Continuous and Discrete Time Signals and Systems (Mandal & Asif) Solutions - Chap03

Chapter 3: Time Domain Analysis of LTIC Systems Problem 3.1 Linearity: For x3(t) = α x1(t) + β x2(t) applied as the input, the output y3(t) is given by

d n y3 dt n

+ an −1

d n −1 y3

+ " + a1

dt n −1

dy3 d m −1 (αx1 (t ) + βx2 (t )) d m (αx1 (t ) + βx2 (t )) + a0 y3 (t ) = bm + b m −1 dt dt m −1 dt m . d (αx1 (t ) + βx2 (t )) + " + b1 + b0 (αx1 (t )βx2 (t )) dt

Rearranging the terms on the right hand side of the equation, we get d n y3 dt n

d n −1 y3

+ an −1

dt n −1

+ " + a1

 d mx  dy3 d m −1 x dx + a0 y3 (t ) = α bm m1 + bm −1 m −11 + " + b1 1 + b0 x1 (t ) dt dt dt dt   m m − 1  d x2  d x dx + β bm + bm −1 m −12 + " + b1 2 + b0 x2 (t ) m dt dt dt  

Expressing the higher order derivatives of x1(t) and x2(t) in terms of y1(t) and y2(t), we get d n y3 dt n

+ an −1

d n −1 y3 dt n −1

+ " + a1

d ny  dy3 d n −1 y dy + a0 y3 (t ) = α  n1 + an −1 n −11 + " + a1 1 + a0 y1 (t )  dt dt dt  dt 

dny  d n −1 y dy + β  n1 + an −1 n −11 + " + a1 1 + a0 y1 (t ) dt dt  dt 

d n y3 or,

dt n

+ an −1

d n y3 dt n

+ " + a1

dy3 d n (αy1 + β y2 ) d n −1 (αy1 + β y2 ) a + a0 y3 (t ) = + n − 1 dt dt n dt n −1 d (αy1 + β y2 ) + " + a1 + a0 (αy1 (t ) + βy2 (t ) ) dt

y 3 (t ) = αy1 (t ) + βy 2 (t ) .

which implies that

The system is therefore linear. Time-invariance: For x(t – t0) applied as the input, the output y1(t) is given by

d m x (t − t 0 ) d m −1 x(t − t0 ) d n y1 d n −1 y1 dy1 + a + " + a + a y ( t ) = b + b n −1 m m −1 1 0 1 dt dt n dt n −1 dt m dt m −1 dx(t − t0 ) + " + b1 + b0 x(t − t0 ) dt Substituting τ = t – t0 (which implies that dt = dτ), we get d n y1 (τ + t0 ) dτ n

+ an −1

d n −1 y1 (τ + t0 ) dτ n −1

+ " + a1

dy1 ( τ + t0 ) d m x(τ) d m −1 x(τ) + a0 y1 ( τ + t0 ) = bm + b m −1 dτ dτ m dτ m −1 dx(τ) + b0 x(τ) + " + b1 dτ

Comparing with the original differential equation representation of the system, we get

78

Chapter 3 y (τ) = y1 (τ + t0 ) or,

y1 (τ) = y ( τ − t0 ) ,

proving that the system is time-invariant. Note that the time invariance property is only valid if the coefficients ar’s and br’s are constants. If ar’s and br’s are functions of time, then the substitution (τ = t – t0) will also affect them. Clearly, y(τ) ≠ y1(τ + t0) in such a case and the system will NOT be timeinvariant. ▌ Problem 3.2

x (t ) = e −4 t u(t ), y (0) = 0, and y (0) = 0.

(i)

 y (t ) + 4 y (t ) + 8 y (t ) = x (t ) + x (t ) with

(a)

Zero-input response of the system: The characteristic equation of the LTIC system (i) is s 2 + 4s + 8 = 0 , which has roots at s = −2 ± j2. The zero-input response is given by

y zi (t ) = Ae −2t cos(2t ) + Be −2t sin(2t ) for t ≥ 0, with A and B being constants. To calculate their values, we substitute the initial conditions y (0 − ) = 0 and y (0 − ) = 0 in the above equation. The resulting simultaneous equations are

