Ultimo 22 supch15.pdf

CHAPTER 15 - LAPLACE TRANSFORM

List of topics for this chapter:Definition and Properties of the Laplace TransformInverse Laplace TransformApplication to CircuitsTransfer FunctionsConvolution IntegralApplication to Integrodifferential EquationsApplications

DEFINITION AND PROPERTIES OF THE LAPLACE TRANSFORM

The Laplace transform is an integral transformation of a function f(t) from the time domain intothe frequency domain, giving F(s).

Problem 15.1 Find the Laplace transform for(a) )t(u)e1010( t3-(b) )1t(u)e1010( 1)-(t-3 (c) )1t(u)e1010( t3-

(a) =+

3s10

s

10

++++ )3s(s3

10

(b) =

+

=

+ 3s

1s

1e103s

e10s

e10s-

s-s-

++++ )3s(s3

e10 s-

(c) Manipulate this to match a transform pair using the time-shifting property.)1t(u)e1010( t3- )1t(u)e1010( )11t(3- = +

)1t(u)ee1010( 3-1)-(t-3 =

The Laplace transform is

3see10

s

e10 s-3-s-

+

+

= 3se

s

1e10

3-s-

=

++++

++++

)3s(s3s)e1(

e10-3

s-

Problem 15.2 Find the Laplace transform for(a) )t(u]etet[ t2-2t3- +(b) )t(u]ttt[ 32 ++(c) )1t(u]tt[ 2 +(d) )t(u]ete)4t(et[ t-2t-t- +++

(a) 32 )2s(2

)3s(1

++++++++

++++

(b) =++ 432 s6

s

2s

14

2

s

6s2s ++++++++

(c) Manipulate this to match a transform pair using the time-shifting property.)1t(u]tt[ 2 + )1t(u])11t()11t([ 2 +++=

)1t(u]1)1t(2)1t(1)1t([ 2 ++++=)1t(u])2)1t(2)1t()1t([ 2 +++=

)1t(u]2)1t(3)1t([ 2 ++=


=

++=++s

3s

3s

2e

s

e3s

e3s

e223

s-

s-

2

s-

3

s-

]2s3s3[s

e 23

s-

++++++++

(d) First, expand the third term. Then, combine like terms.)t(u]ete)4t(et[ t-2t-t- +++ )t(u]ete4etet[ t-2t-t-t- +++=

)t(u]etete3t[ t-2t-t- ++=


322 )1s(2

)1s(1

)1s(3

s

1++++

++++++++

++++++++

INVERSE LAPLACE TRANSFORM

Finding the inverse Laplace transform of F(s) involves two steps:1. Decomposing F(s) into simple terms using partial fraction expansion, and2. Finding the inverse of each term by matching entries in a Laplace transform table.

Partial Fraction Expansion

Suppose F(s) has the general form )s(D)s(N)s(F = . Consider the three possible forms F(s) may

take and how to perform partial fraction expansion of F(s) for each form.

Real Poles

If F(s) has only real poles, then the denominator becomes a product of factors so that

)ps()ps)(ps()s(N)s(F

n21 +++= ,

where n21 p-,,p-,p-s = are the real poles and ji pp for all ji .

Assuming the degree of N(s) is less than the degree of D(s), partial fraction expansion is used todecompose F(s) as

n

n

2

2

1

1

psk

psk

psk)s(F

+++

++

+=

There are many ways to find the expansion coefficients, n,21 k,k,k , or residues of F(s). Oneway is the residue method. If both sides of the equation are multiplied by )ps( 1+ , then

n

1n

2

1211 ps

)ps(kps

)ps(kk)s(F)ps(

+

+++

+

++=+

Since ji pp , set 1p-s = to find 1k .

Hence, 1p-s1 k)s(F)ps( 1 =+ =

Thus, in general,ip-sii

)s(F)ps(k=

+=

When the residues of F(s) are known, the inverse of F(s) can be found. Since the inversetransform of each term is

)t(ukeas

kat-1-

=

+

L

thentp-

ntp-

2tp-

1n21 ekekek)t(f +++=

Repeated Poles

Now, suppose that F(s) has repeated poles at p-s = . Then,

)s(Fps

k)ps(

k)ps(

k)ps(

k)s(F 11221n1nnn +++++++++= ,

where F1(s) is the remaining part of F(s) that does not have a pole at p-s = .

The expansion coefficient nk is determined as

p-sn

n )s(F)ps(k =+=similar to the method for real poles. To determine 1nk , multiply F(s) by n)ps( + , differentiateand evaluate the result at p-s = to isolate 1nk . Thus,

p-sn

1n )]s(F)ps[(dsd

k=

+=

and p-sn

2

2

2n )]s(F)ps[(dsd

!21

k=

+=

In general, the mth term is p-sn

m

m

m )]s(F)ps[(dsd

!m1

k=

+=

where 1n,,2,1m = .

