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 t≥0
,
[e
− αt
]
− e −βt . .
88
Chapter 3
h(τ)
x(τ) 1
1
e−ατu(τ) τ
0
e−βτu(τ)
(a) Waveform for x(τ)
(a) Waveform for h(τ) h(t−τ)
h(−τ) 1
eβτu(−τ)
τ
t
τ
0
(d) Waveform for h(t − τ)
(c) Waveform for h(−τ) x(τ) h(t−τ)
Case 2: 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<3)
(h) Overlap btw x(τ) and x(t−τ) for (3≤t<4)
y1(t) = x(t)*x(t)
0.5 0 -0.5 -1 -1.5 0
1
2
(j) Convolution output
3
4
y1 (t )
Fig. S3.6.1: Convolution of x(t) with x(t) in Problem 3.6(i). Case I (t < 0): Since there is no overlap,
y1 (t ) = 0 . t
∫
y1 (t ) = 1.1dτ = t .
Case II (0 ≤ t < 1):
0
t −1
y1 (t ) =
∫
1
∫
t
∫
(−1).1dτ + 1.1dτ + 1.(−1)dτ t −1
0
1
= −(t − 1) + (1 − t + 1) − (t − 1) = 4 − 3t. y1 (t ) =
t −1
t
(i) Overlap btw x(τ) and x(t−τ) for (t>4)
1
-2
2 t −2
−1
−1
−1
Case IV (2 ≤ t < 3):
2
τ 1 t −2 2 t −1
2 t
Case III (1 ≤ t < 2):
1 t
(f) Overlap btw x(τ) and x(t−τ) for (1≤t<2)
x(τ) x(t−τ)
τ t −2 1 t −1
t −1 −1
(d) Overlap btw x(τ) and x(t−τ) for (t < (e) Overlap btw x(τ) and x(t−τ) for (0≤t<1) 0) 1
x(τ) x(t−τ)
τ 1
−1
x(τ) x(t−τ)
τ
0
t
x(τ) x(t−τ)
0 t
t −1
x(t−τ)
−1
(b) Waveform for x(−τ)
x(τ) x(t−τ)
0
t
1
t −1
−1
(a) Waveform for x(τ)
t −1
1
0
−1
−1
t −2
x(−τ)
1
t −1
2
t −2
1
t −1
∫ (−1).1dτ + ∫ (−1).(−1)dτ + ∫ 1.(−1)dτ
= −(1 − t + 2) + (t − 1 − 1) − (2 − t + 1) = 3t − 8.
Solutions
2
Case V (3 ≤ t < 4):
y1 (t ) =
∫ (−1).(−1)dτ = (2 − t + 2) = 4 − t .
t −2
Case VI (t > 4): Since there is no overlap, y1 (t ) = 0. Combining all the cases, the result of the convolution y1 (t ) = x(t ) ∗ x(t ) is given by t (0 ≤ t < 1) (4 − 3t ) (1 ≤ t < 2) y1 (t ) = (3t − 8) (2 ≤ t < 3) (4 − t ) (3 ≤ t < 4) 0 elsewhere. The output is y1(t) plotted in Fig. S3.6.1(j). (ii)
Using the graphical approach, the convolution of x(t) with z(t) is shown in Fig. S3.6.2, where we consider six different cases for different values of t. Case I (t < −1): Since there is no overlap, y2 (t ) = 0 . t
∫
2
y 2 (t ) = 1.τdτ = t2 − 12 .
Case II (−1 ≤ t < 0):
−1
t −1
Case III (0 ≤ t < 1):
∫
y2 (t ) =
t
∫
(−1).τdτ + 1.τdτ
−1
t −1
2 2 (t −1) 2 (t −1) 2 = − 2 − 12 + t2 − 2 = − t2 + 2t − 12 .
t −1
Case IV (1 ≤ t < 2):
y2 (t ) =
∫
1
∫
(−1).τdτ + 1.τdτ
t −2
(t −1) 2 = − 2 −
t −1 (t − 2) 2 2
1
Case V (2 ≤ t < 3):
y2 (t ) =
∫
(−1).τdτ =
t −2
+ 1 − (t −1) 2 = − t 2 + 3 . 2 2 2 2 (t − 2) 2 2
− 12 =
t2 2
− 2t + 32 .
Case VI (t > 4): Since there is no overlap, y2 (t ) = 0. Combining all the cases, the result of the convolution y2 (t ) = x(t ) ∗ z (t ) is given by t2 − 1 22 2 − t2 + 2t − 12 2 y2 (t ) = − t2 + 32 t2 3 2 − 2t + 2 0 The output is y2(t) plotted in Fig. S3.6.2(j).
(−1 ≤ t < 0) (0 ≤ t < 1) (1 ≤ t < 2) (2 ≤ t < 3) elsewhere.
93
94
Chapter 3
1
z(τ)
τ 0
−1
x(τ)
1
1
τ
0
1
1
−2
2
−1
−1
(a) Waveform for z(τ)
(b) Waveform for x(τ)
(c) Waveform for x(−τ)
x(t−τ)
1
z(τ) x(t−τ) 1
t −2
t
t −1
0
t −1
−1
t −2
(e) Overlap btw z(τ) and x(t−τ) for (t<−1)
z(τ) x(t−τ)
1
0 t
t −1 −1 t
1
(f) Overlap btw z(τ) and x(t−τ) for (−1≤t<0)
z(τ) x(t−τ)
1
z(τ) x(t−τ)
τ −1 t −2
1
−1
t −1 1
τ
t
t −2 1 t −1
−1
−1
(g) Overlap btw z(τ) and x(t−τ) for (0≤t<1)
0 −1
τ t −2 −1 t −1
τ
1
−1
(d) Waveform for x(t−τ) 1
z(τ) x(t−τ)
τ
τ t −1
τ
0
−1
−1
1
t −2
x(−τ)
t
−1
(h) Overlap btw z(τ) and x(t−τ) for (1≤t<2)
(i) Overlap btw z(τ) and x(t−τ) for (2≤t<3)
y 2(t) = x (t)*z (t)
1
0.5
0
-0.5 -1
0
1
2
(j) Convolution output
3
y2 (t )
Fig. S3.6.2: Convolution of x(t) with z(t) in Problem 3.6(ii). (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, y3 (t ) = 0 . t
∫
y3 (t ) = 1.(1 + τ) dτ =
Case II (−1 ≤ t < 0):
−1 t −1
Case III (0 ≤ t < 1):
y3 (t ) = =
∫
0
∫
(1+ t ) 2 2
=
t2 2
+ t + 12 . t
∫
(−1).(1 + τ)dτ + 1.(1 + τ)dτ + 1.(1 − τ)dτ
−1 2 − t2
(
) (
−0 +
1 2
t −1
−
t2 2
)
0
2 (1− t ) 2 − 2 − 12 = − 32t + t + 12 .
