Instructors Solution for Assignment 3 Chapter 3: Time ...asif/3451fall10/solution-Assignment3.pdf · 2 Solution to Assignment 3 To compute the values of constants C 1 and C 2, we

Instructors Solution for Assignment 3 Chapter 3: Time Domain Analysis of LTIC Systems

Problem 3.2

(i) y (t )+ 4 y (t )+8 y (t ) = x (t )+ x (t ) with x (t ) = e −4tu (t ), y (0) = 0, and y (0) = 0.

(a) Zero-input response of the system: The characteristic equation of the LTIC system (i) is

0842 =++ ss ,

which has roots at s = −2 ± j2. The zero-input response is given by

)2sin()2cos()( 22 tBetAety ttzi

−− +=

for t ≥ 0, with A and B being constants. To calculate their values, we substitute the initial conditions 0)0( =−y and 0)0( =−y in the above equation. The resulting simultaneous equations are

0220

=+−=

BAA

that has the solution, A = 0 and B = 0. The zero-input response is therefore given by

.0)( =tyzi

Because of the zero initial conditions, the zero-input response is also zero.

(b) Zero-state response of the system: To calculate the zero-state response of the system, the initial conditions are assumed to be zero. Hence, the zero state response yzs(t) can be calculated by solving the differential equation

)(842

2tx

dtdx

dtdy

dtyd

+=++

with initial conditions, 0)0( =−y and 0)0( =−y , and input x(t) = exp(−4t)u(t). The homogenous solution of system (i) has the same form as the zero-input response and is given by

)2sin()2cos()( 22

21

)( teCteCty tthzs

−− +=

for t ≥ 0, with C1 and C2 being constants. The particular solution for input x(t) = exp(−4t)u(t) is of the form

)(4)( tu Ke(t) y tpzs

−= .

Substituting the particular solution in the differential equation for system (i) and solving the resulting equation gives K = −3/8. The zero-state response of the system is, therefore, given by

( ) )()2sin()2cos()( 4832

22

1 tueteCteCty tttzs

−−− −+= .

2 Solution to Assignment 3

To compute the values of constants C1 and C2, we use the initial conditions, y(0−) = 0 and 0)0( =−y assumed for the zero-state response. Substituting the initial conditions in yzs(t) leads to the following simultaneous equations

0220

23

21

83

1

=++−

=−

CCC

with solution C1 = 3/8 and C2 = −3/8. The zero-state solution is given by

( ) )()2sin()2cos()( 42283 tuetetety ttt

zs−−− −−= .

(c) Overall response of the system: The overall response of the system is obtained by summing up the zero-input and zero-state responses, and is given by

( ) )()2sin()2cos()( 42283 tuetetety ttt −−− −−= .

(d) Steady state response of the system: The steady state response of the system is obtained by applying the limit, t → ∞, to y(t) and is given by

( ) 0)()2sin()2cos(lim)( 42283 =−−= −−−

∞→tuetetety ttt

t.

(iii) y (t )+ 2 y (t )+ y (t ) = x (t ) with x (t ) = cos(t )+ sin(2t )!" #$u (t ), y (0) = 3, and y (0) =1.

(a) Zero-input response of the system: The characteristic equation of the LTIC system (iii) is

0122 =++ ss ,

which has roots at s = −1, −1. The zero-input response is given by tt

zi BteAety −− +=)(

for t ≥ 0, with A and B being constants. To calculate their values, we substitute the initial conditions 3)0( =−y and 1)0( =−y in the above equation. The resulting simultaneous equations are

13

=+−=

BAA

that has a solution, A = 3 and B = 4. The zero-input response is therefore given by

( ) )(43)( tuteety ttzi

−− += .


)(122

2tx

dtdy

dtyd =++

with initial conditions, 0)0( =−y and 0)0( =−y , and input x(t) = [cos(t) + sin(t)]u(t). The homogenous solution of system (iii) has the same form as the zero-input response and is given by

tthzs teCeCty −− += 21)( )(

Solutions 3

for t ≥ 0, with C1 and C2 being constants. The particular solution for input x(t) = [cos(t) + sin(t)]u(t) is of the form

)2sin()2cos()sin()cos( 4321)( tKtKtKt K(t) y p

zs +++= .

Substituting the particular solution in the differential equation for system (iii) and solving the resulting equation gives

( ) () ( ) )2sin(4)cos()2sin()2cos()sin()cos(1)2cos(2

)2sin(2)cos()sin(2)2sin(4)2cos(4)sin()cos(43214

3214321tttKtKtKtKtK

tKtKtK tKtKtKtK−−=+++++

−+−+−−−−

Collecting the coefficients of the cosine and sine terms, we get

( ) ( )( ) ( ) 0)2sin(444)2cos(44

)sin(2)cos(12434343

212121=++−−+++−

++−−++++−tKKK tKKK

tKKK tKKK

which gives K1 = 0, K2 = −0.5, K3 = 0.64, and K4 = 0.48. The zero-state response of the system is

( ) )()2sin(48.0)2cos(64.0)sin(5.0)( 21 tutttteCeCty ttzs ++−+= −− .