Once the values n,21 k,k,k are obtained using partial fraction expansion, apply the inversetransform

)!1n(et

)as(1 at-1n

n

1-

=

+

L

to each term in order to obtain

)t(fet)!1n(k

et!2k

etkek)t(f 1pt-1nnpt-23pt-2pt-1 +

++++=

Complex Poles

A pair of complex poles may be handled the same as real poles, but the complex algebra may bequite cumbersome. Even so,

)js(K

)js(K)s(F

*

11

++++=and

+=+= j-s1 )s(F)js(Kwhere = 11 KK and = -KK 1*1 .

Then, t)j(-j-1t)j(-j

1 eeKeeK)t(f + +=]ee[eK)t(f )t(j-)t(jt-1 ++ +=

)tcos(eK2)t(f t-1 +=

An easier approach is a method known as completing the square. Since N(s) and D(s) alwayshave real coefficients and complex roots occur in conjugate pairs, F(s) may have the general form

)s(Fbass

AsA)s(F 12 21 ++++

= ,

where F1(s) is the remaining part of F(s) that does not have complex poles.

Begin completing the square by letting2222 s2sbass +++=++

Also, let ++=+ 1121 B)s(AAsA

Then, )s(F)s(B

)s()s(A)s(F 1221221 +++

+++

+=

and the inverse transform is )t(f)tsin(eB)tcos(eA)t(f 1t-1t-1 ++=

Problem 15.3 Find the inverse Laplace transform for

(a) 6s

1+

(b) 9s

12 +

(c) 1s2s

12 ++

(a) t6-e

(b) Manipulate this to match a transform pair.

+

=

+ 9s3

31

9s1

22

The inverse Laplace transform is )t3sin()31(

(c) Manipulate this to match a transform pair.

22 )1s(1

1s2s1

+=

++

The inverse Laplace transform is t-te


(a) s6s

12 +

(b) 3s4s

2s2 ++

+

(c) 4s4ss

1s2s23

2

+++

++

None of the above match a transform pair. Manipulate these problems, perform partial fractionexpansion, and then use a table of Laplace transform pairs to find the inverse Laplace transforms.

Manipulate and perform partial fraction expansion to get the following.

(a) 6s61-

s

61)6s(s

1s6s

12 +

+=+

=

+

(b) 3s21

1s21

)3s)(1s(2s

3s4s2s

2 ++

+=

++

+=

++

+

(c)

+

++

=

+

+=

++

+=

+++

++

4s2

21

4ss

4s1s

)4s)(1s()1s(

4s4ss1s2s

2222

2

23

2

Now, the inverse Laplace transforms are

(a) t-e61)t(u6

1

(b) =+ t3-t- e21

e21 ]ee[2

1 t3-t- ++++

(c) )t2sin(5.0)t2cos( ++++


(a) 4s4ss

2s2s23

2

+++

++

(b) 1s3s3s

123 +++

The inverse Laplace transforms are

(a) )87.36t2cos(e51 t-

++++ (b) )et(21 t-2

APPLICATION TO CIRCUITS

Begin by transforming the circuit from the time domain to the s-domain. Solve the circuit usingnodal analysis, mesh analysis, source transformation, superposition, or any circuit analysistechnique with which we are familiar. Take the inverse Laplace transform of the solution andthus obtain the solution in the time domain.

Solving circuits with initial conditions is a straightforward process following the same basicapproach as circuits without initial conditions. Let us start with a capacitor and see how weactually solve such circuits. Start with the defining equation for a capacitor.

dt)t(dvC)t(i CC =

Taking the Laplace transform gives)0(vC)s(VCs])0(v)s(Vs[C)s(I CCCCC ==

or

s)0(v)s(I)Cs1()s(V CCC += .

We now can use the following circuit model for capacitors with initial conditions.

Now let us look at the inductor. The defining equation for the inductor is

dt)t(diL)t(v LL = .

Taking the Laplace transform gives)0(iL)s(ILs])0(i)s(sI[L)s(V LLLLL ==

or

s)0(iLs)s(V)s(I LLL += .

We can use the following circuit model for inductors with initial conditions.

Problem 15.6 Solve a first-order capacitive circuit with an initial condition using Laplacetransforms. In Figure 15.1, solve for )t(vC for all 0t > . The initial value of 50)t(vC = volts,or 50)0(vC = volts, and the value of )t(u20)t(vS = volts.

