CIRCUIT ANALYSIS IV - MWEBhome.mweb.co.za/rf/rfitch/eicam4/CamEval.pdf · Circuit Analysis EICAM4 Unit 2 First Assessment 4 October 2007 Page 1 Question 1 Refer to the circuit in

VUT Vaal University of Technology

CIRCUIT ANALYSIS IV

Previous Evaluation and Assessment

Module Code: EICAM4

Circuit Analysis EICAM4 Unit 1 First Assessment 30 August 2007 Page 1

Question 1 For the circuit in Figure 1, determine v(t) and i(t) for all t, if switch S is opened when t = 0. (12) Question 2 For the circuit shown in Figure 2, find an expression for i(t) for all t, if vs(t) is given by:

vs(t) =

0 for t1

0 for t1-

(10) Question 3 Refer to the circuit in Figure 3 and determine the zero state step response v(t). (10)

Question 4 The switch in the circuit of Figure 4, is opened at time t = 0. Determine i(t) and v(t) for all time t. (18) ---ooo000ooo--- Total: 50

Appendix

Trigonometric identities: Acost + Bsint = cos (t – tan-1(B/A)) cos = sin( + /2)

= sin (t + tan-1(A/B)) sin = cos( – /2)

Euler’s rule: e j = cos + jsin cos = ½ (e j + e – j) sin = ½ j (e j – e – j)

Laplace transforms: L[f(t)] = F(s) =

f(t) F(s) f(t) F(s)

df(t)/dt sF(s)–f(0) te-at u(t) 1/(s+a)2

t0 f(t)dt F(s)/s sint u(t) /(s2+2)

(t) 1 cost u(t) s/(s2+2) u(t) 1/s e-at sint u(t) /[(s+a)2+2]

e-atu(t) 1/(s+a) e-at cost u(t) (s+a)/[(s+a)2+2]

tu(t) 1/s2 2Ae-at cos(t+)u(t)

jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1i

101 F

t = 0

v

2 3 10 V

S

v vs(t) = u(t)

1

1 F31

1

Figure 3

1 F

1 H

t=0

2 i

v 1 V

Figure 4

– 1 t

vs(t)

0

1

1

1 vs 1 H

i Figure 2

Stroombaananalise IV EICAM4 Eenheid 1 Eerste Evaluasie 30 Augustus 2007 Memorandum Bladsy 1

1. t < 0: v(t) = 10 V.[1] en i(t) = 0 [1] t 0: v+2i+3i = 0 5i+v = 0 en i = (1/10)dv/dt (1/2)dv/dt + v = 0 dv/dt + 2v = 0 [5] v = Ae-2t [1] en v(0) = 10 A= 10 [1] v(t) = 10e-2t [1] en i(t) = (1/10)dv/dt = -2e-2t [2] 2. t < 0: -1 – i – 0 = 0 i = -1 A [1] t 0: 1 – v’ = v’ + i 1 = 2v’ + i en v’ = 1di/dt 1 = 2di/dt +i di/dt +(1/2)i = (1/2) [5] i = 1 + Ae-t/2 [1] Maar i(0)= -1 -1 = 1 + A A = -2 [1] i = 1 – 2e-t/2 [2] 3. t < 0: v(t) = 0 [1] (zero toestand) t 0: (1-v’)/1 = v’/1 + (v’ – v)/1 1 = 3v’ - v en (v’ – 0)/1 = (1/3)d/dt(0-v) v’ = -(1/3)dv/dt 1 = 3(-(1/3)dv/dt – v 1 = -dv/dt – v dv/dt + v = -1 [5] v = -1 + Ae-t [1] Maar v(0) = 0 0 = -1 + A A= 1 [1] v(t) = -(1 – e-t) [2] vir t>= 0 4. t < 0: i(t) = 1/2A [1] en v(t) = 0 V [1] t 0: 2i + di/dt + v = 1 en i = dv/dt 2dv/dt + d2v/dt2 + v = 1 d2v/dt2 + 2dv/dt + v = 1 [6] = 1 en n = 1 krities gedemp v(t) = 1 + Ae-t + Bte-t [2] v(0) = 0 A = -1 [2] dv/dt = – Ae-t + Be-t – Bte-t dv(0)/dt = -A+B [1] maar i(0) = dv(0)/dt dv(0)/dt = ½ [1] -A + B = ½ B = -½ [2] v(t) = 1 – e-t – ½tet [1] vir t>0 en i(t) = dv/dt = e-t – ½e-t + ½te-t [1] = ½e-t + ½te-t = ½e-t(1+t)

[12]

1

1 1 V 1 H

i v’

1 -1 V

i

0V

i

t < 0 t 0

[10]

[10]

[18]

i

10

1 F

v

2 3

i v

2 3

10V

t < 0 t 0

1

v u(t)

1

1

1/3 F

v’

1 F

1 H 2 i

v 1 V

2 i

v1 V

t < 0

t 0

Circuit Analysis EICAM4 Unit 2 First Assessment 4 October 2007 Page 1 Question 1 Refer to the circuit in Figure 1.

a) Find the steady state response i(t) in the time domain, solving the differential equation for i, if the supply voltage vs, is given by the real sinusoid vs(t) = cost. Give your answer in the form, Acos(t + ). (10)

b) Find the steady state response i(t) in the time domain, solving the differential equation for i, if the supply voltage is the complex sinusoid vs = e jt. Give your answer in the form Ae j(t + ). (6)

c) Use frequency domain analysis to determine the phasor current I, if the supply voltage is the phasor Vs = 10. Reconstruct the real form of i(t), Acos(t + ), again from the phasor current I. (4)

Question 2 Determine the resonance frequency of the circuit in Figure 2, from the perspective that the imaginary part of the impedance Z, must vanish. (12) Question 3 Refer to the circuit in Figure 3. The circuit is supplied from a voltage source vs(t) = cost u(t) volt. The initial current through the inductor is i(0) = 0 amp. and the initial voltage across the capacitor is v(0) = 1 volt. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (18) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

vs

Figure 1 2

cost e jt

101 H

i

1 F

1 1 H

Figure 2

2 Z

i(0) = 0

v(0) = 1 31 F

i 1 H 4

vvs = cost u(t)

Figure 3

Stroombaananalise IV EICAM4 Eenheid 2 Eerste Evaluasie 4 Oktober 2007 Memorandum Bladsy 1

1. a) vs – 2i – di/dt = 0 di/dt + 2i = vs With vs = cost: di/dt + 2i = cost [4] But the steady state current must be of the form: i(t) = Acost + Bsint – Asint + Bcost + 2Acost + 2Bsint = cost (2A + B)cost + (– A + 2B)sint = cost 2A + B = 1 and – A + 2B = 0 A = 2/5 [2] and B = 1/5 [2] iss(t) = 2/5cost + 1/5sint [1] = 0.4472cos(t – 0.4636) amp [1] (10)

b) For vs(t) = e jt di/dt + 2i = e jt i(t) must have the form Ae j(t + ) [1] d/dt[Ae j(t + )] + 2Ae j(t + ) = e jt jAe j(t + ) + 2Ae j(t + ) = e jt (2 + j)Ae j(t + ) = e jt (2 + j)Ae jt×e j = e jt (2 + j)Ae j = 1 Ae j = 1/(2 + j) = 0.4472 – 0.4636 = 0.4472e – j0.4636 Ae j = 0.4472e –j 0.4636 [3] A = 0.4472 and = – 0.4636r iss(t) = 0.4472e j(t – 0.4636) amp [2] (6)

c) Z = 2 + 190 = 2.23626.57 I = V/Z = 10/2.23626.57) = 0.4472-26.57 A [3] v(t) = 0.4472cos(t – 0 4637r) [1] (4) 2. Z = 1 + j + 2//(1/j) [2] = 1 + j + (2/j)/(2 + 1/j) = 1 + j + 2/(j2 + 1) = 1 + j + (2 – j4)/(1 + 42) = [1 + 2/(1 + 42)] + j[ – 4/(1 + 42)] = [(1 + 42 + 2)/(1 + 42)] + j[(1 + 42 – 4)/(1 + 42)] [6] At resonance, Im{Z} = 0 1 + 42 – 4 = 0 42 = 3 r = 0.866 r/s.[4] 3. [s/(s2 + 1)] – 4I(s) – sI(s) – (1/s) – (3/s)I(s) = 0 [2] [4 + s + (3/s)]I(s) = [s/(s2+1)] – (1/s) = [(s2 – (s2 + 1))/s(s2+1)] = –1/s(s2 + 1) [(4s + s2 + 3)/s]I(s) = –1/s(s2 + 1) (s2 + 4s + 3)I(s) = –1/(s2 + 1) I(s)(s + 1)(s + 3) = –1/(s2 + 1) I(s) = –1/(s + 1)(s + 3)(s2 + 1) [4] I(s) = –1/(s + 1)(s + 3)(s – 11.571)(s – 1– 1.571) I(s) = –¼/(s + 1) + (1/20)/(s + 3) + 0.11180.4635/(s – 11.571) + 0.1118– 0.4635/(s – 1– 1.571)

[4] I(s) = – ¼/(s + 1) + (1/20)/(s + 3) + 0.11180.4635/(s – 0 – j) + 0.1118– 0.4635/(s – 0 + j) i(t) = – ¼e – t u(t) + (1/20)e – 3t u(t) + 2×0.1118cos(t + 0.4635) u(t) [4] = [– 0.25e – t + 0.05e – 3t + 0.2236cos(t + 0.4635)]u(t)

(cost, e jt)

vs

2

1 H

i

I

10 V

Vs

2

190 = 1 r/s

1 H

[20]

1/j

1 j

2 Z

[12]

s3

4 I(s)

12s

s

s

V(s)

Model 4 marks

s1

[18]

Circuit Analysis EICAM4 Unit 1 Final Assessment 1 November 2007 Page 1

Question 1 Refer to the circuit in Figure 1. The switch S moves from position a to position b, at time t = 0. Determine v(t) and i(t) for all time t. Question 2 Find the zero state response v(t), for the circuit in Figure 2, if vs(t) = u(t). Question 3 Determine the complete solution for i(t), from the following differential equation:

4i4dtdi4

dtid2

2

, with the initial conditions, i(0) = 0 and dt

di(0)= 0. (6)

Question 4 For the circuit shown in Figure 3, the switch S was open for a long time. The switch is closed at time t = 0. Find v(t) and i(t) for all time t.

---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

(12)

(10)

(22)

i

51 F

S

t = 0

5 V

Figure 3

51 H

v

a

S

3

3

i

t = 0

6 A v 3 b

101

F

Figure 1

v vs(t) = u(t) v 1 F

1

1

Figure 2

85

Stroombaananalise IV EICAM4 Eenheid 1 Finale Evaluasie 1 November 2007 Memorandum Bladsy 1

1. t < 0: i3+3 = (3/9)6 = 2A v(t) = 6V [2] and i(t) = 0 A [1] t 0: v/6 + i + v/3 = 0 en i = (1/10)dv/dt dv/dt+5v = 0 [4] v=Ae-5t v(0) = 6 A = 6 v(t) = 6e-5t [3] i(t) = (1/10)dv/dt = (1/10)(-30e-5t) = -3e-5t [2]

2. (vs – v)/1 = ic + iR = dv/dt + vR and v + v – vR = 0 vR = 2v vs – v = dv/dt + 2v dv/dt + 3v = vs t < 0: dv/dt + 3v = 0 v(t) = 0 [1] (or zero state) t 0 dv/dt + 3v = 1 [5] v(t) = 1/3 + Ae-3t [1] en v(0) = 0 A = -1/3 [1] v(t) = 1/3 – 1/3e-6t [2]

3. d2i/dt2 + 4di/dt + 4i = 4 = 2 [1] en n = 2 [1] stelsel is krities gedemp i(t) = Ae-2t + Bte-2t + 1 [1] i(0) = 0 A + 1 = 0 A = -1 [1] di/dt = -2Ae-2t – Bte-2t + Be-2t [1] di(0)/dt = -2A + B = 0 2 + B = 0 B = –2 [1] i(t) = – e-2t – 2te-2t + 1

4. t < 0: i(t) = 5/(5/8) = 8 A [1] en v(0) = 0 [1]. t 0: 5 - (1/5)di/dt - v = 0 …....…….(1) en i = v/(5/8) + (1/5)dv/dt ……(2) (2) in (1): 5 – (1/5)d/dt[v/(5/8) + (1/5)dv/dt] – v = 0 5 – (8/25)dv/dt – (1/25)d2v/dt2 – v = 0 (1/25)d2v/dt2 + (8/25)dv/dt + v = 5 d2v/dt2 + 8di/dt + 25i = 125 [6] = 4 [1] en n = 5 [1] ondergedemp met d = 3 r/s [1] v(t) = Ae-4tcos3t + Be-4tsin3t + 5 [1] v(0) = 0 A + 5 = 0 A = -5 [1] en i(0) = v(0)/(5/8) + (1/5)dv(0)/dt 8 = 0 + (1/5)dv(0)/dt dv(0)/dt = 40 [1] dv/dt = -4Ae-4tcos3t – 3Ae-4tsin3t – 4Be-4tsin3t + 3Be-4tcos3t dv(0)/dt = -4A + 3B [1] 40 = -4(-5) + 3B B = 6.667 [1] v(t) = -5e-4tcos3t + 6.667e-4tsin3t + 5 [2] dv/dt = 20e-4tcos3t + 15e-4tsin3t – 26.67e-4tsin3t + 20e-4tcos3t = 40e-4tcos3t – 11.67e-4tsin3t i(t) = (8/5)v + (1/5)dv/dt = (8/5)[-5e-4tcos3t + 6.667e-4tsin3t + 5] + (1/5)[40e-4tcos3t – 11.67e-4tsin3t] = 8 – 8e-4tcos3t + 8e-4tcos3t + 10.67e-4tsin3t – 2.334e-4tsin3t = 8 + 8.336e-4tsin3t [4]

85

3

101 F

[12]

3

i

3 v

3

3 i

6 A v

3

t < 0

t 0

v

1 F v

1

1 vs=u(t)

iC iR

vR

[10]

i

51

F

5 V 51

H v

[22]

85

i

5 V v

t < 0

t 0

[6]

Circuit Analysis EICAM4 Unit 2 Final Assessment 1 November 2007 Page 1

Question 1 Refer to the circuit in Figure 1.

a) Find the steady state response v(t) in the time domain, solving the differential equation for v, if the supply voltage vs, is given by the real sinusoid vs(t) = cost. Give your answer in the form, Acos(t + ).

b) Find the steady state response v(t) in the time domain, solving the differential equation for v, if the supply voltage is the complex sinusoid vs = e jt. Give your answer in the form Ae j(t + ). (6)

c) Use frequency domain analysis to determine the phasor voltage V, if the supply voltage is the phasor Vs = 10. Reconstruct the real form of v(t), Acos(t + ), again from the phasor voltage V. (4)

Question 2 Determine the resonance frequency of the circuit in Figure 2, from the perspective that the imaginary part of the impedance Z, must vanish. Question 3 Refer to the circuit in Figure 3. The circuit is supplied from a voltage source vs(t) = cost u(t) volt. The initial current through the inductor is i(0) = 1 amp. and the initial voltage across the capacitor is v(0) = 1 volt. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t).


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

vs 1 F

2 H

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

v 1

Cost ejt 10

(12)

Figure 2

Z 2 F

1 H

1

i(0) = 1

v(0) = 1 31 F

i 1 H 4

vvs = cost u(t)

Figure 3

(10)

(18)

Stroombaananalise IV EICAM4 Eenheid 2 Finale Evaluasie 1 November 2007 Memorandum Bladsy 1

1. a) vs – 2di/dt – v = 0 and i = v + dv/dt vs – 2d/dt[v + dv/dt] – v = 0 vs – 2dv/dt – 2d2v/dt2 – v = 0 With vs = cost: d2v/dt2 + dv/dt + ½v = ½cost [6] But the steady state voltage must be of the form: v(t) = Acost + Bsint – Acost – Bsint – Asint + Bcost + ½Acost + ½Bsint = ½cost (– ½A + B)cost + (– A – ½B)sint = cost – ½A + B = 1 and – A – ½B = 0 A = – 1/5 [2] and B = 2/5 [2] vss(t) = – (1/5)cost + (2/5)sint [1] = 0.4472cos(t – 2.0344) volt [1] (12)

b) For vs(t) = e jt d2v/dt2 + dv/dt + ½v = ½e jt v(t) must have the form Ae j(t + ) [1] d2/dt2[Ae j(t + )] + d/dt[Ae j(t + )] + ½[Ae j(t + )]v = ½e jt j2Ae j(t + ) + jAe j(t + ) + ½Ae j(t + ) = ½e jt ( – 1 + j + ½)Ae j(t + ) = ½e jt ( – ½ + j)Ae j(t + ) = ½e jt ( – ½ + j)Ae jt×e j = ½e jt ( – ½ + j)Ae j = ½ Ae j = ½/( – ½ + j) = 0.4472– 2.0344 = 0.4472e – j2.0344 [4] A = 0.4472 and = – 2.0344r vss(t) = 0.4472e j(t – 2.0344) volt [1] (6)

c) Z1F//1 = (1×1-90)/(1 + 1-90) = 0.7071-45 Zt = 290 + 0.7071-45 = 1.58171.57 V = (Z1F//1/Zt)×Vs = (0.7071-45/1.58171.57)×10 = 0.4472-116.6 volt [3] v(t) = 0.4472cos(t – 2.035r) [1] (4) 2. Z = j + 1//(1/j2) [1] = j + (1/j2)/(1 + 1/j2) = j + 1/(j2 + 1) = j + (1 – j2)/(1 + 42) = [1/(1 + 42)] + j[1 – 2/(1 + 42)] [5] At resonance, Im{Z} = 0 1 – 2/(1 + 42) = 0 2/(1 + 42) = 1 1 + 42 = 2 42 = 1 2 = 1/4 r = 0.5 r/s.[4] 3. [s/(s2 + 1)] – 4I(s) + 1 – sI(s) – (1/s) – (3/s)I(s) = 0 [2] [4 + s + (3/s)]I(s) = [s/(s2+1)] + 1 – (1/s) = [(s2 + s(s2 + 1) – (s2 + 1))/s(s2+1)] = (s3 + s – 1)/s(s2 + 1) [(4s + s2 + 3)/s]I(s) = (s3 + s – 1)/s(s2 + 1) (s2 + 4s + 3)I(s) = (s3 + s – 1)/(s2 + 1) I(s)(s + 1)(s + 3) = (s3 + s – 1)/(s2 + 1) I(s) = (s3 + s – 1)/[(s + 1)(s + 3)(s2 + 1)] [4] I(s) = (s3 + s – 1)/[(s + 1)(s + 3)(s – 11.571)(s – 1– 1.571)] I(s) = –(3/4)/(s + 1) + (31/20)/(s + 3) + 0.11180.4635/(s – 11.571) + 0.1118– 0.4635/(s – 1– 1.571)

[4] I(s) = – (3/4)/(s + 1) + (31/20)/(s + 3) + 0.11180.4635/(s – 0 – j) + 0.1118– 0.4635/(s – 0 + j) i(t) = – (3/4)e – t u(t) + (31/20)e – 3t u(t) + 2×0.1118cos(t + 0.4635) u(t) [4] = [– 0.75e – t + 1.55e – 3t + 0.2236cos(t + 0.4635)]u(t)

I(s)

[22]

i

(cost, e jt)

vs

1 F

2 H

v1

I

10 V

Vs

= 1 r/s 1 F

2 H

V1

[10]

1/j2

j

1 Z

s3

4

1

12s

s

s V(s)

Model 4 marks

s1

[18]

Circuit Analysis EICAM4 Unit 1 First Assessment 28 August 2008 Page 1 Question 1 For the circuit shown in Figure 1, find the zero state step response, i(t). (10) Question 2 For the circuit in Figure 2, the switch S is closed at time t = 0. Calculate v(t) for all time t. (10) Question 3 For the circuit in Figure 3, the switch S is opened at time t = 0. Determine a solution for v(t) for all time t. (15)

Question 4 Refer to the active circuit in Figure 4 and determine the zero state step response, v(t). (Zero state implies that v(0) = 0 and e(0) = 0). (15)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

vs(t) = u(t)

i 1

1 1 H

Figure 1

t=0

S

v 1 V

1

1 1 F

Figure 2

21

F

S

t=0

3 i v

1 H

1 V

Figure 3

e

4

161

F

251

F

4

v vs = u(t)

Figure 4


1. t<0: i(t) = 0 [1] t>=0: 1 – v = v + i 1 = 2v + i

en v = 1di/dt 1 = 2di/dt +i di/dt +(1/2)i = (1/2) [6] i = 1 + Ae-t/2 [1] Maar i(0)= 0 0 = 1 + A A = -1 [1] i = 1 – e-t/2 [1] 2. t<0: v(t) = 1 V [2] t>=0: 1 – v = dv/dt + v dv/dt + 2v = 1 [5] v = Ae-2t + ½ [1] v(0) = 1 A = ½ [1] v(t) = ½(e-2t + 1) [1] 3. t<0: v(t) = 0 V [1] en i(t) = 1/3 A [1] t>=0: 1 = 3i + di/dt + v en i = (1/2)dv/dt 1 = (3/2)dv/dt + (1/2)d2v/dt2 + v d2v/dt2 + 3dv/dt + 2v = 2 [5] = 1.5 en n = 2 v(t) = Ae-t + Be-2t + 1 [4] v(0) = 0 A + B = -1 i(0) = ½dv(0)/dt = 1/3 dv(0)/dt=2/3 en dv/dt = -Ae-t – 2Be-2t -A – 2B = 2/3 v(t) = 1 – (4/3)e-t + (1/3)e-2t [4] 4. t<0: v(t) = 0 en e(t) = 0 t>=0: (1 – e)/4 = (e – v)/4 + (1/16)d/dt(e – v) 4(1 – e) = 4(e – v) + de/dt – dv/dt 4 – 4e = 4e – 4v + de/dt – dv/dt 4 = 8e + de/dt – 4v – dv/dt ….... (1) en (e – v)/4 = (1/25)dv/dt e – v = (4/25)dv/dt e = v + (4/25)dv/dt ……..…….. (2) (2) in (1): 4 = 8[v + (4/25)dv/dt] + d/dt[v + (4/25)dv/dt] – 4v – dv/dt 4 = 8v + (32/25)dv/dt] + dv/dt + (4/25)d2v/dt2 – 4v – dv/dt 4 = 4v + (32/25)dv/dt + (4/25)d2v/dt2 d2v/dt2 + 8dv/dt + 25v = 25 [7] = 4 en n = 5 v(t) = Ae-4tcos3t + Be-4tsin3t + 1 [4] ……. (3) v(0) = 0 A + 1 = 0 A = -1 van (2): e(0) = v(0) + (4/25)dv(0)/dt dv(0)/dt = (25/4)(e(0) – v(0)) = 0 en van (3): dv/dt = -4Ae-4tcos3t – 3Ae-4tsin3t – 4Be-4tsin3t + 3Be-4tcos3t dv(0)/dt = -4A + 3B dv(0)/dt = 0 -4A + 3B = 0 3B = 4A = 4(-1) = -4 B = -(4/3) v(t) = –e-4tcos3t – (4/3)e-4tsin3t + 1 [4] = 1 + 1.667e-4tcos(3t - 4.069)

1 V

1 v i

1 H 1

[10]

21

F

3 i

v

1 H

1 V

[10]

v 1 V

1

1 1 F

[15]

e

v 4

161 F

251 F

4

v 1

[15]

Circuit Analysis EICAM4 Unit 2 First Assessment 25 September 2008 Page 1


a) Find the steady state response i(t) in the time domain, solving the differential equation for i, if the supply voltage vs, is given by the real sinusoid vs(t) = cost. Give your answer in the form, Acos(t + ).

b) Find the steady state response i(t) in the time domain, solving the differential equation for i, if the supply voltage is the complex sinusoid vs = e jt. Give your answer in the form Ae j(t + ). (6)

c) Use frequency domain analysis to determine the phasor current I, if the supply voltage is the phasor Vs = 10. Reconstruct the real form of i(t), Acos(t + ), again from the phasor current I. (6)

Question 2 Determine the resonance frequency of the circuit in Figure 2. (12) Question 3 Refer to the circuit in Figure 3. The circuit is supplied from a voltage source vs(t) = e

– 2t

u(t) volt. The initial current through the inductor is i(0) = 0 amp. and the initial voltage across the capacitor is v(0) = 0 volt. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (16) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

1

1 H

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

i(0) = 0

v(0) = 0 31 F

i 1 H 4

vvs = e –

2t

u(t)