A=0 − 2 A + 2B = 0 that has the solution, A = 0 and B = 0. The zero-input response is therefore given by y zi (t ) = 0. 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 d2y dy dx +4 +8= + x(t ) 2 dt dt dt

with initial conditions, y (0 − ) = 0 and y (0 − ) = 0 , 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 (h) (t ) = C1e −2t cos(2t ) + C2e −2t sin(2t ) y zs

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

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

(

)

y zs (t ) = C1e −2t cos(2t ) + C2 e −2t sin(2t ) − 83 e −4t u (t ) . To compute the values of constants C1 and C2, we use the initial conditions, y(0−) = 0 and y (0 − ) = 0 assumed for the zero-state response. Substituting the initial conditions in yzs(t) leads to the following simultaneous equations

Solutions

79

C1 − 83 = 0

− 2C1 + 2C2 +

3 2

=0

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

y zs (t ) = (c)

(e

−2t

)

cos(2t ) − e −2t sin(2t ) − e −4t u (t ) .

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

(

)

y (t ) = 83 e −2t cos(2t ) − e −2t sin(2t ) − e −4t u (t ) . (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 3 t →∞ 8

y (t ) = lim

(e

−2 t

)

cos(2t ) − e −2t sin(2t ) − e −4t u (t ) = 0 .

(ii)

y (t ) + 6 y (t ) + 4 y (t ) = x (t ) + x (t ) with

x (t ) = cos(6t )u(t ), y (0) = 2, and y (0) = 0.

(a)

Zero-input response of the system: The characteristic equation of the LTIC system (ii) is s 2 + 6s + 4 = 0 , which has roots at s = −3 ± 2.2361 = −5.2361 and −0.7639. The zero-input response is given by

y zi (t ) = Ae −5.2361t + Be −0.7639t for t ≥ 0 with A and B being constants. To calculate their values, we substitute the initial conditions y (0 − ) = 2 and y (0 − ) = 0 in the above equation. The resulting simultaneous equations are A+ B = 2 − 5.2361A − 0.7639 B = 0 that has a solution, A = −0.3416 and B = 2.3416. The zero-input response is therefore given by

(

)

y zi (t ) = − 0.3416e −5.2361t + 2.3416e −0.7639t u (t ) . (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

d2y dy dx +6 +4= + x(t ) 2 dt dt dt with initial conditions, y (0 − ) = 0 and y (0 − ) = 0 , and input x(t) = cos(6t)u(t). The homogenous solution of system (ii) has the same form as its zero-input response and is given by (h) y zs (t ) = C1e −5.2361t + C2e −0.7639t

for t ≥ 0, with C1 and C2 being constants. The particular solution for input x(t) = cos(6t)u(t) is of the form ( p) y zs (t) = K1 cos(6t ) + K 2 sin(6t ) .

80

Chapter 3 Substituting the particular solution in the differential equation for system (ii) and solving the resulting equation gives

(− 36 K1 cos(6t ) − 36 K 2 sin(6t ) ) + 6(− 6 K1 sin(6t ) + 6 K 2 cos(6t ) ) + 4(K1 cos(6t ) + K 2 sin(6t ) ) = −6 sin(6t ) + cos(6t ) Collecting the coefficients of the cosine and sine terms, we get

(− 36 K1 + 36 K 2 + 4 K1 − 1) cos(6t ) + (− 36 K 2 − 36 K1 + 4 K 2 + 6)sin(6t ) = 0 or, − 32 K1 + 36 K 2 = 1 − 36 K1 − 32 K 2 = −6

which has the solution, K1 = 0.0793 and K2 = 0.0983. The zero-state response of the system is

(

)

y zs (t ) = C1e −5.2361t + C2e −0.7639t + 0.0793 cos(6t ) + 0.0983 sin(6t ) u (t ) . To compute the values of constants C1 and C2, we use the zero initial conditions, y(0−) = 0 and y (0 − ) = 0 assumed for the zero-state response. Substituting the initial conditions in yzs(t) leads to the following simultaneous equations C1 + C2 + 0.0793 = 0 − 5.2361C1 − 0.7639C2 + 6(0.0983) = 0 with solution C1 = 0.1454 and C2 = −0.2247. The zero-state solution is given by

(

)

y zs (t ) = 0.1454e −5.2361t − 0.2247e −0.7639t + 0.0793 cos(6t ) + 0.0983 sin(6t ) u (t ) . (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

( + (0.1454e

)

y (t ) = − 0.3416e − 5.2361t + 2.3416e − 0.7639t u (t )

or, (d)

− 5.2361t

)

− 0.2247e − 0.7639t e − 0.7639t + 0.0793 cos(6t ) + 0.0983 sin(6t ) u (t )

(

)

y (t ) = − 0.1962e − 5.2361t + 2.1169e − 0.7639t + 0.0793 cos(6t ) + 0.0983 sin(6t ) u (t ) .