Figure 15.1

+

vC(0)/s

1/Cs+

VC(s)

Ls

IL(s)

+

VL(s)

iL(0)/s

5

+

0.1 F+

vC(t)

vS(t)

10

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.In this circuit, we are expected to find )t(vC for all 0t > using Laplace transforms. Weneed to set up a circuit in the frequency domain that will lead us to the desired solution. Itshould be noted that the initial condition on the capacitor must be included so that the factthat we do not have initial conditions is not violated. This is easy to do. Initial conditionscan be represented by impulse functions in the time domain. An impulse function transformsinto a constant. Thus, the capacitor can be represented by a capacitor in series with whatappears to be a constant DC source. In reality, the Laplace transform of an impulse functionis a constant in the frequency domain. Thus, we have the following circuit.

Please note that the actual value of )s(VC is the voltage across both the initial conditionsource and the s-domain capacitor. This is in agreement with the model developed earlier.Once we have found ),s(VC then we can take the inverse Laplace transform to find )t(vC .

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.Because we have accounted for the initial condition in the frequency domain equivalentcircuit, we can use nodal analysis, mesh analysis, or basic circuit analysis to find the voltageacross the capacitor for any time greater than zero. Looking at the frequency domain circuitabove, it is evident that nodal analysis is the best technique. Because there is only onenonreference node, this could be considered an application of KCL and Ohm's law.

ATTEMPT a problem solution.Summing the currents flowing out of the node gives us

0s1.01

s50)s(V5

0)s(V10

s20)s(V CCC=

+

+

Simplifying and collecting terms,5s2)s(V]10s51101[ C +=++

Multiplying both sides by 10,50)s20()s(V)3s( C +=+

5

+

20/s

10+

50/s

1/0.1s+

VC(s)

VC(s)

Thus,

3s50

)3s(s20)s(VC +++=

Let us simplify the first term, using partial fraction expansion.

3s320-

s

320)3s(s

20+

+=+

Finally,

3s50

3s320-

s

320)s(VC+

++

+=

Taking the inverse Laplace transform produces)t(u}e50]e1[)320({)t(v t3-t3-C += volts

This can also be written as )t(u}e3130320{)t(v t3-C += volts.

EVALUATE the solution and check for accuracy.How does this compare with our earlier approach, where we used a generic solution for afirst-order differential equation?

The value for += )tt(-C 0BeA)t(v

where 0t0 = sec and 31)1.0()510(

)5)(10(RC =

+== ,

320)(vC = volts, and 50)0(vC = volts.

Thus,320A = volts

313032050A)0(vB C === volts

Therefore, )t(u}e3130320{)t(v t3-C += volts, and our answer is the same as what wegot using the Laplace transform approach!

Our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=)t(vC V)t(ue3130

320 3t-

++++

Problem 15.7 Given the circuit in Figure 15.1, solve for )t(iL where )t(u)e10()t(v -2t=volts and -2)0(iL = amps.

Figure 15.1

We now convert the above circuit into its Laplace equivalent as shown below.

Write a mesh equation, where -2)0(iL = amps.0)]s2-()s(I[s5)s(I5)s(V- LL =++

10)s(V)s(I)1s(5 L =+

)1s(510)s(V)s(IL

+

=

Since )t(u)e10()t(v -2t= volts, )2s(10)s(V += . Now, substitute V(s) into the equation.

)1s(510)2s(10)s(IL +

+=

1s2-

)2s)(1s(2)s(IL ++++=

Simplifying the first term using partial fraction expansion,

1s2-

2s2-

1s2)s(IL

++

++

+=

2s2-)s(IL+

=

Taking the inverse Laplace transform of )s(IL gives=)t(iL A)t(u}e2-{ -2t

5

5 H

+

iL(t)

v(t)

5s

5

+

V(s) iL(0)/s

IL(s)

+

VL(s)

iL(0)/sIL(s)

The answer can be easily checked by again summing the voltages around the loop using thisvalue of )t(iL .

0 voltagesof =0)t(udt

)e2-(d5)e2-)(5(e10-

t2-t2-t2-

=

++

0)t(u}]e)2-)(2-[()5()e10-(e10-{ t2-t2-t2- =++0)t(u}e20e20-{ t2-t2- =+

The sum does equal zero.

Problem 15.8 Find the current, )t(iL for all t, given )t(u)e1010()t(v -2t= volts for thecircuit in Figure 15.1. Note that there are no initial conditions.

Figure 15.1

This problem is most easily solved using Laplace transforms. The first thing to do is to write amesh equation since we are looking for a current.