Figure 3

(10)

vs 1 F

Figure 1

Cost ejt 10

i 1

2 H

6

Figure 2

361 F

Stroombaananalise IV EICAM4 Eenheid 2 Eerste Evaluasie 25 September 2008 Memorandum Bladsy 1

1. a) (vs –v) = dv/dt + i en v = i + di/dt vs – i – di/dt = di/dt + d2i/dt2 + i d2i/dt2 + 2di/dt + 2i = vs d2i/dt2 + 2di/dt + 2i = cost [5] We insist that i = Acost + Bsint , therefore di/dt = -Asint + Bcost and d2i/dt2 = –Acost – Bsint –Acost – Bsint – 2Asint + 2Bcost + 2Acost + 2Bsint = cost [A + 2B]cost + [B – 2A]sint = cost A + 2B = 1 en B – 2A = 0 A = 0.2 en B = 0.4 i(t) = 0.2cost + 0.4sint [4] = 0.4472cos(t – 1.107) ampere (9) b) For vs(t) = e jt d2i/dt2 + 2di/dt + 2i = e jt i(t) must have the form Ae j(t + ) [1] d2/dt2[Ae j(t + )] + 2d/dt[Ae j(t + )] + 2[Ae j(t + ) = e jt j2Ae j(t+)

+ j2Ae j(t+) +2Ae j(t+) = e jt (–1 + j2 + 2)Ae j(t+) = e jt (1 + j2)Ae jt×e j = e jt (1 + j2)Ae j = 1 Ae j = 1/(1+j2) Aej = 0.4472–1.107 Aej = 0.4472e-j1.107 [3] A = 0.4472 and = – 1.107r iss(t) = 0.4472e j(t – 1.107) amp [2] (6) c) Z = 1 + 1-90//(1+190) = 1 + [1-90×(1+190)/[1-90 + 1 + 190) = 1 + 1.4142-45/1 = 2.236-26.56 [2] It = Vs/Z = 10/2.236-26.56) = 0.447226.56 A [2] I = [1-90/(1-90 + 1 + 1-90)]×It = [1-90/(1)]×0.447226.56 = 0.4472-63.44 [2] i(t) = 0.4472cos(t – 1.107r) [1] (7) 2. 1/Z=1/(1/j(1/36))+1/(6+j2) = j(1/36)+(6-j2)/(36+42) = 6/(36+42)+j[/36-2/(36+42)]

[4] Vir resonansie moet Im{Z} = 0 [2] /36 - 2/(36+42) = 0 1/36 = 2/(36+42) 36+42 = 72 42 = 36 2 = 9 = 3 rad/sek [6] 3. [1/(s + 2)] – 4I(s) – sI(s) – (3/s)I(s) = 0 [2] [4 + s + (3/s)]I(s) = [1/(s+2)] [(4s + s2 + 3)/s]I(s) = 1/(s + 2) (s2 + 4s + 3)I(s) = s/(s + 2) I(s)(s + 1)(s + 3) = s/(s + 2) I(s) = s/(s + 1)(s + 2)(s + 3) [4] I(s) = –½/(s + 1) + 2/(s + 2) – 1½/(s + 3) [3] i(t) = – ½e – t u(t) + 2e – 2t u(t) – 1½ e – 3t [3]

cost ejt

[16]

1

1 vs

1

i 1

v

[22]

Vs

It = 1 r/s

1-90

I

10 V

190

1 1

[12]

s3

4 I(s)

2s

1

s

V(s)

Model 4 marks

Circuit Analysis EICAM4 Unit 1 Final Assessment 30 October 2008 Page 1

Question 1 For the circuit shown in Figure 1, find an expression for i(t) for all t,

if the supply current, is(t), is given by:

0 for t1

0 for t1- (t)is (10)

Question 2 Determine the zero state step response v(t) for the circuit in Figure 2. (10)

Question 3 For the circuit in Figure 3, determine the zero state step response v(t). (12) Question 4 The switch S, in the circuit of Figure 4, is opened at time t = 0. Determine i(t) and v(t) for all t. (18) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2 3 i

i is

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

1

1

-1

t

is(t)

0

1

1 H

Figure 2

vs(t) = u(t) 3 H 2i v

Figure 4

t = 0 251 F

1 H 6 i

v 6 V

vs = u(t) 1 F

3 1 H

Figure 3

v 1

S

Stroombaananalise IV EICAM4 Eenheid 1 Finale Evaluasie 30 October 2008 Memorandum Bladsy 1

i

1. t < 0: i(t) = -½ A [2] t 0: 1 = v’ + i en v’ = i + di/dt 1 = i +di/dt + i di/dt + 2i = 1 [5] i = ½ + Ae-2t Maar i(0) = -½ , -½ = ½ + A A = -1 i(t) = ½ - e-2t [3] vir t 1

2. t < 0: i(t) = v(t) = iL(t) = 0 [1] (zero state) t 0: i + 2i = iL 3i = iL

v = 3diL/dt 1 = 3i + 2iL + v 1 = iL + 2iL + 3diL/dt diL/dt + iL = 1/3 [5] iL = 1/3 + Ae-t i(0) = 0 A = -1/3 iL (t) = (1/3)(1 - e-t) [2] v(t) = 3diL/dt = 3(1/3)e-t = e-t [2]

3. t < 0: v(t) = 0 [½] , i(t) = 0 [½] t 0: 3i + di/dt = 1 – v en i = v + dv/dt 3(v + dv/dt) +d/dt(v + dv/dt) = 1 – v 3v + 3dv/dt +dv/dt + d2v/dt2 = 1 – v d2v/dt2 + 4dv/dt + 4v = 1 [5] = 2 en n = 2 kritiek gedemp v(t) = Ate-2t + Be-2t + ¼ [2] v(0) = 0 B + ¼ = 0 B = -¼ and dv/dt = Ae-2t – 2Ate-2t – 2Be-2t dv(0)/dt = A – 2B but dv(0)dt = i(0) – v(0) A – 2B = 0 A – 2(-1/4) = 0 A = -1/2 v(t) = -½te-2t – ¼e-2t + ¼ [4] 4. t < 0: i(t) = 6/6 = 1A [1] en v(t) = 0 [1] t 0: 6 = 6i + di/dt + v en i =(1/25)dv/dt (6/25)dv/dt +(1/25)dv2/dt2 + v = 6 d2v/dt2 + 6dv/dt + 25v = 150 [6] = 3 en n = 5 ondergedempte stelsel met d = 4 r/s v(t) = 6 + Ae-3tcos4t + Be-3tsin4t [4] v(0) = 0 0 = 6 + A A = -6 dv/dt = -3Ae-3tcos4t – 4Ae-3tsin4t –3Be-3tsin4t + 4Be-3tcos4t = (4B-3A)e-3tcos4t – (4A+3B)e-3tsin4t …………........(i) dv(0)/dt = -3A + 4B = 18 + 4B maar i(0) = (1/25)dv(0)/dt dv(0)/dt = 25 18 + 4B = 25 4B = 7 B = 1.75 v(t) = 6 – 6e-3tcos4t + 1.75e-3tsin4t [3] en i(t) = (1/25)dv/dt = (1/25)[25e-3tcos4t + 18.75e-3tsin4t] van (i) = e-3tcos4t + 0.75e-3tsin4t [3]

i

v’

1 H is

i 1

1

[10]

v

iL

2i 1 V

2 3

3 H

[10]

[12]

1 1 F

3 1 H

v 1

251 F

1 H 6 i

v 6 V

251 F

1 H 6 i

v 6 V

[18]


Question 1 The supply current to the circuit in Figure 1, is given by is(t) = cos t. a) Determine the time domain

solution for the steady state current response, i(t), by writing down and solving a differential equation for i(t). Give your answer in the form Acos(t+). (10)

b) Use frequency domain analysis to determine the phasor current I. Reconstruct i(t) again from the phasor current I. (6)

Question 2

a) Determine the resonance frequency of the circuit in Figure 2. (8)

b) Determine an equivalent parallel structure for the circuit in Figure 3. Calculate the resonance frequency r, the quality factor Q, the bandwidth BW and the half power frequencies 1 and 2, for the equivalent circuit. (8)

Question 3 Refer to the circuit in Figure 4. The circuit is supplied from a voltage source vs(t) = cost u(t) volt. The initial current through the inductor is i(0) = 0 amp. and the initial voltage across the capacitor is v(0) = 0 volt. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (18)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

1 F

is(t) = cost

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

i(0) = 0

v(0) = 0

i 1 Hvvs = cost u(t)

Figure 4

i 1

1 F 1 H

1

2

1 H

1

2 F

Figure 2

Figure 3

0.8 H 0.05 F

0.1


1. a) is = dvc/dt + i en vc = i + di/dt is = di/dt + d2i/dt + i d2i/dt2 + di/dt + i = cost [5] With i(t) = Acost + Bsint -Acost – Bsint – Asint + Bcost +Acost + Bsint = cost Bcost -Asint = cost B=1 en A = 0 i(t) = sint = cos(t-1.571) ampere [5] (10) b) I = [1-90/(1-90+1+190)]10 = 1-90 A i(t) = cos(t-1.571) A (6) 2. a) Z = 1/j2 + 1//(2+j) = -j(1/2) + (2+j)/(3+j) = -j(1/2) + (2+j)(3-j)/(9+2) = -j(1/2) + (6+j+2)/(9+2) = (6+2)/(9+2) + j[/(9+2)-(1/2)] [4] Vir resonansie Im{Z} = 0 [1] /(9+2) = 1/2 9+2 = 22 2 = 9 =3 r/s [3] (8) b) RP = L/RC = 0.8/0.10.05 = 160 [1] r = (1/LC) = 5 r/s [1] Q = RC/L = 160(0.05/0.8) = 40 [1] BW = r/Q = 5/40 = 0.125 r/s [1] 12 = 25 and 2 - 1 = 0.125 1 = 4.9375 en 2 = 5.0625 [4] (8) 3. [s/(s2 + 1)] – sI(s) – V(s) = 0 [2] ……..…. (1) I(s) = V(s)/1 + V(s)/(1/s) [2] ……………. (2) From (1): V(s) = [s/(s2 + 1)] – sI(s) …………..…… (3) (3) in (2): I(s) = [s/(s2 + 1)] – sI(s) + s{[s/(s2 + 1)] – sI(s)} = [s/(s2 + 1)] – sI(s) + s2/(s2 + 1)] – s2I(s) I(s) + sI(s) + s2I(s) = s/(s2 + 1)] + s2/(s2 + 1)] (s2 + s + 1)I(s) = (s2 + s)/(s2 + 1) I(s) = (s2 + s)/(s2 + 1)(s2 + s + 1) [5] = (s2 + s)/(s – j)(s + j)(s + 0.5 – j0.866)(s + 0.5 + j0.866) I(s) = (s2 + s)/(s – 11.571)(s – 1– 1.571)(s – 12.094)(s – 1–2.094) I(s) = 0.7078–0.7853/(s – 11.571) + 0.70780.7853/(s – 1– 1.571) + 0.5782.618/(s – 12.094) + 0.578–2.618/(s – 1–2.094) [3] I(s) = 0.7078–0.7853/(s + 0 – j) + 0.70780.7853/(s + 0 + j) + 0.5782.618/(s +0.5 – j0.866) + 0.578–2.618/(s + 0.5 + j0.866) i(t) = 2×0.7078×cos(t – 0.7853) + 2×0.578×e-0.5tcos(0.866t – 2.618) = 1.416cos(t – 0.7853) + 1.156e-0.5tcos(0.866t – 2.618) [3]

[16]

[18]

160 0.05 F 0.8 H

2

1 H

1

2 F

Model 3 marks

I(s)

s1

12s

s

s V(s)

1

i

cost

1

1 F 1 H vc

[16]

Circuit Analysis EICAM4 Unit 1 First Assessment 27 August 2009 Page 1

Question 1 Determine the zero state step response i(t) for the circuit in Figure 1. (10) Question 2 For the circuit shown in Figure 2, find an expression for i(t) for all t, if the supply vs(t), is given

by: vs(t) =

0 for t1

0 for t1

Question 3 For the circuit in Figure 3, determine the zero state step response i(t) and v(t). (13)

Question 4 For the circuit shown in Figure 4, the switch S was open for a long time. The switch is closed at t = 0. Find i(t), for all time t. (15)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

1 H

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

vs = u(t)

1

1 Figure 3

1 F

i v

1 i

Figure 2

1

1 1 H vs

– 1 t

vs(t)

0

1

(12)

1 H

1 i

vs(t) = u(t) 1

Figure 11

Figure 4

v

85

H 51

F 51

t = 0 i

5 V

S


1. Vir t 0: i(t) = 0 (zero toestand respons) Vir t > 0: 1 - v’ = i + v’ en v’ = i + di/dt 1 - (i + di/dt) = i + (i + di/dt) 2di/dt + 3i = 1 di/dt + 3/2i = ½ [6] i(t) = 1/3+Ae-3/2t [2] Maar i(0) = 0 A=-1/3 [1] i(t) = 1/3(1-e-3/2t) vir t>=0 [1] 2. t < 0: -1 – v’ = v’ + i and v’ = i + 0 = i -1 – i = i + i i = -1/3 A [2] t 0: 1 – v’ = v’ + i 1 = 2v’ + i en v’ = i + di/dt 1 = 2(i + di/dt) +i 2di/dt + 3i = 1 di/dt + (3/2)i = (1/2) [6] i = (1/3) + Ae-3t/2 [2] Maar i(0)= -(1/3) -(1/3) = (1/3) + A A = -(2/3) i = (1/3) – (2/3)e-3t/2 [2] 3. t < 0: i(t) = 0 and v(t) = 0 (zero state) t 0: 1 – v = v + i + dv/dt and v = di/dt 1 – di/dt = di/dt + i + d2/dt2 d2i/dt2 + 2di/dt + i = 1 [6] = 1 and n = 1 critically damped: i(t) = Ae-t + Bte-t + 1 [3] i(0) = 0 0 = A + 1 A = -1 v(t) = di/dt = -Ae-t + Be-t – Bte-t v(0) = 0 -A + B = 0 B = A = -1 i(t) = –e-t – te-t + 1 [2] v(t) = e-t – e-t + te-t v(t) = te-t [2] 4. Vir t < 0: i(t)=0 [1] en v(t)=5 [1] Vir t 0: (5-v)/(5/8) = (1/5)dv/dt +i and v = (1/5)di/dt (1/25)d2i/dt2 + (8/25)di/dt + i = 8 d2i/dt2 +8di/dt + 25i = 200 [6] = 4 en n = 5 ondergedemp met d = 3 r/s i(t) = Ae-4tcos3t + Be-4tsin3t + 8 [4] i(0) = 0 A + 8 = 0 A = -8 en (1/5)di(0)/dt = v(0) di(0)/dt = 25 di/dt = -4Ae-4tcos3t – 3Ae-4tsin3t –4Be-4tsin3t + 3Be-4tcos3t di(0)/dt = -4A + 3B 25 = -4(-8) + 3B B = -2.333 i(t) = -8e-4tcos3t – 2.33e-4tsin3t + 8 [5]

1F

i

v’

1

i

1 H

1 1 V

[10]

[12]

1H

1

1 Vi v’ t 0

1 -1V

i

0V

i t < 0 v’

1

1

1

i

1

di/dt

1V

1

1 1H v

[13]

[15]

v

8

5

H 5

1 F

5

1

i

5V

8

5

F 5

1

5V

i = 0

v = 5

Circuit Analysis EICAM4 Unit 2 First Assessment 8 October 2009 Page 1


a) Find the steady state response v(t) in the time domain, solving the differential equation for v, if the supply current is, is given by the real sinusoid is(t) = 10sin2t. Give your answer in the form, Acos(t + ).

b) Use frequency domain analysis to determine the phasor voltage V, if the supply current is the phasor Is = 10–90. Reconstruct the real form of v(t), Acos(t + ), again from the phasor voltage V. (8)

Question 2 Refer to the circuit in Figure 2. a) Calculate the impedance Z(),

seen by the source. b) Determine the resonance

frequency r of the circuit in Figure 2, from the perspective that the impedance must become real.

Question 3 Refer to the circuit in Figure 3. The supply current to the circuit is given by is(t) = 10sin2t u(t) ampere. The initial voltage across the capacitor is v(0) = 0 volt. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for v(t). (18) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

(10)

is 21

F

Figure 1

10sin2t (10cos(2t–/2))

10–/2r

(10–90) 4

v

(10)

(4) Figure 2

Z

1

1 F

1 F

1 H

Figure 3

21

F is = 10sin2t u(t)

4 v

v(0) = 0


1. a) is = ½dv/dt + v/4 dv/dt + ½v = 2is dv/dt + ½v = 20sin2t [4] We insist that v = Acos2t + Bsin2t , therefore dv/dt = -2Asint + 2Bcost –2Asin2t + 2Bcos2t + ½Acos2t + ½Bsint = 20sin2t [½A + 2B]cos2t + [– 2A + ½B]sin2t = 20sin2t ½A + 2B = 0 and – 2A + ½B = 20 A = -9.412 en B = 2.353 v(t) = -9.412cos2t + 2.353sin2t [4] = 9.702cos(2t – 2.897) volt [2] (10) b) Ic = [4/(4 + 1–90)]×10–90 = 9.7014–75.96 A [3] V = Ic×1–90 = 9.7014–75.96×1–90 = 9.7014–165.96 V [3] v(t) = 9.7014cos(2t – 2.897) [2] (8) 2. a) Z1 = 1 + 1/j = 1 – j(1/)

Z2 = (j)(1/j)/[j + (1/j)] = 1/[j + (1/j)] = j/(1 – 2) Z = Z1 + Z2 = [1 – j(1/)] + [j/(1 – 2)] = 1 + j[/(1–2) – (1/)] Z = 1 + j{(22 –1)/[(1-2)]} (10)

b) For Im{Z} to vanish: 22 –1 = 0 = 0.7071 rad./sec. (4)

3. 20/(s2 + 4) = V(s)/(2/s) + V(s)/4 [2] sV(s)/2 + V(s)/4 = 20/(s2 + 4) sV(s) + 0.5V(s) = 40/(s2 + 4) (s + 0.5)V(s) = 40/(s2 + 4) V(s) = 40/(s2 + 4)(s + 0.5) [4] V(s) = 40/(s + 0.5)(s – 21.571)(s – 2-1.571) V(s) = 9.412/(s + 0.5) + 4.851-2.897/(s – 21.571) + 4.8512.897/(s – 2-1.571) = 9.412/(s + 0.5) + 4.851-2.897/(s – j2) + 4.8512.897/(s – j2) [4] v(t) = 9.412e – 0.5t u(t) + 2×4.851×e – 0t×cos(2t – 2.897) u(t) = 9.412e – 0.5t u(t) + 9.702cos(2t – 2.897) u(t) [4]

[18]

Model 4 marks

is

21

10sin2t( =10cos(2t–/2) )

4 v

Is 10–90 = 2r/s

4 V

1-90

Ic

Z

1

1/j

1/j

j

[14]

s

2

42s

20

4

V(s)

[18]


Question 1 For the circuit shown in Figure 1, find an expression for v(t) for all t,


0 for t1

0 for t2 (t)is (10)

Question 2 Determine the zero state step response v(t) for the circuit shown in Figure 2. (10)

Question 3 Determine the zero state step response v(t), for the circuit shown in Figure 3. (15)

Question 4 The switch S in the circuit of Figure 4, is opened at time t = 0. Determine i(t) and v(t), for all time t. (15)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

t

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

1F 5 v

is

-1

is(t)

0

2

2 Figure 2

4

81 F

v vs(t) = u(t)

1 F 41

Figure 3

v vs = u(t) 21 H

i

Figure 4 1 F

1 H

t=0

2 i

v 1 V S


1. For t < 0: v(t) = 10 [1] For t 0: is = v/5 + dv/dt -1 = v/5 + dv/dt dv/dt + v/5 = -1 [5] v = [-1/(1/5)] + Ae-t/5 [1] v(0) = 10 10 = -5 + A A = 15 [1] v(t) = -5 + 15e-t/5 [2] , t 0 2. For t < 0: v(t) = 0 [1] ;zero toestand For t 0: (1 - v)/4 = (1/8)dv/dt + v/2 2(1-v) = dv/dt + 4v dv/dt + 6v = 2 [5] v(t) = (1/3) + Ae-6t [1] maar v(0) = 0 0 = (1/3) + A A = -(1/3) [1] v(t) = (1/3)(1 -e-6t ) [2] vir t>0 3. Vir t < 0: v(t) = i(t) = 0 [1] zero toestand Vir t 0: i = v + (1/4)dv/dt en (1-v) = (1/2)di/dt v = 1 – (1/2)di/dt i = 1 – (1/2)di/dt + (1/4)d/dt[1 – (1/2)di/dt] i = 1 – (1/2)di/dt – (1/8)d2i/dt2 d2i/dt2 + 4di/dt + 8i = 8 [6] = 2 en n = 8 ondergedemp met d = (8-4) = 2 r/s i(t) = Ae-2tcos2t + Be-2tsin2t + 1 [4] i(0) = 0 A + 1 = 0 A = -1 en (1/2)di(0)/dt = 1 - v(0) di(0)/dt = 2 di/dt = -2Ae-2tcos2t – 2Ae-2tsin2t – 2Be-2tsin2t + 2Be-2tcos2t di(0)/dt = -2A + 2B 2 = -2(-1) + 2B B = 0 i(t) = 1 - e-2tcos2t v(t) = 1 - ½di/dt = 1 - ½[2e-2tcos2t + 2e-2tsin2t] = 1 - e-2tcos2t - e-2tsin2t [4] 4. Vir t < 0: i(t) = 1/2A [1] en v(t) = 0 V [1] Vir t 0: 2i + di/dt + v = 1 en i = dv/dt 2dv/dt + d2v/dt2 + v = 1 d2v/dt2 + 2dv/dt + v = 1 [6] = 1 en n = 1 krities gedemp v(t) = 1 + Ae-t + Bte-t [3] v(0) = 0 A = -1 dv/dt = -Ae-t + Be-t –Bte-t dv(0)/dt = -A+B maar i(0) = dv(0)/dt dv(0)/dt = ½ -A + B = ½ B = -½ v(t) = 1 – e-t – ½tet vir t > 0 [2] en i(t) = dv/dt = e-t – ½e-t + ½te-t t > 0 [2]

1F5 v

2A1F5

v

is

t < 0

t 0 [10]

[10]

2

4

81 F

v 1 V

1 41 F

v 1 V 21 H

i

[15]

1 F

1 H

2 i

v 1 V

1 F

1 H

2 i

v 1 V

t < 0

t 0

[15]


Question 1 The supply voltage to the circuit in Figure 1, is given by vs(t) = cos t. a) Determine the time domain



Question 2 Determine the resonance frequency for the circuits in Figure 2 a), (4) and Figure 2 b). (8)

Question 3 Refer to the circuit in Figure 3. The supply voltage is given by vs(t) = 10 u(t) volt. The initial current through the inductor is i(0) = 2 amp. and the initial voltage across the capacitor is v(0) = 4 volt. Draw an equivalent Laplace network model for the circuit and use this model to find expressions for i(t) and v(t). (18)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

41 F

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

1 F

Figure 1

2 H

i 1

vs = cos t

2 H

1 H 6

361 F

1

1 F

i

v

2 H 6

vs = 10u(t)

i(0) = 2

v(0) = 4

Figure 3

Figure 2(a) Figure 2(b)


1. a) vs –vc = dvc/dt + i and vc = 2di/dt vs – 2di/dt = 2d2i/dt2 + i d2i/dt2 + di/dt + i/2 = (1/2)cost [6] Let i = Acost + Bsint , then di/dt = -Asint + Bcost and d2i/dt2 = -Acost – Bsint

-Acost – Bsint –Asint +Bcost +(A/2)cost +(B/2)sint = (1/2)cost [B – A/2]cost + [-A-B/2]sint = (1/2)cost A = -1/5 [2] en B = 2/5 [2] i(t) = -(1/5)cost + (2/5)sint = 0.4472cos(t-2.0344) [2] (12) b) Zt = 1+(-j//j2) = 1-j2 = 2.236-63.435 [2] It = 10/2.236-63.435 = 0.447263.435 A [2] I = [-j/(-j+j2)]0.447263.435 = (1-90/190)0.447263.435 A = 0.4472-116.57 A [2] i(t) = 0.4472cos(t-2.0344) A [2] (8)