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 y (t ) = (0.0793 cos(6t ) + 0.0983 sin(6t ) ) u (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 s 2 + 2s + 1 = 0 , which has roots at s = −1, −1. The zero-input response is given by y zi (t ) = Ae −t + Bte −t

Solutions

81

for t ≥ 0, with A and B being constants. To calculate their values, we substitute the initial conditions y (0 − ) = 3 and y (0 − ) = 1 in the above equation. The resulting simultaneous equations are A=3 − A+ B =1 that has a solution, A = 3 and B = 4. The zero-input response is therefore given by

(

)

y zi (t ) = 3e −t + 4te −t u (t ) . (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 d2y dy + 2 + 1 = x(t ) 2 dt dt

with initial conditions, y (0 − ) = 0 and y (0 − ) = 0 , 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 (h) y zs (t ) = C1e −t + C2te −t

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 ( p) y zs (t) = K1 cos(t ) + K 2 sin(t ) + K 3 cos(2t ) + K 4 sin(2t ) .

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

(− K1 cos(t ) − K 2 sin(t ) − 4 K 3 cos(2t ) − 4 K 4 sin(2t )) + 2(− K1 sin(t ) + K 2 cos(t ) − 2 K 3 sin(2t ) + 2 K 4 cos(2t ) ) + 1(K1 cos(t ) + K 2 sin(t ) + K 3 cos(2t ) + K 4 sin( 2t ) ) = − cos(t ) − 4 sin(2t ) Collecting the coefficients of the cosine and sine terms, we get

(− K1 + 2 K 2 + K1 + 1) cos(t ) + (− K 2 − 2 K1 + K 2 ) sin(t ) + (− 4 K 3 + 4 K 4 + K 3 ) cos(2t ) + (− 4 K 4 − 4 K 3 + K 4 + 4)sin(2t ) = 0 which gives K1 = 0, K2 = −0.5, K3 = 0.64, and K4 = 0.48. The zero-state response of the system is

(

)

y zs (t ) = C1e −t + C2te −t − 0.5 sin(t ) + 0.64 cos(2t ) + 0.48 sin(2t ) u (t ) . To compute the values of constants C1 and C2, we use the initial conditions, y(0−) = 0 and y (0 − ) = 0 . Substituting the initial conditions in yzs(t) leads to the following simultaneous equations C1 + 0.64 = 0 − C1 + C2 − 0.5 + 0.48 = 0 with solution C1 = −0.64 and C2 = −1.1. The zero-state solution is given by

(

)

y zs (t ) = − 0.64e −t − 1.1te −t − 0.5 sin(t ) + 0.64 cos(2t ) + 0.48 sin( 2t ) u (t ) . (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

(

)

(

)

or, y (t ) = 3e −t + 4te −t u (t ) + − 0.64e −t − 1.1te −t − 0.5 sin(t ) + 0.64 cos(2t ) + 0.48 sin(2t ) u (t )

82

Chapter 3

or, (d)

(

)

y (t ) = 2.36e −t + 2.9te −t − 0.5 sin(t ) + 0.64 cos(2t ) + 0.48 sin(2t ) u (t ) .

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

(

)

y (t ) = lim 2.36e −t + 2.9te −t − 0.5 sin(t ) + 0.64 cos(2t ) + 0.48 sin( 2t ) u (t ) t →∞

y (t ) = (− 0.5 sin(t ) + 0.64 cos(2t ) + 0.48 sin(2t ) ) u (t ) .

or,

x (t ) = 4te− t u(t ), y (0) = −2, and y (0) = 0.