To compute the values of constants C1 and C2, we use the initial conditions, y(0−) = 0 and 0)0( =−y . Substituting the initial conditions in yzs(t) leads to the following simultaneous equations

048.05.0064.0

211

=+−+−=+

CCC

with solution C1 = −0.64 and C2 = −1.1. The zero-state solution is given by

( ) )()2sin(48.0)2cos(64.0)sin(5.01.164.0)( tutttteety ttzs ++−−−= −− .


or, ( ) ( ) )()2sin(48.0)2cos(64.0)sin(5.01.164.0)(43)( tutttteetuteety tttt ++−−−++= −−−−

or, ( ) )()2sin(48.0)2cos(64.0)sin(5.09.236.2)( tutttteety tt ++−+= −−.


( ) )()2sin(48.0)2cos(64.0)sin(5.09.236.2lim)( tutttteety ttt

++−+= −−

∞→

or, ( ) )()2sin(48.0)2cos(64.0)sin(5.0)( tutttty ++−= .

(v) .1)0(,0)0()0()0(),(2)()()(2 2

2

4

4

======++ yandyyytutxwithtxtydtyd

dtyd

(a) Zero-input response of the system: The characteristic equation of the LTIC system (v) is

0124 =++ ss ,

which has roots at s = ±j1, ±j1. The zero-input response is given by

jtjtjtjtzi DteCeBteAety −− +++=)( ,


for t ≥ 0, with A and B being constants. To calculate their values, we substitute the initial conditions in the above equation. The resulting simultaneous equations are

03302210

=−+−−=−−+−=+−+=+

DjCBjADjCBjA

DjCBjACA

that has a solution, A = −j0.75 Β = −0.25, C = j0.75 and D = −0.25. The zero-input response is

( ) )(25.075.025.075.0)( tuteejteejty jtjtjtjtzi

−− −+−−= ,

which reduces to

( ) )(cos5.0sin5.1)( tuttttyzi −= .


)()(2 2

2

4

4

txtydtyd

dtyd =++

with all initial conditions set to 0 and input x(t) = 2u(t). The homogenous solution of system (v) has the same form as its zero-input response and is given by

jtjtjtjthzs teCeCteCeCty −− +++= 4321)( )(

where Ci’s are constants. The particular solution for input x(t) = 2 u(t) is of the form

)()( t Ku(t) y pzs = .

Substituting the particular solution in the differential equation for system (v) and solving the resulting equation gives

2)0(20 =++ K , or, K = 2.

The zero-state response of the system is given by

2)( 4321 ++++= −− jtjtjtjtzs teCeCteCeCty ,

for (t ≥ 0). To compute the values of constants Ci’s, we use zero initial conditions. Substituting the initial conditions in yzs(t) leads to the following simultaneous equations

03302202

=−+−−=−−+−=+−+

−=+

DjCBjADjCBjA

DjCBjACA

with solution C1 = −1, C2 = j0.5, C3 = −1, and C4 = −j0.5. The zero-state solution is given by

( ) )(5.05.0)( tutejetejety jtjtjtjtzs

−− −−+−= ,

which reduces to

( ) )(sincos2)( tuttttyzi −−= .


Solutions 5

or, ( ) ( ) )(sincos2)(cos5.0sin5.1)( tuttttutttty −−+−=

or, ( ) )(2cos5.0sincos2sin5.1)( tuttttttty +−−−= .


( ) ∞→+−−−=∞→

)(2cos5.0sincos2sin5.1lim)( tutttttttyt

. ▌

Problem 3.5

(i) The output y(t) is given by

∫∫∞∞

∞−

ττ−=ττ−τ=∗=0

)()()()()()( dtudtuutututy .

Recall that ⎩⎨⎧

>τ≤τ=τ− ).(if0)(if1)( tttu

Therefore, the output y(t) is given by

).()()0(if0)0(if)( trttut

ttty ==⎩⎨⎧

<≥=

The aforementioned convolution can also be computed graphically.

(iii) The output y(t) is given by

[ ] [ ])1()1()2()1(2)()( −−+∗−+−−= tutututututy

Using the properties of the convolution integral, the output is expressed as

[ ] [ ] [ ][ ] [ ] [ ])1()2()1()2()1()1(2

)1()1(2)1()()1()()(−∗−−+∗−+−∗−+

+∗−−−∗−+∗=tutututututu

tututututututy

Based on the results of part (i), i.e., u(t) * u(t) = r(t), the overall output is given by

)3()1()2(2)(2)1()1()( −−−+−+−−−+= trtrtrtrtrtrty .

(vi) Considering the two cases (t < 0) and (t ≥ 0) separately

Case I (t < 0): ∫∫∫∞

τ−τ−τ−τ

∞−

τ−−τ τ+τ+τ=0

)(520

)(52)(52)( deedeedeety t

t

tt

t

which reduces to ∫∫∫∞

τ−−τ−

∞−

τ− τ+τ+τ=0

750

3575)( deedeedeety t

t

tt

t

or, ( ) tttttttt eeeeeeeety 8315

2142

71

7153

3157

715 1)( −−−−− +−=×+−×+×=

Case II (t ≥ 0): ∫∫∫∞

τ−τ−τ−−τ−

∞−

τ−−τ τ+τ+τ=t

tt

tt deedeedeety )(52

0

)(520

)(52)(


which reduces to ∫∫∫∞

τ−−τ−

∞−

τ− τ+τ+τ=t

tttt deedeedeety 75

0

35075)(

or, ( ) ttttttttt eeeeeeeeety 12715

312

315

717

7153

3155

71 1)( −−−−−− +−+=×+−×+= .