0dt)t(id5)t(i5)t(v- LL =++)t(v)t(i5dt)t(id5 LL =+

where)t(u)e1010()t(v t2-= volts

Taking the Laplace transform of both sides, noting that 0)0(iL = ,

2s10-

s

10)s(I5)0)s(Is)(5( LL ++=+or

2s2-

s

2)s(I)1s( L ++=+and

++

++

=

+

+

+= )2s)(1s(1-

)1s(s1

21s

12s

2-s

2)s(IL

Now perform a partial fraction expansion of both terms.

+

++

+

+=2s

1-1s

11s

1-s

12)s(IL

5

5 H

+

v(t) iL(t)

+

++

+= 2s1

1s2-

s

12)s(IL

Now to find )t(iL all we have to do is to take the inverse transform of )s(IL .=)t(iL A)t(u}ee21{2 t2-t- ++++

It is interesting to check and see if our answer is correct. To do that, we need to place this intothe original equation.

)t(v)t(i5dt)t(id5 LL =+where )t(u}e20e20{)t(u}e2)]e2)(1-(0)[2({5dt)t(id5 t2-t-t2-t-L ==

)t(u}e10e2010{)t(u}ee21){2)(5()t(i5 t2-t-t2-t-L +=+=

Clearly,)t(v)t(u}e1010{)t(u}e10e2010{)t(u}e20e20{ t2-t2-t-t2-t- ==++

and our answer checks.

Problem 15.9 Given the circuit in Figure 15.1, = 10R1 , = 10R 2 , F101C = , and)t(u10)t(v = volts. Calculate )t(i 2R .

Figure 15.1

)s(V])Cs1(||R[R)Cs1(||R)s(V

21

22R +

=

where1CsR

R)Cs1(R

CsR)Cs1(||R2

2

2

22 +

=

+=

R2C

R1

+

iR2(t)

v(t)

R2C

R1

+

iR2(t)

v(t)

Time domain

R21/Cs

R1

+

IR2(s)

V(s)

Frequency domain

Then,

)s(VR)1CsR(R

R)s(V])1CsR(R[R)1CsR(R)s(V

221

2

221

222R ++

=

++

+=

where 10)1.0)(10)(10(CRR 21 ==

)s(V2s

1)s(V20s10

10)s(V1010s10

10)s(V 2R+

=

+=

++=

Taking the Laplace transform of )t(v , s

10)s(V = .Then,

)2s(s10

s

102s

1)s(V 2R +=

+

=

The partial fraction expansion is

2s5-

s

5)s(V 2R+

+=

The inverse Laplace transform is)t(u)e1)(5()t(v t2-2R =

Now,

10)t(u)e1)(5(

R)t(v)t(i

t2-

2

2R2R

==

Finally,=)t(i 2R A)t(u)e1)(5.0( t2-

Problem 15.10 Given )t(u)e1()t(i -t1 = amps, )t(u)e1()t(i -t2 = amps, 0)0(v1 = volts,0)0(v2 = volts, and the circuit in Figure 15.1, find )t(v1 and )t(v2 .

Figure 15.1

0.1 F

10

0.1 F

v1(t) v2(t)

i1(t) i2(t)

Since there are no initial conditions, the above circuit is easily changed into the s-domain circuitshown below.

Using nodal analysis,At node 1 : At node 2 :

010

)s(V)s(V)s(Is10

0)s(V 211

1=

+ 0

10)s(V)s(V)s(I

s100)s(V 12

22

=

+

which leads to

=

+

+

)s(I10)s(I10

)s(V)s(V

1s1-1-1s

2

1

2

1

+

+

+

=

)s(I10)s(I10

s2s1s1

11s

)s(V)s(V

2

12

2

1

Hence,

)s(Is2s

10)s(Is2s

)1s)(10()s(V 22121 ++++

=

)s(Is2s

)1s)(10()s(Is2s

10)s(V 22122 ++

++

=

Taking the Laplace transform of )t(i1 and )t(i 2 , we find that)1s(1)s(I)s(I 21 +==

Substitute these currents into )s(V1 and )s(V2

)1s)(2s(s10

)1s)(2s(s)1s)(10()s(V1 +++++

+=

)1s)(2s(s)1s)(10(

)1s)(2s(s10)s(V2 ++

++

++=

)2s)(1s(s10

)2s(s10)s(V)s(V 21 ++++==

10/s

10

10/s

V1(s) V2(s)

I1(s) I2(s)

+

++

++

+

+==2s

51s

10-s

52s

5-s

5)s(V)s(V 21

1s10-

s

10)s(V)s(V 21+

+==

Therefore,=)t(v1 V)t(u}e1010{ t- and =)t(v2 V)t(u}e1010{ t-

Problem 15.11 Develop the matrix representation for the circuit shown in Figure 15.1, usingnodal analysis with three nodes.