(iv)

 y (t ) + 4 y (t ) = 5 x (t ) with

(a)

Zero-input response of the system: The characteristic equation of the LTIC system (iv) is s2 + 4 = 0 , which has roots at s = ±j2. The zero-input response is given by

y zi (t ) = A cos(2t ) + B sin(2t ) for t ≥ 0, with A and B being constants. To calculate their values, we substitute the initial conditions y (0 − ) = −2 and y (0 − ) = 0 in the above equation. The resulting simultaneous equations are A = −2 2B = 0 that has a solution, A = −2 and B = 0. The zero-input response is therefore given by y zi (t ) = −2 cos(2t )u (t ) (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 d2y + 4 = 5 x(t ) dt 2 with initial conditions, y (0 − ) = 0 and y (0 − ) = 0 , and input x(t) = 4t exp(−t) u(t). The homogenous solution of system (iv) has the same form as the zero-input response and is given by (h) y zs (t ) = C1 cos(2t ) + C2 sin(2t )

where C1 and C2 are constants. The particular solution for input x(t) = 4t exp(−t) u(t) is of the form ( p) y zs (t) = K1e −t + K 2te −t .

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

(K e 1

−t

) (

Collecting the coefficients of exp(−t) and texp(−t), we get

(K e 1

−t

)

− K 2 e −t − K 2 e −t + K 2te −t + 4 K1e −t + K 2te −t = 20te −t

) (

)

− K 2 e −t − K 2 e −t + 4 K1e −t + K 2te −t + 4 K 2te −t = 20te −t

which gives K1 = 1.6 and K2 = 4. The zero-state response of the system is given by

Solutions

(

83

)

y zs (t ) = C1 cos(2t ) + C 2 sin( 2t ) + 1.6e −t + 4te −t . To compute the values of constants C1 and C2, we use the initial conditions, y(0−) = 0 and y (0 − ) = 0 . Substituting the initial conditions in yzs(t) leads to the following simultaneous equations C1 + 1.6 = 0 2C2 − 1.6 + 4 = 0

with solution C1 = −1.6 and C2 = −1.2. The zero-state solution is given by

(

)

y zs (t ) = − 1.6 cos(2t ) − 1.2 sin( 2t ) + 1.6e −t + 4te −t u (t ) . (c)

(d)

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

(

)

or,

y (t ) = −2 cos(2t ) u (t ) + − 1.6 cos(2t ) − 1.2 sin(2t ) + 1.6e −t + 4te −t u (t )

or,

y (t ) = − 3.6 cos(2t ) − 1.2 sin(2t ) + 1.6e −t + 4te −t u (t ) .

(

)

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

(

)

y (t ) = lim − 2.32 cos(2t ) − 0.24 sin(2t ) + 0.32e −t + 0.8te −t u (t ) = (− 2.32 cos(2t ) − 0.24 sin(2t ) ) u (t ) . t →∞

(v) (a)

d4y dt 4

+2

d2y dt 2

+ y (t ) = x(t ) with

x(t ) = 2u (t ), y (0) = y(0) = y(0) = 0, and y (0) = 1.

Zero-input response of the system: The characteristic equation of the LTIC system (v) is s 4 + 2s + 1 = 0 , which has roots at s = ±j1, ±j1. The zero-input response is given by y zi (t ) = Ae jt + Bte jt + Ce − jt + Dte − jt , 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 A jA +B − A + j 2B − jA − 3B

+C − jC −C + jC

= +D = − j 2D = − 3D =

0 1 0 0

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

(

)

y zi (t ) = − j 0.75e jt − 0.25te jt + j 0.75e − jt − 0.25te − jt u (t ) , which reduces to y zi (t ) = (1.5 sin t − 0.5t cos t ) u (t ) . (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

84

Chapter 3

d4y dt 4

+2

d2y dt 2

+ y (t ) = x(t )

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 (h) y zs (t ) = C1e jt + C 2 te jt + C 3 e − jt + C 4 te − jt

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

Substituting the particular solution in the differential equation for system (v) and solving the resulting equation gives 0 + 2(0) + K = 2 , or, K = 2. The zero-state response of the system is given by

y zs (t ) = C1e jt + C 2 te jt + C 3 e − jt + C 4 te − jt + 2 , 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 A jA +B − A + j 2B − jA − 3B

+C − jC −C + jC

= +D = − j 2D = − 3D =

−2 0 0 0

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

(

)

y zs (t ) = − e jt + j 0.5te jt − e − jt − j 0.5te − jt u (t ) , which reduces to y zi (t ) = (− 2 cos t − t sin t ) u (t ) . (c)

(d)

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 or,

y (t ) = (1.5 sin t − 0.5t cos t ) u (t ) + (− 2 cos t − t sin t ) u (t )

or,

y (t ) = (1.5 sin t − 2 cos t − t sin t − 0.5t cos t + 2) u (t ) .