Solutions
1
w(τ)
1
τ 0
−1
τ 1
t −1
t −1
(b) Waveform for x(τ)
w(τ) x(t−τ)
0
1
(c) Waveform for x(t−τ)
w(τ) x(t−τ)
1
t −1 −1
t
0
1
1
w(τ) x(t−τ)
τ
t
t −2
−1
−1
1 t −1
τ
t
1 t −2
−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)
y3(t) = x(t)*w(t)
0.4 0.2 0 -0.2 -0.4 -0.6 0
1 (iii)
2
(j) Convolution output
3
y3 (t )
Fig. S3.6.3: Convolution of x(t) with w(t) in Problem 3.6(iii). t −1
0
y3 (t ) =
∫
(−1).(1 + τ)dτ +
t −2
= − 12 −
∫
1
∫
(−1).(1 − τ) dτ + 1.(1 − τ)dτ
0 (t −1) 2 ( t − 2) 2 + 2 2
− 12 − 0 −
1
Case V (2 ≤ t < 3):
y3 (t ) =
∫
(−1).(1 − τ) dτ = 0 −
t −2
t
(i) Overlap btw w(τ) and x(t−τ) for (t>3)
0.6
-1
t −1
−1
−1
Case IV (1 ≤ t < 2):
1
(f) Overlap btw w(τ) and x(t−τ) for (0≤t<1)
w(τ) x(t−τ)
τ 1
0t −1
(e) Overlap btw w(τ) and x(t−τ) for (−1≤t<0)
w(τ) x(t−τ)
t −1
w(τ) x(t−τ)
τ t −2 −1t −1
1
−1
(d) Overlap btw w(τ) and x(t−τ) for (t<−1)
−1 t −2
t
τ t −2
1
−1
1
t −1
−1
τ t −2
τ t −2
2
−1
(a) Waveform for z(τ)
x(t−τ)
1
0
1
−1
1
x(τ)
95
(3− t ) 2 2
t −1 (2 −t ) 2 2
=
3t 2 2
− 5t + 72 .
2
= − t2 + 3t − 92 .
Case VI (t > 4): Since there is no overlap, y3 (t ) = 0. Combining all the cases, the result of the convolution y3 (t ) = x(t ) ∗ w(t ) is given by
96
Chapter 3
t2 + t + 1 2 22 3t 1 t − + + 2 2 2 y3 (t ) = 32t − 5t + 72 t2 9 2 + 3t − 2 0
(−1 ≤ t < 0) (0 ≤ t < 1) (1 ≤ t < 2) (2 ≤ t < 3) elsewhere.
The output is y3(t) plotted in Fig. S3.6.3(j). (iv)
Using the graphical approach, the convolution of x(t) with v(t) is shown in Fig. 3.6.4, where we consider six different cases for different values of t.
1
v(τ)
1
x(τ)
τ 0
−1
1
(b) Waveform for x(τ)
v(τ) x(t−τ)
t −1
1
(c) Waveform for x(t−τ)
v(τ) x(t−τ)
t −1 −1 t
−1
1
t −2
1 −1
(d) Overlap btw v(τ) and x(t−τ) for (t<−1)
1
1
τ t
1
v(τ) x(t−τ)
1
v(τ) x(t−τ)
τ
t
t −2 1 t −1
−1
−1
τ
t
1 t −2
−1
(h) Overlap btw v(τ) and x(t−τ) for (2≤t<3)
y 4(t) = v (t)*v (t)
0
-0.5 0
t
(i) Overlap btw v(τ) and x(t−τ) for (t>3)
0.5
-1
t −1
−1
−1
(g) Overlap btw v(τ) and x(t−τ) for (1≤t<2)
−1 t −1
(e) Overlap btw v(τ) and x(t−τ) for (−1≤t<0) (f) Overlap btw v(τ) and x(t−τ) for (0≤t<1)
v(τ) x(t−τ)
t −1
v(τ) x(t−τ)
−1
τ −1t −2
t
τ t −2
1
1
t −1
−1
τ t −1
τ t −2
2
−1
(a) Waveform for v(τ) 1
τ
0
1
−1
t −2
x(t−τ)
1
e−2τ
e2τ
1 (i )
(j) Convolution output
2
3
y4 (t )
Fig. S3.6.4: Convolution of x(t) with v(t) in Problem 3.6(iv). Case I (t < −1): Since there is no overlap, y4 (t ) = 0 .
Solutions
t
∫
y4 (t ) = 1.e 2τ dτ =
Case II (−1 ≤ t < 0):
−1
(e
1 2
)
− e−2 .
2t
0 t −1 t y4 (t ) = (−1).e 2 τ dτ + 1.e 2 τ dτ + 1.e − 2 τ dτ 0 −1 t −1 = − 12 e 2(t −1) − e − 2 + 12 1 − e 2(t −1) + 12 1 − e − 2t Case III (0 ≤ t < 1): = −e 2(t −1) + 12 e − 2 + 1 − 12 e − 2t .
∫
∫
(
∫
) (
) (
)
0 1 t −1 y4 (t ) = (−1).e 2 τ dτ + (−1).e − 2 τ dτ + 1.e − 2 τ dτ 0 t −2 t −1 Case IV (1 ≤ t < 2): = − 12 1 − e 2(t − 2) + 12 e − 2(t −1) − 1 + ( −12) e − 2 − e − 2(t −1) = 12 e 2(t − 2) − 1 − 12 e − 2 + e − 2(t −1) .
∫
∫
(
) (
1
Case V (2 ≤ t < 3):
y4 (t ) =
∫
∫ (−1).e
− 2τ
dτ =
t −2
)
1 2
(e
−2
(
)
)
− e − 2(t − 2) .