Figure 15.1

Transforming Figure 15.1 to the frequency domain,

Using nodal analysis,

0s5VV

200V

10)s(VV 2111

=

+

+

0s155

VV10

0Vs5VV 32212

=

+

+

+

0)s(I20

0Vs155

VV 323=

++

Simplifying these equations,0Vs4Vs4V)s(V2V2 2111 =++

0)VV()2(V)1s3()VV()]1s3)(s2[( 32212 =++++0)s(I)1s3)(20(V)1s3()VV)(4( 323 =+++

20

1/5 F10

+

v(t) 10 20

15 H5

i(t)

20

5/s10

+

V(s) 10 20

15s5

I(s)

V1 V3V2

Combining like terms,)s(V2Vs4V)s412( 21 =++

0V2V]2)1s3()1s3)(s2[(V)]1s3)(s2(-[ 321 =++++++)s(I)1s3)(20(V)]1s3(4[V4- 32 +=+++

Simplifying further,)s(V2Vs4V)3s4( 21 =+

0V2V)3s5s6(V)s2s6-( 32212 =+++)s(I)1s3)(20(V)5s3(V4- 32 +=++

Therefore, the matrix equation is

++++

====

++++

++++++++

++++

)s(I)1s3)(20(0

)s(V2

VVV

5s34-02-3s5s6s2s6-

0s4-3s4

3

2

122

Problem 15.12 Given the circuit in Figure 15.1, write the s-domain equations for 1V , 2V ,3V , and 4V . DO NOT SOLVE.

Figure 15.1

Converting this circuit to the s-domain yields,


At node 1 : 0s10VV

100V

10)s(VV 2111

=

+

+

10

0.1 F10

+

v(t) 5 ix(t)

ix(t)

10

0.1 F

10

0.1 F

10

v1 v3v2 v4

10

10/s10

+

V(s) 5 Ix(s)

Ix(s)

10

10/s

10

10/s

10

V1 V3V2 V4

At node 2 : 0s10VV

100V

s10VV 32212

=

+

+

At node 3 : 0s10VV

100V

s10VV 43323

=

+

+

At node 4 : 0)s(I510

0Vs10VV

x

434=

+

where10

V)s(V)s(I 1x

=

Simplifying, )s(VVsV)2s( 21 =+0VsV)1s2(Vs- 321 =++0VsV)1s2(Vs- 432 =++

)s(V5V)1s(VsV5 431 =++

Finally, collect the equations and place them into a matrix form.

====

++++

++++

++++

++++

)s(V500

)s(V

VVVV

1ss-05s-1s2s-0

0s-1s2s-00s-2s

4

3

2

1

Problem 15.13 [15.49] For the circuit in Figure 15.1 find )t(vo for 0t > .

Figure 15.1

Consider the circuit shown below.

s

2I2s2s

1

+

6/s I1

1 H

2 1 H2 H

1

+

6 u(t)+

vo

For mesh 1,

21 IsI)s21(s6

++= (1)

For mesh 2,21 I)s2(Is0 ++=

21 Is2

1-I

+= (2)

Substituting (2) into (1) gives

2

2

22 Is2)5s(s-

IsIs

212s)-(1

s

6 ++=+

++=

or 2s5s6-

I 22 ++=

4.561)0.438)(s(s12-

2s5s12-

I2V 22o ++=

++==

Since the roots of 02s5s2 =++ are 0.438- and 4.561- ,

561.4sB

438.0sA

Vo +++=

-2.914.123

12-A == , 91.24.123-

12-B ==

561.4s91.2

0.438s2.91-)s(Vo +++=

=)t(vo V)t(u]ee[91.2 t438.0-4.561t

Problem 15.14 Given the circuit shown in Figure 15.1 and )t(u)e1010()t(v t-in += volts,find )t(vout , t .

Figure 15.1

10

10 H+

vin(t)

1/10 F10

+

vout(t)

10

=)t(vout V)et5e10( t-t-

Problem 15.15 Given the circuit shown in Figure 15.1 and )}t(ue5{)t(v -2tout = volts.Find )t(v .

Figure 15.1

=)t(v V)t(u20

TRANSFER FUNCTIONS

The transfer function )s(H is the ratio of the output response )s(Y to the input excitation )s(X ,assuming all initial conditions are zero.

Problem 15.16 [15.51] The transfer function of a system is

1s3s)s(H

2

+=

Find the output when the system has an input of )t(ue4 3t- .

