15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients...

15: Transients (A)

• Differential Equation

• Piecewise steady stateinputs

• Step Input

• Negative exponentials

• Exponential Time Delays

• Inductor Transients

• Linearity

• Transient Amplitude

• Capacitor VoltageContinuity

• Summary

E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 1 / 11

Differential Equation

15: Transients (A)

• Step Input

• Linearity

• Summary

To find y(t):

15: Transients (A)

• Step Input

• Linearity

• Summary

To find y(t):x(t) constant: Nodal analysis

15: Transients (A)

• Step Input

• Linearity

• Summary

To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysis

15: Transients (A)

• Step Input

• Linearity

• Summary

To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

General Solution: Particular Integral + Complementary Function

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

Particular Integral: Any solution to RC dydt + y = x

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution for y(t).

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

Complementary Function: Solution to RC dydt + y = 0

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

Does not depend on x(t), only on the circuit.

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

Does not depend on x(t), only on the circuit.Solution is y(t) = Ae−

where τ = RC is the time constant of the circuit.

15: Transients (A)

• Step Input

• Linearity

• Summary

i(t) = C dydt =

x−yR ⇒ RC dy

dt + y = x

Does not depend on x(t), only on the circuit.Solution is y(t) = Ae−

where τ = RC is the time constant of the circuit.

The amplitude, A, is determined by the initial conditions at t = 0.

Piecewise steady state inputs

15: Transients (A)

• Step Input

• Linearity

• Summary

We will consider input signals that are sinusoidal or constant for a particulartime interval and then suddenly change in amplitude, phase or frequency.

15: Transients (A)

• Step Input

• Linearity

• Summary

Output is the sum of the steady state and a transient:y(t) = ySS(t) + yTr(t)

15: Transients (A)

• Step Input

• Linearity

• Summary

Steady state, ySS(t), is the same frequency as theinput; use phasors + nodal analysis.

15: Transients (A)

• Step Input

• Linearity

• Summary

Transient is always yTr(t) = Ae−t

τ at each change.

15: Transients (A)

• Step Input

• Linearity

• Summary

Transient is always yTr(t) = Ae−t

τ at each change.

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy

dt + y = x= 4Time Const: τ = RC = 1ms

-1 0 1 2 3

t (ms)

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

Transient (Complementary Function)yTr(t) = Ae−

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

Steady State + Transienty(t) = ySS + yTr = 4 +Ae−

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

To find A, use capacitor property:Capacitor voltage never changes abruptly

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

y(0+) = 4 +A and y(0−) = 1

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

So transient: yTr(t) = −3e−t/τ

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

So transient: yTr(t) = −3e−t/τ and total y(t) = 4− 3e−t/τ

Step Input

15: Transients (A)

• Step Input

• Linearity

• Summary

-1 0 1 2 3

t (ms)

-1 0 1 2 3

t (ms)

y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3

So transient: yTr(t) = −3e−t/τ and total y(t) = 4− 3e−t/τ

Transient amplitude ⇐ capacitor voltage continuity: vC(0+) = vC(0−)

Negative exponentials

15: Transients (A)

• Step Input

• Linearity

• Summary

Positive exponentials grow to ±∞:et

0 1 2 3 4 5

15: Transients (A)

• Step Input

• Linearity

• Summary

Positive exponentials grow to ±∞:et, 3et/4

0 1 2 3 4 5

10 et3e¼t

15: Transients (A)

• Step Input

• Linearity

• Summary

Positive exponentials grow to ±∞:et, 3et/4, −2et/2

0 1 2 3 4 5

10 et3e¼t

-2e½t

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

Negative exponentials decay to 0:2e−t

0 2 4 6 8-2

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

Negative exponentials decay to 0:2e−t, e

−t/4

0 2 4 6 8-2

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

0 2 4 6 8-2

-2e-½t

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

Transients are negative exponentials.

0 2 4 6 8-2

-2e-½t

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

0 2 4 6 8-2

-2e-½t

Decay rate of e−t/a

a 2a 3a 4a 5a0

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

0 2 4 6 8-2

-2e-½t

37% after 1 time constant

a 2a 3a 4a 5a0

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

0 2 4 6 8-2

-2e-½t

37% after 1 time constant5% after 3

a 2a 3a 4a 5a0

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

0 2 4 6 8-2

-2e-½t

37% after 1 time constant5% after 3, <1% after 5

a 2a 3a 4a 5a0

0.05 0.01

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

0 2 4 6 8-2

-2e-½t

a 2a 3a 4a 5a0

0.05 0.01

Gradient of e−t/a

Gradient at t hits zero at t+ a.

a 2a 3a0

15: Transients (A)

• Step Input

• Linearity

• Summary

0 1 2 3 4 5

10 et3e¼t

-2e½t

−t/4, −2e−t/2

0 2 4 6 8-2

-2e-½t

a 2a 3a 4a 5a0

0.05 0.01

Gradient of e−t/a

Gradient at t hits zero at t+ a.True for any t.

a 2a 3a0

Exponential Time Delays

15: Transients (A)

• Step Input

• Linearity

• Summary

Negative exponential with a finalvalue of F .

y(t) = F + (A− F ) e−(t−TA)/τ

15: Transients (A)

• Step Input

• Linearity

• Summary

y(t) = F + (A− F ) e−(t−TA)/τ

How long does it take to go from A to B ?