Case VI (t > 4): Since there is no overlap, y4 (t ) = 0. Combining all the cases, the result of the convolution y 4 (t ) = x(t ) ∗ v(t ) is given by 1 2t e − 12 e −2 2 2 (t −1) + 12 e − 2 + 1 − 12 e − 2t −e y4 (t ) = 12 e 2(t − 2) − 1 − 12 e − 2 + e − 2(t −1) 1 −2 e − 12 e − 2(t − 2) 2 0
(−1 ≤ t < 0) (0 ≤ t < 1) (1 ≤ t < 2) (2 ≤ t < 3) elsewhere.
The output is y4(t) plotted in Fig. S3.6.4(j). (v)
Using the graphical approach, the convolution of z(t) with z(t) is shown in Fig. 3.6.5, where we consider four different cases for different values of t. Case I (t < −2): Since there is no overlap, y5 (t ) = 0 . t +1 t +1 y5 (t ) = τ(t − τ)dτ = 12 tτ 2 − 13 τ3 −1 −1 Case II (−2 ≤ t < 0): = 12 t (t + 1) 2 − 12 t − 13 (t + 1)3 + 13 =
∫
[
[
][
]
] (t 1 6
3
)
− 6t − 4 .
97
Chapter 3
1 1 y5 (t ) = τ(t − τ)dτ = 12 tτ 2 − 13 τ3 t −1 1 t − Case III (0 < t < 2): . 2 3 1 1 1 1 1 3 = 2 t − 2 t (t − 1) − 3 − 3 (t − 1) = − 6 t + 6t + 4 .
∫
[
[
][
]
]
(
)
z(τ)
1
τ 0
−1
1
−1
(a) Waveform for z(τ) 1
z(−τ)
z(t−τ)
1
1
z(τ) z(t−τ)
τ 0
−1
1
t
t−1
t+1
0
−1
(b) Waveform for z(−τ) 1
1
t−1
−1
−1
z(τ) z(t−τ)
1
t
t+1
0
−1
1
−1
(d) Overlap btw z(τ) and z(t−τ) for (t<−2)
(c) Waveform for z(t−τ) z(τ) z(t−τ)
1
z(τ) z(t−τ)
τ t−1 −1
t
t+1
t
−1 t−1
1
−1
1t+1
1 t−1
−1
(e) Overlap btw z(τ) and z(t−τ) for (f) Overlap btw z(τ) and z(t−τ) for (0≤t<2) (−2≤t<0)
0 -0.2 -0.4 -0.6 -1
t+1
(g) Overlap btw z(τ) and z(t−τ) for (t>2)
0.2
-2
t
−1
−1
y 5(t) = z (t)*z (t)
98
0
(h) Convolution output
1
2
y5 (t )
Fig. S3.6.5: Convolution of z(t) with z(t) in Problem 3.6(v). Case IV (t > 2): Since there is no overlap, y5 (t ) = 0. Combining all the cases, the result of the convolution y5 (t ) = z (t ) ∗ z (t ) is given by
Solutions
1 t3 − t − 2 3 6 y5 (t ) = − 16 t 3 + t + 23 0
99
(−2 ≤ t < 0) ( 0 ≤ t < 2) elsewhere
The output is y5(t) shown in Fig. S3.6.5(h). (vi)
Using the graphical approach, the convolution of w(t) with z(t) is shown in Fig. 3.6.6, where we consider six different cases for different values of t. w(τ)
1
z(τ)
1
1
τ 0
−1
t
1
t−1
(a) Waveform for w(τ)
(b) Waveform for z(τ)
t+1
1
w(τ) z(t−τ)
0
−1
1
1
(c) Waveform for z(t−τ)
w(τ) z(t−τ) 1
t −1 t+1
t−1 −1 t
1
w(τ) z(t−τ)
1
t+1
1
−1
(e) Overlap btw w(τ) and z(t−τ) for (−2≤t<−1)
w(τ) z(t−τ)
(f) Overlap btw w(τ) and z(t−τ) for (−1≤t<0)
w(τ) z(t−τ)
1
τ t
0
−1
−1
(d) Overlap btw w(τ) and z(t−τ) for (t<−2)
−1 t−1
t+1
τ
t−1
−1
1
t
−1
−1
τ t−1
τ
τ 0
−1
1
−1
1
z(t−τ)
w(τ) z(t−τ)
τ
τ
1 t+1
−1
t−1
1 t
t+1
−1
−1
(g) Overlap btw w(τ) and z(t−τ) for (0≤t<1)
(h) Overlap btw w(τ) and z(t−τ) for (1≤t<2)
1 t−1
−1
(i) Overlap btw w(τ) and z(t−τ) for (t>2)
0.3 y6(t) = w(t)*z(t)
0.2 0.1 0 -0.1 -0.2 -0.3 -1
t+1
−1
Problem 3.6(vi)
-2
t
0 (vi)
(j) Convolution output
1
2
y6 (t )
Fig. S3.6.6: Convolution of w(t) with z(t) in Problem 3.6(vi). Case I (t < −2): Since there is no overlap, y6 (t ) = 0 .
100
Chapter 3
t +1 t +1 y6 (t ) = (1 + τ)(t − τ)dτ = tτ + 12 tτ 2 − 12 τ 2 − 13 τ3 −1 −1 Case II (−2 ≤ t < −1): = t (t + 1) + 12 t (t + 1) 2 − 12 (t + 1) 2 − 13 (t + 1)3 − − t + 12 t − 12 + 13 = 16 t 3 + 12 t 2 − 23 .
[
∫
]
[
][
t +1 0 y (t ) = (1 + τ)(t − τ) dτ + (1 − τ)(t − τ)dτ 6 −1 0 Case III (−1 < t < 0): 0 2 2 = tτ + 12 tτ − 12 τ − 13 τ3 + tτ − 12 tτ 2 − 12 τ 2 + 13 τ3 −1 = 12 t + 16 + − 16 t 3 + 12 t 2 + 12 t − 16 = − 16 t 3 + 12 t 2 + t.