)s(X)s(H)s(Y = , 1s312

31s4)s(X

+=

+=

22

2

)1s3(34s8

34

)1s3(s12)s(Y

+

+=

+=

22 )31s(1

274

)31s(s

98

34)s(Y

+

+=

Let 2)31s(s

98-)s(G

+=

10 20

+

10 +

vout(t)

v(t) 5 H

Using the time differentiation property,

+

==3t-3t-3t- eet3

1-98-)et(dt

d98-)t(g

3t-3t- e98

et278)t(g =

Hence,3t-3t-3t- et

274

e98

et278)t(u

34)t(y +=

=)t(y 3t-3t- et274

e98)t(u3

4++++

Problem 15.17 [15.59] Refer to the network in Figure 15.1. Find the following transferfunctions:

(a) )s(V)s(V)s(H so1 =(b) )s(I)s(V)s(H so2 =(c) )s(I)s(I)s(H so3 =(d) )s(V)s(I)s(H so4 =

Figure 15.1

(a) Consider the following circuit.

At node 1,

s

VVVs

1VV o1

11s +=

o1s Vs1

Vs

1s1V

++= (1)

s

1

1

+

Vs 1/s

V1 Vo Io

+

Vo

1/s

Is

1 H

1

1

+

vs 1 F

io

+

vo

1 F

is

At node o,

ooo

o1 V)1s(VVss

VV+=+=

o2

1 V)1ss(V ++= (2)

Substituting (2) into (1)oo

2s Vs1V)1ss)(s11s(V ++++=

o23

s V)2s3s2s(V +++=

==

s

o

1 VV)s(H

2s3s2s1

23 ++++++++++++

(b) o2o231ss V)1ss(V)2s3s2s(VVI +++++==o

23s V)1s2ss(I +++=

==

s

o

2 IV)s(H

1s2ss1

23 ++++++++++++

(c) 1

VI oo =

==== )s(HIV

II)s(H 2

s

o

s

o

3 1s2ss1

23 ++++++++++++

(d) ==== )s(HVV

VI)s(H 1

s

o

s

o

4 2s3s2s1

23 ++++++++++++

CONVOLUTION INTEGRAL

The convolution of two signals consists of time-reversing one signal, shifting it, multiplying itpoint by point with the second signal, and integrating the product.

Steps to evaluate the convolution integral :1. Folding : Take the mirror image of )(h about the ordinate axis to obtain )-(h .2. Displacement : Shift or delay )-(h by t to obtain )t(h .3. Multiplication : Find the product of )t(h and )(x .4. Integration : For a given time t, calculate the area under the product )(x)t(h for

t0

Problem 15.18 [15.69] Given that )1s(s)s(F)s(F 221 +== , find [ ])s(F)s(F 211-L byconvolution.

)tcos()t(f)t(f 21 ==

[ ] = t021

1- d)tcos()cos()s(F)s(FL

[ ])BAcos()BAcos(21)Bcos()Acos( ++=

[ ] = t021

1- d)tcos()2tcos(21)s(F)s(FL

[ ] t0t0211- 2-)2tsin(

21)tcos(

21)s(F)s(F +=L

[ ] =)s(F)s(F 211-L )tsin(21)tcos(t21 ++++

APPLICATION TO INTEGRODIFFERENTIAL EQUATIONS

Problem 15.19 Given the following is a matrix representation of a circuit in the frequencydomain. Determine the value of )t(v1 and )t(v2 .

=

+

+

0s8

)s(V)s(V

2ss-s-2s

2

1

+

+

+

=

++

+

+

=

0s8

)1s)(4(2ss

s2s

0s8

s4s4s2ss

s2s

)s(V)s(V

222

1

Hence,

+

++

=

+

+

+

=

+

+

+

++

+

=

1s2

1s2-

s

4

)1s)(s4(s8

)1s)(s4()2s)(8(

0s8

)1s)(4()2s(

)1s)(4(s

)1s)(4(s

)1s)(4()2s(

)s(V)s(V

2

1

Therefore, =)t(v1 V)t(u}e24{ t- and =)t(v2 V)t(u}e2{ t- .

Problem 15.20 Solve for )t(v1 and )t(v2 given the following matrix equation.

+=

+

+

)3s(4s2

)s(V)s(V

2s3-s-1s

2

1

++

+

+

=

+++

+

+

=

)3s(4s2

2s1s3

s2s

)3s(4s2

s3)2s)(1s(1s3

s2s

)s(V)s(V

22

1

+

+

+

+

++

+

=

3s4s

2

2s1s

2s3

2ss

2s2s

)s(V)s(V

22

22

2

1

)3s)(2s(s12s10s6

)3s)(2s(s)s)(s4()3s)(2s)(2(

3s4

2ss

s

22s2s)s(V 2

2

2221 ++

++=

++

+++=

+

+

+

+

+=

)3s)(2s(s18s10s4

)3s)(2s(s)s)(1s)(4()3s)(6(

3s4

2s1s

s

22s

3)s(V 22

2222 ++

++=

++

+++=

+

+

++

+

=

Now, we find the inverse Laplace transforms of )s(V1 and )s(V2 .