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

y (t ) = lim (1.5 sin t − 2 cos t − t sin t − 0.5t cos t + 2) u (t ) → ∞ . t →∞

Problem 3.3

(i)

To evaluate the impulse response, set x(t) = δ(t). The resulting equation is t

∫

h(t ) = 2δ(t ) ⇒ h(t ) = 2 δ(t )dt + C = 2u (t ) + C −∞

▌

Solutions

85

where C is a constant that can be evaluated from the initial condition. Since the initial condition is 0, then C = 0. To solve parts (ii)-(vi), we make use of the following theorem. Theorem S2.1: The impulse response of an LTIC system initially at rest and described by the differential equation n

∑ a p ddt y = x(t ) p

p

p =0

n

∑ a p ddt h = 0

is given by

p

p

p =0

with initial condition

(ii)

d n −1 h dt n −1

(0 + ) =

1 an

. The remaining lower order initial conditions are all zero.

Based on Theorem S2.1, the impulse response of system (ii) is given by h(t ) + 6h(t ) = 0 with initial condition h(0+) = 1. The characteristic equation for the homogenous equation is

s+6=0 which has a root at s = −6. The impulse response is given by h(t ) = C1e −6t +

for t ≥ 0, with C1 being a constant. Use the initial condition h(0 ) = 1, the value of C1 = 1. The impulse response of system (ii) is given by

h(t ) = e −6t u (t ) . (iii) Assume w(t) = dx/dt. System (iii) can, therefore, be represented as a cascaded combination of two systems w(t ) =

System S1(iii):

dx (t ) dt

2 y (t ) + 5 y (t ) = w(t )

System S2(iii):

Based on Theorem S2.1, the impulse response of system S2(iii) is given by 2h2 (t ) + 5h2 (t ) = 0 +

with initial condition h2(0 ) = 1/2. The characteristic equation for the homogenous equation is (2s + 5) = 0 which has a root at s = −5/2. The impulse response is given by h2 (t ) = C1e −5t / 2

86

Chapter 3 +

for t ≥ 0, with C1 being a constant. Use the initial condition h2(0 ) = 1/2, the value of C1 = 1/2. The impulse response of system S2(iii) is h2 (t ) = 12 e −5t / 2u (t ) . for t ≥ 0. Combining the cascaded configuration, the impulse response of the overall system is

h(t ) = (iv)

dh2 (t ) dt

=

(

d 1 −5t / 2 e u (t ) dt 2

)=

1 −5t / 2 e δ(t ) − 12 × 52 e −5t / 2u (t ) 2

= 12 δ(t ) − 54 e −5t / 2u (t ) .

System (iv) is represented as a cascaded combination of two systems w(t ) = 2

System S1(iv):

dx (t ) dt

+ 3x(t )

y (t ) + 3 y (t ) = w(t )

System S2(iv):

Based on Theorem S2.1, the impulse response of system S2(iv) is given by h2 (t ) + 3h2 (t ) = 0 +

with initial condition h2(0 ) = 1. The characteristic equation for the homogenous equation is ( s + 3) = 0 which has a root at s = −3. The impulse response is given by h2 (t ) = C1e −3t +

for t ≥ 0, with C1 being a constant. Use the initial condition h2(0 ) = 1, the value of C1 = 1. The impulse response of system S2(ii) is h2 (t ) = e −3t u (t ) . for t ≥ 0. Combining the cascaded configuration, the impulse response of the overall system is

h(t ) = 2

dh2 (t ) dt − 3t

= 2e (v)

(

)

+ 3h2 (t ) = 2 dtd e −3t u (t ) + 3e −3t u (t )

δ(t ) − 6e

− 3t

u (t ) + 3e

− 3t

u (t ) = 2δ(t ) − 3e − 3t u (t )

.