Hence, the overall expression for y(t) is given by

⎪⎩

⎪⎨⎧

≥

<== − ).0(

)0()( 3

51

251

tete

ty t

t

Problem 3.6

(iii) Using the graphical approach, the convolution of x(t) with w(t) is shown in Fig. S3.6.3, where we consider six different cases for different values of t.

Case I (t < −1): Since there is no overlap, 0)(3 =ty .

Case II (−1 ≤ t < 0): .)1.(1)( 21

22)1(

13

22

++==ττ+= +

−∫ tdty ttt

Case III (0 ≤ t < 1):

( ) ( ) .0

)1.(1)1.(1)1).(1()(

21

23

21

2)1(

221

2

0

0

1

1

13

2222++−=⎟

⎠⎞⎜

⎝⎛ −−−+−−=

ττ−+ττ++ττ+−=

−−

−

−∫∫∫

t

dddty

tttt

t

t

t

τ

1

−1

w(τ)

0 1−1

τ

1

−1

w(τ)

0 1−1 0 1−1

τ

1

−1

1 20

x(τ)

τ

1

−1

1 20

x(τ)1

−1

1 20 1 20

x(τ)

τt −2 t −1 t

x(t−τ)1

−1

τt −2 t −1 tt −2 t −1 t

x(t−τ)1

−1

(a) Waveform for z(τ) (b) Waveform for x(τ) (c) Waveform for x(t−τ)

τt −2 t −1 t

w(τ) x(t−τ)1

−1

0 1−1

τt −2 t −1 tt −2 t −1 t

w(τ) x(t−τ)1

−1

0 1−1 0 1−1

τt −2 t −1 t

w(τ) x(t−τ)1

−1

0 1−1

τt −2 t −1 tt −2 t −1 t

w(τ) x(t−τ)1

−1

0 1−1 0 1−1

τt −2 t −1 t

w(τ) x(t−τ)1

−1

0 1−1

τt −2 t −1 tt −2 t −1 t

w(τ) x(t−τ)1

−1

0 1−1 0 1−1

(d) Overlap btw w(τ) and x(t−τ) for (t<−1)

(e) Overlap btw w(τ) and x(t−τ) for (−1≤t<0) (f) Overlap btw w(τ) and x(t−τ) for (0≤t<1)

τt −2 t −1 t

w(τ) x(t−τ)1

−1

1−1

τt −2 t −1 tt −2 t −1 t

w(τ) x(t−τ)1

−1

1−1 1−1

τt −2 t −1 t

w(τ) x(t−τ)1

−1

1−1

τt −2 t −1 tt −2 t −1 t

w(τ) x(t−τ)1

−1

1−1 1−1

τt −2 t −1 t

w(τ) x(t−τ)1

−1

1−1

τt −2 t −1 tt −2 t −1 t

w(τ) x(t−τ)1

−1

1−1 1−1

(g) Overlap btw w(τ) and x(t−τ) for (1≤t<2)

(h) Overlap btw w(τ) and x(t−τ) for (2≤t<3) (i) Overlap btw w(τ) and x(t−τ) for (t>3)

Solutions 7

0 1 2 3 4-2

-1.5

-1

-0.5

0

0.5

1

y1(t)

= x

(t)*x

(t)

Problem 3.6(i)

(i)-1 0 1 2 3

-0.5

0

0.5

1

y2(t)

= x

(t)*z

(t)

Problem 3.6(ii)

(ii)-1 0 1 2 3

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

y3(t)

= x

(t)*w

(t)

Problem 3.6(iii)

(iii)

-1 0 1 2 3-0.5

0

0.5

y4(t)

= v

(t)*v

(t)

Problem 3.6(iv)

(iv)-2 -1 0 1 2

-0.6

-0.4

-0.2

0

0.2

y5(t)

= z

(t)*z

(t)

Problem 3.6(v)

(v)-2 -1 0 1 2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3y6

(t) =

w(t)

*z(t)

Problem 3.6(vi)

(vi)

-2 -1 0 1 2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

y7(t)

= v

(t)*z

(t)

Problem 3.6(vii)

(vii)-2 -1 0 1 2

0

0.1

0.2

0.3

0.4

0.5

0.6

y8(t)

= w

(t)*w

(t)

Problem 3.6(viii)

(viii)-2 -1 0 1 2

0

0.1

0.2

0.3

0.4

0.5

y9(t)

= w

(t)*w

(t)

Problem 3.6(ix)

(ix)

(j) Convolution output 3( )y t

Fig. S3.6.3: Convolution of x(t) with w(t) in Problem 3.6(iii).

Case IV (1 ≤ t < 2):

.50

)1.(1)1).(1()1).(1()(

27

23

2)2(

21

2)2(

2)1(

21

1

1

1

0

0

23

2222

+−=⎟⎠⎞⎜

⎝⎛ −−⎟

⎠⎞⎜

⎝⎛ −+⎟

⎠⎞⎜

⎝⎛ −−=

ττ−+ττ−−+ττ+−=

−−−−

−

−∫∫∫

t

dddty

ttttt

t

t

Case V (2 ≤ t < 3): 29

22)3(

1

23 30)1).(1()(

22

−+−=−=ττ−−= −

−∫ tdty tt

t

.

Case VI (t > 4): Since there is no overlap, )(3 ty = 0.

Combining all the cases, the result of the convolution )()()(3 twtxty ∗= is given by

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

<≤−+

<≤+−

<≤++−

<≤−++

=

elsewhere.0)32(3)21(5)10()01(

)(

29

2

27

23

21

23

21

2

32

2

2

2

tttttttt

tyt

t

t

t

The output is y3(t) plotted in Fig. S3.6.3(j).