Partial fraction expansion yields

3s0909.1-

j2sj9642.04545.0-

j2sj9642.04545.0-

s

2)s(V1+

++

++

+=

3s7273.0-

j2sj6428.01364.1-

j2sj6428.01364.1-

s

3)s(V2+

++

++

+=

The inverse Laplace transform ist3-t2j-j24.115t2jj24.115-

1 e0909.1ee066.1ee066.1)t(u2)t(v ++=t3-t2j-j51.150t2jj51.150-

2 e7273.0ee3056.1ee3056.1)t(u3)t(v ++=

or t3-)24.115t2(j-)24.115t2(j1 e0909.1)ee(066.1)t(u2)t(v ++= t3-)51.150t2(j-)51.150t2(j

2 e7273.0)ee(3056.1)t(u3)t(v ++= Finally,

=)t(v1 t3-e0909.1)24.115t2cos(132.2)t(u2 ++++

=)t(v2 t3-e7273.0)51.150t2cos(611.2)t(u3 ++++

Problem 15.21 Given the circuit in Figure 15.1 and the following s-domain matrix nodalequations, determine the values for R, C, and L.

Figure 15.1

The matrix representation of the s-domain nodal equations is

=

++

+

)s(I)s(I

VV

s204s2s

20s-

20s-

204s5

2

1

2

12

So, the two equations are

)s(IV20

s-V20

4s5121 =+

+

)s(IVs20

4s2sV20

s-22

2

1 =++

+

Draw the s-domain circuit.

Writing the node equations gives,)s(IVCsV)Cs5sR1( 121 =++

and)s(IV)Ls1101Cs(VCs- 221 =+++

Clearly,51

4s

204s5Cs

5s

R1

+=+

=++

R L0.2 F

Cv1 v2

i1(t) i2(t)10

R Ls5/s

1/CsV1 V2

I1(s) I2(s)10

Thus, 51R1 = or =R 5)41(s)C2.0(s =+ or == 2.025.0C F05.0

Also,s20

4s2sLs1101Cs2 ++

=++

Finally, s511.0s05.0Ls11.0s05.0 ++=++ or =L H5

APPLICATIONS

Problem 15.22 Given the op-amp circuit in Figure 15.1, determine the value of )t(vout ,where the value of )t(vC is equal to 2 volts at t = 0. In other words, 2)0(vC = volts.

Figure 15.1

Begin by transforming the circuit from the time domain to the frequency domain.

+ +

vout

5

+

5et u(t)

+ vC(t)

0.1 F

+ +

Vout

5

+

5/(s+1)

10/s

V(s)

Vb

Va

+

2/s

+Vc(s)

Writing a node equation at a,

0s/10

)]s(V)s/2[(V5

)s(VV outaa=

++

which is only one equation with two unknowns.

We need a constraint equation.0VV ba == .

Then,

0s/10

)]s(V)s/2([-5

)s(V- out=

++

0)]s(Vs2[)s(V2- out =+

s

2-)1s(s

10-s

2-1s

5s

2-s

2-s

)s(V2-)s(Vout ++=+

+

=+=

Partial fraction expansion of the first term leads to

1s10

s

12-s

2-1s

10s

10-)s(Vout+

+=++

+=

Taking the inverse Laplace transform of both sides produces=)t(vout V)t(u}e1012-{ t-++++

Problem 15.23 In Figure 15.1, find )t(vo and )t(io , given that )t(u)e10()t(v -t= volts.

Figure 15.1

10 H

+

20

+

v(t) 1

io(t)+

vo(t)

Begin by transforming the time domain circuit to the frequency domain.

At node a :

020

)s(VVs10

)s(VV oaa=

+

We cannot help ourselves by writing any more node equations. So we have one equation and twounknowns. We need another equation without generating any additional unknowns. So we go tothe constraint equation.

0VV ba ==

So, now our node equation becomes

020

)s(V-s10

)s(V- o=+

10sV(s)-

20)s(Vo=

which leads to

V(s)s

2-)s(Vo =

Taking the Laplace transform of )t(v gives

1s10)s(V+

=

Substituting for )s(V yields

)1s(s02-

1s10

s

2-)s(Vo +=

+

=

After partial fraction expansion,

1s02

s

20-)s(Vo+

+=

Taking the inverse Laplace transform produces=)t(vo V)t(u)e2020-( t-++++

10s

+

20

+

V(s) 1

Io(s)+

Vo(s)

Va

Vb

Since Riv = in either domain, we do not need to find )s(Io .== 1/)t(v)t(i oo A)t(u)e2020-( t-++++

Problem 15.24 Given the circuit in Figure 15.1, calculate )s(Vout in terms of )s(Vin .