2. a) Z = 1+j-j(1/) = 1+j(-1/) [2] Vir resonansie: Im{Z} = 0 -1/ = 0 2 = 1 = 1 rad/sec [2] (4) b) 1/Z = 1/(1/j(1/36)) + 1/(6+j2) = j(1/36) + (6-j2)/(36+42) = 6/(36 + 42) +j[/36 - 2/(36+42)] [4] Vir resonansie moet Im{Z} = 0 /36 - 2/(36+42) = 0 1/36 = 2/(36+42) 36+42 = 72 42 = 36 2 = 9 = 3 rad/sek [4] (8) 3. 10/s – 6I(s) – 2sI(s) + 4 – (4/s)I(s) – 4/s = 0 [2] 10 – 6sI(s) – 2s2I(s) + 4s – 4I(s) – 4 = 0 I(s)[2s2 + 6s + 4] = 10 + 4s – 4 2I(s)[s2 + 3s + 2] = 4s + 6 I(s)(s + 2)(s + 1) = 2s + 3 I(s) = (2s + 3)/(s + 2)(s + 1) = 1/(s + 2) + 1/(s + 1) [4] i(t) = [e-2t + e-t]u(t) [2] V(s) = (4/s)I(s) + 4/s = (4/s)[(2s + 3)/(s + 2)(s + 1)] + 4/s = [4(2s+3) + 4(s + 2)(s + 1)]/[s(s + 1)(s + 2)] = (8s+12+4s2+12s+8)/s(s+1)(s+2) = (4s2+20s+20)/s(s+1)(s+2) = 10/s - 4/(s+1) – 2/(s+2) [4] v(t) = 10u(t) – 4e-tu(t) – 2e-2tu(t) [2]

36/j

vc

2 H 1 F

i 1

vs = cos t

j 1

1/j

j2

6

[20]

[12]

10/s

4/s

4

2s 6

4/s

I(s)

V(s)

[18]

[4]


Question 1 For the circuit shown in Figure 1, the switch S was closed for a long time. The switch is opened at time t=0. Find v(t) for all time t. (10)

Question 2 For the circuit shown in Figure 2, find an expression for i(t) for all t, if vs(t) is given by:

vs(t) =

0 for t1

0 for t1-

(10)

Question 3 The switch S, in the circuit of Figure 3, is closed at time t = 0. Determine i(t) and v(t) for all time t. (15) Question 4 The switch S in the circuit of Figure 4, is opened at time t = 0. Determine i(t) and v(t) for all time t. (15) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

101 F

v 10V t = 0 3

2

S Figure 1

– 1 t

vs(t)

0

1

1

1 vs 1 H

i Figure 2

41 F

21 H

t=0

1 i

v4 V 1

S

Figure 3

1 F

1 H

t=0

2 i

v1 V

Figure 4

S


1. t < 0: v(t) = 10V [2] t 0: 2i + 3i – v = 0 5i – v = 0 and i = –(1/10)dv/dt 5[–(1/10)dv/dt] – v = 0 dv/dt + 2v = 0 [5] v = Ae-2t [1] v(0) = A, but v(0) = 10 A = 10 [1] v(t) = 10e-2t for t 0 [1]

2. t < 0: -1 – i – 0 = 0 i = -1 A [1] t 0: 1 – v’ = v’ + i 1 = 2v’ + i en v’ = 1di/dt 1 = 2di/dt +i di/dt +(1/2)i = (1/2) [5] i = 1 + Ae-t/2 [1] Maar i(0)= -1 -1 = 1 + A A = -2 [1] i = 1 – 2e-t/2 [2]

3. t < 0: i(t) = 2A [1] en v(t) = 2 V [1] t 0: (1/2)di/dt + v = 0 en i = (1/4)dv/dt + v (1/2)d/dt[(1/4)dv/dt + v] + v = 0 (1/8)d2v/dt2 + (1/2)dv/dt + v = 0 d2v/dt2 + 4dv/dt + 8v = 0 [6] = 2 en n = 8 ondergedemp met d = 2 r/s v(t) = Ae-2tcos2t + Be-2tsin2t [3] v(0) = 2 A = 2 dv/dt = -2Ae-2tcos2t – 2Ae-2tsin2t –2Be-2tsin2t +2Be-2tcos2t dv(0)/dt = -2A+2B maar i(0)=(1/4)dv(0)/dt + v(0) 2 = (1/4)dv(0)/dt +2 dv(0)/dt = 0 -2(2) + 2B=0 B=2 v(t) = 2e-2tcos2t + 2e2tsin2t vir t>0 [2] en dv/dt = -4e-2tcos2t – 4e-2tsin2t –4e-2tsin2t +4e-2tcos2t = -8e-2tcos2t i(t) = (1/4)dv/dt + v = (1/4)[-8e-2tsin2t] + 2e-2tcos2t+2e-2tsin2t i(t) = 2e-2tcos2t vir t>0 [2]

4. t < 0: i(t) = 1/2A [1] en v(t) = 0 V [1] t 0: 2i + di/dt + v = 1 en i = dv/dt 2dv/dt + d2v/dt2 + v = 1 d2v/dt2 + 2dv/dt + v = 1 [6] = 1 en n = 1 krities gedemp v(t) = 1 + Ae-t + Bte-t [3] v(0) = 0 A = -1 dv/dt = – Ae-t + Be-t – Bte-t dv(0)/dt = -A+B maar i(0) = dv(0)/dt dv(0)/dt = ½ -A + B = ½ B = -½ v(t) = 1 – e-t – ½te-t [2] vir t>0 en i(t) = dv/dt = e-t – ½e-t + ½te-t = ½e-t + ½te-t = ½e-t(1+t) [2]

[10]

[10]

[15]

1 F

1 H 2 i

v 1 V

2 i

v1 V

t < 0

t 0

4

1 2

1

1 i

v 4 1

4

1 2

1

i

v

1

t < 0

t 0

[15]

101 v 10

3

2

t < 0

101 v

3

2

t 0 i

1

1 1 V 1 H

i v’

1 -1 V

i

0V

i t < 0 t 0

Circuit Analysis EICAM4 Unit 2 First Assessment 7 October 2010 Page 1


a) Find the steady state response i(t) in the time domain, solving the differential equation for i, if the supply current is, is given by is(t) = cost. Give your answer in the form, Acos(t + ).

b) Use frequency domain analysis to determine the phasor current I. Reconstruct the real form of i(t), Acos(t + ), again from the phasor voltage V. (6)

Question 2 Refer to the circuit in Figure 2. a) Calculate the admittance Y(),


frequency r of the circuit in Figure 2, from the perspective that the admittance must become real.

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = 5sin2t u(t) volt. The initial current through the inductor is i(0) = 0 amp. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (16) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

(12)

(12)

(4) Figure 2

Y

3

1 H

v

i 1

2 F 1 His(t) = cost

Figure 1

1 F

2

Figure 3

i 1

1 H vs = 5sin2t u(t)

i(0) = 0


1. a)) is = 2dv/dt + i en v = i + di/dt is = 2d/dt(di/dt + i) + i 2d2i/dt2 + 2di/dt + i = cost d2i/dt2 + di/dt + ½i = ½cost [6] i(t) = Acost + Bsint -Acost–Bsint–Asint+Bcost+½Acost+½Bsint=½cost (-½A+B)cost + (-A- ½B)sint = ½cost -½A+B=½ en -A- ½B = 0 A=-1/5 [2] en B = 2/5 [2] i(t) = -0.2cost + 0.4sint = 0.447214cos(t-2.0344) A [2] (12) b) I = [0.5-1.5708r/(0.5-1.5708+1+11.5708)]10 = 0.44721-2.0344r A [4] i(t) = 0.44721cos(t – 2.0344) [2] (6) 2. a) Y1 = 1/(3 + j) = (3 – j)/(9 + 2) [3] Y2 = 1/(2 + 1/j) = j/(1 + j2) = j(1 – j2)/(1 + 42) = (22 + j)/(1 + 42) [3] Y = Y1 + Y2 = (3 – j)/(9 + 2) + (22 + j)/(1 + 42) = [(3 – j)(1 + 42) +(22 + j)(9 + 2)]/[(9 + 2)(1 + 42)] = [(3 – j + 122 – j43] + (182 + j9 + 24 + j3)]/[(9 + 2)(1 + 42)] = [(3 + 302 + 24) + j(– + 9 – 43 + 3)]/[(9 + 2)(1 + 42)] = [(3 + 302 + 24) + j(8 – 33)]/[(9 + 2)(1 + 42)] [6] (12) b) Im[Y] = 0 8 – 33 = 0 (8 – 32) = 0 = 0 (The circuit does not resonate although Y is real (1/3 S) at = 0 as L = SC & C = OC) or 8 – 32 = 0 2 = 8/3 = 1.633 r/s (4) 3. 10/(s2 + 4) = I(s) + sI(s) [2] (s + 1)I(s) = 10/(s2 + 4) I(s) = 10/(s2 + 4)(s + 1) [4] I(s) = 10/(s + 1)(s – 21.571)(s – 2-1.571) I(s) = 2/(s + 1) + 1.118-2.678/(s – 21.571) + 1.1182.678/(s – 2-1.571) = 2/(s + 1) + 1.118-2.678/(s + 0 – j2) + 1.1182.678/(s + 0 + j2) [4] i(t) = 2e –t u(t) + 2×1.118×e – 0t×cos(2t – 2.678) u(t) = 2e –t u(t) + 2.236cos(2t – 2.678) u(t) [2]

[18]

Model 4 marks

2

Y

1/jj

[16]

v

i 1

2 F 1 H cost

3

[16]

I(s) 1

s 42s

10


Question 1 For the circuit in Figure 1, the switch S, is opened at time t = 0. Calculate i(t) for all time t. (8)

Question 2 For the circuit shown in Figure 2, find an expression for i(t) for all time t,

if the supply voltage, vs(t), is given by:

0 for t1

0 for t1- (t)vs (12)

Question 3 For the circuit shown in Figure 3, the switch S, was open for a long time. The switch is closed at time t = 0. Find i(t), for all time t. (15) Question 4 For the circuit shown in Figure 4, the switch S was closed for a long time. The switch is opened at time t = 0. Find v(t), for all time t. (15)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

1

2 vs

-1

t

vs(t)

0

1

2 H

i

Figure 2

Figure 4

85

H 51

F 51

t = 0 i

5 V

Figure 3

v

S

i

2

5H

9V

t=0

3

S

21 F

t=0 2

i

v 21 H

6 VS


1. t < 0: i(t) = 9/3 = 3A [1] t 0: 5di/dt + 5i = 0 di/dt + i = 0 [4] i = Ae-t [1] i(0) = 3 3 = A [1] i(t) = 3e-t [1] 2. t < 0: i(t) = -1 A [2] t 0: 1 – v’ = (v’/2) + i en v’ = 2di/dt 1 – 2di/dt = di/dt + i 3di/dt + i = 1 di/dt + i/3 = 1/3 [5] i = Ae-t/3 + (1/3)/(1/3) i = Ae-t/3 + 1 [2] en i(0) = -1 -1 = A + 1 A = -2 [1] i(t) = -2e-t/3 + 1 [2] 3. t < 0: i(t)=0 [1] en v(t)=5 [1] t 0: (5-v)/(5/8) = (1/5)dv/dt +i en v = (1/5)di/dt (1/25)d2i/dt2 + (8/25)di/dt + i = 8 d2i/dt2 +8di/dt + 25i = 200 [6] = 4 en n = 5 Dus ondergedemp met d = 3 r/s i(t) = Ae-4tcos3t + Be-4tsin3t + 8 [4] i(0) = 0 A + 8 = 0 A = -8 en (1/5)di(0)/dt = v(0) di(0)/dt = 25 di/dt = –4Ae-4tcos3t – 3Ae-4tsin3t – 4Be-4tsin3t + 3Be-4tcos3t di(0)/dt = -4A + 3B 25 = -4(-8) + 3B B = -2.333 i(t) = -8e-4tcos3t – 2.33e-4tsin3t + 8 [3] 4. Vir t < 0: i(t) = 3 A [1] en v(t) = 0 V [1] Vir t 0: 2i + ½di/dt + v = 6 en i = ½dv/dt dv/dt + ¼d2v/dt2 + v = 6 d2v/dt2 + 4dv/dt + 4v = 24 [6] = 2 en n = 2 krities gedemp v(t) = 6 + Ae-2t + Bte-2t [4] v(0) = 0 A = –6 dv/dt = –2Ae-2t + Be-2t – 2Bte-2t dv(0)/dt = –2A+B maar i(0) = ½dv(0)/dt dv(0)/dt = 6 –2A + B = 6 B = 6 + 2A B = 6 + 2(–6) = –6 v(t) = 6 – 6e-2t – 6te-2t vir t > 0 [3]

i

2 9V 3

i

2

5H

3

t < 0

t 0

i 1

2 1 V

i 1

2

1 V 2 H

v’

t < 0

t 0

[8]

[12]

8

5 i

5 V

8

5

H 5

1 F

5

1

i

5 V

v

t < 0

t 0

v

[15]

[15] ½ F

½ H

2 i

v 6 V

2 i

v 6 V

t < 0

t 0



solution for the steady state voltage response, v(t), by writing down and solving a differential equation for v(t). Give your answer in the form Acos(t+).

b) Use frequency domain analysis to determine the phasor voltage V. Reconstruct v(t) again from the phasor voltage V. (6)

Question 2 Refer to the circuit in Figure 2. a) Calculate the impedance Z(),


frequency r of the circuit in Figure 2, from the perspective that the impedance must become real.

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = 5sin2t u(t) volt. The initial current through the inductor is i(0) = 0 amp and the initial voltage across the capacitor is v(0) = 0 volt. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for v(t). (16)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

2 H 1

vs = cos t

(12)

(4) Figure 2

Z 1 H

1 F

21

2 v

21 F

(12)

Figure 3

i 1

1 H

vs = 5sin2t u(t)

i(0) = 0

v

v(0) = 0

1 F


1. a) vs –v = i+2di/dt and i = v/2 + (1/2)dv/dt vs – v = v/2 + (1/2)dv/dt + dv/dt + d2v/dt2 d2v/dt2 + (3/2)dv/dt + (3/2)v = cost [6] v = Acost + Bsint , dv/dt = -Asint + Bcost and d2v/dt2 = -Acost – Bsint -Acost – Bsint – (3/2)Asint + (3/2)Bcost + (3/2)Acost + (3/2)Bsint = cost [0.5A + 1.5B]cost + [0.5B – 1.5A]sint = cost A + 3B = 2 en B – 3A = 0 A = 0.2 en B = 0.6 v(t) = 0.2cost + 0.6sint = 0.6325cos(t – 1.249) volt [6] (12) b) V = (ZV/ZT)10 ZV = 2//[1/j(1/2)] = 2//–j2 = –j4/(2 – j2) = –j2/(1 – j) ZV/ZT = [–j2/(1 – j)]/[–j2/(1 – j) + 1 + j2] = –j2/[–j2 + (1 + j2)(1 – j)] = –j2/(–j2 + 1 + j + 2) = –j2/(3 – j) V = [–j2/(3 – j)]10 = 0.6325–1.249 r volt [4] v(t) = 0.6325cos(t – 1.249) [2] (6)

2. a) Z = j//(½ + 1/j) [2] = [j(½ + 1/j)]/(j + ½ + 1/j) [4] = (j½ + 1)/(j + ½ + 1/j) = (–½2 + j)/[(1 – 2) + j½] = {(–½2 + j)[(1 – 2) – j½]}/[(1 – 2)2 + ¼2] [2] = {[(–½2 )(1 – 2) + ½2] + j[(1 – 2) + ¼3]}/[(1 – 2)2 + ¼2] = [(–½2 + ½4 + ½2) + j( – 3 + ¼3)]/(1 – 22 + 4 + ¼2) [4] = [½4/(1 – 1¾2 + 4)] + j[( – ¾3)/(1 – 1¾2 + 4)] (12) b) Im{Z} = 0 – ¾3 = 0 ¾2 = 1 2 = (4/3) = 1.155 r/s (4) 3. [10/(s2 + 4)] – V(s) = I(s) + sV(s) [1] .... (1) and I(s) = V(s)/s [1] ............................... (2) (2) in (1): 10/(s2 + 4) – V(s) = V(s)/s + sV(s) s2V(s) +sV(s) + V(s) = 10s/(s2 + 4) (s2 + s + 1)V(s) = 10s/(s2 + 4) V(s) = 10s/[(s2 + 4)(s2 + s + 1) [4] V(s) = 10s/[(s – 21.571)(s – 2–1.571)(s – 12.094)( s – 1–2.094)] V(s) = 1.387–2.554/(s – 21.571) + 1.3872.554/(s – 2–1.571) + 1.6020.7655/(s – 12.094) + 1.602–0.7655/(s – 1–2.094) [4] = 1.387–2.554/[s – (0 + j2)] + 1.3872.554/[s – (0 – j2)] + 1.6020.7655/[s – (–0.5 + j0.8662)] + 1.602–0.7655/[s – (–0.5 – j0.8662)] = 1.387–2.554/(s – 0 – j2) + 1.3872.554/(s – 0 + j2) + 1.6020.7655/(s + 0.5 – j0.8662) + 1.602–0.7655/(s + 0.5 + j0.8662) v(t) = 2×1.387×e – 0t×cos(2t – 2.554) u(t) + 21.602e–0.5tcos(0.8662t + 0.7655)u(t) = 2.774cos(2t – 2.554) u(t) + 3.204e–0.5tcos(0.8662t + 0.7655)u(t) [2]

½Z

1/jj

[16]

2 H 1

vs = cos t 2

v

2

1 F

i

[18]

[16]

Model 4 marks

I(s) 1

s 42s

10

1/s

V(s)

Zv

Circuit Analysis EICAM4 Unit 1 First Assessment 1 September 2011 Page 1Question 1 For the circuit shown in Figure 1, the switch S was closed for a long time. The switch is opened at time t = 0. Find v(t) for all time t. (10) Question 2 Determine the zero state step response i(t) for the circuit in Figure 2. (10) Question 3 For the circuit shown in Figure 3, determine the response v(t) for all t, if vs(t) is given by:

0 for t1

0 for t 2 (t)vs .

(15) Question 4 Refer to the active circuit in Figure 4 and determine the zero state step response, v(t). (15)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 21

vs(t) = u(t) 1 H 1

1 i

1 F v 10V

t = 0 1

1

S Figure 1

1

Figure 4

1 F

1

v vs=u(t)

1

1 F

t

1 H 1

–1

v 1 F

vs

i

1

vs(t)

0

2

Figure 3


1. t < 0: v(t) = 5V [2] t 0: i + i – v = 0 2i – v = 0 and i = –dv/dt 2[–dv/dt] – v = 0 dv/dt + ½v = 0 [5] v = Ae–½t [1] v(0) = A, but v(0) = 5 A = 5 [1] v(t) = 5e–½t for t 0 [1] 2. t < 0: i(t) = 0 [1] (zero toestand respons) t 0: 1 – v’=i + v’ en v’ = i + di/dt 1 – (i + di/dt) = i + (i + di/dt) 2di/dt + 3i = 1 di/dt + 3/2i = ½ [6] i(t) = 1/3 + Ae–3/2t [1] Maar i(0) = 0 A = –1/3 [1] i(t) = 1/3(1 – e–3/2t) vir t 0 [1] 3. t < 0: v(t) = 1 V [1] en i(t) = 1 A [1] t 0: i = v + dv/dt en vs – i – di/dt – v=0 –1 – (v + dv/dt) – d/dt(v + dv/dt) – v = 0 –1 – 2v – 2dv/dt – d2v/dt2 = 0 d2v/dt2 + 2dv/dt + 2v = –1 [6] = 1 en n = 2 ondergedemp met d = (2 – 1) = 1 v(t) = –½ + Ae–tcost + Be–tsint [3] v(0)=1 1= –½ +A A = 1.5 en i(0) = v(0) + dv(0)/dt dv(0)/dt = 0 dv/dt=–Ae–tcost–Ae–tsint–Be–tsint+Be–tcost 0 = –A + B B = A = 1.5 v(t) = -0.5 + 1.5e–tcost + 1.5e–tsint [4] 4. t < 0: v = 0 [½] en v1 = 0 [½] (zero state, given) t 0: (1 – v1)/1=(v1 – v)/1 + d/dt(v1 – v) 1 – v1 = v1 – v + dv1/dt – dv/dt 1 = 2v1 + dv1/dt – dv/dt – v ….…. (1) en (v1 – v)/1 = dv/dt v1 = dv/dt + v …..…….........…… (2) (2) in (1): 1 = 2(dv/dt + v) + d/dt(dv/dt + v) – dv/dt – v 1 = 2dv/dt + 2v + d2v/dt2 + dv/dt – dv/dt – v d2v/dt2 + 2dv/dt + v = 1 [6] = 1 en n = 1 krities gedemp v(t) = Ae–t + Bte–t + 1 [2] v(0) = 0 A + 1 = 0 A = –1 dv/dt = –Ae–t + Be–t – Bte–t Van (2): v1(0) = dv(0)/dt + v(0) dv(0)/dt = 0 –A + B = 0 B = –1 v(t) = –e–t – te–t + 1 [6]

[10]

1 v 10

1

1

t < 0 1

1 v

1

1

t 0 i

v’

1

i

1 H

1

1

[10]

i

1

1

v 1 F

-1

1 H

i

1

1

v 2

t < 0

t 0

[15]

v v1

1

1

1

1

1

v

[15]



solution for the steady state voltage response, v(t), by writing down and solving a differential equation for v(t). Give your answer in the form Acos(t+). (12)


Question 2 Refer to the circuit in Figure 2. a) Determine the resonance

frequency of the circuit from the perspective that the impedance seen by the source, must become real. (12)

b) Calculate the impedance of the circuit at resonance. (2)

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = 10sin2t u(t) volt. The initial current through the inductor is i(0) = 1 amp. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (18)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 3

i 1

1 H vs = 10sin2t u(t)

i(0) = 1

Figure 1

2 H 1

vs = cos t 1

v 2 F

1 F

2

1

1 H

Figure 2


1. a) vs – i – 2di/dt – v = 0 and i = v + 2dv/dt vs – (v + 2dv/dt) – 2d/dt(v + 2dv/dt) – v = 0 vs – v – 2dv/dt – 2dv/dt – 4d2v/dt2 – v = 0 4d2v/dt2 + 4dv/dt + 2v = vs d2v/dt2 + dv/dt + 0.5v = 0.25cost [6] v = Acost + Bsint, dv/dt = –Asint + Bcost and d2v/dt2 = –Acost – Bsint –Acost – Bsint – Asint + Bcost + 0.5Acost + 0.5Bsint = 0.25cost [–0.5A + B]cost + [–A – 0.5B]sint = 0.25cost –0.5A + B = 0.25 en –A – 0.5B = 0 A = –0.1 en B = 0.2 v(t) = –0.1cost + 0.2sint = 0.2236cos(t – 2.0344) volt [6] (12) b) ZV = 1//(1/j2) = (1/j2)/[1 + (1/j2)] = 1/(1 + j2) = 0.4472–1.1072 ZT = 1 + j2 + Zv = 1 + j2 + 0.4472–1.1072 20.92731 ZV/ZT = 0.4472–1.1072/20.92731 = 0.2236–2.0345 V = (ZV/ZT)10 = 0.2236–2.034510 = 0.2236–2.0345 volt [4] v(t) = 0.2236cos(t – 2.0345) [2] (6)

2. a) Z = 1 + 1/j + 2//(j) [2] = 1 – j(1/) + (j2)/(2 + j) = 1 – j(1/) + j2(2 – j)/(4 + 2) = 1 – j(1/) + (j4 + 22)/(4 + 2) = [1 + 22/(4 + 2)] + j[4/(4 + 2) – 1/] Vir resonansie moet Im{Z} = 0 4/(4 + 2) – 1/ = 0 [2] 4/(4 + 2) = 1/ 42 = 4 + 2 32 = 4 r = 2/3 = 1.154701 r/s [6] (10) b) Zr = 1 + 2r

2/(4 + r2) = 1 + 8/3/(4 + 4/3) = 1.5 (4)

3. 20/(s2 + 4) – I(s) – sI(s) + 1 = 0 [2] (s + 1)I(s) = 20/(s2 + 4) + 1 (s + 1)I(s) = (20 + s2 + 4)/(s2 + 4) I(s) = (s2 + 24)/(s2 + 4)(s + 1) [2] I(s) = (s2 + 24)/(s + 1)(s – 21.571)(s – 2–1.571) I(s) = 5/(s + 1) + 1.118–2.678/(s – 21.571) + 1.1182.678/(s – 2–1.571) = 5/(s + 1) + 2.2362–2.678/(s + 0 – j2) + 2.23622.678/(s + 0 + j2) [4] i(t) = 5e –t u(t) + 2×2.2362×e – 0t×cos(2t – 2.678) u(t) = 5e –t u(t) + 4.472cos(2t – 2.678) u(t) [4]

[18]

[18]

2 H 1

vs = cos t 1

v

2 F

Zv

i

1/j

2

1

j

[14]

I(s) [1] 1 [1]

s [1]

42s

20

[2]

1 [1]


Question 1 For the circuit shown in Figure 1, the switch S was closed for a long time. The switch is opened at time t = 0. Find i(t) for all time t.