(vii) Using the graphical approach, the convolution of v(t) with z(t) is shown in Fig. 3.6.7, where we consider six different cases for different values of t.

τ

1

−1

v(τ)

0 1−1

e−2τe2τ

τ

1

−1

v(τ)

0 1−1 1−1

e−2τe2τ

τ

1

−1

z(τ)

0 1−1

τ

1

−1

z(τ)

0 1−1

τ

1

−1

z(t−τ)

t t+1t−1 0 1−1

τ

1

−1

z(t−τ)

t t+1t−1 0 1−1

(a) Waveform for v(τ) (b) Waveform for z(τ) (c) Waveform for z(t−τ)


τ

1

−1

v(τ) z(t−τ)

t t+1t−1 0 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 0 1−1 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 0 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 t t+1t−1 0 1−1 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 t t+1t−1 1−1 1−1

(d) Overlap btw v(τ) and z(t−τ) for (t<−2) (e) Overlap btw v(τ) and z(t−τ) for (−2≤t<−1) (f) Overlap btw v(τ) and z(t−τ) for (−1≤t<0)

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 0 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 t t+1t−1 0 1−1 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−10 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 t t+1t−10 1−1 1−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−11−1

τ

1

−1

v(τ) z(t−τ)

t t+1t−1 t t+1t−11−1 1−1

(g) Overlap btw v(τ) and z(t−τ) for (0≤t<1) (h) Overlap btw v(τ) and z(t−τ) for (1≤t<2) (i) Overlap btw v(τ) and z(t−τ) for (t>2)

0 1 2 3 4-2

-1.5

-1

-0.5

0

0.5

1

y1(t

) =

x(t

)*x(

t)

Problem 3.6(i)

(i)-1 0 1 2 3

-0.5

0

0.5

1

y2(t

) =

x(t

)*z(

t)

Problem 3.6(ii)

(ii)-1 0 1 2 3

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

y3(t

) =

x(t

)*w

(t)

Problem 3.6(iii)

(iii)

-1 0 1 2 3-0.5

0

0.5

y4(t

) =

v(t

)*v(

t)

Problem 3.6(iv)

(iv)-2 -1 0 1 2

-0.6

-0.4

-0.2

0

0.2

y5(t

) =

z(t

)*z(

t)

Problem 3.6(v)

(v)-2 -1 0 1 2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

y6(t

) =

w(t

)*z(

t)

Problem 3.6(vi)

(vi)

-2 -1 0 1 2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

y7(t

) =

v(t

)*z(

t)

Problem 3.6(vii)

(vii)-2 -1 0 1 2

0

0.1

0.2

0.3

0.4

0.5

0.6

y8(t

) =

w(t

)*w

(t)

Problem 3.6(viii)

(viii)-2 -1 0 1 2

0

0.1

0.2

0.3

0.4

0.5

y9(t

) =

w(t

)*w

(t)

Problem 3.6(ix)

(ix)

(j) Convolution output 7 ( )y t

Fig. S3.6.7: Convolution of v(t) with z(t) in Problem 3.6(vii).


Case II (−2 ≤ t < −1):

[ ]

[ ] [ ]

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−−−=

++−+−=

+τ−=ττ−=

−−+

−−++

+

−ττ

+

−

τ∫

.

)1(

)()()(

2432

21)1(2

41

2412

21)1(2

41)1(2

21

1

12

412

21

1

1

27

etee

eetee

eetdtety

t

tt

tt

Solutions 9

Case III (−1 < t < 0): [ ] [ ][ ] [ ]

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

+−−=

−+−+−−++=

+τ−−++τ−=

ττ−+ττ−=

−−+−

+−−−

+τ−τ−−

ττ

+τ−

−

τ ∫∫

.

)()(

)()()(

2432

21)1(2

43

41

21)1(2

412

432

21

41

21

1

02

412

210

12

412

21

1

0

20

1

27

tetee

teetet

eeteet

dtedtety

t

t

t

t

Case IV (0 < t < 1): [ ] [ ][ ] [ ]

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

++−−=

−++−+−+=

+τ−−++τ−=

ττ−+ττ−=

−−−

−−−

τ−τ−−

ττ

τ−

−

τ ∫∫

.

)()(

)()()(

2432

21)1(2

43

41

212

432

21)1(2

43

41

21

1

02

412

210

12

412

21

1

0

20

1

27

tetee

teteet

eeteet

dtedtety

t

t

t

t

Case V (1 ≤ t < 2):

[ ]

[ ] [ ]

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

+−=

++−+−=

+τ−−=ττ−=

−−−−

−−++

−τ−τ−

−

τ−∫

.

)1(

)()()(

2432

21)1(2

41

2412

21)1(2

41)1(2

21

1

12

412

21

1

1

27

etee

eetee

eetdtety

t

tt

tt

Case VI (t > 2): Since there is no overlap, 0)(7 =ty .

Combining all the cases, the result of the convolution )()()(7 tvtzty ∗= is given by

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

<≤+−

<≤++−−

<≤−+−−

−<≤−−−−

=−−−−

−−−

−−+−

−−+

elsewhere.0)21()10()01()12(

)(2

432

21)1(2

41

2432

21)1(2

43

2432

21)1(2

43

2432

21)1(2

41

7

teteetteteetteteetetee

tyt

t

t

t

The output is y7(t) shown in Fig. S3.6.7(j) at the end of the solution of this problem.