Figure 15.1



0R

)s(VVR

)s(VV2

outa

1

ina=

+

This is one equation with two unknowns. Thus, we need the following constraint equation.0VV ba ==

Then the node equation becomes

0R

)s(V-R

)s(V-2

out

1

in=+

1

in

2

out

R)s(V-

R)s(V=

+

R2

+

vin(t) RL+

vout(t)

R1

+

R2

+

Vin(s) RL+

Vout(s)

R1 Va

Vb

Therefore,

=)s(Vout )s(VRR-

in1

2

Problem 15.25 Given the circuit in Figure 15.1, calculate )s(Vout in terms of )s(Vin .

Figure 15.1



0Cs1

)s(VVR

)s(VV outa1

ina=

+

This is one equation with two unknowns. Thus, we need the following constraint equation.0VV ba ==

Then, the node equation becomes

0Cs1

)s(V-R

)s(V- out1

in=+

+

+

vin(t) RL+

vout(t)

R1

C

1/Cs

+

+

Vin(s) RL+

Vout(s)

R1 Va

Vb

1inout

R)s(V-

Cs1)s(V=

Therefore,

=)s(Vout )s(VCsR1-

in1

Problem 15.26 Given = k100R1 . What value of C, in Figure 15.1, yieldsdt)t(V-)t(V inout =

Taking the Laplace transform,

)s(Vs

1-)s(V inout =

From Problem 15.25, we know that

)s(VCsR1-)s(V in

1out =

Thus,

1CR1 = or ====5-

31

10110100

1R1

C F10

Problem 15.27 Given = k10R1 , = 100R L , F50C = , and )t(ue10)t(V t2-in = .Calculate the total energy that inv delivers to the circuit shown in Figure 15.1. Also, find thetotal energy delivered to LR .

== dt)t(vR1dt)t(p)t(w 2To find the total energy that inv delivers to the circuit,= dt)t(vR1)t(w 2in1vin

= t0

2-2-v d)e10)(e10(000,101)t(w in

= t0

4- de000,10100

)e41-)(1001( t04- =)41e41-)(1001( t4- +=

= )e1)(4001( t4-

To find the total energy delivered to LR , = dt)t(vR1)t(w 2outLR LFirst, find )t(vout . From Problem 15.25, we know that

)s(VCsR1-)s(V in

1out =

where 5.010500)1050)(1010(CR 3-6-31 ===

Then,

)s(Vs

2-)s(VCsR1-)s(V inin

1out ==

Taking the Laplace transform of )t(vin ,

2s10)s(Vin+

=

Substitute )s(Vin into the equation for )s(Vout .

)2s(s20-

2s10

s

2-)s(Vs

2-)s(V inout +=

+

==

Take the partial fraction expansion of )s(Vout .

2s10

s

10-)s(Vout ++=

Hence,)t(u)e1010-()t(V t2-out +=

Now, = dt)t(vR1)t(w 2outLR L ++= t

02-2- d)e1010-)(e1010-(1001

+= t0

2-4- d)100e200e100(1001

+= t0

t

02-t

04- d1de2de

t]21e)21(-)[2(]41e)41-[( t2-t4- +++== t)1e-()1e-)(41( t2-t4- ++++++++++++++++

Problem 15.28 [15.89] A gyrator is a device for simulating an inductor in a network. Abasic gyrator is shown in Figure 15.1. By finding )s(I)s(V oo , show that the inductanceproduced by the gyrator is 2RCL = .

Figure 15.1

The gyrator is equivalent to two cascaded inverting amplifiers. Let 1V be the voltage atthe output of the first op amp.

ii1 -VVRR-

V ==

i1o VsCR1

VRsC1-

V ==

CsRV

RV

I 2oo

o ==

CsRIV 2

o

o=

CRLwhen,sLIV 2

o

o========

+

R

R

+

Vi

+

C

R

R

Io

SearchHelpEWB Help PageWe want your feedbacke-Text Main MenuTextbook Table of ContentsProblem Solving WorkbookWeb LinksTextbook WebsiteOLC Student Center WebsiteMcGraw-Hill Website

PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformDefinition of the Laplace TransformInverse Laplace TransformApplication to CircuitsTransfer FunctionsConvolution Integral Application to Integrodifferential EquationsApplications

Chapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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