Question 2 For the circuit shown in Figure 2, find an expression for v(t) for all time t,

if the supply voltage, vs(t), is given by:

0 for t1

0 for t1- (t)vs

Question 3 For the circuit shown in Figure 3, the switch S was closed for a long time. The switch is opened at time t = 0. Find v(t), for all time t.

Question 4 Refer to the active circuit in Figure 4 and determine the zero state step response, v(t). (Hint: for zero state, v(0) = 0 and and v1(0) = 0) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

1 H 3 V

t = 0 1

1

S

Figure 1

1 i

1

1 vs

-1

t

vs(t)

0

1

1 F v

Figure 2

(16)

(14)

(10)

Figure 3

251 F

t=0 6

i

v

1 H

6 VS

Figure 4

1 F v vs=u(t)

1

1 F 21

v1

(10)

Stroombaananalise IV EICAM4 Eenheid 1 Finale Evaluasie 27 Oktober 2011 Memorandum Bladsy 1

1. t < 0: i(t) = 1A [2] t 0: – i – i – v = 0 – 2i – v = 0 and v = di/dt –2i – di/dt = 0 di/dt + 2i = 0 [5] i = Ae–2t [1] i(0) = A, but i(0) = 1 A = 1 [1] i(t) = e–2t for t 0 [1]

2. t < 0: v(t) = –½ V [2] t 0: 1 – v = v + dv/dt dv/dt + 2v = 1 [5] v = Ae-2t + ½ [1] v(0) = A + ½ and v(0) = –½ A + ½ = –½ A = –1 [1] v(t) = –e-2t + ½ [1]

3. t < 0: i(t) = 6/6 = 1A [1] en v(t) = 0 [1] t 0: 6 = 6i + di/dt + v en i =(1/25)dv/dt (6/25)dv/dt +(1/25)dv2/dt2 + v = 6 d2v/dt2 + 6dv/dt + 25v = 150 [6] = 3 en n = 5 ondergedempte stelsel met d = 4 r/s v(t) = 6 + Ae-3tcos4t + Be-3tsin4t [4] v(0) = 0 0 = 6 + A A = –6 dv/dt =–3Ae-3tcos4t – 4Ae-3tsin4t –3Be-3tsin4t + 4Be-3tcos4t = (4B-3A)e-3tcos4t – (4A+3B)e-3tsin4t dv(0)/dt = –3A + 4B = 18 + 4B maar i(0) = (1/25)dv(0)/dt dv(0)/dt = 25 18 + 4B = 25 4B = 7 B = 1.75 v(t) = 6 – 6e-3tcos4t + 1.75e-3tsin4t [4]

4. t < 0: v(t)=0 and v1(t)=0 (zero state given) t 0: (1–v1)/(1/2) = (v1–v)/1 + 1d/dt(v1–v) 2 – 2v1 = v1 – v + dv1/dt – dv/dt 2 = 3v1 + dv1/dt – v – dv/dt .................(1) and (v1 – v) = dv/dt v1 = v + dv/dt ......(2) (2) in (1): 2 = 3(v + dv/dt) + d/dt(v + dv/dt) – v – dv/dt d2v/dt2 + 3dv/dt + 2v = 2 [6] = 3/2 en n = 2 stelsel oorgedemp met s1,2 = [–3 (9–8)]/2 = [–3 1]/2 s1 = –2 en s2 = –1 v(t) = 1 + Ae-2t + Be-t [4] v(0) = 0 1 + A + B = 0 ...............................(i) dv/dt = –2Ae-2t – Be-t dv(0)/dt = –2A – B Maar van (2): v1(0) = v(0) +dv(0)/dt dv(0)/dt = 0 –2A – B = 0 .................................................(ii) Van (i) en (ii): A = 1 en B = –2 v(t) = 1 + e-2t – 2e-t [4]

[10]

1

i

3

1

1

t < 0 1

1

i 1

1

t 0

v

[10]

v

1

1 1 V

1

1

1 V

1 F

t < 0 t 0

v

25

1 F

6 i

v

1 H

6 V

6 i

v 6 V

[16]

t < 0

t 0

v v1

1F

½

1

1

1F

v

[14]


Question 1 The supply voltage to the circuit in Figure 1, is given by vs(t) = cos 2t. a) Determine the time domain




frequency of the circuit in Figure 2, from the perspective that the impedance must become real.

b) Calculate the impedance of the circuit at resonance

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = 10sin2t u(t) volt. The initial current through the inductor is i(0) = 1 amp and the initial voltage across the capacitor is v(0) = 0 volt . Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (18)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

1 H

v

Figure 1

vs = cos 2t v 1 F

21 H

21

(12)

(3) Figure 2

Z 1 H

1 F 21

Figure 3

i 1

vs = 10sin2t u(t)

i(0) = 1

v(0) = 01 F


1. a) vs – ½i – ½di/dt – v = 0 and i = dv/dt vs – ½dv/dt – ½d2v/dt2 – v = 0 ½d2v/dt2 + ½dv/dt + v = vs d2v/dt2 + dv/dt + 2v = 2cos2t [6] v = Acos2t + Bsin2t dv/dt = –2Asin2t + 2Bcos2t and d2v/dt2 = –4Acos2t – 4Bsin2t –4Acos2t – 4Bsin2t – 2Asin2t + 2Bcos2t + 2Acos2t + 2Bsin2t = 2cos2t [–2A + 2B]cos2t + [–2A – 2B]sin2t = 2cos2t –2A + 2B = 2 and –2A – 2B = 0 A = –0.5 en B = 0.5 v(t) = –0.5cos2t + 0.5sin2t = 0.7071cos(2t – 2.356) volt [6] (12) b) Z = ½ + 190˚ + 0.5–90˚ = 0.707145˚ I = 10/0.707145˚ = 1.4142–45˚ A V = 0.5-901.4142-45˚ = 0.7071-135˚ V v(t) = 0.7071cos(2t – 2.356) volt (5)

2. a) Z = (1/j)//(½ + j) [2] = [(1/j)(½ + j)]/(1/j + ½ + j) [2] = (½ + j)/[(1 – 2) + j½] = {(½ + j)[(1 – 2) – j½]}/[(1 – 2)2 + ¼2] = {[½(1 – 2) + j(1 – 2) – ¼j + ½2]}/[(1 – 2)2 + ¼2] = (½ – ½2 + j – j3 – ¼j + ½2)/(1 – 22 + 4 + ¼2) = [½ + j(¾ – 3)]/(1 – 1¾2 + 4) [6] Im{Z} = 0 ¾ – 3 = 0 2 = ¾ = 0.866 r/s [2] (12) b) Z = ½/(1 – 1¾2 + 4) = 2 (3) 3. 20/(s2 + 4) – I(s) – sI(s) + 1 – V(s) = 0 [2] and V(s) = I(s)/s [1] 20/(s2 + 4) – I(s) – sI(s) + 1 – I(s)/s = 0 s2I(s) + sI(s) + I(s) = 20s/(s2 + 4) + s (s2 + s + 1)I(s) = [20s + s(s2 + 4)]/(s2 + 4) I(s) = (s3 + 24s)/(s2 + s + 1)(s2 + 4) [4] I(s) = (s3 + 24s)/(s – 12.0944)(s – 1–2.0944)(s – 21.571)(s – 2–1.571) I(s) = 3.76560.72933/(s – 12.0944) + 3.7656–0.72933/(s – 1–2.0944) + 2.774–2.5538/(s – 21.571) + 2.7742.5538/(s – 2–1.571) [4] = 3.76560.72933/(s + 0.5 – 0.866) + 3.7656–0.72933/(s + 0.5 + j0.866) + 2.774–2.5538/(s + 0 – j2) + 2.7742.5538/(s + 0 – j2) i(t) = 23.7656e –0.5t cos(0.866t + 0.72933)u(t) + 22.774cos(2t – 2.5538)u(t) = 7.531e –0.5t cos(0.866t + 0.72933)u(t) + 5.548cos(2t – 2.5538)u(t) [2]

[17]

½ H½

vs = cos 2t v

1 F

i

190˚ ½

10 V0.5–90˚

I

½Z

1/j j

[15]

[18]

I(s) 1 [1] s [1]

42s

20

[1] V(s)

1 [1]

1/s [1]


Question 1 For the circuit shown in Figure 1, the switch S was closed for a long time. The switch is opened at time t = 0. Find v(t) for all time t. (10) Question 2 Determine the zero state step response i(t) for the circuit in Figure 2. (10)

Question 3 For the circuit shown in Figure 3, determine the current response i(t) in the 1 H inductor, for all t, if vs(t) is given by:

0 for t2

0 for t 1 (t)vs .

Question 4 Refer to the active circuit in Figure 4 and determine the zero state step response, v(t). (15)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

1F 1 v 1 A t = 0

S

1

Figure 1

1 H 1 i

1 vs(t)=u(t)

Figure 2

2i

Figure 4

1

v vs=u(t)

1

1 F

21 F

1 H

1

v 1 F

vs i

1

t

–2

vs(t)

0

1Figure 3

(15)


1. t < 0: v(t) = 1½ = ½V [2] t 0: v + i = 0 and i = dv/dt v + dv/dt = 0 dv/dt + v = 0 [5] v = Ae–t [1] v(0) = A But v(0) = ½ A = ½ [1] v(t) = ½e–t for t 0 [1]

2. t < 0: i(t) = 0 [1] (zero toestand respons) t 0: 1 – v’ = 2i + i and v’ = i + di/dt 1 – (i + di/dt) = 3i di/dt + 4i = 1 [5] i(t) = 1/4 + Ae–4t [1] But i(0) = 0 A = –1/4 [1] i(t) = 1/4(1 – e–4t) for t 0 [2]

3. t < 0: v(t) = 1 V [1] en i(t) = 1 A [1] t 0: Loop containing 1H and 1F: –2 – di/dt – v = 0 v = –2 – di/dt ................................ (1) node N: i + di/dt = dv/dt + v .......................... (2) (1) in (2): i + di/dt = d/dt(–2 – di/dt) + (–2 – di/dt) i + di/dt = –d2/dt2 – 2 – di/dt) d2i/dt2 + 2di/dt + i = –2 [6] = 1 en n = 1 krities gedemp: i(t) = –2 + Ae–t + Bte–t [3] i(0) = –2 + A But i(0) = 1 1= –2 +A A = 3 and di/dt = –Ae–t – Bte–t + Be–t di(0)/dt = –A + B = –3 + B. But from (1): v(0) = –2 – di(0)/dt 1 = –2 – di(0)/dt di(0)/dt = –3 –3 + B = –3 B = 0 i(t) = –2 + 3e–t [4]

4. t < 0: v = 0 [½] en v1 = 0 [½] (zero state, given) t 0: (1 – v1)/1=(v1 – v)/1 + d/dt(v1 – v) 1 – v1 = v1 – v + dv1/dt – dv/dt 1 = 2v1 + dv1/dt – dv/dt – v …...…. (1) en (v1 – v)/1 = ½dv/dt v1 = ½dv/dt + v …..…….........…… (2) (2) in (1): 1 = 2(½dv/dt + v) + d/dt(½dv/dt + v) – dv/dt – v 1 = dv/dt + 2v + ½d2v/dt2 + dv/dt – dv/dt – v ½d2v/dt2 + dv/dt + v = 1 d2v/dt2 + 2dv/dt + 2v = 2 [6] = 1 en n = 2 ondergedemp met d = 1: v(t) = 1 + Ae–tcost + Be–tsint [2] v(0) = 0 1 + A = 0 A = –1 dv/dt = –Ae–tcost –Ae–tsint – Be–tsint + Be–tcost dv(0)/dt = –A + B From (2): v1(0) = ½dv(0)/dt + v(0) dv(0)/dt = 0 –A + B = 0 B = A = –1 v(t) = 1 – e–tcost – e–tsint [6] [15]

1 v

1 1 1 v 1 A

1

t < 0 t 0

[10]

i

i

[10]

[15]

1 1 i

2i 1 1 v’

t < 0

v i

1 1V

t 0

1 H

1

v 1 F

i 1

–2

di/dt N

v v1

½

1

1

1

1

v

Circuit Analysis IV EICAM4A Unit 2 First Assessment 27 September 2012 Page 1


solution for the steady state voltage response, i(t), by writing down and solving a differential equation for i(t). Give your answer in the form Acos(t+). (12)





Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = sin2t u(t) volt. The initial current through the inductor is i(0) = 1 amp and the initial voltage across the capacitor is v(0) = 1 volt . Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (18)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

5

5 1 H

Figure 2

1 H

1

vs = cos 2t

1

1 F

i

1 F

1 H

v

Figure 3

i 1

vs = sin2t u(t)

i(0) = 1

v(0) = 11 F

Stroombaananalise IV EICAM4A Eenheid2 Eerste Evaluasie 27 September 2012 Memorandum Bladsy 1

1. a) Node N: vs – v = dv/dt + i ........... (1) and loop L: v – i – di/dt = 0 v = i + di/dt .............................. (2) (2) in (1): vs – (i + di/dt ) = d/dt(i + di/dt ) + i vs – i – di/dt = di/dt + d2i/dt2 + i d2i/dt2 + 2di/dt + 2i = vs d2i/dt2 + 2di/dt + 2i = cos2t [6]

i = Acos2t + Bsin2t, di/dt = –2Asin2t + 2Bcos2t, and d2v/dt2 = –4Acos2t – 4Bsin2t –4Acos2t – 4Bsin2t – 4Asin2t + 4Bcos2t + 2Acost + 2Bsint = cos2t [–2A + 4B]cos2t + [–4A – 2B]sin2t = cos2t –2A + 4B = 1 and –4A – 2B = 0 A = –0.1 en B = 0.2 i(t) = –0.1cos2t + 0.2sin2t = 0.2236cos(2t – 2.0344) amp [6] (12) b) Z1 = 0.5–90˚//(1+290˚) = [0.5–90˚(1+290˚)]/(0.5–90˚+1+290˚) = (0.5–90˚+10˚)/1.8027856.31˚ = 1.11803–26.565˚/1.8027856.31˚ Z1 = 0.62017–82.875˚ ZT = 1+ 0.62017–82.875˚=1.24035–29.7338˚ IT = 10/1.24035–29.7338˚ =0.8062229.7338˚ I = [0.5–90˚/(0.5–90˚ +1+290˚)]0.8062229.7338˚ = 0.27735–146.31˚0.8062229.7338˚ I = 0.22361–116.576˚ [4] = 0.22361–2.035r i(t) = 0.2236cos(2t–2.035) ampere [2] (6) 2. a) Z = 5 + j + 5//(1/j) [2] = 5 + j + {(5/j)/[5 + (1/j)]} = 5 + j + [5/(1 + j5)] = 5 + j + [5(1 – j5)/(1 + 252)] = [5 + 5/(1 + 252)] + j{1 – [25/(1 + 252)]} For resonance: Im{Z} = 0 1 – [25/(1 + 252)] = 0 [2] 1 = 25/(1 + 252) 1 + 252 = 25 252 = 24 2 = 24/25 r = (24)/5 = 0.979796 r/s [6] (10) b) Zr = 5 + 5/(1 + 25r

2) = 5 + 5/[1 + 25(24/25)] = 5 + 5/(1 + 24) = 5 + 1/5 = 5.2 (4) 3. 2/(s2 + 4) – I(s) – sI(s) + 1 – I(s)/s – 1/s = 0 [3] 2s/(s2 + 4) – sI(s) – s2I(s) + s – I(s) – 1 = 0 s2I(s) + sI(s) + I(s) = 2s/(s2 + 4) + s – 1 (s2 + s + 1)I(s) = [2s + s(s2+4) – (s2+4)]/(s2+4) (s2 + s + 1)I(s) = (2s + s3 + 4s – s2 – 4)/(s2 + 4) (s2 + s + 1)I(s) = (s3 – s2 + 6s – 4)/(s2 + 4) I(s) = (s3 – s2 + 6s – 4)/[(s2 + s + 1)(s2 + 4)] [4] I(s) = (s3 – s2 + 6s – 4)/[(s – 12.0944)(s – 1–2.0944)(s – 21.571)(s – 2–1.571)] I(s) = 1.3109150.979219/(s – 12.0944) + 1.310915–0.979219/(s – 1–2.0944) + 0.27739–2.5538/(s – 21.571) + 0.277392.5538/(s – 2–1.571) [4] = 1.3109150.979219/(s + 0.5 – j0.866) + 1.310915–0.979219/(s + 0.5 + j0.866) + 0.27739–2.5538/(s + 0 – j2) + 0.277392.5538/(s + 0 + j2) i(t) = 21.310915e –0.5t cos(0.866t + 0.979219)u(t) + 20.27739cos(2t – 2.5538)u(t) = 2.6218e –0.5t cos(0.866t + 0.979219)u(t) + 0.55478cos(2t – 2.5538)u(t) [2] {or:I(s) = [(As+B)/(s2+s+1)]+[(Cs+D)/(s2+4)] = [(A+C)s3+(B+C+D)s2+(4A+C+D)s+(4B+D)]/(s2+s+1)(s2+4) A+C=1, B+C+D=–1, 4A+C+D=6 & 4B+D=–4 I(s)={[(19/13)s–(15/13)]/(s2+s+1)}+{[–(6/13)s+(8/13)]/(s2+4)} I(s)={(19/13)(s+½)/[(s+½)2+(3/2)2]}–{[(49/133)[(3/2)/((s+½)2+(3/2)2)]}+{–(6/13)[s/(s2+22)]}+{(4/13)[2/(s2+22]}

i(t) = (19/13)e–½tcos(3/2)t – (49/133)e–½tsin(3/2)t – (6/13)cos2t + (4/13)sin2t i(t) = e–½t[1.4615cos0.866t–2.1762sin0.866t]+[–0.4615cos2t+0.30769sin2t] i(t) = 2.6214e–½tcos(0.866 + 0.9794) + 0.5547cos(2t – 2.554) }

[18]

[14]

290˚

1

10

1

V

0.5–90˚

I

Z1

IT

[18]

1 H

1

vs = cos 2t

1

v

1 F

i dv/dt

di/dt

(vs-v)/1 N

L

i

5

5 j

I(s) 1 s [1]

42s

2

[1]

V(s)

1 [1]

1/s [1]

1/s [1]

I(s) sI(s) 1

1/s [1] s

I(s)

s

1

1/j


Question 1 For the circuit shown in Figure 1, the switch S was closed for a long time. The switch is opened at time t = 0. Find v(t) for all time t. (10)

Question 2 For the circuit shown in Figure 2, determine the zero state step response v(t). (10) Question 3 For the circuit shown in Figure 3, determine the current response i(t), in the 1 H inductor, for all time t, if vs(t) is given by:

0 for t2

0 for t 2 (t)vs

Question 4 For the circuit shown in Figure 4, find the zero state step response v(t), if the input to the circuit is the step current source, is(t) = u(t) ampere.


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

(15)

Figure 1

3 A t = 0 1

1

S

v1 F 1

(15)

1 H

1

i

1 F vs

1

t

–2

vs(t)

0

2

Figure 3

1 1

vs = u(t)

1 F

v

Figure 2

1

Figure 4

v 1 F 1 H

is = u(t) 1 1


1. t < 0: v(t) = 1V [2] t 0: v – i = 0 and i = – dv/dt v – (–dv/dt) = 0 dv/dt + v = 0 [5] v = Ae–t [1] v(0) = A, but v(0) = 1 A = 1 [1] i(t) = e–t for t 0 [1]

2. t < 0: v(t) = 0 [1] zero toestand t 0: (1 – v’)/1 = i + v’/1 .............. (1) and v’ = i + v ................................. (2) also i = dv/dt .................................. (3) (3) in (2): v’ = dv/dt + v ................ (4) (4) and (3) in (1): 1 – (dv/dt + v) = dv/dt + (dv/dt + v) 3dv/dt + 2v = 1 dv/dt + (2/3)v = (1/3) [6] v(t) = ½ + Ae-(2/3)t [1] v(0) = ½ + A = 0 A = –½ [1] v(t) = ½[1 – e-(2/3)t] [1]

3. t < 0: v(t) = 1 V [1] en i(t) = 1 A [1] t 0: Loop L: –2 – i – di/dt – v = 0 v = –2 – i – di/dt ................................. (1) node N: i = v + dv/dt .............................. (2) (1) in (2): i = (–2 – i – di/dt) + d/dt(–2 – i – di/dt) i = –2 – i – di/dt – di/dt –d2/dt2 d2i/dt2 + 2di/dt + 2i = –2 [6] = 1 en n = 2 Ondergedemp met d = 1: i(t) = –1 + Ae–tcost+ Be–tsint [3] i(0) = –1 + A But i(0) = 1 1= –1 +A A = 2 and di/dt = –Ae–tcost – Ae–tsint –Be–tsint + Be–tcost di(0)/dt = –A + B = –2 + B. But from (1): v(0) = –2 – i(0) – di(0)/dt { also for v(t): d2v/dt2 + 2dv/dt + 2v = –2 } 1 = –2 – 1 – di(0)/dt di(0)/dt = –4 { with v(0)=1 and dv(0)/dt=i(0)-v(0)=1–1=0 } –2 + B = –4 B = –2 { or from v(t) = –2 – i(t) – di(t)/dt } i(t) = –1 + 2e–tcost – 2e–tsint [4] { is v(t) = –1 + 2e–tcost + 2e–tsint }

4. t < 0: i(0) = 0A en v(0) = 0 [1] zero state t 0: Node N: 1 = i + dv/dt i = 1 – dv/dt ......... (1) Loop L: di/dt + i – dv/dt – v = 0 ................... (2) (1) in (2): d/dt(1 – dv/dt) + (1 – dv/dt) – dv/dt – v = 0 0 – d2v/dt2 + 1 – dv/dt – dv/dt – v = 0 d2v/dt2 + 2dv/dt + v = 1 [6] = 1 en n = 1 critically damped v(t) = 1 + Ae-t + Bte-t [4] v(0) = 1 + A and v(0) = 0 1 + A = 0 A = –1 dv/dt = –Ae-t – Bte-t + Be-t dv(0)/dt = –A + B = 1 + B (as A = –1) and from (1): i(0) = 1 – dv(0)/dt dv(0)/dt = 1 – i(0) = 1 – 0 1 + B = 1 B = 0 v(t) = 1 – e-–t [4]

[10]

[10]

[15]

t 0 1 H 1

v

1 F

i

1

–2Vdi/dt

N

i dv/dtv

L

v 3

1

1

t < 0 i t 0

1 1 F v

1 i

1

1 1 V

1 F v

1

i

v’

t < 0

v i

1 2V

1

[15]

v

1 A i

1

1F

1H

1

dv/dt

di/dt L

i dv/dt

N

Circuit Analysis IV EICAM4A Unit 2 Final Assessment 25 October 2012 Page 1







Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = cos2t u(t) volt. The initial current through the inductor is i(0) = 1 amp and the initial voltage across the capacitor is v(0) = 1 volt . Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (18) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

vs = cos 2t v

1 H 3

(10)

(4) Figure 2

2

1 1 H

1 F

1 H

v

Figure 3

i 3

vs = cos2t u(t)

i(0) = 1

v(0) = 1

21 F

21 F

Stroombaananalise IV EICAM4A Eenheid 2 Finale Evaluasie 25 Oktober 2012 Memorandum Bladsy 1

1. a) vs – 3i – di/dt – v = 0 and i = ½dv/dt vs – (3/2)dv/dt – ½d2v/dt2 – v = 0 d2v/dt2 + 3dv/dt + 2v = 2vs d2v/dt2 + 3dv/dt + 2v = 2cos2t [6] v = Acos2t + Bsin2t dv/dt = –2Asin2t + 2Bcos2t and d2v/dt2 = –4Acos2t – 4Bsin2t –4Acos2t – 4Bsin2t – 6Asin2t + 6Bcos2t + 2Acos2t + 2Bsin2t = 2cos2t [–2A + 6B]cos2t + [–6A – 2B]sin2t = 2cos2t –2A + 6B = 2 and –6A – 2B = 0 A = –0.1 en B = 0.3 vss(t) = –0.1cos2t + 0.3sin2t = 0.316228cos(2t – 1.89255) volt [6] (12) b) Z = 30˚+290˚+1–90˚ = 3.1622818.43495˚ I = 10/3.1622818.43495˚ = 0.316228–18.435˚ A V = 0.316228–18.435˚1–90˚ = 0.316228–108.435˚ volt vss(t) = 0.316228cos(2t – 1.89255) volt (6) {For comparison with Question 3: iss(t) = ½dvss(t)/dt = ½(0.6cos2t+0.2sin2t) = 0.3cos2t+0.1sin2t}