∫
[ [
∫
] [
] [
]
0 1 y (t ) = (1 + τ)(t − τ)dτ + (1 − τ)(t − τ)dτ 6 t −1 0 Case IV (0 < t < 1): 0 = tτ + 12 tτ 2 − 12 τ 2 − 13 τ3 + tτ − 12 tτ 2 − 12 τ 2 + 13 τ3 t −1 = − 16 t 3 + 12 t 2 − 12 t − 16 + 12 t − 16 = − 16 t 3 − 12 t 2 + t.
∫
[
]
t +1 0
∫
]
[
][
[
]
]
1 0
1 1 y6 (t ) = (1 − τ)(t − τ)dτ = tτ − 12 tτ 2 − 12 τ 2 + 13 τ3 t −1 t −1 = t − 12 t − 12 + 13 − t (t − 1) − 12 t (t − 1) 2 − 12 (t − 1) 2 + 13 (t − 1)3 Case V (1 ≤ t < 2): = 16 t 3 − 12 t 2 + 23 .
[
∫
[
]
]
] [
]
Case VI (t > 2): Since there is no overlap, y6 (t ) = 0. Combining all the cases, the result of the convolution y6 (t ) = z (t ) ∗ w(t ) is given by 1 t 3 + 1 t 2 − 2 (−2 ≤ t < −1) 61 3 21 2 3 ( −1 ≤ t < 0) − 6 t + 2 t + t 1 3 1 2 y6 (t ) = − 6 t − 2 t + t (0 ≤ t < 1) 1 3 1 2 2 (1 ≤ t < 2) 6t − 2t + 3 0 elsewhere. The output is y6(t) shown in Fig. S3.6.6(j) at the end of the solution of this problem. (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.
Solutions
1
v(τ)
1
z(τ)
1
τ 0
−1
τ 0
−1
1
1
t−1
(a) Waveform for v(τ) 1
(b) Waveform for z(τ)
v(τ) z(t−τ)
0
−1
1
1
t −1 t+1
0
t−1 −1 t
1
1
t+1
1
(f) Overlap btw v(τ) and z(t−τ) for (−1≤t<0)
(e) Overlap btw v(τ) and z(t−τ) for (−2≤t<−1)
v(τ) z(t−τ)
0t
v(τ) z(t−τ)
−1
v(τ) z(t−τ)
1
τ −1 t−1
1
(c) Waveform for z(t−τ)
v(τ) z(t−τ)
−1
(d) Overlap btw v(τ) and z(t−τ) for (t<−2)
0
−1
τ t−1
1
−1
1
t+1
−1
τ t+1
t
−1
−1
t
z(t−τ)
e−2τ
e2τ
t−1
101
v(τ) z(t−τ)
τ 0 t−1 1
−1
1 t+1
t
−1
t+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)
y 7(t) = v (t)*z (t)
0.2 0.1 0 -0.1 -0.2 -0.3 -1
0
(j) Convolution output
1
2
y7 (t )
Fig. S3.6.7: Convolution of v(t) with z(t) in Problem 3.6(vii). Case I (t < −2): Since there is no overlap, y7 (t ) = 0 . t +1 t +1 y7 (t ) = e 2 τ (t − τ) dτ = 12 (t − τ)e 2τ + 14 e 2 τ −1 −1 = − 12 e 2(t +1) + 14 e 2(t +1) − 12 (t + 1)e −2 + 14 e −2 Case II (−2 ≤ t < −1): = − 14 e 2(t +1) − 12 te −2 − 34 e −2 .
∫
[
t
t+1
(i) Overlap btw v(τ) and z(t−τ) for (t>2)
0.3
-2
1 t−1 −1
[
]
][
]
102
Chapter 3
t +1 0 y (t ) = e 2 τ (t − τ)dτ + e − 2 τ (t − τ)dτ 7 −1 0 0 = 12 (t − τ)e 2 τ + 14 e 2 τ + − 12 (t − τ)e − 2 τ + 14 e − 2 τ −1 Case III (−1 < t < 0): = + 12 t + 14 − 12 te − 2 − 43 e − 2 + − 14 e − 2(t +1) + 12 t − 14 = 34 e − 2(t +1) − 12 te − 2 − 34 e − 2 + t.
∫
∫
[ [
] [ ][
0 1 y (t ) = e 2 τ (t − τ)dτ + e − 2τ (t − τ)dτ 7 0 t −1 0 = 12 (t − τ)e 2 τ + 14 e 2 τ + − 12 (t − τ)e − 2 τ + 14 e − 2 τ t −1 Case IV (0 < t < 1): = 12 t + 14 − 43 e 2(t −1) + − 12 te − 2 + 34 e − 2 + 12 t − 14 = − 34 e 2(t −1) − 12 te − 2 + 43 e − 2 + t.
∫
[ [
] ]
t +1 0
∫
]
[
][
]
]
1 0
1 1 y7 (t ) = e − 2 τ (t − τ)dτ = − 12 (t − τ)e − 2 τ + 14 e − 2τ t −1 t −1 Case V (1 ≤ t < 2): = − 12 e 2(t +1) + 14 e 2(t +1) − 12 (t + 1)e − 2 + 14 e − 2 = 14 e − 2(t −1) − 12 te − 2 + 34 e − 2 .
∫
[
[
]
][
]
Case VI (t > 2): Since there is no overlap, y7 (t ) = 0 . Combining all the cases, the result of the convolution y7 (t ) = z (t ) ∗ v(t ) is given by − 1 e 2(t +1) − 1 te −2 − 3 e −2 (−2 ≤ t < −1) 3 −42(t +1) 1 2 − 2 3 4− 2 (−1 ≤ t < 0) − 2 te − 4 e + t 4e 3 2(t −1) 1 − 2 3 − 2 y7 (t ) = − 4 e (0 ≤ t < 1) − 2 te + 4 e + t 1 − 2(t −1) 1 − 2 3 − 2 (1 ≤ t < 2) − 2 te + 4 e 4e 0 elsewhere. The output is y7(t) shown in Fig. S3.6.7(j) at the end of the solution of this problem. (viii) Since w(t) = 1 − |t|, therefore, the expression for w(t – τ) is 1 − (t − τ) if τ < t w(t − τ) = 1 − t − τ = 1 − (τ − t ) if τ > t. Using the graphical approach, the convolution of w(t) with w(t) is shown in Fig. 3.6.8, where we consider six different cases for different values of t.