(ix) Using the graphical approach, the convolution of v(t) with w(t) is shown in Fig. 3.6.9, where we consider six different cases for different values of t.


τ

1

−1

v(τ)

0 1−1

e−2τe2τ

τ

1

−1

v(τ)

0 1−1 1−1

e−2τe2τ

τ

1

−1

w(τ)

0 1−1

τ

1

−1

w(τ)

0 1−1 0 1−1

τ

1

−1

w(t−τ)

0 1−1t t+1t−1τ

1

−1

w(t−τ)

0 1−1t t+1t−1 t t+1t−1

(a) Waveform for v(τ) (b) Waveform for w(τ) (c) Waveform for w(t−τ)

τ

1

−1

v(τ) w(t−τ)

t t+1t−1 1−1

τ

1

−1

v(τ) w(t−τ)

t t+1t−1 t t+1t−1 1−1 1−1

t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 11−1t t+1t−1 t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 11−1 1−1

t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 1−1 t t+1t−1 t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 1−1 1−1

(d) Overlap btw v(τ) and w(t−τ) for (t<−2) (e) Overlap btw v(τ) and w(t−τ) for (−2≤t<−1) (f) Overlap btw v(τ) and w(t−τ) for (−1≤t<0)

t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 1−1 t t+1t−1 t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 1−1 1−1

t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 1−1 t t+1t−1 t t+1t−1τ

1

−1

v(τ) w(t−τ)

0 1−1 1−1

τ

1

−1

v(τ) w(t−τ)

t t+1t−11−1

τ

1

−1

v(τ) w(t−τ)

t t+1t−11−1 1−1

(g) Overlap btw v(τ) and w(t−τ) for (0≤t<1) (h) Overlap btw v(τ) and w(t−τ) for (1≤t<2) (i) Overlap btw v(τ) and w(t−τ) for (t>2)

0 1 2 3 4-2

-1.5

-1

-0.5

0

0.5

1

y1(t

) =

x(t

)*x(

t)

Problem 3.6(i)

(i)-1 0 1 2 3

-0.5

0

0.5

1

y2(t

) =

x(t

)*z(

t)

Problem 3.6(ii)

(ii)-1 0 1 2 3

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

y3(t

) =

x(t

)*w

(t)

Problem 3.6(iii)

(iii)

-1 0 1 2 3-0.5

0

0.5

y4(t

) =

v(t

)*v(

t)

Problem 3.6(iv)

(iv)-2 -1 0 1 2

-0.6

-0.4

-0.2

0

0.2

y5(t

) =

z(t

)*z(

t)

Problem 3.6(v)

(v)-2 -1 0 1 2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3y6

(t)

= w

(t)*

z(t)

Problem 3.6(vi)

(vi)

-2 -1 0 1 2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

y7(t

) =

v(t

)*z(

t)

Problem 3.6(vii)

(vii)-2 -1 0 1 2

0

0.1

0.2

0.3

0.4

0.5

0.6

y8(t

) =

w(t

)*w

(t)

Problem 3.6(viii)

(viii)-2 -1 0 1 2

0

0.1

0.2

0.3

0.4

0.5

y9(t

) =

w(t

)*w

(t)

Problem 3.6(ix)

(ix)

(j) Convolution output 9 ( )y t

Fig. S3.6.9: Convolution of v(t) with w(t) in Problem 3.6(ix).

Since w(t) = 1 − |t|, therefore, the expression for w(t – τ) is

⎩⎨⎧

>τ−τ−<ττ−−=τ−−=τ− . if)(1

if)(11)( ttttttw


Solutions 11

Case II (−2 ≤ t < −1): ( ) ( )

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−−=

++−+−−+=

ττ−τ+=τ+τ−=

−−+

−−++−+

+

−

τ+

−

τ+

−

τ ∫∫∫

.

)1()1(

)1()1()(

2452

21)1(2

41

2412

21)1(2

41)1(2

212)1(2

21

1

1

21

1

21

1

29

etee

eeeeteet

dedetdtety

t

ttt

ttt

Case III (−1 < t < 0):

[ ] [ ] [ ] [ ]

[ ] [ ]

⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪

⎨

⎧

++++−=

−τ−−−++

−τ−++−τ+−=

ττ−++ττ−++ττ+−=

−−+−

+τ−τ−+τ−

τττ−

ττ−

τ

+τ−τ

−

τ ∫∫∫

.1

)1(

)1()1(

)1()1()1()(

2412

212

21)1(2

41

1

02

412

211

02

21

02412

2102

21

12

412

21

12

21

1

0

202

1

29

teteee

eeet

eeeteeet

dtedtedtety

tt

tt

tt

tt

t

t

t

.

Case IV (0 < t < 1):

[ ] [ ] [ ] [ ]

[ ] [ ]

⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪

⎨

⎧

+−+−−=

−τ−−−++

−τ+−+−τ+−=

ττ−++ττ+−+ττ+−=

−−−−

τ−τ−τ−

τττ−

ττ−

τ

τ−τ

−

τ ∫∫∫

.1

)1(

)1()1(

)1()1()1()(

2412

212

21)1(2

41

12412

2112

21

0

02

412

210

02

210

12

412

210

12

21

12

0

20

1

29

teteee

eeet

eeeteeet

dtedtedtety

tt

tt

tt

t

t

t

Case V (1 ≤ t < 2): ( ) ( )

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−+=

+−+−−+−−−=

ττ+τ−=ττ+−=

−−−−

−−−−−−−−−

−

τ−

−

τ−

−

τ− ∫∫∫

.