2. a) Z = 1 + j + 2//(1/j) [2] = 1 + j + {(2/j)/[2 + (1/j)]} = 1 + j + [2/(1 + j2)] = 1 + j + [2(1 – j2)/(1 + 42)] = [1 + 2/(1 + 42)] + j{1 – [4/(1 + 42)]} For resonance: Im{Z} = 0 1 – [4/(1 + 42)] = 0 [2] 1 = 4/(1 + 42) 1 + 42 = 4 42 = 3 2 = 3/4 r = (3)/2 = 0.866025 r/s [6] (10) b) Zr = 1 + 2/(1 + 4r

2) = 1 + 2/[1 + 4(3/4)] = 1 + 2/4 = 1.5 (4)

3. s/(s2 + 4) – 3I(s) – sI(s) + 1 – (2/s)I(s) – 1/s = 0 [3] s2/(s2 + 4) – 3sI(s) – s2I(s) + s – 2I(s) – 1 = 0 s2I(s) + 3sI(s) + 2I(s) = s2/(s2 + 4) + s – 1 (s2 + 3s + 2)I(s) = [s2 + s(s2+4) – (s2+4)]/(s2+4) (s2 + 3s + 2)I(s) = (s2 + s3 + 4s – s2 – 4)/(s2 + 4) (s2 + 3s + 2)I(s) = (s3 + 4s – 4)/(s2 + 4) I(s) = (s3 + 4s – 4)/[(s2 + 3s + 2)(s2 + 4)] [4] I(s) = (s3 + 4s – 4)/[(s + 1)(s + 2)(s – 21.5708)(s – 2–1.5708)] I(s) = –1.8/(s + 1) + 2.5/(s + 2) + 0.158113–0.321757/(s – 21.5708) + 0.1581130.321757/(s – 2–1.5708) [4] = –1.8/(s + 1) + 2.5/(s + 2) + 0.158113–0.321757/(s + 0 – j2) + 0.1581130.321757/(s + 0 + j2) i(t) = –1.8e –t u(t) + 2.5e –2t u(t) + 20.158113cos(2t – 0.321757)u(t) = –1.8e –t u(t) + 2.5e –2t u(t) + 0.316226cos(2t – 0.321757)u(t) [2] {Check i(0) 1} {Compare steady state term 0.316226cos(2t–0.321757) with Question 1b: I=0.316228–18.435˚ } {or:I(s)=A/(s+1)+B/(s+2)+(Cs+D)/(s2+4)=[A(s+2)(s2+4)+B(s+1)(s2+4)+(Cs+D)(s+1)(s+2)]/(s+1)(s+2)(s2+4) = [A(s3+2s2+4s+8)+B(s3+s2+4s+4)+(Cs3+3Cs2+2Cs+Ds2+3Ds+2D)]/(s+1)(s+2)(s2+4) = [(A+B+C)s3+(2A+B+3C+D)s2+(4A+4B+2C+3D)s+(8A+4B+2D)]/(s+1)(s+2)(s2+4) A+B+C=1, 2A+B+3C+D=0, 4A+4B+2C+3D=4, 8A+4B+2D=–4 I(s)=–1.8/(s+1)+2.5/(s+2)+(0.3s+0.2)/(s2+4) I(s) = –1.8/(s+1) + 2.5/(s+2) + 0.3[s/(s2+4)] + 0.1[2/(s2+4)] i(t) = –1.8e–t + 2.5e–2t + 0.3cos2t + 0.1sin2t = –1.8e–1 + 2.5e–2 + 0.316228cos(2t – 0.32175) }

[18]

1 H 3

vs = cos 2t v

½ F

i

290˚ 3

10 V1–90˚

I

[14]

2

1 j

1/j

[18]

I(s) 3 s [1]

42s

s

[1]

V(s)

1 [1]

1/s [1]

1/s [1]

I(s) sI(s) 1

2/s [1] s

I(s)

s

1

Circuit Analysis IV EICAM4A Unit 1 First Assessment 29 August 2013 Page 1

Question 1 For the circuit shown in Figure 1, the switch S was closed for a long time. The switch is opened at time t = 0. Find v(t) for all time t. (10) Question 2 Determine the zero state step response v(t) for the circuit in Figure 2. (12) Question 3 For the circuit shown in Figure 3, the switch S was open for a long time, and it is then closed at time t = 0. Determine the current i(t) in the ½ H inductor and the voltage v(t) across the ¼ F capacitor, for all time t. (16) Question 4 Refer to the circuit in Figure 4 and determine the zero state step response of the voltage v(t), across the capacitor. (12)

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

1F 2 v 2 V

t = 0

S

2

1

Figure 1


41 F

21 H

Figure 3

1 i

v2 V t=0 1

1 F

i

vs = u(t) v

Figure 431

21 H

Figure 2

1 v vs = u(t)

1

1 F 1

Stroombaananalise IV EICAM4A Eenheid 1 Eerste Evaluasie 29 Augustus 2013 Memorandum Bladsy 1

1. t < 0: v(t) = (1/2)2 = 1V [2] t 0: i1 + iC + i2 = 0 (v/2) + 1dv/dt + (v/2) = 0 dv/dt + v = 0 [5] v = Ae–t [1] v(0) = A But v(0) = 1 A = 1 [1] v(t) = e–t for t 0 [1] 2. t < 0: v(t) = 0 [1] (zero state) t 0: Node N1: (v’ – 0)/1 = (0 – v)/1 v’ = –v ............ (1) Node N2: (1 – v’)/1 = (v’ – 0)/1 + d/dt(v’ – v) 1 = 2v’ + d/dt(v’ – v) .................................................... (2) (1) in (2): 1 = 2(–v) + d/dt(–v – v) 2dv/dt + 2v = –1 dv/dt + v = –½ [6] v(t) = –½ + Ae–t [1] but v(0) = 0 A = ½ [1] v(t) = ½(e–t – 1) for t 0 [3] 3. t < 0: v(t) = 1 V [1] en i(t) = 1 A [1] t 0: Loop containing ½1H and ¼F: vL + v = 0 ½di/dt + v = 0 v = –½di/dt ............................................ (1) Node N: i = iC + iR i = ¼dv/dt + v/1 i = ¼dv/dt + v ........................................ (2) (1) in (2): {or (2) in (1)} i = ¼d/dt(–½di/dt) + (–½di/dt) {v = –½d/dt(¼dv/dt+v} (1/8)d2i/dt2 + ½di/dt + i = 0 {(1/8)d2v/dt2+½dv/dt+v=0} d2i/dt2 + 4di/dt + 8i = 0 [6] {d2v/dt2+4dv/dt+8v=0 [6] } = 2 and n = 8 underdamped with d = 2 r/s i(t) = Ae-2tcos2t+Be-2tsin2t [2] {v(t)=Ae-2tcos2t+Be-2tsin2t [2]} i(0) = 1 A = 1 {v(0)=1 A=1} di/dt = –2Ae–2tcos2t – 2Ae–2tsin2t – 2Be–2tsin2t + 2Be–2tcos2t di(0)/dt = –2A + 2B {dv/dt=–2Ae–2tcos2t–2Ae–2tsin2t–2Be–2tsin2t+2Be–2tcos2tdv(0)/dt=–2A+2B} but v(0) = –½di(0)/dt (from (1)) 1 = –½di(0)/dt di(0)/dt = –2 {but i(0)=¼dv(0)/dt+v(0) (from (2)) 1=¼dv(0)/dt+1 dv(0)/dt=0} –2(1) + 2B = –2 B = 0 i(t) = e–2tcos2t for t 0 [4] {–2(1) + 2B = 0 B = 1 v(t) = e–2tcos2t + e–2tsin2t [4] } and di/dt = –2e–2tcos2t–2e–2tsin2t {and dv/dt = –2e–2tcos2t–2e–2tsin2t–2e–2tsin2t+2e–2tcos2t = –4e–2tsin2t} v(t) = –½di/dt = –½(–2e–2tcos2t – 2e–2tsin2t) = e–2tcos2t + e–2tsin2t [2] {i(t)=¼dv/dt+v=¼[–4e–2tsin2t] + e–2tcos2t + e–2tsin2t = e–2tcos2t [2] } 4. t < 0: v(t) = 0 V and i(t) = 0 A [1] zero state. t 0: Node N: i = 3v + dv/dt ................................ (1) Loop L: 1 – ½di/dt – v = 0 v = 1 – ½di/dt ........... (2) (1) in (2): {or (2) in (1)} v = 1 – ½d/dt(3v + dv/dt) = 0 {i=3(1–½di/dt)+d/dt(1–½di/dt) ½d2v/dt2 + (3/2)dv/dt + v = 1 {i=3–(3/2)di/dt–½d2i/dt2} d2v/dt2 + 3dv/dt + 2v = 2 [6] {d2i/dt2 + 3di/dt + 2i = 6 [6]} = 1.5 and n = 1.414 overdamped with s1 = –1 and s2 = –2 v(t) = 1 + Ae–t + Be–2t [2] {i(t) = 3 + Ae–t + Be–2t [2]} v(0) = 0 1 + A + B = 0 .............................................. (3) {i(0) = 0 A + B + 3 = 0}} and dv/dt = –Ae–t – 2Be–2t but from (1): i(0)=3v(0)+dv(0)/dt 0 = 0 + dv(0)/dt dv(0)/dt = 0 –A – 2B = 0 ......... (4) {from (2): di(0)/dt=2 –A–2B = 2 } From (3) and (4): A = –2 and B = 1 v(t) = 1 – 2e–t

+ e–t [3] {A= –4 and B= 1i(t)=3–4e–t+e–2t} {v(t)=1–½di/dt=1 – 2e–t + e–2t [3] }

[10]

[12]

1 2

v 2 V

2

1 t < 0

1 2

v

2

t 0 i1 iC i2

[16]

41

21 i

v 1

vL iC iR

t 0N

[12]

1

i

1 v 31

21

dv/dt 3v

½di/dt

1

i

v 2 V 1

t < 0

1 v

1 1

1 1

v’

N1 N2

N

L

Circuit Analysis IV EICAM4A Unit 2 First Assessment 3 October 2013 Page 1

Question 1 The supply current to the circuit in Figure 1, is given by is(t) = cos t. a) Determine the time domain






Question 3 Refer to the circuit in Figure 3. The supply current to the circuit is given by is(t) = sin2t u(t) amp. The initial current through the inductor is i(0) = 0 amp and the initial voltage across the capacitor is v(0) = 0 volt . Draw an equivalent Laplace network model for the circuit and use this model to find an expression for v(t). (18)


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

2

1 H

Figure 2

1 F

1 H

v

Figure 3

i is = sin2t u(t)

i(0) = 0

v(0) = 0 1 F

Figure 1

1 H

1

is = cos t

1

2 F

i

1

Stroombaananalise IV EICAM4A Eenheid2 Eerste Evaluasie 3 Oktober 2013 Memorandum Bladsy 1

1. a) Node N: is = 2dv/dt + i ............... (1) and loop L: v – i – di/dt = 0 v = i + di/dt .............................. (2) (2) in (1): is = 2d/dt(i + di/dt ) + i is = 2d2i/dt2 + 2di/dt + i d2i/dt2 + di/dt + ½i = ½is d2i/dt2 + di/dt + ½i = ½cost [6]

i(t) = Acost + Bsint, di/dt = –Asint + Bcost, and d2v/dt2 = –Acost – Bsint –Acost – Bsint – Asint + Bcost + ½Acost + ½Bsint = ½cost [–½A + B]cost + [–A – ½B]sint = ½cost –½A + B = ½ and –A – ½B = 0 A = –0.2 and B = 0.4 i(t) = –0.2cost + 0.4sint = 0.447214cos(t – 2.0344) amp [6] (12)

b) I = [0.5–90˚/(0.5–90˚ +1+190˚)]Is = 0.4472136–116.56505˚10˚ A I = 0.4472136–116.56505˚ A = 0.4472136–2.034444r A i(t) = 0.4472136cos(t – 2.034444) A (6)

2. a) Z = j + 2//(1/j) [2] = j + {(2/j)/[2 + (1/j)]} = j + [2/(1 + j2)] = j + [2(1 – j2)/(1 + 42)] = [2/(1 + 42)] + j{1 – [4/(1 + 42)]} For resonance: Im{Z} = 0 1 – [4/(1 + 42)] = 0 [2] 1 = 4/(1 + 42) 1 + 42 = 4 42 = 3 2 = 3/4 r = 0.886025 r/s [6] (10) b) Zr = 2/(1 + 4r

2) = 2/[1 + 4(3/4)] = 2/(1 + 3) = 2/4 = 0.5 (4)

3. 2/(s2 + 4) = IR(s) + I(s) + IC(s) 2/(s2 + 4) = V(s)/1 + V(s)/s + V(s)/(1/s) 2s/(s2 + 4) = sV(s) + V(s) + s2V(s) s2Vs) + sV(s) + V(s) = 2s/(s2 + 4) (s2 + s + 1)V(s) = 2s/(s2+4) V(s) = 2s/[(s2 + s + 1)(s2 + 4)] [4] {or I(s) = V(s)/s = 2/[(s2 + s + 1)(s2 + 4)]} V(s) = 2s/[(s – 12.0944)(s – 1–2.0944)(s – 21.5708)(s – 2–1.5708)] [2] V(s) = 0.3202570.766165/(s – 12.0944) + 0.320257–0.766165/(s – 1–2.0944) + 0.27735–2.55359/(s – 21.571) + 0.277352.55359/(s – 2–1.571) [4] = 0.3202570.766165/(s + 0.5 – j0.866025) + 0.320257–0.766165/(s + 0.5 + j0.866025) + 0.27735–2.55359/(s + 0 – j2) + 0.277352.55359/(s + 0 + j2) v(t) = 20.320257e –0.5t cos(0.866025t + 0.766165)u(t) + 20.27735cos(2t – 2.55359)u(t) = 0.640514e –0.5t cos(0.866025t + 0.766165)u(t) + 0.5547cos(2t – 2.55359)u(t) [4] {or:V(s) = [(As+B)/(s2+s+1)]+[(Cs+D)/(s2+4)] = [(A+C)s3+(B+C+D)s2+(4A+C+D)s+(4B+D)]/(s2+s+1)(s2+4) A+C=0, B+C+D=0, 4A+C+D=2 & 4B+D=0 V(s)={[(6/13)s–(2/13)]/(s2+s+1)}+{[–(6/13)s+(8/13)]/(s2+4)} V(s)={(6/13)(s+½)/[(s+½)2+(3/2)2]}–{[(10/133)[(3/2)/((s+½)2+(3/2)2)]}+{–(6/13)[s/(s2+22)]}+{(4/13)[2/(s2+22]}

v(t) = (6/13)e–½tcos(3/2)t – (10/133)e–½tsin(3/2)t – (6/13)cos2t + (4/13)sin2t v(t) = e–½t[0.461538cos0.866025t – 0.444116sin0.866025t] + [–0.461538cos2t + 0.307692sin2t] v(t) = 0.640513e–½tcos(0.866025t + 0.766164) + 0.5547cos(2t – 2.55359) }

[18]

[14]

[18]

2

j

1/j

1 H

1

is = cos t

1

v

2 F

i 2dv/dt

di/dt

is N

L

i

190˚

1

10

1

0.5–90˚

I Is

I(s)

s [1]

42s

2

[1]

1/s [1]

V(s) 1 [1]

IR(s) IC(s)


Question 1 Determine the zero state step response of the current i(t) through the 1H inductor, for the circuit in Figure 1. (10) Question 2 For the circuit shown in Figure 2, find an expression for v(t) for all t, if is(t) is given by:

0 for t1

0 for t 1(t)is

Question 3 For the circuit shown in Figure 3, determine the zero state step response of the voltage v(t) across the 1F capacitor. Question 4 The switch S in the circuit of Figure 4, is closed at time t = 0. Determine i(t) and v(t) for all time t. (16) ---ooo000ooo--- Total: 50

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 2

is

1

v1 F 1

–1

t

is(t) (amp)

0

1

41 F

21 H

t = 0

1 i

v 4 V 1

Figure 4

S

1 H

1 i

vs = u(t) (volt)

1

Figure 1

1

1 H

1 F 1 vvs = u(t)

(volt)

3

Figure 3(14)

(10)


1. Vir t 0: i(t) = 0 [1] (zero toestand respons) Vir t > 0: 1 – v’ = i + v’ en v’ = i + di/dt 1 - (i + di/dt) = i + (i + di/dt) 2di/dt + 3i = 1 di/dt + 3/2i = ½ [5] i(t) = 1/3+A e–3/2t [1] But i(0) = 0 A = –1/3 [1] i(t) = 1/3(1 – e–3/2t) for t 0 [2] 2. t < 0: v(t) = –1V [2] t 0: 1 = v + dv/dt dv/dt + v = 1 [5] v = 1 + Ae–t [1] v(0) = –1 A = –2 [1] v(t) = 1 – 2e–t for t 0 [1] 3. t < 0: v(t), i(t) = 0 [1] t 0: 1 – 3i – di/dt – v = 0 en i = v + dv/dt 1 – 3(v + dv/dt) –d/dt(v + dv/dt) – v = 0 1 – 3v – 3dv/dt – dv/dt – d2v/dt2 – v = 0 d2v/dt2 + 4dv/dt + 4v = 1 [6] = 2 en n = 2 kritiek gedemp v(t) = Ate-2t + Be-2t + ¼ [3] v(0) = 0 B + ¼ = 0 B = –¼ dv/dt = Ae-2t – 2Ate-2t – 2Be-2t dv(0)/dt = A – 2B en dv(0)/dt = i(0) – v(0) dv(0)/dt = 0 A – 2B = 0 A – 2(–1/4) = 0 A = –1/2 v(t) = -½te-2t – ¼e-2t + ¼ for t 0 [4] 4. t < 0: i(t) = 2A [1] en v(t) = 2 V [1] t 0: (1/2)di/dt + v = 0 and i = (1/4)dv/dt + v (1/2)d/dt[(1/4)dv/dt + v] + v = 0 (1/8)d2v/dt2 + (1/2)dv/dt + v = 0 d2v/dt2 + 4dv/dt + 8v = 0 [6] = 2 en n = 8 Thus underdamped system with d = 2 r/s v(t) = Ae-2tcos2t + Be-2tsin2t [2] v(0) = 2 A = 2 dv/dt = –2Ae-2tcos2t – 2Ae-2tsin2t –2Be-2tsin2t +2Be-2tcos2t dv(0)/dt = –2A+2B but i(0)=(1/4)dv(0)/dt + v(0) 2 = (1/4)dv(0)/dt +2 dv(0)/dt = 0 –2(2) + 2B = 0 B = 2 v(t) = 2e-2tcos2t + 2e2tsin2t for t 0 [3] and dv/dt = -4e-2tcos2t – 4e-2tsin2t –4e-2tsin2t +4e-2tcos2t = –8e-2tcos2t i(t) = (1/4)dv/dt + v = (1/4)[–8e-2tsin2t] + 2e-2tcos2t + 2e-2tsin2t i(t) = 2e-2tcos2t for t 0 [3]

v’

1

i

1 H

1 1 V

[10]

[10]

–1

1

v 1

1

1

v

1 F 1

t 0 t < 0 dv/dt v

di/dt 3i i

1 H

1 F 1

v 1 V

3

v

[14]

[16]

1 i

v 4 V

1 t < 0

4

1 F

2

1 H

i

v 1

t 0

½di/dt

¼dv/dt v

1 v’


Question 1 The supply current to the circuit in Figure 1, is given by is(t) = sin2t ampere. a) Determine the time domain

solution for the steady state current response, i(t), by writing down and solving a differential equation for i(t). Give your answer in the form Acos(t+).

b) Use frequency domain analysis to determine the phasor current I. Reconstruct i(t) again from the phasor current I.

Question 2 a) Determine the resonance



Question 3 Refer to the circuit in Figure 3. The supply current to the circuit is given by is(t) = sin2t u(t) amp. The initial current through the inductor is i(0) = 0 amp and the initial voltage across the capacitor is v(0) = 0 volt. Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t).


Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

1 H

v

Figure 3

i is = sin2t u(t)

i(0) = 0

v(0) = 0 1 F 1

Figure 1

1 H

v

i

is = sin2t

1 F 1

(10)

(4) Figure 2

1

1 2 F

1 H

(18)

(12)

(6)

Stroombaananalise IV EICAM4A Eenheid2 Finale Evaluasie 24 Oktober 2013 Memorandum Bladsy 1

1. a) is = iR + i + iC is = v/1 + i + 1dv/dt ........ (1) And v = 1di/dt .................... (2) (2) in (1): is = di/dt + i + d2i/dt2 d2i/dt2 + di/dt + i = sin2t [6] i = Acos2t + Bsin2t di/dt = –2Asin2t + 2Bcos2t and d2i/dt2 = –4Acos2t – 4Bsin2t –4Acos2t – 4Bsin2t – 2Asin2t + 2Bcos2t + Acos2t + Bsin2t = sin2t [–3A + 2B]cos2t + [–2A – 3B]sin2t = sin2t –3A + 2B = 0 and –2A – 3B = 1 A = –0.1538462 and B = –0.2307692 i(t) = –0.1538462cos2t – 0.2307692sin2t A [5] i(t) = 0.2773501cos(2t – 4.124386) A [1] {=0.2773501cos(2t + 2.158700)} (12) b) sin2t = cos(2t – /2) Is = 1–90˚ A YT = (1/1)+(1/290˚)+(1/0.5–90˚) = 1.80277656.30993˚ S V = Is/YT = 1–90˚/1.80277656.30993˚ = 0.5547001–146.30993˚ V I = 0.5547001–146.3099˚/290˚ = 0.2773501–236.3099˚ A [5] = 0.2773501–4.124386 r i(t)=0.2773501cos(2t–4.124386)A [1] (6)

2. a) Z = 1 + 1/j2 + 1//j [2] = 1 – j/2 + j/(1 + j) = 1 – j/2 + [j(1 – j)/(1 + 2)] = 1 – j/2 + (j + 2)/(1 + 2)] = {1 + [2/(1 + 2)]} + j{[/(1 + 2)] – 1/2} For resonance: Im{Z} = 0 [/(1 + 2)] – 1/2 = 0 [2] /(1 + 2) = 1/2 22 = 1 + 2 2 – 1 0 = 1 rad/sec [6] (10) b) Zr = 1 + 1/(1 + 1) = 1.5 (4)

3. 2/(s2 + 4) = IR(s) + I(s) + IC(s) 2/(s2 + 4) = V(s)/1 + V(s)/s + V(s)/(1/s) 2/(s2 + 4) = V(s) + V(s)/s + sV(s) .............. (1) and V(s) = sI(s) .......................................... (2) (2) in (1): 2/(s2 + 4) = sI(s) + I(s) + s2I(s) s2I(s) + sI(s) + I(s) = 2/(s2 + 4) (s2 + s + 1)I(s) = 2/(s2+4) I(s) = 2/[(s2 + s + 1)(s2 + 4)] [4] I(s) = 2/[(s – 12.0944)(s – 1–2.0944)(s – 21.5708)(s – 2–1.5708)] [2] I(s) = 0.320257–1.328235/(s – 12.0944) + 0.3202571.328235/(s – 1–2.0944) + 0.13867512.158795/(s – 21.571) + 0.1386751–2.158795/(s – 2–1.571) [4] = 0.320257–1.328235/(s + 0.5 – j0.866025) + 0.3202571.328235/(s + 0.5 + j0.866025) + 0.13867512.158795/(s + 0 – j2) + 0.1386751–2.158795/(s + 0 + j2) i(t) = 20.320257e –0.5t cos(0.866025t – 1.328235)u(t) + 20.1386751cos(2t + 2.158795)u(t) = 0.640514e –0.5t cos(0.866025t – 1.328235)u(t) + 0.2773502cos(2t + 2.158795)u(t) [4] {or:I(s) = [(As+B)/(s2+s+1)]+[(Cs+D)/(s2+4)] = [(A+C)s3+(B+C+D)s2+(4A+C+D)s+(4B+D)]/(s2+s+1)(s2+4) A+C=0, B+C+D=0, 4A+C+D=0 & 4B+D=2 I(s)={[(2/13)s+(8/13)]/(s2+s+1)}+{[–(2/13)s–(6/13)]/(s2+4)} I(s)={(2/13)(s+½)/[(s+½)2+(3/2)2]}+{[(14/133)[(3/2)/((s+½)2+(3/2)2)]}+{–(2/13)[s/(s2+22)]}–{(3/13)[2/(s2+22]}

i(t) = (2/13)e–½tcos(3/2)t + (14/133)e–½tsin(3/2)t – (2/13)cos2t – (3/13)sin2t i(t) = e–½t[0.1538462cos0.866025t+0.6217618sin0.866025t]+[–0.1538461cos2t–0.2307692sin2t] i(t) = 0.640513e–½tcos(0.866025t – 1.328232) + 0.27735cos(2t – 4.124386) = 0.640513e–½tcos(0.866025t – 1.328232) + 0.27735cos(2t + 2.158799) }