Solutions
w(τ)
1
w(−τ)
1
1
103
w(t−τ)
τ
τ
0
−1
(a) Waveform for w(τ)
1
t−1
t
t+1
0
−1
1
−1
−1
(b) Waveform for w(−τ)
(c) Waveform for w(t−τ)
−1
1
0
−1
1
w(τ) w(t−τ)
w(τ) w(t−τ)
1
1
w(τ) w(t−τ)
τ
τ
t−1
t
t+1
0
−1
t −1
t−1
1
t−1−1
1
−1
−1
(d) Overlap btw w(τ) and w(t−τ) for (t<−2) 1
t+1 0
t
0 t+1 1
−1
(e) Overlap btw w(τ) and w(t−τ) for (−2≤t<−1) (f) Overlap btw w(τ) and w(t−τ) for (−1≤t<0)
w(τ) w(t−τ)
w(τ) w(t−τ)
1
1
w(τ) w(t−τ)
τ
τ
−1 t−1
0 t
1 t+1
0 t−1 1
−1
t
−1
t+1
−1
−1
(g) Overlap btw w(τ) and w(t−τ) for (0≤t<1)
1 t−1
(i) Overlap btw w(τ) and w(t−τ) for (t>2)
0.6 y 8 (t ) = w (t)*w (t )
0.5 0.4 0.3 0.2 0.1 -1
0
1
(j) Convolution output
2
y8 (t )
Fig. S3.6.8: Convolution of w(t) with w(t) in Problem 3.6(viii). Case I (t < −2): Since there is no overlap, y8 (t ) = 0 . t +1 y8 (t ) = (1 + τ)(1 + t − τ)dτ −1 t +1 t +1 2 Case II (−2 ≤ t < −1): = (1 − τ ) dτ + t (1 + τ) dτ = τ − 13 τ3 + 12 t (1 + τ) 2 −1 −1 = 16 t 3 + t 2 + 2t + 43 .
∫ ∫
t+1
−1
(h) Overlap btw w(τ) and w(t−τ) for (1≤t<2)
0 -2
t
∫
[
]
t +1 −1
104
Chapter 3
Case III (−1 < t < 0): t +1
0
t
∫
∫
y8 (t ) = (1 + τ)(1 − t + τ)dτ + (1 + τ)(1 + t − τ)dτ + −1 t
∫
t
t
∫
2
0
∫
0
∫ (1 − τ)(1 + t − τ)dτ 0
t +1
∫
2
∫ (1 − τ)
= (1 + τ) dτ − t (1 + τ)dτ + (1 − τ )dτ + t (1 + τ)dτ + −1
=
−1
t
3
1 3
2
1 2
1 3 3
1 2 2
t +1
∫
dτ + t (1 − τ)dτ
0
t
[ (1 + t ) ]− t [ (1 + t ) ]− [t − t ]− t [t + t ]+ [−
2
1 3 t 3
0
] [
− 13 + t (t + 1) − 12 (t + 1) 2
]
= 23 − t 2 − 12 t 3 . Case IV (0 < t < 1): 0
y8 (t ) =
∫ (1 + τ)(1 − t + τ)dτ + ∫ (1 − τ)(1 + t − τ)dτ + ∫ (1 − τ)(1 + t − τ)dτ
t −1 0
=
∫ (1 + τ)
t −1
=
1
t
[
1 3
2
0
0
∫
t
∫
t
t
1
∫
2
∫
1
∫
2
dτ − t (1 + τ)dτ + (1 − τ) dτ + t (1 − τ)dτ + (1 − τ) dτ + t (1 − τ) dτ
] [
− 13 t 3 − t
1 2
t −1
][
0
] [
0
t
t
] [ (1 − t ) ]+ t[ (1 − t ) ]
− 12 t 2 + − 13 (1 − t )3 + 13 + t t − 12 t 2 +
1 3
3
2
1 2
= 23 − t 2 + 12 t 3 . 1 y8 (t ) = (1 − τ)(1 − t + τ)dτ t −1 1 1 Case V (1 ≤ t < 2): = (1 − τ 2 )dτ − t (1 − τ)dτ = τ − 13 τ3 − t (τ − 12 τ 2 ) t −1 t −1 = − 16 t 3 + t 2 − 2t + 43
∫ ∫
∫
[
]
1 t −1
.
Case VI (t > 2): Since there is no overlap, y8 (t ) = 0. Combining all the cases, the result of the convolution y8 (t ) = w(t ) ∗ w(t ) is given by 1 t 3 + t 2 + 2t + 4 6 2 2 1 3 3 3 −t − 2t y8 (t ) = 23 − t 2 + 12 t 3 1 3 2 4 − 6 t + t − 2t + 3 0
(−2 ≤ t < −1) ( −1 ≤ t < 0) (0 ≤ t < 1) (1 ≤ t < 2) elsewhere.
The output is y8(t) shown in Fig. S3.6.8(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.
Solutions
1
v(τ)
1
w(τ)
1
τ
τ 0
−1
0
−1
1
τ
1
t−1
t
t+1
0
−1
1
−1
−1
−1
(a) Waveform for v(τ)
(b) Waveform for w(τ)
(c) Waveform for w(t−τ)
1
v(τ) w(t−τ)
1
v(τ) w(t−τ)
1
t+1
−1
v(τ) w(t−τ)
τ
τ t
w(t−τ)
e−2τ
e2τ
t−1
105
t−1
1
t −1
−1
t+1 0
τ t−1−1
1
t
0 t+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) 1
v(τ) w(t−τ)
1
v(τ) w(t−τ)
1
τ
τ −1 t−1
0 t
v(τ) w(t−τ)
−1
1 t+1
0 t−1 1
t
τ −1
t+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)
1 t−1
(i) Overlap btw v(τ) and w(t−τ) for (t>2)
y 9 (t) = w (t)*w (t)
0.4 0.3 0.2 0.1 -1
t+1
−1
0.5
0 -2
t
0
(j) Convolution output
1
2
y9 (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 1 − (t − τ) if τ < t w(t − τ) = 1 − t − τ = 1 − (τ − t ) if τ > t. Case I (t < −2): Since there is no overlap, y9 (t ) = 0 .