)1()1(

)1()1()(

2452

21)1(2

41

)1(241)1(2

212

412

21)1(22

21

1

1

21

1

21

1

29

etee

eeteeeet

dedetdtety

t

ttt

ttt

.

Case VI (t > 2): Since there is no overlap, 0)(9 =ty .

Combining all the cases, the result of the convolution )()()(9 twtvty ∗= is given by


⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

<≤−+

<≤+−+−−

<≤−++++−

−<≤−−−

=−−−−

−−−−

−−+−

−−+

elsewhere.0)21()10(1)01(1)12(

)(2

452

21)1(2

41

2412

212

21)1(2

41

2412

212

21)1(2

41

2452

21)1(2

41

9

teteetteteeetteteeetetee

tyt

tt

tt

t

The output is y9(t) shown in Fig. S3.6.9(j).

Problem 3.12

(iii) System h3(t) is NOT memoryless since h3(t) ≠ 0 for t ≠ 0. System h3(t) is causal since h3(t) = 0 for t < 0. System h3(t) is BIBO stable since

5 5 5

0 0 1

5 51 1 15 5 50

0

| 3( ) | sin(2 ) ( ) sin(2 ) sin(2 )

0 ( )

t t t

t t

h t dt e t u t dt e t dt e t dt

e dt e

π π π∞ ∞ ∞ ∞

− − −

−∞ −∞ ≤

∞∞− −

= = =

⎡ ⎤≤ = − = − − = < ∞⎡ ⎤⎣ ⎦⎣ ⎦

∫ ∫ ∫ ∫

∫

14 2 43.

(vii) System h7(t) is NOT memoryless since h7(t) ≠ 0 for t ≠ 0. System h7(t) is causal since h7(t) = 0 for t < 0. System h7(t) is NOT BIBO stable since

∞==∫ ∫∞

∞−

∞

0

)5cos(|)(7| dttdtth .

Consider the bounded input signal cos(5 )t . If this signal is applied to the system, the output can be calculated as:

0

( ) ( ) ( ) cos(5 5 )cos(5 ) ( ) cos(5 5 )cos(5 )y t x t h d t u d t dτ τ τ τ τ τ τ τ τ τ∞ ∞ ∞

−∞ −∞

= − = − = −∫ ∫ ∫ .

The output at t=0 is given by,

( )

{

2 12

0 0 0

1 12 2

0 0=

(0) cos( 5 )cos(5 ) cos (5 ) 1 cos(10 )

cos(10 )

finite value

y d d d

d d

τ τ τ τ τ τ τ

τ τ τ

∞ ∞ ∞

∞ ∞

∞ =

= − = = +

= + = ∞

∫ ∫ ∫

∫ ∫1 44 2 4 43

It is observed that the output becomes unbounded at t=0 even if the input is always bounded. This proves that the system is not BIBO stable.

(viii) System h8(t) is NOT memoryless since h8(t) ≠ 0 for t ≠ 0. System h8(t) is NOT causal since h8(t) ≠ 0 for t < 0. System h8(t) is BIBO stable since

Solutions 13

[ ]

ln(0.95) ln(0.95)

00 0

2| 8( ) | 0.95 2 0.95 2ln(0.95)

2 2 0 1 39ln(0.95) ln(0.95)

t t t th t dt dt dt e dt e∞ ∞ ∞ ∞

∞

−∞ −∞

⎡ ⎤= = = = ⎣ ⎦

= − = − = < ∞

∫ ∫ ∫ ∫


Solutions to Problems of Chapter 6 Laplace Transform

Problem 6.1

(b) By definition

∫ ∫∫∫∫∫∞

∞−

∞+−

∞−

−∞

−−

∞−

−−−∞

∞−

− +=+===

II

ts

I

tssttsttsttst dtedtedteedteedteedtetxsX0

)3(0

)3(

0

30

33)()( .

Integral I reduces to

[ ]00 (3 )

(3 )1

1 11 0 Re{(3 )} 0 :Re{ } 3(3 ) (3 ) 3

s ts t eI e dt provided s ROC R s

s s s

−−

−∞ −∞

−= = = − = − > ⇒ <

− − −∫

while integral II reduces to

[ ]( 3)

( 3)1

0 0

1 10 1 Re{( 3)} 0 :Re{ } 3( 3) ( 3) 3

s ts t eII e dt provided s ROC R s

s s s

∞∞ − +− + −

= = = − = + > ⇒ > −− + + +∫

The Laplace transform is therefore given by

1 22

1 1 6( ) : : ( 3 Re{ } 3)3 3 9

X s I II with ROC R R R or R ss s s

−= + = − = = − < <

+ − −I .

(e) By definition

∫ ∫∫∞

∞−

∞−−−

∞

∞−

− ===0

)7(7 )9cos()()9cos()()( dttedtetutedtetxsX tssttst .

The above expression reduces to

[ ] ∞−−−−∞

−− +−−+−

== ∫ 0)7()7(

220

)7( )9sin(9)9cos()7(9)7(

1)9cos()( tetess

dttesX tststs

or,

[ ] .7}Re{0)}7Re{(provided9)7()7()0)7(()00(

9)7(1)( 2222 >⇒>−

+−

−=+−−−+

+−= ss

sss

ssX

(f) We derive the Laplace transform for two cases: (s = 0) and (s ≠ 0).