[18]

1 H v

i

is = sin2t 1 F 1

iR iC

V

I

1 290˚ 0.5–90˚

1–90˚

1

1

j

1/j2

[14]

[18]

I(s)

s [1]

42s

2

[1]

1/s [1]

V(s) 1 [1]

IR(s) IC(s)


Question 1 For the circuit shown in Figure 1, the switch S was closed for a long time. The switch S is opened at time t = 0. Find v(t) for all time t. (10) Question 2 For the circuit shown in Figure 2, find an expression for i(t) for all t, if the source current, is(t), is given by:

0 for t1

0 for t 1 (t)is

Question 3 Refer to the circuit in Figure 3 and determine the zero state step response of the voltage v(t), across the capacitor. (12) Question 4 For the circuit shown in Figure 4, switch S1 was closed for a long time and switch S2 was open for a long time. Switch S1 is opened at time t = 0 and switch S2 is closed at time t = 0. Determine the inductor current i(t) and the capacitor voltage v(t), for all time t. (16)

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)


2 V

t = 0

S2

1

1

v

1F 1 V

t = 0

S1

i

Figure 4

21 H

(12)

1 H

1 i

1 is

–1

t

is(t) amp

0

1

Figure 2

6 V

1

t = 0 S

v

1F

1

1

Figure 1

vs = u(t)

3

1 1 F

v1 H i

Figure 3


1. t < 0: v(t) = (2/3)6 = 4V [1] t 0: v + v1 + v2 = 0 v + 1iC + 1iC = 0 v+dv/dt+dv/dt = 0 2dv/dt + v = 0 dv/dt + ½v = 0 [6] v = Ae–½t [1] v(0) = A But v(0) = 4 A = 4 [1] v(t) = 4e–½t for t 0 [1] 2. t < 0: i(t) = ½ A [1] t 0: Node N: –1 = i+v’... (1) Loop L: v’ – i – di/dt = 0 v’ = i + di/dt ................. (2) (2) in (1): –1 = i + i + di/dt di/dt + 2i = –1 [6] i(t) = (–1/2) + Ae–2t [2] but i(0) = ½ A = 1 [1] i(t) = –½ + e–2t amp, for t 0 [2] 3. t < 0: v(t) = 0 [½] , i(t) = 0 [½] zero state t 0: 1 – 3i – di/dt – v = 0 en i = v + dv/dt 1 – 3(v + dv/dt) – d/dt(v + dv/dt) – v = 0 1 – 3v – 3dv/dt – dv/dt – d2v/dt2 – v = 0 d2v/dt2 + 4dv/dt + 4v = 1 [6] = 2 en n = 2 kritiek gedemp v(t) = Ate-2t + Be-2t + ¼ [2] v(0) = 0 B + ¼ = 0 B = -¼ dv/dt = Ae-2t – 2Ate-2t – 2Be-2t dv(0)/dt = A – 2B dv(0)dt = i(0) – v(0) = 0 A – 2B = 0 A – 2(-1/4) = 0 A = -1/2 v(t) = -½te-2t – ¼e-2t + ¼ [3] 4. t < 0: v(t) = 1V [1] i(t) = 0A[1] t 0: 2 + ½di/dt + i – v = 0 ... (1) and i = –dv/dt ........................ (2) (2) in (1): 2 – ½d2v/dt2 – dv/dt – v = 0 d2v/dt2 + 2dv/dt + 2v = 4 [6] = 1, n = 2 under damped with d = 1 v(t) = Ae–tcost + Be–tsint + 2 [2] v(0) = 1 A + 2 = 1 A = –1 v(t) = –e–tcost + Be–tsint + 2 i(t) = –dv/dt = –[e–tcost + e–tsint – Be–tsint + Be–tcost] = –e–tcost – e–tsint + Be–tsint – Be–tcost and i(0) = 0 –1 – B = 0 B = –1 v(t) = –e–tcost – e–tsint + 2 [3] and i(t) = –dv/dt = –[e–tcost + e–tsint + e–tsint – e–tcost] i(t) = –2e–tsint [3]

[16]

6 V

1

v

1

1 v

1

1

1

t < 0 t 0

[10]

iC v1 v2

1

i

1 v’

v’

1

t < 0

1

i

1

1 v’

v’t 0

–1

[12]

N

L

di/dt

i

1V

3

1 1

v

1 i

[12]

2 V

½

1

v

1

i 1 1

v1 V

i

t < 0 t 0

½di/dt

i

3i di/dt

v dv/dt

i


Question 1 Refer to the circuit in Figure 1. a) Find the steady state current

response i(t) in the time domain, solving the differential equation for i, if the supply voltage vs, is given by the real sinusoid vs(t) = cos2t volt. Give your answer in the form, Acos(t + ).

b) Find the steady state current response i(t) in the time domain, solving the differential equation for i, if the supply voltage is the complex sinusoid vs = e j2t volt. Give your answer in the form Ae j(t + ). (6)

c) Use frequency domain analysis to determine the phasor current I, if the supply voltage is the phasor Vs = 10 volt. Reconstruct the time form of i(t), Acos(t + ), again from the phasor current I. (6)

Question 2 Refer to the circuit in Figure 2. a) Determine the resonance frequency

of the circuit from the perspective that the impedance seen by the source, must become real. (10)


Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = cos2t u(t) volt. The initial current through the inductor is i(0) = 0 amp and the initial voltage across the capacitor is v(0) = 0 volt . Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t). (16)


Appendix Trigonometric identities: Network Models: Acost + Bsint = cos (t – tan-1(B/A)) cos = sin( + /2)




f(t) F(s) f(t) F(s)






jasA

jasA

(10)

Figure 3

1

1 Hvs = cos2t u(t)

1 F

i

v

i(0) = 0

v(0) = 0

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)

V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

1

1 Hvs

1 F

cos2t ej2t 10

i

2

1 H

Figure 2

1 F

1

Stroombaananalise IV EICAM4A Eenheid2 Eerste Evaluasie 25 September 2014 Memorandum Bladsy 1

1. a) (vs –v)/1 = i + dv/dt ........... (1) and v = di/dt ...................... (2) vs – di/dt = i + d2i/dt2 d2i/dt2 + di/dt + i = vs d2i/dt2 + di/dt + i = cos2t [5] we insist that i = Acos2t + Bsin2t , therefore di/dt = –2Asin2t + 2Bcos2t and d2i/dt2 = –4Acos2t – 4Bsin2t –4Acos2t –4 Bsin2t – 2Asin2t + 2Bcos2t + Acos2t + Bsint = cos2t [–3A + 2B]cost + [–3B – 2A]sint = cost –3A + 2B = 1 en –3B – 2A = 0 A = –(3/13) = –0.230769231 and B = (2/13) = 0.153846154 i(t) = –(3/13)cos2t + (2/13)sin2t [4] = 0.277350098cos(2t – 2.55359) ampere [1] (10) b) For vs(t) = e j2t d2i/dt2 + di/dt + i = e j2t we again insist that i(t) must have the form, i(t) = Ae j(2t + ) [1] therefore di/dt = j2Ae j(2t + ) and d2/dt2 = j24Ae j(2t + ) = –4Ae j(2t + ) –4Ae j(2t + ) + j2Ae j(2t + ) + Ae j(2t + ) = e j2t (–4 + j2 + 1)Ae j(2t + ) = e j2t (–3 + j2)Aejej2t = ej2t (–3 + j2)Aej = 1 Aej = 1/(–3 + j2) = 0.277350098–2.553590 Aej = 0.277350098e –j2.553590 A = 0.277350098 [2] and = –2.553590r [2] i(t) = 0.277350098 e j(2t – 2.55359) amp [1] (6) c) Z = 1+ 0.5–90//290) = 1 + [(0.5–90×290)/(0.5–90+290)] = 1 + (10/1.590) = 1+ (2/3)–90= 1.20185043–33.6900676 [2] It = Vs/Z = 10/1.20185043–33.6900676 = 0.83205029133.6900676 A [2] I = [0.5–90/(0.5–90 + 290)]×It = [0.5–90/(1.590]×0.83205029133.6900676 = 0.333333333–180×0.83205029133.6900676 = 0.277350097–146.309932 A [1] i(t) = 0.277350097cos(2t – 2.55359005r) [1] (6) {or V = [(2/3)–90/(1+(2/3)–90)]10=0.5547002–56.309932 I = 0.5547002–56.309932/290=0.2773501–146.309932}

2. a) Z = 1 + j + 2//(1/j) [2] = 1 + j + [2(1/j)]/[2 + (1/j)] = 1 + j + [2/(1 + j2)] = 1 + j + [2(1 – j2)/(1 + 42)] = 1 + j + [2/(1 + 42)] – j[4/(1 + 42)] = {1 + [2/(1 + 42)]} + j{ – [4/(1 + 42]} [4] { = [(1 + 42 + 2)/(1 + 42)] + j{[(1 + 42) – 4]/(1 + 42)} = [(3 + 42)/(1 + 42)] + j[(42 – 3)/(1 + 42)] } For resonance: Im{Z} = 0 – [4/(1 + 42] = 0 4/(1 + 42) = 1 1 + 42 = 4 42 = 3 = 3/2 (0.866925 r/a) [4] (10) b) Z0 = {1 + [2/(1 + 42)]}| = 3/2 = 1 + 2/(1 + 3) = 1 + 0.5 = 1.5 (2)

3. [s/(s2 + 4) – V(s)]/1 = I(s) + V(s)/(1/s) s/(s2 + 4) = V(s) + I(s) + sV(s) But V(s) = sI(s) s/(s2 + 4) = sI(s) + I(s) + s2I(s) (s2 + s + 1)I(s) = s/(s2 + 4) I(s) = s/(s2 + s + 1)(s2 + 4) [4] = s/[(s – 12.0944)(s – 1–2.0944)(s – 21.5708)(s – 2–1.5708)] [2] = 0.1601290.766165/(s – 12.0944) + 0.160129–0.766165/(s – 1–2.0944) + 0.138675–2.55359/(s – 21.571) + 0.1386752.55359/(s – 2–1.571) [4] = 0.1601290.766165/(s + 0.5 – j0.866025) + 0.160129–0.766165/(s + 0.5 + j0.866025) + 0.138675–2.55359/(s + 0 – j2) + 0.1386752.55359/(s + 0 + j2) i(t) = 20.160129e-0.5tcos(0.866025t + 0.766165) + 20.138675e0cos(2t – 2.55359) = 0.320258e-0.5tcos(0.866025t + 0.766165) + 0.27735cos(2t – 2.55359) [2]

[12]

cos2tej2t

1

1 vs

1

i

v

[22]

2

j

1/j

1

1 [1]

s [1]

42s

s

[1]

1/s [1]

I(s)

V(s)

[16]

Vs

It = 2 r/s

0.5–90

I

10 V 290

1

V


Question 1 Refer to the circuit in Figure 1, and determine the zero state step response of the output voltage v(t). (10) Question 2 For the circuit shown in Figure 2, find an expression for v(t) for all t, if is(t) is given by:

0 for t2

0 for t 1 (t)is

Question 3 For the circuit shown in Figure 3, determine the zero state step response of the current i(t), in the 1 H inductor. Question 4 For the circuit shown in Figure 4, the switch S was open for a long time Switch S is closed at time t = 0. Find i(t) for all time t. ---ooo000ooo--- Total: 50

Appendix Trigonometric identities: Network models Acost + Bsint = cos (t – tan-1(B/A)) cos = sin( + /2)




f(t) F(s) f(t) F(s)






jasA

jasA

(12) Figure 2

is

1

v1 F 1

t

is(t) (amp)

0

2

1 1

(14)

(14)

S

1

21 F

1 H

t = 0 i

1 V

Figure 4

v vs(t) = u(t)

1

1 F31

1 Figure 1

2B2A

2B2A

0 dtst-f(t)e

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)

1

v

1 F

vs = u(t)

i

1 Figure 3

1 H


1. t < 0: v(t) = 0 [1] (zero state) t 0: (1 – v’)/1 = (v’ – 0)/1 + (v’ – v)/1 1 = 3v’ – v ....................................................... (1) and (v’ – 0)/1 = (1/3)d/dt(0 – v) v’ = – (1/3)dv/dt ............................................... (2) (2) in (1): 1 = 3[–(1/3)dv/dt] – v 1 = –dv/dt – v dv/dt + v = –1 [5] v = –1 + Ae-t [1] But v(0) = 0 0 = –1 + A A= 1 [1] v(t) = – (1 – e-t) [2] for t 0

2. t < 0: i’ = (1/3)1 = ⅓ amp v(t) = ⅓ volt [2] t 0: 2 = v’ + (v’ – v)/1 = 2v’ – v 2v’ = 2 + v v’ = 1 + ½v .............. (1) and (v’ – v)/1 = v + dv/dt v’ = 2v + dv/dt ......... (2) (2) in (1): 2v + dv/dt = 1 + v/2 dv/dt + (3/2)v = 1 [6] v = (2/3) + Ae–(3/2)t [1] and v(0) = 1/3 (1/3) = (2/3) + A A = –(1/3) [1] v(t) = (2/3) – (1/3)e–(3/2)t for t 0 [2] {or: Thevenin VT=⅓2=⅔ & RT=1//2=⅔ v=⅔+Ae-(3/2)t=⅔–⅓e-1.5t}

3. t < 0: v(t) = 0 V [½] en i(t) = 0 A [½] (zero state) t 0: Loop L: 1 – v – di/dt = 0 v = 1 – di/dt ................................... (1) Node N: v + dv/dt = i + di/dt ............. (2) (1) in (2): i – di/dt + d/dt(1 – di/dt) = i + di/dt i – di/dt + 0 – d2/dt2 = i + di/dt) d2i/dt2 + 2di/dt + i = 1 [6] = 1 and n = 1 Therefore the system is critically damped, and thus i(t) = 1 + Ae–t + Bte–t [3] so i(0) = 1 + A and with i(0) = 0, this implies that 1 + A = 0 A = –1 and di/dt = –Ae–t – Bte–t + Be–t , so di(0)/dt = –A + B. But from (1): v = 1 – di/dt, which must also be true for t = 0 v(0) = 1 – di(0)/dt di(0)/dt = 1 – v(0) = 1 – 0 = 1. Therefore di(0)/dt = –A + B = 1 –(–1) + B = 1 B = 0 i(t) = 1 – e–t [4]

4. t < 0: i(t)=0 [1] v(t) = 1 [1]. t 0: (1 – v)/1 = (1/2)dv/dt +i 2 – 2v = dv/dt + 2i ..................... (1) and v = di/dt ............................. (2) (2) in (1): 2 – 2(di/dt) = d/dt(di/dt) + 2i 2 – 2di/dt = d2i/dt2 + 2i d2i/dt2 + 2di/dt + 2i = 2 [6] = 1 and n = 2 thus underdamped system with d = 1 r/s i(t) = Ae-tcost + Be-tsint + 1 [3] i(0) = 0 A + 1 = 0 A = –1 and from (2): di(0)/dt = v(0) = 1 and di/dt = –Ae-tcost – Ae-tsint – Be-tsint + Be-tcost di(0)/dt = –A + B = –(–1) + B = 1 + B but di(0)/dt = 1 1 + B = 1 B = 0 i(t) = –e– tcost + 1 [3]

[10]

[12]

[14]

[14]

1

1

1

1

N

v1

i

di/dt

di/dt

v

dv/dt

L

1

v 1

1

1

1/3

v’ 0 V

1 1

v 1

i’ 2

1

v

1 1 1

t < 0 t 0

v’ v dv/dt

VT=2/3V v RT=2/3

1 F

Thevenin equivalent circuit

1

1

1

21

v

i

1 V

1

1

21

v

i

1 V

t < 0

t 0 ½dv/dt



a) Find the steady state response v(t) in the time domain, solving the differential equation for v, if the supply voltage vs, is given by the real sinusoid vs(t) = cost. Give your answer in the form, Acos(t + ).

b) Find the steady state response v(t) in the time domain, solving the differential equation for v, if the supply voltage is the complex sinusoid vs = e jt. Give your answer in the form Ae j(t + ). (6)

c) Use frequency domain analysis to determine the phasor voltage V, if the supply voltage is the phasor Vs = 10. Reconstruct the real form of v(t), Acos(t + ), again from the phasor voltage V. (4)

Question 2 Determine the resonance frequency of the circuit in Figure 2, from the perspective that the imaginary part of the impedance Z, must vanish.

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = cos2t u(t) volt. The initial current through the inductor is i(0) = 1 amp and the initial voltage across the capacitor is v(0) = 0 volt . Draw an equivalent Laplace network model for the circuit and use this model to find an expression for i(t).






f(t) F(s) f(t) F(s)






jasA

jasA

Figure 3

1

1 Hvs = cos2t u(t)

1 F

i

v

i(0) = 1 A

v(0) = 0

2B2A

2B2A

0 dtst-f(t)e

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)

vs 1 F

2 H

Figure 1

v 1

cost ejt 10

(12)

(10)

(18)

Figure 2

Z 2 F

1 H

1


1. a) vs – 2di/dt – v = 0 and i = v + dv/dt vs – 2d/dt[v + dv/dt] – v = 0 vs – 2dv/dt – 2d2v/dt2 – v = 0 With vs = cost: d2v/dt2 + dv/dt + ½v = ½cost [6] But the steady state voltage must be of the form: v(t) = Acost + Bsint – Acost – Bsint – Asint + Bcost + ½Acost + ½Bsint = ½cost (– ½A + B)cost + (– A – ½B)sint = cost – ½A + B = 1 and – A – ½B = 0 A = –1/5 [2] and B = 2/5 [2] vss(t) = –(1/5)cost + (2/5)sint

[1] = 0.4472cos(t – 2.0344) volt

[1] (12) b) For vs(t) = e jt d2v/dt2 + dv/dt + ½v = ½e jt and v(t) must have the form Ae j(t + ) [1] d2/dt2[Ae j(t + )] + d/dt[Ae j(t + )] + ½[Ae j(t + )]v = ½e jt j2Ae j(t + ) + jAe j(t + ) + ½Ae j(t + ) = ½e jt ( – 1 + j + ½)Ae j(t + ) = ½e jt ( – ½ + j)Ae j(t + ) = ½e jt ( – ½ + j)Ae jt×e j = ½e jt ( – ½ + j)Ae j = ½ Ae j = ½/( – ½ + j) = 0.4472– 2.0344 = 0.4472e – j2.0344 [4] A = 0.4472 and = – 2.0344r vss(t) = 0.4472e j(t – 2.0344) volt [1] (6) c) Z1F//1 = (1×1–90)/(1 + 1–90) = 0.7071–45 Zt = 290 + 0.7071–45 = 1.58171.57 V = (Z1F//1/Zt)×Vs = (0.7071–45/1.58171.57)×10 = 0.4472–116.6 volt [3] v(t) = 0.4472cos(t – 2.035r) volt [1] (4)

2. Z = j + 1//(1/j2) [1] = j + (1/j2)/(1 + 1/j2) = j + 1/(j2 + 1) = j + (1 – j2)/(1 + 42) = [1/(1 + 42)] + j[1 – 2/(1 + 42)] [5] At resonance, Im{Z} = 0 1 – 2/(1 + 42) = 0 2/(1 + 42) = 1 1 + 42 = 2 42 = 1 2 = 1/4 r = 0.5 r/s.[4]

3. [s/(s2 + 4) – V(s)]/1 = I(s) + V(s)/(1/s) s/(s2 + 4) = V(s) + I(s) + sV(s) [2] ..... (1) But I(s) = 1/s + V(s)/s V(s) = sI(s) – 1 [1] ............................. (2) (2) in (1): s/(s2

+ 4) = [sI(s) – 1] + I(s) + s[sI(s) – 1] s/(s2 + 4) = s2I(s) +sI(s) + I(s) – s – 1 (s2 + s + 1)I(s) = s/(s2 + 4) + s + 1 (s2 + s + 1)I(s) = [s + s(s2

+ 4) + s2 + 4]/(s2

+ 4) I(s) = (s3 + s2 + 5s + 4)/(s2 + 4) I(s) = (s3 + s2 + 5s + 4)/(s2 + s + 1)(s2 + 4) [2] = (s3 + s2 + 5s + 4)/[(s – 12.0944)(s – 1–2.0944)(s – 21.5708)(s – 2–1.5708)] [2] = A/(s – 12.0944) + A*/(s – 1–2.0944) + B/(s – 21.5708) + B*/(s – 2–1.5708) Now (s3+s2+5s+4)|s=12.0944 = 41.0472 and (s3+s2+5s+4)|s=21.5708 = 21.57077 A = [I(s)(s–12.0944)]s=12.0944 = 0.640515–0.281035 and B = [I(s)(s–21.5708)]s=21.5708 = 0.138675–2.55359

I(s) = 0.640515– 0.281035/(s – 12.0944) + 0.6405150.281035/(s – 1–2.0944) + 0.138675–2.55359/(s – 21.5708) + 0.1386752.55359/(s – 2–1.5708) [4] = 0.640515– 0.281035/(s + 0.5 – j0.866025) + 0.6405150.281035/(s + 0.5 + j0.866025) + 0.138675–2.55362/(s + 0 – j2) + 0.1386752.55362/(s + 0 + j2) i(t) = 20.640515e-0.5tcos(0.866025t – 0.281035) + 20.138675e0cos(2t – 2.55359) = 1.28103e-0.5tcos(0.866025t – 0.281035) + 0.27735cos(2t – 2.55359) [2] {or: I(s) = [(As+B)/(s2+s+1)]+[(Cs+D)/(s2+4)]={[(16/13)s+(12/13)]/[(s+½)2+(3/2)2]}+{[–(3/13)s+(4/13)]/(s2+22)} = (16/13){(s+½)/[(s+½)2+(3/2)2]} + (8/133){(3/2)/[(s+½)2+(3/2)2]} – (3/13){s/(s2+22)} + (2/13){2/(s2+22)}

i(t) = (16/13)e-0.5tcos(3/2)t + (8/133)e-0.5tsin(3/2)t – (3/13)cos2t + (2/13)sin2t = 1.28103e-0.5tcos[(3/2)t – 0.281035] + 0.27735cos(2t – 2.5536) }

[18]

[22]

[10]

i

(cost, e jt)

vs

1 F

2 H v

1

or: [s/(s2+4)–V(s)]/1=I(s)+sV(s) ... (1) and V(s) – sI(s) + 1 = 0

V(s) = sI(s) – 1 ............ (2) (2) in (1): s/(s2+4) – [sI(s) – 1]

= I(s)+s[sI(s) – 1] (s2+s+1)I(s) = s/(s2+4) + s + 1

= (s3+s2+5s+4)/(s2+4) I(s)= (s3+s2+5s+4)/(s2+4)(s2+s+1)

42s

s

I(s)

1

s sI(s)

1/s V(s)

sV(s)

1 [1]

s [1]

42s

s

[1]

1/s [1]

I(s)

V(s) 1/s [1]

1

I

Vs =10 V

= 1 r/s 1–90

290 V

1

1/j2

j Z

1


Question 1 Referring to the circuit in Figure 1, determine the zero state step response of the voltage v(t) across the 1 F capacitor. Question 2 For the circuit shown in Figure 2, find an expression for the current i(t) through the 1 H inductor, for all time t, if the voltage supply, vs(t), is given by:

0 for t1

0 for t 1 (t)vs

Question 3 For the circuit in Figure 3, the switch S was in position x for a long time. At time t = 0, the switch is moved to position y. Determine expressions for i(t) and v(t) for t 0 Question 4 For the circuit shown in Figure 4, determine the zero state step response of the capacitor voltage v(t).