106
Chapter 3
t +1 t +1 t +1 y9 (t ) = e 2τ (1 − τ + t )dτ = (1 + t ) e 2 τ dτ − τe 2 τ dτ −1 −1 −1 = 12 (1 + t ) e 2(t +1) − e − 2 − 12 (t + 1)e 2(t +1) − 14 e 2(t +1) + 12 e − 2 + 14 e − 2 Case II (−2 ≤ t < −1): = 14 e 2(t +1) − 12 te − 2 − 54 e − 2 .
∫
∫
(
∫
) (
)
0 t t +1 2τ 2τ y9 (t ) = e (1 − t + τ)dτ + e (1 + t − τ)dτ + e − 2 τ (1 + t − τ) dτ 0 −1 t 0 t t = (1 − t ) 12 e 2 τ + 12 τe 2 τ − 14 e 2 τ + (1 + t ) 12 e 2 τ − 12 τe 2 τ − 14 e 2 τ −1 −1 t Case III (−1 < t < 0): t +1 t +1 + (1 + t ) − 12 e − 2 τ − − 12 τe − 2 τ − 14 e − 2τ 0 0 = 14 e − 2(t +1) − 12 e 2t + 12 te − 2 + 14 e − 2 + t + 1.
∫
∫
∫
[ ] [ [
]
]
[ ] [
[
∫
[ ] [
0 t
.
]
0 1 t 2τ 2τ y9 (t ) = e (1 − t + τ)dτ + e (1 − t + τ)dτ + e − 2 τ (1 + t − τ)dτ 0 t −1 t 0 0 0 = (1 − t ) 12 e 2 τ + 12 τe 2 τ − 14 e 2 τ + (1 − t ) 12 e 2 τ + 12 τe 2 τ − 14 e 2 τ t −1 t −1 0 Case IV (0 < t < 1): 1 1 + (1 + t ) − 12 e − 2 τ − − 12 τe − 2 τ − 14 e − 2 τ t t = 14 e 2(t −1) − 12 e − 2t − 12 te − 2 + 14 e − 2 − t + 1.
∫
]
∫
[
]
] [
[ ] [
]
0 0
]
1 1 1 y9 (t ) = e − 2 τ (1 − t + τ)dτ = (1 − t ) e − 2 τ dτ + τe − 2 τ dτ t −1 t −1 t −1 Case V (1 ≤ t < 2): = − 12 (1 − t ) e − 2 − e − 2(t −1) + − 12 e − 2 − 14 e − 2 + 12 (t − 1)e − 2(t −1) + 14 e − 2(t −1) . = 14 e − 2(t −1) + 12 te − 2 − 54 e − 2 .
∫
∫
(
Case VI (t > 2): Since there is no overlap,
∫
) (
y9 (t ) = 0 .
Combining all the cases, the result of the convolution y9 (t ) = v(t ) ∗ w(t ) is given by
)
Solutions
107
1 2( t +1) − 1 te −2 − 5 e −2 ( −2 ≤ t < −1) e 1 − 2(t +14) 1 2t 21 − 2 4 1 − 2 − 2 e + 2 te + 4 e + t + 1 (−1 ≤ t < 0) 4 e 1 2(t −1) 1 − 2t 1 − 2 1 − 2 y 9 (t ) = 4 e − 2 e − 2 te + 4 e − t + 1 (0 ≤ t < 1) 5 −2 1 − 2 ( t −1) 1 −2 + 2 te − 4 e (1 ≤ t < 2) e 4 0 elsewhere.
The output is y9(t) shown in Fig. S3.6.9(j) at the end of the solution of this problem. Using the graphical approach, the convolution of v(t) with v(t) is shown in Fig. 3.6.10, where we consider six different cases for different values of t.
(x)
v(τ) e−2τ 0
−1
1
Case 1: (t + 1) < −1
0
Case 2: (t + 1) > −1 (t + 1) < 0
y(t) e−2τ
t +1 −1
1
0
t −1
(d) Overlap btw v(τ) and v(t−τ) for (t<−2) Case 4: (t + 1) > 1 (t – 1) < 0
t −1 t +1 0
−1 t −1 0
y(t) 1
τ
1
t −1 −1 t
1
Case 6: (t − 1) > 1 y(t)
y(t) 1
e−2τ
e2τ −1
t +1
(g) Overlap btw v(τ) and v(t−τ) for (0≤t<1)
e−2τ
e2τ
0 t −1 1
τ
t +1
(h) Overlap btw v(τ) and v(t−τ) for (1≤t<2)
−1
0
v(t)*v(t)
2 1.5 1 0.5
-2
0
(j) Convolution output
2
4
y10 (t )
Fig. S3.4.10: Convolution of v(t) with v(t) in Problem 3.6(x). Case I (t < −2): Since there is no overlap, Case II (−2 ≤ t < −1):
y10 (t ) = 0 .
1t −1
t +1
τ
(i) Overlap btw v(τ) and v(t−τ) for (t>2)
2.5
0 -4
τ
0 t +1 1
(f) Overlap btw v(τ) and v(t−τ) for (−1≤t<0)
1 e−2τ
e−2τ
e2τ
(e) Overlap btw v(τ) and v(t−τ) for (−2≤t<−1)
1 e2τ
τ
t +1
e−2τ
Case 5: (t − 1) > 0 (t – 1) < 1
y(t)
t
Case 3: (t + 1) > 0 (t + 1) < 1
y(t) e2τ
1
t −1
(c) Waveform for v(t−τ)
1
e2τ
e−2(t−τ)
e2(t−τ) τ
(b) Waveform for v(−τ)
1
t
e−2τ
e2τ
(a) Waveform for v(τ)
t −1
1
1
e2τ −1
v(t −τ)
v(−τ)
1
108
Chapter 3
t +1
y10 (t ) =
∫
2 τ − 2(t − τ)
e e
dτ = e
− 2t
t +1
−1
∫
4τ
e dτ = e
− 2t e
−1
∫
]
∫
(
)
(
)
)
t 0 1 2 τ 2(t − τ) − 2 τ 2(t − τ) y10 (t ) = e e dτ + e e dτ + e − 2 τ e − 2 ( t − τ ) dτ t −1 t 0 = (1 − t )e 2t + 14 e 2t 1 − e − 4t + (1 − t )e − 2t = 54 − t e 2t + 34 − t e − 2t .