Case I: (s = 0). By definition

1

120 0

0

( ) ( ) 2 (1 ) 2 0.5 1sX s x t dt t dt t t∞

=−∞

⎡ ⎤= = − = − =⎣ ⎦∫ ∫

Solutions 15

Case II: (s ≠ 0). By definition

0 1

01 0

( ) ( ) (1 ) (1 )st st sts

I II

X s x t e dt t e dt t e dt∞

− − −

≠−∞ −

= = + + −∫ ∫ ∫1 44 2 4 43 1 44 2 4 43

.

Integral I is given by

( ) ( )

2

2 2 2

0 0 00 0

1 11 1

1 1 1

1 1 1 1

(1 ) ( 1)

1 1 ( 1) 1 ( 1) 1

st st st st sts s

s s s s ss s s s

I t e dt e dt te dt e e st

e e s se s e s e s

− − − − −−

− −− − −

⎡ ⎤ ⎡ ⎤= + = + = + − −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= − + − − − = − − − − = − −⎣ ⎦ ⎣ ⎦

∫ ∫ ∫

while Integral II is given by

( ) ( )

2

2 2 2

1 1 11 1

1 10 0

0 0 0

1 1 1 1

(1 ) ( 1)

1 ( 1) 1 ( 1) 1 1

st st st st sts s

s s s s ss s s s

II t e dt e dt te dt e e st

e e s s se e s e s

− − − − −−

− − − − −

⎡ ⎤ ⎡ ⎤= − = − = − − −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= − − − − + = − + + − = + −⎣ ⎦ ⎣ ⎦

∫ ∫ ∫

The Laplace transform is therefore given by

( )21

1 0( ) : Entire s-plane

2 0s ss

sX s ROC

e e s−

=⎧⎪= ⎨

+ − ≠⎪⎩

Note that the (s = 0) case can also be derived from the other result by applying the limit, s → 0, and the L’Hopital’s rule

Problem 6.3

(b) Using partial fraction expansion and associating the ROC to individual terms, gives

X (s ) = s 2+2s+1(s+1)(s 2+5s+6)

= s+1(s+2)(s+3) =

As+2

ROC:Re{s}<−2

+ Bs+3

ROC:Re{s}<−3

where constants A, and B were computed in part (a) as A = −1, and B = 2.

Taking the inverse transform of X(s), gives

( ) )(2)( 32 tueetx tt −−= −−

Note that the same rational fraction for X(s) gives different time domain representations if the associated ROC is changed.

(d) X (s ) = s 2+3s−4(s+1)(s 2+5s+6)

= s 2+3s−4(s+1)(s+2)(s+3) =

As+1

ROC:Re{s}<−1 + B

s+2

ROC:Re{s}<−2 + C

s+3

ROC:Re{s}<−3

where constants A, B, and C were computed in part (c) as A = −3, B = 6, and C = −2.


( )2 3( ) 3 6 2 ( )t t tx t e e e u t− − −= − + − .

(f) Using partial fraction expansion and associating the ROC to individual terms, gives


4}Re{

)4(

3}Re{

)3(

2}Re{

)2(2}Re{

)2()4)(3()2(1

22)(−>

+

−>

+

−>

+

−>

+++++ +++==

s

sD

s

sC

s

sB

s

sA

sssssX

where

[ ] [ ]

[ ] [ ]

[ ] [ ] .43)4(and

2)3(

21)2(

4)3()2(1

4)4)(3()2(1

3)4()2(1

1)4)(3()2(1

2)4)(3(1

1

2)4)(3()2(

1

22

22

2

==+=

−==+=

−==+=

−=+++

−=++++

−=+++

−=++++

−=+++

−=++++

ssss

sssss

ssss

sssss

ssss

sssss

sD

sC

sB

To evaluate A, expand X(s) as

)3()2()4()2()4)(3()4)(3)(2(1 22 ++++++++++++=+ ssDssCssBsssAs

and compare the coefficients of s3. We get

DCA ++=0

which has a solution A = 5/4. The Laplace transform may be expressed as

4}Re{

)4(43

3}Re{

)3(2

2}Re{

)2(21

2}Re{

)2(45

2)(

−>

+

−>

+

−>

+

−>

+ +−−=

s

s

s

s

s

ss

ssX


).()(2)()()( 44332

212

45 tuetuetutetuetx tttt −−−− +−−=

(g) Using partial fraction expansion and associating the ROC to individual terms, gives

0}:Re{ROC

)16()(

1}:Re{ROC

)1(

1}:Re{ROC

)1(1}:Re{ROC

)1()16()1(12

23223

2)(

<

+

+

−<

+

−<

+

−<

++++− +++==

s

sEDs

s

sC

s

sB

s

sA

sssssX

where

[ ] [ ] .174)1(

2)16(12

1

3)16()1(

122

2

23

2==+=

−=++−

−=+++−

ssss

sssss sC

To evaluate A, B, and C expand X(s) as

322222 )1)(()16()16)(1()16()1(12 ++++++++++=+− sEDssCssBssAss

and compare the coefficients of s4, s3 , s2, and s. We get

) of nts(coefficie316322) of nts(coefficie33171) of nts(coefficie320) of nts(coefficie0

2

3

4

sEDBAsEDCBAsEDBAsDA

+++=−++++=

+++=+=

, or,

EDBAEDBA

EDBADA

3163223317

3200

1713

+++=−+++=+++=

+=

.

which has a solution of A = 0.0206, B = −0.2076, D = −0.0206, and E = 0.2282.