Appendix





f(t) F(s) f(t) F(s)






jasA

jasA

Network Models:

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)V(s)

V(s)sL

Cv(0)

I(s)

V(s)

---ooo000ooo---

(12)

1 H

1

i 1

vs

–1

t

vs(t) volt

0

1

Figure 2

(10)

1

vs = u(t) (volt) 1

Figure 1

1

v

1 F

1

x

y 1 A 3

t = 0

i

⅓ F

v ¼ H

6

Figure 3 (14)

1 H 1

v 1 F

vs = u(t) 1

Figure 4 (14)

Total: 50

S


1. For t 0: v(t) = 0 [1] (zero toestand respons) For t > 0: 1 – v’ = i + v’ en v’ = i + v 1 – (i + v) = i + (i + v) 3i + 2v = 1 But i = dv/dt 3dv/dt + 2v = 1 dv/dt + ⅔v = ⅓ [5] v(t) = ½ + Ae–⅔t [1] But v(0) = 0 A = –½ [1] v(t) = ½(1 – e–⅔t) for t 0 [2]

2. t < 0: iT = 1/1.5 = ⅔ i(t) = ⅓ A [2] t 0: Node N: –1 – v’ = i+v’ v’ = – ½ – ½i................. (1) Loop L: v’ – i – di/dt = 0 v’ = i + di/dt ................. (2) (1) in (2): – ½ – ½i = i + di/dt di/dt + (3/2)i = –½ [5] i(t) = [(–1/2)/(3/2)] + Ae–(3/2)t = –⅓ + Ae–(3/2)t [2] but i(0) = ⅓ A = ⅔ [1] i(t) = –⅓ + ⅔e–(3/2)t amp, for t 0 [2]

3. t < 0: I = (3/(3+6))×1 = ⅓ A v(t) = ⅓×6 = 2 V [1] (or v(t) = 1×(6//3) = 1×2 = 2 V ) and i(t) = 0 A [1] t 0: KCL:i = i1 + i2 = 0 i2 = i – i1 ………..…. (1) KVL: v1 – v2 = 0 6i1 – 3i2 = 0 6i1 – 3(i – i1) = 0 from (1) 9i1 – 3i = 0 i1 = ⅓i ……………………...….. (2) KVL: v + vL + v1 = 0 v + ¼di/dt + 6i1 = 0 v + ¼di/dt + 6(⅓i) = 0 from (2) v + ¼di/dt + 2i = 0 …………………………….. (3) But i = ⅓dv/dt …………………………..……..… (4) (4) in (3): v + (1/12)d2v/dt2 + ⅔dv/dt = 0 d2v/dt2 + 8dv/dt + 12v = 0 [6]

{ or combine the 3 and 6 resistors in parallel to form a single 2 resistor: v + ¼di/dt + 2i = 0 d2v/dt2 + 8dv/dt + 12v = 0 }

= 4 and n = 12 = 3.464 system overdamped: s2 + 8s + 12 = 0 (s + 2)(s + 6) = 0 s1 = – 2 and s2 = – 6 v(t) = Ae -2t + Be -6t With v(0) = 2 A + B = 2 …...……….… (5) dv(0)/dt = – 2A – 6B = 0 (i(0) = ⅓dv(0)/dt = 0 ) 2A + 6B = 0 ………….…………….. (6) From (5) and (6): A = 3 and B = –1 v(t) = 3e -2t – e -6t [3] and i(t) = ⅓dv/dt = ⅓d/dt(3e -2t – e -6t) i(t) = ⅓( – 6e -2t + 6e -6t) = – 2e -2t + 2e -6t [3]

4. t < 0: v(t) = 0 V [½] and i(t) = 0 A [½] t 0: i = v + dv/dt and 1 – i – di/dt – v=0 1 – (v + dv/dt) – d/dt(v + dv/dt) – v = 0 1 – 2v – 2dv/dt – d2v/dt2 = 0 d2v/dt2 + 2dv/dt + 2v = 1 [6] = 1 en n = 2 underdamped with d = (2 – 1) = 1 v(t) = ½ + Ae–tcost + Be–tsint [3] , v(0) = 0 0= ½ +A A = –½ and i(0) = v(0) + dv(0)/dt dv(0)/dt = 0 dv/dt = –Ae–tcost – Ae–tsint – Be–tsint + Be–tcost 0 = –A + B B = A = – 0.5 v(t) = 0.5 – 0.5e–tcost – 0.5e–tsint [4]

1 1 1 V

[10]

1

i

v 1

v’

i v’

[12]

1

i

1

1

t < 0

1

i

1

1 v’

v’t 0

–1

N

L

di/dt

i

1 1 iT

2i

vL

i

⅓ F

v ¼ H

2

v1

v2

i2

i1

vL

1 A

3

t < 0

i

⅓ F

v ¼ H

6

3

t 0

i

⅓ F

v ¼ H

6

I

[14]

[14]

i

1

1

v 1 F

1

1 H


Question 1 The supply current to the circuit in Figure 1, is given by is(t) = sin 2t ampere. a) Determine the time domain




of the circuit from the perspective that the impedance seen by the source, must become real. (10)


Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = cos2t u(t) volt. The initial current through the inductor is i(0) = 0 amp and the initial voltage across the capacitor is v(0) = 1 volt . Draw an equivalent Laplace network model for the circuit and use this model to find the voltage v(t) across the 1 F capacitor. (18)

---ooo000ooo--- Total: 50 Appendix Trigonometric identities: Network Models: Acost + Bsint = cos (t – tan-1(B/A)) cos = sin( + /2)




f(t) F(s) f(t) F(s)






jasA

jasA

2B2A

2B2A

0 dtst-f(t)e

sC1

s

i(0)

I(s)

I(s) I(s)

sL

Li(0)

V(s)

sC1

s

v(0)

V(s)

V(s)sL

Cv(0)

I(s)

V(s)

Figure 1

2 Hv

i

is = sin 2t 1 F

1

Figure 2

1 H 1 F

4 2

Figure 3

1

1 H

vs = cos2t u(t) 1 F

i

v

i(0) = 0 A

v(0) = 1 V

Stroombaananalise IV EICAM4A Eenheid2 Eerste Evaluasie 17 September 2015 Memorandum Bladsy 11. a) is = i + iC is = i + 1dv/dt .................................. (1) and v – 2di/dt – i = 0 v = 2di/dt + i ................. (2) (2) in (1): is = i+d/dt(2di/dt+i) 2d2i/dt2+di/dt+i = sin2t d2i/dt2+½di/dt+½i = ½sin2t [6] We now insist that i=Acos2t+Bsin2t , therefore di/dt = –2Asin2t + 2Bcos2t

and d2i/dt2 = –4Acos2t – 4Bsin2t –4Acos2t –4Bsin2t–Asin2t+Bcos2t+½Acos2t+½Bsint=½sin2t (–3.5A+B)cost+(–3.5B–A)sint=0.5sin2t–3.5A+B=0 and –3.5B–A=0.5A=–0.037736 and B=–0.132076 i(t)= –0.037736cos2t–0.132076sin2t

[5] =0.13736cos(2t+1.8491) A [1] {or i(t)=0.13736cos(2t–4.4341)} (12)

b) sin2t = cos(2t – /2) Is = 1–90˚ A [1] I = {(0.5–90˚)/[(0.5–90˚)+10+490˚)]}1–90˚ = 0.13736105.945˚ A [3] (current division rule) {or ZT=0.566352-88.0909 VT=0.566353-178.0909 I=0.13736105.945}

I = 0.13736105.945˚ = 0.137361.8491r A [1] i(t)=0.13736cos(2t+1.8491)A

[1] {or I=0.13736(105.945˚–360˚)=0.13736–254.055˚i(t)=0.13736cos(2t-4.4341)}

2. a) Z = (4 + j)//[2 + (1/j)] [2] = {(4 + j)[2 + (1/j)]}/{4 + j + 2 + (1/j)} = [8 + j2 + (4/j) + 1]/[6 + j + (1/j)] = [9 + j2 + (4/j)]/[6 + j + (1/j)] = [(4–22)+j9]/[(1 – 2)+j6]={[(4–22)+j9][(1–2)–j6]}/[(1–2)2+362] = {(4 – 22)(1 – 2) + j9(1 – 2) – j6(4 – 22) + 542]/[(1 – 2)2

+ 362] = {[4 – 22

– 42 + 24

+ 542]/[(1 – 2)2 + 362]} + j{[9 – 93

– 24 + 123]/[(1 – 2)2 + 362]}

= {[4 + 482 + 24]/[(1 – 2)2 + 362]} + j{[33 – 15]/[(1 – 2)2

+ 362]} [6] For resonance: Im{Z} = 0 33 – 15 = 0 2 = 5 0 = 5 = 2.236 r/s [2] (10) {or Y = [1/(4 + j)]+[1/(2+(1/j))] = [(4–j)/(16+2)]+[j/(1+j2)] = [(4–j)/(16+2)]+[j(1–j2)/(1+42)] = [(4–j)/(16+2)]+[(22+j)/(1+42)] = [4(1+42)–j(1+42)+22(16+2)+j(16+2)]/(16+2)(1+42) = [(4+162+322+24)+j(16+2–1–42)/(16+2)(1+42) = [(4+482+24)+j(15–32)/(16+2)(1+42) = {(4 + 482 + 24)/[(16 + 2)(1 + 42)]} + j{(15 – 32)/[(16 + 2)(1 + 42)]} 0 = 5 = 2.236 r/s}

b) Z0 = [4 + 485 + 225]/[(1 – 5)2 + 365] = 294/(16 + 180) = 1.5 (4)

{or Y0 = (4 + 485 + 225)/(16 + 5)(1 + 45) = 294/(2121) = 0.666667 = 1/Z0} 3. [s/(s2 + 4) – V(s)]/1 = I(s) + IC(s) [2] s/(s2 + 4) = V(s) + I(s) + IC(s) .............. (1) But I(s) = V(s)/s and IC(s) = –1 + sV(s) Substitute I(s) and IC(s) in (1): s/(s2+4) = V(s)+[V(s)/s]+[–1+sV(s)] s2/(s2 + 4) = sV(s) + V(s) – s + s2V(s) (s2 + s + 1)V(s) = s2/(s2 + 4) + s = [s2 + s(s2 + 4)]/(s2 + 4) = (s3 + s2 + 4s)/(s2 + 4) V(s)=(s3+s2+4s)/(s2+s +1)(s2+4) [4] = (s3+s2+4s)/[(s–12.0944)(s–1–2.0944)(s–21.5708)(s–2–1.5708)] [2] = A/(s – 12.0944) + A*/(s – 1–2.0944) + B/(s – 21.5708) + B*/(s – 2–1.5708) Now (s3+s2+4s)|s=12.0944 = 32.0944 and (s3+s2+4s)|s=21.5708 = –4 A=[V(s)(s–12.0944)]s=12.0944 = 0.480390.76616 and B=[V(s)(s–21.5708)]s=21.5708 = 0.27735–0.98279 V(s)=0.480390.76616/(s–12.0944)+0.48039–0.76616/(s–1–2.0944)+0.27735–0.9828/(s–21.5708)+0.277350.9828/(s–2–1.5708)[4] =0.480390.76616/(s+0.5–j0.866025)+0.48039–0.76616/(s+0.5+j0.866025)+0.27735–0.9828/(s+0–j2)+0.277350.9828/(s+0+j2)

v(t) = 20.48039e-0.5tcos(0.866025t+0.76616) + 20.27735e0cos(2t–0.9828) with the focus on the –j terms = 0.96078e-0.5tcos(0.866025t+0.76616) + 0.5547cos(2t–0.9828) [1]

[18]

[14]

[18]

or: [s/(s2+4)–V(s)]/1=I(s)+IC(s)s/(s2+4)=V(s)+I(s)+IC(s) ............................................. (1) and V(s) = sI(s) I(s) = V(s)/s ......................................................................... (2) and V(s) – [IC(s)/s)]–(1/s) = 0 IC(s) = sV(s)–1 .............................................. (3) (2) and (3) in (1): s/(s2+4) = V(s)+V(s)/s+{sV(s)–1} s2/(s2+4) = sV(s)+V(s)+s2V(s)–s (s2+s+1)V(s) = s2/(s2+4)+s = [s2+s(s2+4)]/(s2+4) (s2+s+1)V(s) = [s2+s(s2+4)]/(s2+4) V(s) = (s3+s2+4s)/[(s2+s+1)(s2+4)

42s

s

I(s)

s sI(s)

1/s

V(s)

1

1 [1]

s [1]

42s

s

[1] 1/s

[1]

I(s)

V(s)

IC(s)

1 [1]

IC(s)

1/s IC(s)/s

0.5–90˚

I

1

Is 1–90˚

490˚

or: s/(s2+4)4 – V(s) = I(s) + sV(s) – 1 s/(s2+4) – sI(s) = I(s) + s2I(s) – 1 I(s)(s2+s+1) = (s2+s+3)/(s2+4) I(s) = (s2+s+4)/[(s2+4)(s2+s+1)] = V(s)/s of course I(s) = 0.48039–1.3282/(s–12.0944) + 0.480391.3282/(s–1–2.0944) + 0.138675–2.5536/(s–21.5708) + 0.1386752.5536/(s–2–1.5708) i(t) = 0.96078e-½tcos(0.86603t–1.3282) + 0.27735cos(2t–2.5536) and v(t) = di(t)/dt = –0.48039e-½tcos(0.86603t – 1.3282) – 0.83206e-½tsin(0.86603t – 1.3282) – 0.5547sin(2t – 2.5536) = 0.96078e-½tcos(0.86603t – 1.3282 + 2.0944) – 0.5547cos(2t – 2.5536 – 1.5708) v(t) = 0.96078e-½tcos(0.86603t+07662) – 0.5547cos(2t-4.1244) {or i=0.96078e-½tcos(0.86603t+07662)+0.5547cos(2t-0.9828)}

2H

1F

i

v is = sin2t

iC = dv/dt

1 i

2di/dt

4 2

1/j j

(6)


Question 1

For the circuit in Figure 1, the switch S was closed for t < 0 and is opened at time t = 0. Calculate v(t) for all time t.

Question 2

Determine the zero-state step response v(t), for the circuit shown in Figure 2.

Question 3 Refer to the active circuit in Figure 3 and determine the zero state step response, v(t).

Question 4 For the circuit in Figure 4, the switch S was closed for t < 0 and is opened at time t = 0. Determine a solution for v(t) and i(t) for all time t.






f(t) F(s) f(t) F(s)






jasA

jasA

2B2A

2B2A

0 dtst-f(t)e

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)

(10)

(10)

(14)

(16) Figure 4

S

1 F

t = 0 1 i

v

1 H

1 V

Figure 1

S

1 F

1

v 1 V

t = 0

4 V

2

1

1 F

vs = u(t) 1

v

1

Figure 2

1 F

1

v vs = u(t)

1

1 F

Figure 3


1. t < 0: I = [(4–1)/3] = 1 A v = 4 – 21 = 2 V [2] t 0: v + dv/dt – 1 = 0 dv/dt + v = 1 [5] v = 1 + Ae –t [1] v(0) = 2 2 = 1 + A A = 1 [1] v(t) = 1 + e-t [1]

2. t < 0: v=0 [1] (zero state, given) t 0: (1–v1)/1 = dv/dt + v1/1 1–v1 = dv/dt + v1 …..……. (1) en v1 = dv/dt + v …….…….. (2) (2) in (1): 1–(dv/dt+v) = dv/dt+(dv/dt+v) 3dv/dt + 2v = 1 dv/dt + (2/3)v = (1/3) [6] v = ½ + Ae-(2t/3) [1] v(0) = 0 0 = ½ + A A = –½ [1] v(t) = ½(1 – e-0.6667t) [1]

3. t < 0: v = 0 [½] en v1 = 0 [½] (zero state, given) t 0: (1 – v1)/1=(v1 – v)/1 + d/dt(v1 – v) 1 – v1 = v1 – v + dv1/dt – dv/dt 1 = 2v1 + dv1/dt – dv/dt – v ….. (1) en (v1–v)/1 = dv/dt v1 = dv/dt + v …..…......……… (2) (2) in (1): 1 = 2(dv/dt+v) + d/dt(dv/dt+v) – dv/dt – v 1 = 2dv/dt + 2v + d2v/dt2 + dv/dt – dv/dt – v d2v/dt2 + 2dv/dt + v = 1 [6] =1 and n=1 critically damped v(t) = Ae-t + Bte-t + 1 [3] v(0) = 0 A+1 = 0 A = –1 dv/dt = -Ae-t + Be-t – Bte-t From (2): v1(0) = dv(0)/dt + v(0) dv(0)/dt = 0 –A + B = 0 B = –1 v(t) = –e-t – te-t + 1 [4]

4. t < 0: v = 0 [1] and i = 1 [1] t 0: 1 – i – di/dt – v = 0 ……....……… (1) and i = dv/dt …………...…...…… (2) (2) in (1): 1 – dv/dt – d2v/dt2 – v = 0 d2v/dt2 + dv/dt + v = 1 [6] =0.5 and n = 1 System underdamped with d = 0.866 v = Ae-0.5tcos0.866t + Be-0.5tsin0.866t + 1 [3] v(0) = 0 A + 1 = 0 A = –1 dv/dt = –0.5Ae-0.5tcos0.866t – 0.866Ae-0.5tsin0.866t – 0.5Be-0.5tsin0.866t + 0.866Be-0.5tcos0.866t dv(0)/dt = –0.5A + 0.866B and i(0) = dv(0)/dt (from 2) dv(0)/dt = 1 –0.5(–1) + 0.866B = 1 B = 0.5774 v(t) = –e-0.5tcos0.866t + 0.5774e-0.5tsin0.866t + 1 [3] i(t) = dv/dt = 0.5e-0.5tcos0.866t + 0.866e-0.5tsin0.866t 0.5– 0.2887e-0.5tsin0.866t + 0.5e-0.5tcos0.866t = e-0.5tcos0.866t + 0.5773e-0.5tsin0.866t [2]

[10]

dv/dt

1 1

1

v 1

1

v1 dv/dt

dv/dt I 1

v

1 4

2 1

v

1

dv/dt

t < 0 t 0

v v1

1

1

1

1

1

v

[10]

[14]

1 i

v 1 1

1 i

v

1

1 V

t < 0 t 0

[16]


Question 1 The supply voltage to the circuit in Figure 1, is given by vs(t) = cos t volt. a) Determine the time domain




of the circuit from the perspective that the admittance seen by the source, must become real.

b) Calculate the admittance of the circuit at resonance.

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = cos 2t u(t) volt. The initial current through the inductor is i(0) = 1 amp and the initial voltage across the capacitor is v(0) = 1 volt . Draw an equivalent Laplace network model for the circuit and use this model to find an expression for v(t). (18)






f(t) F(s) f(t) F(s)






jasA

jasA

2B2A

2B2A

0 dtst-f(t)e

Figure 3

1

1 H

vs = cos 2t u(t)

1 F

i

v

i(0) = 1 A

v(0) = 1 V

(10)

(4)

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)

1 1 i

1 F 1 H vs = cos t

Figure 1

Figure 2

1 H 1 F

7 2


1. a) (vs – v)/1 = dv/dt + i and v = i + di/dt vs – i – di/dt = di/dt + d2i/dt2 + i d2i/dt2 + 2di/dt + 2i = cost [6] i = Acost + Bsint and di/dt = –Asint + Bcost, d2i/dt2 = –Acost – Bsint –Acost – Bsint – 2Asint + 2Bcost + 2Acost + 2Bsint = cost [A + 2B]cost + [B – 2A]sint = cost A + 2B = 1 en B – 2A = 0 A = 0.2 [2] en B = 0.4 [2] i(t) = 0.2cost + 0.4sint = 0.4472cos(t – 1.10715r) ampere [2] (12) b) vs = cost Vs = 10 V Zt = 1 + (–j)//(1+j) = 1 + (–j)(1+j)/[–j+1+j] = 1+(–j+1) = 2 – j ( = 2.23607–26.565˚ ) and It = 10/(2–j) A ( = 0.4472126.565˚ ) I = [(–j)/(–j+1+j)][10/(2–j)] = [–j/(2–j)]10 = 0.4472–1.10715r A = 0.4472–63.435˚ A i(t) = 0.4472cos(t – 1.1072) (6) 2. a) Y = [1/(7 + j)]+[1/(2+(1/j))] [2] = [(7–j)/(49+2)]+[j/(1+j2)] = [(7–j)/(49+2)]+[j(1–j2)/(1+42)] = [(7–j)/(49+2)]+[(22+j)/(1+42)] = {[7/(49+2)] + [22/(1+42)]} + j{[/(1+42)] –[/(49+2)]} [6] For resonance: Im{Y} = 0 [/(1+42)] =[/(49+2)] 1+42 = 49+2 32 = 48 2 = 16 0 = 4 r/s [2] (10) {or: Z = (7+j)//[2+(1/j)] = {(7+j)[(1+j2)/j]}/[7+j+2+(1/j)] = [(7+j)(1+j2)]/(j7–2+j2+1) = (7+j15–22)/(1+j9–2) = [(7–22)+j15]/[(1–2)+j9] = {[(7–22)+j15][(1–2)–j9]/[(1–2)2+812] = {[(7–22)(1–2)+1352]+j[15(1–2)–9(7–22)]}/[(1–2)2+812] Im{Z} =0 15(1–2)–9(7–22)=0 15–152–63+182=0 32=48 0=4 r/s and Z0=[(7–32)(1–16)+2160]/[(1–16)2+1296]=2535/1521=1.6667 }

b) Y0 = [7/(49+16)] + [216/(1+416)] = (7/65) + (32/65) = 39/65 = 0.6 S (1.6667) (4) 3. [s/(s2 + 4) – V(s)]/1 = I(s) + IC(s) [2] s/(s2 + 4) = V(s) + I(s) + IC(s) .............. (1) But I(s) = 1/s + V(s)/s and IC(s) = –1 + sV(s) Substitute I(s) and IC(s) in (1): s/(s2+4) = V(s)+[1/s+V(s)/s]+[–1+sV(s)] s2/(s2 + 4) = sV(s) + 1 + V(s) – s + s2V(s) (s2 + s + 1)V(s) = s2/(s2 + 4) + s – 1 = [s2 + s(s2 + 4) – (s2 + 4)]/(s2 + 4) = (s3 + 4s – 4)/(s2 + 4) V(s)=(s3+4s–4)/(s2+s +1)(s2+4) [4] = (s3+4s–4)/[(s–12.0944)(s–1–2.0944)(s–21.5708)(s–2–1.5708)] [2] = A/(s – 12.0944) + A*/(s – 1–2.0944) + B/(s – 21.5708) + B*/(s – 2–1.5708) Now (s3+4s–4)|s=12.0944 = 6.08282.5357 and (s3+4s–4)|s=21.5708 = –4 A = [V(s)(s–12.0944)]s=12.0944 = 0.974031.2075 and B = [V(s)(s–21.5708)]s=21.5708 = 0.27735–0.9828 V(s) = 0.974031.2075/(s–12.0944) + 0.97403–1.2075/(s–1–2.0944) + 0.27735–0.9828/(s–21.5708) + 0.27735–0.9828/(s–2–1.5708) [4] =0.974031.2075/(s+0.5–j0.866025) + 0.97403–1.2075/(s+0.5+j0.866025) + 0.27735–0.9828/(s+0–j2) + 0.27735–0.9828/(s+0+j2) v(t) = 20.97403e – 0.5tcos(0.866025t+1.2075) +20.27735e0cos(2t–0.9828) =1.9481e – 0.5tcos(0.866025t+1.2075)+0.5547cos(2t–0.9828) [1] or:

or (just as a matter of interest): from I(s)=1/s+V(s)/s and IC(s)=–1+sV(s) it follows that V(s)=sI(s)–1 and IC(s)=–1+s2I(s)–s and so from [s/(s2+4)–V(s)]/1=I(s)+IC(s) then, [s/(s2+4)]–[sI(s)–1]=I(s)+[–1+s2I(s)–s]I(s)(s2+s+1)=[s/(s2+4)]+2+s=(s3+2s2+5s+8)/(s2+4) I(s)(s2+s+1)=(s3+2s2+5s+8)/(s2+4)I(s)=(s3+2s2+5s+8)/[(s2+4)(s2+s+1)] (check V(s)=sI(s)–1={(s4+2s3+5s2+8s)/[(s2+4)(s2+s+1)]}–1 V(s)=[(s4+2s3+5s2+8s)–(s4+s3+5s2+4s+4)]/[(s2+4)(s2+s+1)]=(s3+4s–4)/[(s2+4)(s2+s+1)] ) Now (s3+2s2+5s+8)|s=12.0944=6.082750.4413 and (s3+2s2+5s+8)|s=21.5708=21.5708 and I(s) = (s3+2s2+5s+8)/[(s–12.0944)(s–1–2.0944)(s–21.5708)(s–2–1.5708)] I(s)=0.97402–0.88693/(s+0.5–j0.866025)+0.974020.88693/(s+0.5+j0.866025)+0.13867–2.5536/(s+0–j2)+0.138672.5536/(s+0+j2) i(t) = 1.948e–0.5tcos(0.866025t–0.88693) + 0.27734cos(2t–2.5536) and v(t) = di/dt = –0.97402e–0.5tcos(0.866025t–0.88693) –1.68702e–0.5tsin(0.866025t–0.88693)–0.55468sin(2t–2.5536)=1.948–0.5tcos(0.866025t–0.88693+2.0944)–0.55468cos(2t–2.5536–1.5708) v(t) = 1.948–0.5tcos(0.866025t+1.2075)–0.5547cos(2t–4.1244) = 1.948–0.5tcos(0.866025t+1.2075)+0.5547cos(2t–0.98281)