Based on Theorem S2.1, the impulse response of system (v) is given by h(t ) + 5h(t ) + 4h(t ) = 0 +

+

with initial conditions dh(0 )/dt = 1 and h(0 ) = 0. The characteristic equation for the homogenous equation is ( s 2 + 5s + 4) = 0 which has roots at s = −1 and −4. The impulse response is given by h2 (t ) = C1e −t + C2e −4t for t ≥ 0, with C1 and C2 being constants. Using the initial conditions C1 + C2 = 0 − C1 − 4C2 = 1

which has the solution C1 = 1/3, C2 = −1/3. The impulse response of system (v) is given by

Solutions

h(t ) = (vi)

(e

1 −t 3

)

− 13 e −4t u (t )

87

.

Based on Theorem S2.1, the impulse response of system (vi) is given by h(t ) + 2h(t ) + h(t ) = 0 +

+

with initial conditions dh(0 )/dt = 1 and h(0 ) = 0. The characteristic equation for the homogenous equation is ( s 2 + 2 s + 1) = 0 which has two roots at s = −1. The impulse response is given by

h2 (t ) = C1e −t + C2te −t for t ≥ 0, with C1 and C2 being constants. Using the initial conditions C1 = 0 − C1 + C2 = 1 which has the solution C1 = 0, C2 = 1. The impulse response of system (vi) is given by h(t ) = te −t u (t ) .

▌

Problem 3.4

(i)

Functions x(τ) = exp(−ατ)u(τ), h(τ) = exp(−βτ)u(τ), and is h(−τ) = exp(βτ)u(−τ) are plotted, respectively, in Fig. S3.4(a)-(c). The function h(t − τ) = h(−(τ − t) is obtained by shifting h(−τ) by time t in Fig. S3.4(d). We consider the following two cases of t. Case 1: For t < 0, the waveform h(t − τ) is on the left hand side of the vertical axis. As apparent in the subplot for step 5a in Fig. 3.7, waveforms for h(t − τ) and x(τ) do not overlap. In other words, x(τ)h(t − τ) = 0 for all τ, hence, y(t) = 0. Case 2: For t ≥ 0, we see from the subplot for step 5b in Fig. 3.7 that the nonzero parts of h(t − τ) and x(τ) overlap over the duration t = [0, t]. Therefore, t

t

y (t ) = e − ατ e −β(t − τ) dτ = e −βt e − ( α −β) τ dτ.

∫

∫

0

(S3.4.1)

0

Since (α ≠ β), therefore, the exponential term can be integrated as y (t ) = e

− ( α −β) τ

t

  =   − (α − β)  0

−βt  e

1 e −βt (β − α )

[e

− ( α − β)t

0  Combining the two cases, we get y (t ) =  1 −βt − αt  (β − α ) e − e

[

which is equivalent to

y (t ) =

1 (β − α )

[e

−βt

]

]

−1 =

]

− e − αt u (t ) .

1 (β − α )

t 0

1

t

1

e− β(t−τ) u(t−τ) 0

Case 1: t < 0

τ

0

x(τ) h(t−τ) 1

τ

0

0

(e) Overlap between x(τ) and h(t − τ) for t < 0

t

τ

(f) Overlap between x(τ) and h(t − τ) for t ≥ 0

Fig. S3.4.1: Convolution between x(τ) = exp(−ατ)u(τ) and h(τ) = exp(−βτ)u(τ) in Problem 3.4. (ii)

For α = β, Eq. (S3.4.1) reduces to t

t

y (t ) = e −βt e − ( α −β) τ dτ = e −βt 1 dτ = te −βt .

∫

∫

0

0

The output y(t) is therefore given by y (t ) = te −βt u (t ) = te − αt u (t ) . (iii) Part (ii) is a special case of part (i) as the result for part (ii) can be obtained by applying the limit, α → β, to the solution of part (i). Since applying the limit results in a 0/0 case, we apply the L’Hopital’s rule to get 1 α →β (β − α )

y (t ) = lim

[e

− αt

]

− e −βt u (t ) = lim

1 α →β ( −1)

[− te

− αt

]

− 0 u (t ) = te − αt u (t ) .

Problem 3.5

(i)

The output y(t) is given by ∞

y (t ) = u (t ) ∗ u (t ) =

∫

∞

∫

u (τ)u (t − τ) dτ = u (t − τ) dτ .