15: Transients (A)

• Step Input

• Linearity

• Summary

y(t) = F + (A− F ) e−(t−TA)/τ

At t = TB :y(TB) = B = F + (A− F ) e−(TB−TA)/τ

15: Transients (A)

• Step Input

• Linearity

• Summary

y(t) = F + (A− F ) e−(t−TA)/τ

B−FA−F = e

−(TB−TA)/τ

15: Transients (A)

• Step Input

• Linearity

• Summary

y(t) = F + (A− F ) e−(t−TA)/τ

B−FA−F = e

−(TB−TA)/τ

Hence TB − TA = τ ln(

A−FB−F

= τ ln(

initial distance toFfinal distance toF

15: Transients (A)

• Step Input

• Linearity

• Summary

y(t) = F + (A− F ) e−(t−TA)/τ

B−FA−F = e

−(TB−TA)/τ

Hence TB − TA = τ ln(

A−FB−F

= τ ln(

initial distance toFfinal distance toF

Useful formula - worth remembering.

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

Solution: Particular Integral + Complementary Function

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

Particular Integral: Any solution to LR

dydt + y = L

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

Complementary Function: Solution to LR

dydt + y = 0

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

dydt + y = 0

Does not depend on x(t), only on the circuit.

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

dydt + y = 0

Does not depend on x(t), only on the circuit.Solution is yTr(t) = Ae−

where τ = LR is the time constant of the circuit.

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

dydt + y = 0

1st order transient is always yTr(t) = Ae−t/τ where τ = RC or L

Inductor Transients

15: Transients (A)

• Step Input

• Linearity

• Summary

We know i = x−yR

y(t) = L didt =

d(x−y)dt = L

Rdxdt − L

⇒ LR

dydt + y = L

dydt + y = 0

1st order transient is always yTr(t) = Ae−t/τ where τ = RC or L

RAmplitude A ⇐ no abrupt change in capacitor voltage or inductor current.

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

1st order circuit has only one C or L.

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C .

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae

−t/τ

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

−t/τ

Time constant is τ = RThC

where RTh is the Thévenin resistance.

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

−t/τ

Replace the capacitor with a voltage sourcev(t); all voltages and currents in the circuitwill remain unchanged.

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

−t/τ

Linearity: y = ax+ bv = ax+ bvSS + bvTr

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

−t/τ

Linearity: y = ax+ bv = ax+ bvSS + bvTr = ySS + bvTr

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

−t/τ

All voltages and currents in a circuit have the same transient (but scaled).

Linearity

15: Transients (A)

• Step Input

• Linearity

• Summary

−t/τ

All voltages and currents in a circuit have the same transient (but scaled).

The circuit’s time constant is τ = RThC or LRTh

where RTh is theThévenin resistance of the network connected to C or L.

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

Find Steady State (DC ⇒ ZL = 0)

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Time Constant

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Time ConstantSet x ≡ 0 → RTh = 2k

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

Time ConstantSet x ≡ 0 → RTh = 2kτ = L

RTh= 2µs

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

RTh= 2µs

Result

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

RTh= 2µs

Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

RTh= 2µs

= 3 + (5− 3) e−t/τ

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

RTh= 2µs

= 3 + (5− 3) e−t/τ

= 3 + 2e−t/τ

Transient Amplitude

15: Transients (A)

• Step Input

• Linearity

• Summary

ySS(0−) = 1, ySS(0+) = 3

At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5

RTh= 2µs

= 3 + (5− 3) e−t/τ

= 3 + 2e−t/τ

Capacitor Voltage Continuity

15: Transients (A)

• Step Input

• Linearity

• Summary

Find Steady State (DC ⇒ ZC = ∞)

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x

4R + v8R + v−y

2R = 0

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

KCL @ Y: y−v2R + y−x

6R = 0

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0 -RC 0 RC 2RC 3RC

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

Time Constantτ = RThC = 2RC (from earlier slide)

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

= 134 +

− 54 − 13

e−t/τ

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

= 134 +

− 54 − 13

e−t/τ

= 134 − 18

4 e−t/τ = 3 1

4 − 4 12e

−t/2RC

-RC 0 RC 2RC 3RC-4

15: Transients (A)

• Step Input

• Linearity

• Summary

4R + v8R + v−y

2R = 0

6R = 0

vSS = 34x, ySS = 13

At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)

2R + y−46R = 0

y(0+) = −9+44 = − 5

= 134 +

− 54 − 13

e−t/τ

= 134 − 18

4 e−t/τ = 3 1

4 − 4 12e

−t/2RC

-RC 0 RC 2RC 3RC-4

Summary

15: Transients (A)

• Step Input

• Linearity

• Summary

• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily

continuous unless it equals vC .

Summary

15: Transients (A)

• Step Input

• Linearity

• Summary

• Circuit time constant: τ = RThC or LRTh

◦ RTh is the Thévenin resistance seen by C or L.◦ Same τ for all voltages and currents.

Summary

15: Transients (A)

• Step Input

• Linearity

• Summary

• Output = Steady State + Transient◦ Steady State: use nodal/Phasor analysis when input is piecewise

constant or piecewise sinusoidal. The steady state has the samefrequency as the input signal.

Summary

15: Transients (A)

• Step Input

• Linearity

• Summary

◦ Transient: Find vC(0−) or iL(0−): unchanged at t = 0+Find y(0+) assuming source of vC(0+) or iL(0+)Amplitude never complex, never depends on t.

Summary

15: Transients (A)

• Step Input

• Linearity

• Summary

◦ y(t) = ySS(t) + (y(0+)− ySS(0+)) e−t/τ

Summary

15: Transients (A)

• Step Input

• Linearity

• Summary

◦ y(t) = ySS(t) + (y(0+)− ySS(0+)) e−t/τ

See Hayt Ch 8 or Irwin Ch 7.

15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients...

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