Solutions 17

).()4sin(057.0)()4cos(0206.0)(1176.0)(2076.0)(0206.0)( 2 tuttuttuettutetuetx ttt −−−+−−−+−−= −−−

Problem 6.13

(a) Calculating the Laplace transform of both sides, we get

s 2Y (s )− s y (0− )=0 − y (0− )

=0

"

#$$

%

&''+3 sY (s )− y (0− )

=0

"

#$$

%

&''+ 2Y (s ) =1

which reduces to 1)()23( 2 =++ sYss

or, 21 1 1 1

( 1)( 2) 1 2( 3 2)( ) s s s ss sY s + + + ++ +

= = = − .

Calculating the inverse Laplace transform, we get

).()()( 2 tuetuety tt −− −=

(e) Calculating the Laplace transform of both sides, we get

s4Y (s )− s 3 y (0− )− s 2 y (0− )− sy (0− )− y (0− )=0

"

#$$

%

&''+ 2 s 2Y (s )− sy (0− )− y (0− )

=0

"

#$$

%

&''+Y (s ) = 1

s,

which reduces to ssYss 124 )()12( =++ ,

or, 22224 )1()1()12(1)(

++

++

++++==

sEDs

sCBs

sA

ssssY ,

where [ ] [ ] .10)12(

10)12(

12424 ===

=++=++ ssssssssA

Equating numerator of Y(s) on both sides and setting A = 1, we get

2 2 2

4 3 2

1 ( 1) ( ) ( 1) ( )(1 ) (2 ) ( ) 1s Bs C s s Ds E sB s Cs B D s C E s

= + + + + + +

= + + + + + + + +

Comparing the coefficients of polynomials of different order we get

4

3

2

Coefficients of : 1 0 1Coefficients of : 0Coefficients of : 2 0 1 0 1Coefficients of : 0 0

s B Bs Cs B D D Ds C E E

+ = ⇒ = −

=

+ + = ⇒ + = ⇒ = −

+ = ⇒ =

The partial fraction expansion of Y(s) is given by

2 2 21

1 ( 1)( ) s s

s s sY s

+ += − − .

The inverse transform is therefore given by

[ ]( ) 1 cos( ) 0.5 sin( ) ( )y t t t t u t= − +

where we have used the following transform pair


( )

022 2

0

20sin( ) ( ) sL

st t u t ω

ωω

+←⎯→

which is proved in Problem 6.10(b).

Problem 6.14

Solution:

(a) The Laplace transform of the input and output signals are given by

)2(2

2114

2

2

2)( and )(+++

+ =+==ssss

sss sYsX .

Dividing Y(s) with X(s), the transfer function is given by

)2(42

)()( 2

)( +++== ssss

sXsYsH .

The impulse response is obtained by taking the partial fraction expansion of H(s) as follows

)2(21

41

41

)2(422

)( ++++ −+≡= sssssssH .

Taking the inverse Laplace transform, the impulse response is given by

).()()()( 221

41

41 tuetutth t−−+δ=

In order to calculate the input-output relationship in the form of a differential equation, we represent the transfer function as

)()(

)2(422

)( sXsY

sssssH == +++ .

Cross multiplying, we get )()2()()2(4 2 sXsssYss ++=+

which can be represented as )(2)()()(8)(4 22 sXssXsXsssYsYs ++=+ .

Taking the inverse Laplace transform and assuming zero initial conditions, the differential equation representing the system is given by

)(284 2

2

2

2tx

dtdx

dtxd

dtdy

dtyd

++=+ .

(c) The Laplace transform of the input and output signals are given by

)4(121 3)( and )( 32 +−== sss

sYsX .

Dividing Y(s) with X(s), the transfer function is given by

)4(32

)()( 2

)( +−== ss

ssXsYsH .

The impulse response is obtained by taking the inverse Laplace transform. The impulse response is given by

[ ] .)(3)(2)( 42

2tuetuth t

dtd −−=

In order to calculate the input-output relationship in the form of a differential equation, we represent the transfer function as

Solutions 19

)()(

)4(3)4(2 3

)( sXsY

sssssH == +

−+ .

Cross multiplying, we get )(8)(2)(3)(4)( 32 sXssXsXsssYsYs ++−=+ .

Taking the inverse Laplace transform, the input-output relationship of the system is given by

)(8234 3

3

2

2tx

dtdx

dtxd

dtdy

dtyd

++−=+ .

(e) Note that there is no overlap between the ROC’s of the two terms exp(t)u(−t) and exp(−3t)u(t), therefore, the Laplace transform for y(t) does not exist.

Problem 6.15

(a) 22

2

)1())((

121)(

+

−+

+++ ==

sjsjs

ssssH

Two zeros at s = j, −j.

Two poles at s = 1, −1.

Because both poles are in the left hand side of the s-plane, the system is always BIBO stable.

(c) )6)(3()3/10(

189103 3)( 2 ++

+

+++ == ss

sssssH

One zero at s = −10/3.

Two poles at s = −3, −6.

Because both poles are in the left side of the s-plane, the system is always BIBO stable.

(e) sssssssH 12323

23

2)( ==

++++

The system does not have any zero.

One pole at s = 0.

There is only one pole, which is located on the imaginary axis. Therefore, the system is a marginally stable system.

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