−∞

0

▌

Solutions

1 if (τ ≤ t ) u (t − τ) =  0 if ( τ > t ).

Recall that

Therefore, the output y(t) is given by

t if (t ≥ 0) y (t ) =  = tu (t ) = r (t ). 0  if (t < 0) The aforementioned convolution can also be computed graphically. (ii)

The output y(t) is given by ∞

y (t ) = u ( −t ) ∗ u (−t ) =

∫

0

u (− τ)u (τ − t ) dτ =

−∞

∫ u(τ − t ) dτ .

−∞

The output y(t) is given by 0 if (t ≥ 0)  0 0 if (t ≥ 0) = = −tu (−t ). y (t ) = u (τ − t ) dτ =  u ( τ − t ) dτ if (t < 0) − t if (t < 0)  −∞ t 0

∫

∫

The aforementioned convolution can also be computed graphically. (iii) The output y(t) is given by

y (t ) = [u (t ) − 2u (t − 1) + u (t − 2)] ∗ [u (t + 1) − u (t − 1)] Using the properties of the convolution integral, the output is expressed as y (t ) = [u (t ) ∗ u (t + 1)] − [u (t ) ∗ u (t − 1)] − 2[u (t − 1) ∗ u (t + 1)] + 2[u (t − 1) ∗ u (t − 1)] + [u (t − 2) ∗ u (t + 1)] − [u (t − 2) ∗ u (t − 1)]

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

y (t ) = r (t + 1) − r (t − 1) − 2r (t ) + 2r (t − 2) + r (t − 1) − r (t − 3) . (iv)

The output y(t) is given by y (t ) = e 2t u (−t ) ∗ e − 3t u (t ) =

∞

∫

e 2 τu (−τ)e − 3(t − τ) u (t − τ) dτ = e − 3t

−∞

0

∫e

5τ

u (t − τ) dτ .

−∞

Solving for the two cases (t ≥ 0) and (t < 0), we get t  − 3t  e e 5τ dτ (t < 0) 0  1 e 2t  − 3t 5τ − ∞ y (t ) = e e u (t − τ) dτ =  =  15 − 3t 0  e  − 3t −∞ e 5τ dτ (t ≥ 0)  5 e −∞ 

∫

∫

∫

Therefore, the output y(t) is given by y (t ) = 15 e 2t u (−t ) + 15 e −3t u (t ). (v)

The output y(t) is given by

(t < 0) (t ≥ 0).

89

90

Chapter 3

    u (t − τ ) − u (t − τ − 2) ] dτ − − − sin(2 πτ ) u ( τ 2) u ( τ 5) ∫  

[   −∞ 0 for τ 2, τ 5 = < >   ∞

y (t ) = x(t ) * h(t ) = 5

5

5

= ∫ sin(2πτ ) [u (t − τ ) − u (t − τ − 2) ] dτ = ∫ sin(2πτ )u (t − τ )dτ − ∫ sin(2πτ )u (t − τ − 2)dτ 2 2 2      

=A

=B

Calculating Term A and Term B separately, we get   t≤2 0 t≤2 0 t  1−cos 2π t  A =  ∫ sin(2πτ )dτ 2 ≤ t ≤ 5 =  2π 2≤t ≤5 2  t ≥5  0 5  sin(2πτ )dτ t ≥5 ∫ 2   t−2≤ 2  0 t≤4  0 0 t − 2  1−cos 2π (t − 2)   2π t B =  ∫ sin(2πτ )dτ 2 ≤ t − 2 ≤ 5 =  2π 4 ≤ t ≤ 7 =  1−cos 2π 2    0 t≥7   0  5  sin(2πτ )dτ t −2≥5 ∫  2

t≤4 4≤t ≤7 t≥7

The overall output is given by t ≤ 2, t ≥ 7 0 1 2≤t ≤4  2π (1 − cos 2π t ) Therefore, y (t ) = A − B =  4≤t ≤5 0  1 5≤t ≤7 − 2π (1 − cos 2π t )

(vi)

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

Case I (t < 0):