∫
∫
∫
(
(
)
(
1
Case V (1 ≤ t < 2):
[
e 4(t +1) − e − 4 1 2t + 4 − e − ( 2t − 4 ) . = e − 2t = e 4 4
t t +1 0 2 τ 2(t − τ) 2 τ − 2(t − τ) y10 (t ) = e e dτ + e e dτ + e − 2τ e − 2(t − τ) dτ t 0 −1 . = (t + 1)e 2t + 14 e − 2t 1 − e 4t + (t + 1)e − 2t = t + 34 e 2t + t + 54 e 2t .
(
Case IV (0 < t < 1):
t +1
4 −1
∫
Case III (−1 < t < 0):
4τ
∫e
y10 (t ) =
)
)
− 2 τ 2(t − τ)
e
t −1
(
)
dτ = 14 e 2t e − 4(t −1) − e − 4 . .
y10 (t ) = 0.
Case VI (t > 2): Since there is no overlap,
Combining all the cases, the result of the convolution y10 (t ) = v(t ) ∗ v(t ) is given by y10 (t ) =
(
1 2t + 4 − e − ( 2t − 4 ) e 4
)
(t + 34 ) e 2t + (t + 54 ) e 2t (54 − t ) e2t + (34 − t ) e− 2t 1 2t e 4
(e
− 4(t −1)
− e−4
0
)
( −2 ≤ t < −1) (−1 ≤ t < 0) (0 ≤ t < 1) (1 ≤ t < 2) elsewhere.
The output y10(t) is shown in Fig. S3.6.10(j).
▌
Problem 3.7 ∞
(a)
Distributive property: By definition, x1 (t ) ∗ z (t ) =
∫ x1 (τ) z(t − τ)dτ .
−∞
Substituting z(t) = x2(t) + x3(t), we obtain
Solutions ∞
x1 (t ) ∗ (x2 (t ) + x3 (t ) ) =
∫ x1 (τ)(x2 (t − τ) + x3 (t − τ))dτ
−∞
or,
∞
x1 (t ) ∗ (x2 (t ) + x3 (t ) ) =
∫
∞
x1 (τ) x2 (t − τ)dτ +
− ∞
x1 (t ) ∗ x 2 (t )
∫ x1 (τ) x3 (t − τ)dτ
− ∞
x1 (t ) ∗ x 3 (t )
x1 (t ) ∗ (x2 (t ) + x3 (t ) ) = x1 (t ) ∗ x2 (t ) + x1 (t ) ∗ x3 (t ) .
or,
∞
(b)
∫ x1 (τ)w(t − τ)dτ .
Associative property: By definition, x1 (t ) ∗ w(t ) =
−∞ ∞
w(t ) = x2 (t ) ∗ x3 (t ) =
Substituting
∫ x3 (α) x2 (t − α)dα ,
−∞
we obtain
x1 (t ) ∗ (x2 (t ) ∗ x3 (t ) ) =
∞ x1 (τ) x3 (α) x2 (t − τ − α) dα dτ . −∞ −∞ ∞
∫
∫
Changing the order of integrations, the above equation simplifies to x1 (t ) ∗ (x2 (t ) ∗ x3 (t ) ) =
∞
∞ x3 (α) x1 (τ) x2 (t − τ − α) dτ dα . −∞ −∞
∫
∫
Similarly, expanding ∞
∞ (x1 (t ) ∗ x2 (t )) ∗ x3 (t ) = x3 (τ) x1 (α) x2 (t − τ − α)dα dτ . −∞ −∞
∫
∫
Since the right hand side of the two expressions are equal, we obtain the following.
x1 (t ) ∗ (x2 (t ) ∗ x3 (t ) ) = (x1 (t ) ∗ x2 (t ) ) ∗ x3 (t ) . (c)
Scaling Property: We consider the two cases α > 0 and α < 0 separately. Case I: (α = k) where k is a positive constant. ∞
By definition,
∫ x1 (kτ) z(kt − kτ)dτ
x1 (kt ) ∗ z (kt ) =
−∞ ∞
Substituting p = kτ, we get x1 ( kt ) ∗ z (kt ) =
dp
∫ x1 ( p) z(kt − p) k
−∞
=
1 y (kt ) . k
Case II: (α = −k) where k is a positive constant. ∞
By definition,
x1 (− kt ) ∗ z (− kt ) =
∫ x1 (−kτ) z(−kt + kτ)dτ
−∞
109
110
Chapter 3 −∞
Substituting p = −kτ, we get x1 (− kt ) ∗ z (− kt ) =
dp
∫ x1 ( p) z (−kt − p) (−k ) .
∞
By changing the order of the upper and lower limits 1 x1 (− kt ) ∗ z (− kt ) = − (−k )
∞
1
∫ x1 ( p) z ((−k )t − p)dp = k y(−kt )
−∞
▌
Collectively, Cases I and II prove the scaling property. Problem 3.8
u (t ) → (1 − e −t )u (t ) .
We know that
u (t )
Then
dt N
→
d dt
[(1 − e
−t
]
)u (t ) ,
δ (t )
δ(t ) → δ(t ) + e −t u (t ) − e −t δ(t )
which simplifies to or,
δ(t ) → e −t u (t ) .
Hence, the impulse response is given by
h(t ) = e −t u (t ) .
▌
Problem 3.9
Convolution of two signals that are, respectively, nonzero over the range [tℓ1, tu1] and [tℓ2, tu2] is nonzero over the range [tℓ1 + tℓ2, tu1 + tu2]. Therefore, tℓ1 + tℓ2 = −5 and tu1 + tu2 = 6. By substituting, tℓ1 = −2 and tu1 = 3, the values of tℓ2 = −5 + 2 = −3 and tu2 = 6 – 3 = 3. The possible nonzero range of the impulse ▌ response h(t) is therefore [−3, 3]. Problem 3.10
In Example 3.8, it was shown that
1 − t x(t ) = e − t u (t ) ∗ h(t ) = 0
[
]
0 0 ≤ t ≤ 1 = 2 − t − 2e − t otherwise − (t −1) e − 2e − t
t<0 0 ≤ t ≤1 t > 1.