[18]

[18]

[s/(s2+4)–V(s)]/1=I(s)+IC(s) s/(s2+4) = V(s) + I(s) + IC(s) ........................... (1) and V(s) – sI(s) + 1 = 0 I(s) = [V(s)/s] + (1/s) .............................................. (2) and V(s) – [IC(s)/s)] – (1/s) = 0 IC(s) = sV(s) – 1 .......................................... (3) (2) and (3) in (1): s/(s2+4) = V(s)+{[V(s)/s]+(1/s)}+{sV(s)–1} s2/(s2+4) = sV(s)+V(s)+1+s2V(s)–s (s2+s+1)V(s) = s2/(s2+4)+s–1 (s2+s+1)V(s) = [s2+s(s2+4)–(s2+4)]/(s2+4) V(s) = (s3+4s–4)/[(s2+s+1)(s2+4)

42s

s

I(s)

1

s sI(s)

1/s

V(s)

1

IC(s)

1/s IC(s)/s

[14]

1 1 I

–j j Vs

10

It

v

1 1 i

1 1 vs=cost

1

s [1]

42s

s

[1] 1/s

[1]

I(s)

V(s)

IC(s)

1 [1]

s1

[1]

7 2

1/j j


Question 1 Determine the zero state step response of the current i(t) through the 1 H inductor, for the circuit in Figure 1. (10)

Question 2 For the circuit shown in Figure 2, find an expression for v(t) for all t, if is(t) is given by:

0 for t1

0 for t 1(t)is

Question 3 For the circuit shown in Figure 3, determine the current response i(t), in the 1 H inductor, for all time t, if vs(t) is given by:

0 for t2

0 for t 2 (t)vs

Question 4 For the circuit shown in Figure 4, find the zero state step response v(t), if the input to the circuit is the step current source, is(t) = u(t) ampere.





f(t) F(s) f(t) F(s)






jasA

jasA


(10)

(15)

(15)

2B2A

2B2A

0 dtst-f(t)e

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)

1 H

1 i

vs = u(t) (volt)

1

Figure 1

1

Figure 2

is

1

v1 F 1

–1

t

is(t) (amp)

0

1

1 H

1

i

1 F vs

1

t

–2

vs(t)

0

2

Figure 3

Figure 4

v 1 F 1 H

is = u(t) 1 1


1. Vir t 0: i(t) = 0 [1] (zero toestand respons) Vir t > 0: 1 – v’ = i + v’ en v’ = i + di/dt 1 - (i + di/dt) = i + (i + di/dt) 2di/dt + 3i = 1 di/dt + 3/2i = ½ [5] i(t) = 1/3+A e–3/2t [1] But i(0) = 0 A = –1/3 [1] i(t) = 1/3(1 – e–3/2t) for t 0 [2] 2. t < 0: v(t) = –1V [2] t 0: 1 = v + dv/dt dv/dt + v = 1 [5] v = 1 + Ae–t [1] v(0) = –1 A = –2 [1] v(t) = 1 – 2e–t for t 0 [1] 3. t < 0: v(t) = 1 V [1] en i(t) = 1 A [1] t 0: Loop L: –2 – i – di/dt – v = 0 v = –2 – i – di/dt ................................. (1) node N: i = v + dv/dt .............................. (2) (1) in (2): i = (–2 – i – di/dt) + d/dt(–2 – i – di/dt) i = –2 – i – di/dt – di/dt –d2/dt2 d2i/dt2 + 2di/dt + 2i = –2 [6] = 1 en n = 2 Ondergedemp met d = 1: i(t) = –1 + Ae–tcost+ Be–tsint [3] i(0) = –1 + A But i(0) = 1 1= –1 +A A = 2 and di/dt = –Ae–tcost – Ae–tsint –Be–tsint + Be–tcost di(0)/dt = –A + B = –2 + B. But from (1): v(0) = –2 – i(0) – di(0)/dt { also for v(t): d2v/dt2 + 2dv/dt + 2v = –2 } 1 = –2 – 1 – di(0)/dt di(0)/dt = –4 { with v(0)=1 and dv(0)/dt=i(0)-v(0)=1–1=0 } –2 + B = –4 B = –2 { or from v(t) = –2 – i(t) – di(t)/dt } i(t) = –1 + 2e–tcost – 2e–tsint [4] { is v(t) = –1 + 2e–tcost + 2e–tsint } 4. t < 0: i(0) = 0A en v(0) = 0 [1] zero state t 0: Node N: 1 = i + dv/dt i = 1 – dv/dt ......... (1) Loop L: di/dt + i – dv/dt – v = 0 ................... (2) (1) in (2): d/dt(1 – dv/dt) + (1 – dv/dt) – dv/dt – v = 0 0 – d2v/dt2 + 1 – dv/dt – dv/dt – v = 0 d2v/dt2 + 2dv/dt + v = 1 [6] = 1 en n = 1 critically damped v(t) = 1 + Ae-t + Bte-t [4] v(0) = 1 + A and v(0) = 0 1 + A = 0 A = –1 dv/dt = –Ae-t – Bte-t + Be-t dv(0)/dt = –A + B = 1 + B (as A = –1) and from (1): i(0) = 1 – dv(0)/dt dv(0)/dt = 1 – i(0) = 1 – 0 1 + B = 1 B = 0 v(t) = 1 – e-–t [4]

v’

1

i

1 H

1 1 V

[10]

[10]

–1

1

v 1

1

1

v

1 F 1

t 0 t < 0 dv/dt v

1 v’

[15]

t 0 1 H 1

v

1 F

i

1

–2Vdi/dt

N

i dv/dtv

L

t < 0

v i

1 2V

1

[15]

v

1 A i

1

1F

1H

1

dv/dt

di/dt L

i dv/dt

N

Circuit Analysis IV EICAM4A Unit 2 Second Assessment 20 October 2016 Page 1

Question 1 Refer to the circuit in Figure 1. a) If the supply voltage to the circuit is

given by vs(t) = cos 2t, determine the time domain solution for the steady state capacitor voltage response, v(t), by writing down and solving the differential equation for v(t). Give your answer in the form Acos(t+). (12)

b) Verify your answers for A and in Question 1 a), by solving for the steady state capacitor voltage response v(t) in the form v(t) = Aej(2t+) , by applying the complex sinusoid voltage vs(t) = ej2t as the input voltage source to the circuit. (4)

c) Use frequency domain analysis to determine the phasor voltage V across the capacitor in Figure 1. Reconstruct the time domain expression for v(t), from the phasor voltage V, in the form Acos(t+) (4)


of the circuit from the perspective that the impedance seen by the source, must become real.

b) Calculate the impedance of the circuit at resonance.

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = cos2t u(t) volt. The initial current through the inductor is i(0) = 1 amp and the initial voltage across the capacitor is v(0) = 0 volt.

a) Draw an equivalent Laplace network model for the circuit

b) Use this model to find an expression for the voltage v(t) across the ½ F capacitor. (12)






f(t) F(s) f(t) F(s)






jasA

jasA

Figure 2

Figure 1

vs = cos 2t

vs = ej2t Vs = 10

v

1 H 3

21 F

1

1 H 2

1 F

(10)

(4)

(4)

2B2A

2B2A

0 dtst-f(t)e

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)

i

i(0) = 1 A

v(0) = 0 V

Figure 3

vs = cos 2t u(t) v

1 H 3

21 F

Stroombaananalise IV EICAM4A Eenheid2 Tweede Evaluasie 20 Oktober 2016 Memorandum Bladsy 1

i(t) = [0.31622cos(2t – 0.32175) – 0.8e –t + 1.5e –2t]u(t) v(t) = dt it

0C1 + v(0) = 2 dt i

t

0 + 0

v(t) = 2 t

0[0.31622cos(2t–0.32175)–0.8e –t+1.5e –2t] = 2[0.15811sin(2t–0.32175)+0.8e –t–0.75e –2t]

v(t) = 0.31622cos(2t – 0.32175 – /2) + 1.6e –t – 1.5e –2t = 0.31622cos(2t+1.249 – 1.89255) + 1.6e –t – 1.5e –2t }

1. a) vs – 3i – di/dt – v = 0 and i = ½dv/dt vs – (3/2)dv/dt – ½d2v/dt2 – v = 0 d2v/dt2 + 3dv/dt + 2v = 2vs d2v/dt2 + 3dv/dt + 2v = 2cos2t [6] v = Acos2t + Bsin2t dv/dt = –2Asin2t + 2Bcos2t and d2v/dt2 = –4Acos2t – 4Bsin2t –4Acos2t – 4Bsin2t – 6Asin2t + 6Bcos2t + 2Acos2t + 2Bsin2t = 2cos2t [–2A + 6B]cos2t + [–6A – 2B]sin2t = 2cos2t –2A + 6B = 2 and –6A – 2B = 0 A = –0.1 and B = 0.3 vss(t) = –0.1cos2t+0.3sin2t = 0.316228cos(2t–1.89255) volt [6] (12) b) Assuming vs(t) = ej2t and v(t) = Aej(2t+) : dv/dt = j2Aej(2t+) [1] and d2v/dt2 = –4Aej(2t+) [1] –4Aej(2t+)

+ j6Aej(2t+) + 2Aej2t+)

= 2ej2t [1] Aej2tej[(–4+2)+j6] = 2ej2t Aej(–2+j6) = 2 A6.32461.8925 = 20

[1] 6.3245A=2 and +1.8925=0 A = 0.31623 and = –1.8925 (4)

c) ZL=L90˚=290˚ and ZC=(1/C)–90˚=1–90˚ ZT = 30˚+290˚+1–90˚ = 3.1622818.43495˚ [1] I = (10)/3.1622818.43495˚ = 0.316228–18.435˚ A

[1] V = IZC = 0.316228–18.435˚1–90˚ = 0.316228–108.435˚ volt [1]

vss(t) = 0.316228cos(2t – 1.89255r) volt [1] (4)

2. a) Z = 1 + 1/j + 2//(j) [2] = 1 – j(1/) + (j2)/(2 + j) = 1 – j(1/) + j2(2 – j)/(4 + 2) = 1 – j(1/) + (j4 + 22)/(4 + 2) = [1 + 22/(4 + 2)] + j[4/(4 + 2) – 1/] [2] For resonance Im{Z} = 0 4/(4 + 2) – 1/ = 0 [2] 4/(4 + 2) = 1/ 42 = 4 + 2 32 = 4 r = 2/3 = 1.154701 r/s [4] (10) b) Zr = 1 + 2r

2/(4 + r2) = 1 + (8/3)/[4 + (4/3)] = 1.5 (4)

3. a) b) s/(s2 + 4) – 3I(s) – sI(s) + 1 – V(s) = 0 [2] ...... (1)

But V(s) = 2I(s)/s I(s) = sV(s)/2 ........... (2) (2) in (1): s/(s2+4) – 3sV(s)/2 – s2V(s)/2 + 1 – V(s) = 0 s2V(s)/2 + 3sV(s)/2 + V(s) = s/(s2 + 4) + 1 s2V(s) + 3sV(s) + V(s) = 2s/(s2 + 4) + 2 s2V(s)+3sV(s)+2V(s) = [2s+2(s2+4)]/(s2+4) (4) (s2 + 3s + 2)V(s) = (2s2 + 2s + 8)/(s2 + 4) V(s) = 2(s2 + s + 4)/(s2 + 4)(s2 + 3s + 2) [2] V(s) = 2(s2 + s + 4)/[(s – 21.5708)(s – 2–1.5708)(s + 1)(s + 2)] [2] = A/(s – 21.5708) + A*/(s – 2–1.5708) + B/(s + 1) + C/(s + 2) B = 2(s2+s+4)/(s2 + 4)(s + 2)|s = –1 = 8/5 = 1.6 and C = 2(s2+s+4)/(s2 + 4)(s + 1)|s = –2 = 12/(–8) = –1.5 A = 2(s2 + s + 4)/(s – 2–1.5708)(s + 1)(s + 2)|s = 21.5708 = 0.15811– 1.8925 V(s) = (0.15811–1.8925)/(s–21.5708)+(0.158111.8925)/(s–2–1.5708) +1.6/(s+1) –1.5/(s+2) [4]

= (0.15811–1.8925)/(s+0–j2)+(0.158111.8925)/(s+0+j2) +1.6/(s+1) –1.5/(s+2) v(t) = [20.15811e0tcos(2t – 1.8925) + 1.6e –t – 1.5e –2t]u(t) = [0.31622cos(2t – 1.8925) + 1.6e –t – 1.5e –2t]u(t) [2] (12) { I(s) = sV(s)/2 = (s/2)[2(s2 + s + 4)/(s2 + 4)(s2 + 3s + 2)] = (s3 + s2 + 4s)/(s2 + 4)(s2 + 3s + 2) = A/(s – 21.5708) + A*/(s – 2–1.5708) + B/(s + 1) + C/(s + 2) B = (s3+s2+4s)/(s2 + 4)(s + 2)|s = –1 = –(4/5) = –0.8 , C = (s3 + s2 + 4s)/(s2 + 4)(s + 1)|s = –2 = 12/8 = 1.5 A = (s3 + s2 + 4s)/(s – 2–1.5708)(s + 1)(s + 2)|s = 21.5708 = 0.15811– 0.32175 I(s) = (0.15811–0.32175)/(s+0–j2)+(0.158110.32175)/(s+0+j2) – 0.8/(s+1) + 1.5/(s+2)

[14]

[20]

1 H 3

vs = cos 2t v

½ F

i

1

j 2

1/j

[16]

I(s)

V(s)

s [1] 3

42s

s

[1]

2/s [1]

2I(s)/s

3I(s) sI(s) 1 [1]

290˚ 3

10 V1–90˚

I

= 2


Question 1 For the circuit shown in Figure 1, find an expression for i(t) for all t,


0 for t1

0 for t1 (t)i

s (10)

Question 2 Determine the zero state step response v(t) for the circuit in Figure 2. (10)

Question 3 For the circuit in Figure 3, determine the zero state step response v(t). (12) Question 4 The switch S, in the circuit of Figure 4, is opened at time t = 0. Determine i(t) and v(t) for all t. (18)

---ooo000ooo--- Total: 50 Appendix Trigonometric identities: Network models Acost + Bsint = cos (t – tan-1(B/A)) cos = sin( + /2)




f(t) F(s) f(t) F(s)






jasA

jasA

2 3 i

i is

Figure 1

1

1

–1

t

is(t) ampere

0

1

1 H

Figure 2

vs(t) = u(t) 3 H 2i v

Figure 4

t = 0 251 F

1 H 6 i

v 6 V

vs = u(t) 1 F

3 1 H

Figure 3

v 1

S

2B2A

2B2A

0 dtst-f(t)e

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)


i

1. t < 0: i(t) = -½ A [2] t 0: 1 = v’ + i en v’ = i + di/dt 1 = i +di/dt + i di/dt + 2i = 1 [5] i = ½ + Ae-2t Maar i(0) = -½ , -½ = ½ + A A = -1 i(t) = ½ - e-2t [3] vir t 1

2. t < 0: i(t) = v(t) = iL(t) = 0 [1] (zero state) t 0: i + 2i = iL 3i = iL

v = 3diL/dt 1 = 3i + 2iL + v 1 = iL + 2iL + 3diL/dt diL/dt + iL = 1/3 [5] iL = 1/3 + Ae-t iL(0) = 0 A = –1/3 iL(t) = (1/3)(1 – e-t) [2] v(t) = 3diL/dt = 3(1/3)e-t = e-t [2]

3. t < 0: v(t) = 0 [½] , i(t) = 0 [½] (zero state) t 0: 3i + di/dt = 1 – v en i = v + dv/dt 3(v + dv/dt) +d/dt(v + dv/dt) = 1 – v 3v + 3dv/dt +dv/dt + d2v/dt2 = 1 – v d2v/dt2 + 4dv/dt + 4v = 1 [5] = 2 en n = 2 kritiek gedemp v(t) = Ate-2t + Be-2t + ¼ [2] v(0) = 0 B + ¼ = 0 B = -¼ and dv/dt = Ae-2t – 2Ate-2t – 2Be-2t dv(0)/dt = A – 2B but dv(0)dt = i(0) – v(0) A – 2B = 0 A – 2(–1/4) = 0 A = –1/2 v(t) = – ½te-2t – ¼e-2t + ¼ [4] 4. t < 0: i(t) = 6/6 = 1A [1] en v(t) = 0 [1] t 0: 6 = 6i + di/dt + v en i =(1/25)dv/dt (6/25)dv/dt +(1/25)dv2/dt2 + v = 6 d2v/dt2 + 6dv/dt + 25v = 150 [6] = 3 en n = 5 ondergedempte stelsel met d = 4 r/s v(t) = 6 + Ae-3tcos4t + Be-3tsin4t [4] v(0) = 0 0 = 6 + A A = -6 dv/dt = -3Ae-3tcos4t – 4Ae-3tsin4t –3Be-3tsin4t + 4Be-3tcos4t = (4B-3A)e-3tcos4t – (4A+3B)e-3tsin4t …………........(i) dv(0)/dt = -3A + 4B = 18 + 4B maar i(0) = (1/25)dv(0)/dt dv(0)/dt = 25 18 + 4B = 25 4B = 7 B = 1.75 v(t) = 6 – 6e-3tcos4t + 1.75e-3tsin4t [3] en i(t) = (1/25)dv/dt = (1/25)[25e-3tcos4t + 18.75e-3tsin4t] van (i) = e-3tcos4t + 0.75e-3tsin4t [3]

i

v’

1 H is

i 1

1

[10]

v

iL

2i 1 V

2 3

3 H

1 1 F

3 1 H

v 1

251 F

1 H 6 i

v 6 V

251 F

1 H 6 i

v 6 V

[18]

t < 0

t 0

[10]

[12]

Circuit Analysis IV EICAM4A Unit 2 Third Assessment 10 November 2016 Page 1

Question 1 The supply voltage to the circuit in Figure 1, is given by vs(t) = sin 2t volt.

a) Determine the time domain solution for the steady state current response, i(t) through the ½ H inductor, by writing down and solving a differential equation for i(t). Give your answer in the form i(t) = Acos(t+).

b) Given that vs(t) = sin 2t = cos (2t – /2), use Vs = 1–90 volt as the phasor input supply voltage, and determine the phasor current I, through the ½ H inductor in Figure 1, using frequency domain analysis. Reconstruct the time domain expression for i(t), from the phasor current I, in the form Acos(t+) (8)


of the circuit from the perspective that the admittance Y, seen by the source, must become real.

b) Calculate the admittance of the circuit at resonance.

Question 3 Refer to the circuit in Figure 3. The supply voltage to the circuit is given by vs(t) = cos 2t u(t) volt. The initial current through the inductor is i(0) = 0 amp and the initial voltage across the capacitor is v(0) = 0 volt.

a) Draw an equivalent Laplace network model for the circuit.

b) Use this model to find an expression for the current i(t) through the ½ H inductor. (14)






f(t) F(s) f(t) F(s)






jasA

jasA

2B2A

2B2A

0 dtst-f(t)e

(8)

(4)

I(s)

sL

Li(0)

V(s) s

i(0)

I(s)

V(s)sL

I(s)

sC1

s

v(0)V(s)

sC1 Cv(0)

I(s)

V(s)

Figure 2

2 H

1 F 1

Figure 1

i

1 F vs = sin 2t Vs = 1–90

21 H

31

(12)

(4) Figure 3

vs = cos 2t u(t)

1 F v

v(0) = 0 V

31

i

i(0) = 0 A

21 H

I

Y

Stroombaananalise IV EICAM4A Eenheid2 Derde Evaluasie 10 November 2016 Memorandum Bladsy 1

1. a) (vs – v)/⅓ = dv/dt + i and v = ½di/dt vs – ½di/dt = ⅓d/dt(½di/dt) + ⅓i d2i/dt2+3di/dt+2i=6vs d2i/dt2+3di/dt+2i = 6sin2t [6] i = Acos2t + Bsin2t di/dt = –2Asin2t+2Bcos2t and d2i/dt2 = –4Acos2t–4Bsin2t –4Acos2t–4Bsin2t–6Asin2t+6Bcos2t+2Acos2t+2Bsin2t=6sin2t [–2A+6B]cos2t+[–2B–6A]sin2t = 6sin2t –2A+6B=0 and –2B–6A=6 A= –0.9 and B=–0.3 i(t) = –0.9cos2t – 0.3sin2t ampere [5] = 0.94868cos(2t + 2.8198) A [1] (12) b) vs = sin2t Vs = 1–90 V Zt = ⅓ + (0.5–90)//(190) [1] = ⅓ + (0.50)/(0.590) = ⅓ + 1–90 = 1.0541–71.562 [2] and It = 1–90/1.0541–71.562 = 0.94868–18.438 A [2] I = [0.5–90/(0.5–90+190)]0.94868–18.438 = 1–1800.94868–18.438 = 0.94868–198.438 [2] i(t) = 0.94868cos(2t – 3.4634) = 0.94868cos(2t + 2.8198) [1] (8) 2. a) Y = [1/(1 + j2)] + j [2] = [(1 – j2)/(1 + 42)] + j = [1/(1 + 42)] + {j – [j2/(1+42)]} = [1/(1 + 42)] + j{1 – [2/(1+42)]} For resonance: Im{Y} = 0 1 – [2/(1+42)] = 0 2/(1+42) =1 1+42 = 2 42 = 1 2 = ¼ 0 = ½ rad/sec [6] (8) b) Y0 = [1/(1+1)] = ½ Siemens (2) (4) 3. a) b) [2/(s2 + 4) – V(s)]/⅓ = V(s)/(1/s) + I(s) ... (1) [1] and V(s) = I(s)(s/2) ................................(2) [1] (2) in (1): [2/(s2+4)–(sI(s)/2)] = (s/3)(sI(s)/2) + ⅓I(s) s2I(s)/6 + sI(s)/2 + ⅓I(s) = 2/(s2+4) (s2 + 3s + 2)I(s) = 12/(s2 + 4) (4) I(s) = 12/[(s2 + 4)(s2 + 3s + 2)] [4] = 12/(s–21.5708)(s–2–1.5708)(s+1)(s+2) [2] I(s) = A/(s – 21.5708) + A*/(s – 2–1.5708) + B/(s+1) + C/(s+2) A = 12/(s–2–1.5708)(s+1)(s+2)|s = 21.5708 = 0.474342.8198 B = 12/(s2+4)(s+2)|s=-1 = 2.4 and C=12/(s2+4)(s+1)|s=-2 = –1.5 I(s) = (0.474342.8198)/(s–21.5708)+(0.47434–2.8198)/(s–2–1.5708)+2.4/(s+1)–1.5/(s+2) [4] = (0.474342.8198)/(s + 0 – j2)+(0.47434–2.8198)/(s + 0 + j2) + 2.4/(s + 1) – 1.5/(s + 2) i(t) = 0.94868cos(2t + 2.8198) + 2.4e-t – 1.5e-2t [2] (14)

[20]

s/2 [1]

42s

2

[2] 1/s

[1] I(s)

V(s)

⅓0 I

0.5–90

190 Vs 1–90

It

⅓

v

⅓ i

1

½

vs=sin2t

1

1/j j2

[12]

[18]

Documents

CIRCUIT ANALYSIS IV - MWEBhome.mweb.co.za/rf/rfitch/eicam4/CamEval.pdf · Circuit Analysis EICAM4 Unit 2 First Assessment 4 October 2007 Page 1 Question 1 Refer to the circuit in