∫e

y (t ) =

2 τ − 5( t − τ )

e

0

∫

dτ + e e

−∞

which reduces to

∞

∫

dτ + e − 2 τ e 5 ( t − τ ) dτ 0

t

y (t ) = e

− 5t

t

∫e

7τ

dτ + e

−∞

or,

2 τ 5( t − τ )

5t

0

∫e

− 3τ

dτ + e

− 5t

∫e

− 7τ

dτ

0

t

(

∞

)

4 −5 t y (t ) = e −5t × 17 e 7t + e 5t × 13 e −3t − 1 + e −5t × 17 = 17 e 2t − 21 e + 13 e −8t 0

Case II (t ≥ 0):

y (t ) =

∫e

2 τ − 5( t − τ )

e

−∞

which reduces to

y (t ) = e − 5 t

t

∫

dτ + e

− 2 τ − 5( t − τ )

e

∫

∫

dτ + e − 2 τ e 5(t − τ) dτ t

0

0

∞

t

∞

0

t

∫

∫

e 7 τ dτ + e − 5t e 3τ dτ + e − 5t e − 7 τ dτ

−∞

Solutions

(

91

)

y (t ) = 17 e 5t + e −5t × 13 e 3t − 1 + e −5t × 17 e −7t = 71 e 5t + 13 e −2t − 13 e −5t + 71 e −12t .

or,

Hence, the overall expression for y(t) is given by  1 e 2t y (t ) ==  15 − 3t  5 e

(t < 0) (t ≥ 0).

(vii) Note that

sin(t )u (t ) ∗ cos(t )u (t ) =

1 2j

=

1 4j

+ =

1 4j 1 4j

(e − e ) u(t ) * (e + e ) u(t ) [e u(t ) ∗ e u(t )] − [e u(t ) ∗ e u(t )] [e u(t ) ∗ e u(t )]− [e u(t ) ∗ e u(t )] [e u(t ) ∗ e u(t )]− [e u(t ) ∗ e u(t )] − jt

jt

− jt

jt

1 2

jt

jt

1 4j

− jt

jt

jt

− jt

1 4j

− jt

− jt

jt

jt

− jt

1 4j

− jt

Based on the result of Problem 3.4, we know that e − jt u (t ) ∗ e − jt u (t ) = te − jt u (t ) e jt u (t ) ∗ e jt u (t ) = te jt u (t ) .

and Hence, the output is given by y (t ) =

1 4j

te jt u (t ) −

1 4j

te − jt u (t ) = 12 t sin(t )u (t ) .

Note that the above convolution can also be performed directly as follows: y (t ) = [sin(t )u (t ) ] * [ cos(t )u (t )] =

∞

∞

−∞

0

∫ sin(τ )u(τ ) cos(t − τ )u(t − τ )dτ = ∫ sin(τ ) cos(t − τ )u (t − τ )dτ

t

= ∫ sin(τ ) cos(t − τ )dτ

t >0

[ y (t ) = 0, t < 0]

0 t

t

t

0

0

= ∫ sin(τ ) [ cos(t ) cos(τ ) + sin(t )sin(τ )] dτ = cos(t ) ∫ sin(τ ) cos(τ )dτ + sin(t ) ∫ sin 2 (τ )dτ 0

t

t

τ)  + 0.5sin(t ) τ − sin(22 τ )  = 0.5cos(t ) ∫ sin(2τ )dτ + 0.5sin(t ) ∫ [1 − cos(2τ )] dτ = 0.5cos(t )  − cos(2 2 0

0

t

t

0

0

  = 0.5cos(t ) [ 12 − 12 cos(2t )] + 0.5sin(t ) [t − 12 sin(2t )] = 14 cos(t ) + 12 t sin(t ) − 14 cos(t ) cos(2t ) + sin(t ) sin(2t )   

   = cos(2 t −t ) = cos( t )   = 12 t sin(t ) + 14 cos(t ) − 14 cos(t ) = 12 t sin(t )

▌ Problem 3.6

(i)

Using the graphical approach, the convolution of x(t) with itself is shown in Fig. S3.6.1, where we consider six different cases for different values of t.

92

Chapter 3

1

x(τ)

1

τ

0

1

−2

2

τ

t −2

1

1

τ 1

t −2

2

−1

(c) Waveform for x(t−τ) 1

τ t −2

2

1

x(τ) x(t−τ)

τ 1

t

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