Using the commutative property, Eq. (3.3), we interchange the impulse response and the input to obtain 1 − t y (t ) = x(t ) = 0
0 0 ≤ t ≤1 − t ∗ h(t ) = e u (t ) = 2 − t − 2e − t otherwise e − (t −1) − 2e − t
[
]
t<0 0 ≤ t ≤1 t > 1. ▌
Problem 3.11
By inspection, we note that x´(t) = x (t − 2) and h´(t) = h (t − 4). Using the shifting property, Eq. (3.40), the output
Solutions
111
y ′(t ) = x′(t ) ∗ h′(t ) = x(t − 2) ∗ h(t − 4) = y (t − 6) . Therefore, the convolution output for the shifted input x´(t) = x (t − 2) and shifted impulse response h´(t) = h (t − 4) is given by 0 y ′(t ) = y (t − 6) = 2 − (t − 6) − 2e − (t − 6) − ((t − 6) −1) e − 2e − (t − 6) 0 y ′(t ) = 8 − t − 2e − (t − 6) − (t − 7 ) − 2e − ( t − 6 ) e
or,
(t − 6) < 0 0 ≤ (t − 6) ≤ 1 (t − 6) > 1, t<6
6≤t≤7
▌
t > 7.
Problem 3.12
(i)
System h1(t) is NOT memoryless since h1(t) ≠ 0 for t ≠ 0. System h1(t) is causal since h1(t) = 0 for t < 0. System h1(t) is BIBO stable since ∞
∞
∞
−∞
−∞
−∞
∫ | h1(t ) | dt =
(ii)
∞
−5t −5t ∫ δ (t )dt + ∫ e u (t )dt = 1 + − 15 e 0 =
6 <∞. 5
System h2(t) is NOT memoryless since h3(t) ≠ 0 for t ≠ 0. System h2(t) is causal since h2(t) = 0 for t < 0. System h2(t) is BIBO stable since ∞
∞
∫ | h2(t ) | dt = ∫ e
−∞
−2 t
−∞
∞
∞
u (t )dt = ∫ e−2t dt = − 12 e−2t = 0
0
1 <∞. 2
(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 ∞
∞
−∞
−∞
∫ | h3(t ) | dt = ∫ e
(iv)
− 5t
∫
sin( 2πt )u (t )dt = e − 5t sin( 2πt )dt < ∞ . 0
System h4(t) is NOT memoryless since h4(t) ≠ 0 for t ≠ 0. System h4(t) is NOT causal since h4(t) ≠ 0 for t < 0. System h4(t) is BIBO stable since ∞
∫
0
| h 4(t ) | dt =
−∞
(v)
∞
∫
∞
∫
1
∫
e 2t dt + e − 2t dt + 1 dt = 3 < ∞ .
−∞
0
−1
System h5(t) is NOT memoryless since h5(t) ≠ 0 for t ≠ 0. System h5(t) is NOT causal since h5(t) ≠ 0 for t < 0. System h5(t) is BIBO stable since
112
Chapter 3
∞
∫
4
−∞
(vi)
t2 | h5(t ) | dt = tdt = 2
∫
−4
4
= 16 < ∞ . −4
System h6(t) is NOT memoryless since h6(t) ≠ 0 for t ≠ 0. System h6(t) is NOT causal since h6(t) ≠ 0 for t < 0. System h6(t) is NOT BIBO stable since ∞
∞
−∞
−∞
∫ | h6(t ) | dt = ∫ | sin(10t ) | dt = ∞ .
Consider the bounded input signal sin(10t ) . If this signal is applied to the system, the output can be calculated as: ∞
y (t ) =
∫
x (τ )h (t − τ )dτ =
−∞
∞
∫ sin(10τ )sin(10t − 10τ )dτ
−∞
The output at t=0 is given by, ∞
y (0) =
∞
2 ∫ sin(10τ )sin(−10τ )dτ = − ∫ sin (10τ )dτ = − 12
−∞
−∞
∞
= − 12
∞
∫ (1 − cos(20τ ) ) dτ
−∞
∞
∫ dτ + ∫ cos(20τ )dτ = −∞ 1 2
N
−∞
−∞
=∞
= finite value
It is observed that the output becomes unbounded even if the input is always bounded. This is because the system is not BIBO stable. (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
∫ | h7(t ) | dt = ∫ cos(5t )dt = ∞ .
Consider the bounded input signal cos(5t ) . If this signal is applied to the system, the output can be calculated as: ∞
∫
y (t ) =
x (t − τ )h (τ )dτ =
−∞
∞
∫
−∞
∞
cos(5t − 5τ ) cos(5τ )u (τ )dτ = ∫ cos(5t − 5τ ) cos(5τ )dτ . 0
The output at t=0 is given by, ∞
∞
y (0) = ∫ cos( −5τ ) cos(5τ )dτ = ∫ cos (5τ )dτ = 2
0
0
∞
=
1 2
∞
∫ dτ + ∫ cos(10τ )dτ = ∞ 1 2
0 N =∞
0
= finite value
∞
1 2
∫ (1 + cos(10τ ) ) dτ 0
Solutions
113
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 ∞
∫ | h8(t ) | dt =
−∞
= (ix)
∞
∫
−∞
∞
∞
0
0
0.95 t dt = 2 ∫ 0.95t dt = 2 ∫ et ln(0.95) dt =
∞ 2 et ln(0.95) 0 ln(0.95)
2 2 = 39 < ∞ [0 − 1] = − ln(0.95) ln(0.95)
System h9(t) is NOT memoryless since h8(t) ≠ 0 for t ≠ 0. System h9(t) is NOT causal since h8(t) = 0 for t < 0. System h8(t) is BIBO stable since ∞
1
−∞
−1
∫ | h9(t ) | dt = ∫ 1dt = 2 < ∞ .
▌
Problem 3.13
(i)
y (t ) = x(t ) * h(t ) = x(t ) * [δ (t ) − δ (t − 2)] = x(t ) − x(t − 2) .
From the input-output relationship, we observe that the output at time t depends on the values of the input at time (t − 2). Therefore, the system is NOT memoryless. However, it is causal. (ii)
∞
y (t ) = x(t ) * h(t ) = x(t ) * rect (t / 2) =
∫
1
−∞
rect (τ / 2)x(t − τ )dτ = ∫ x(t − τ )dτ . −1
From the input-output relationship, we observe that the output at time t depends on the values of the input from time (t − 1) to (t + 1). Therefore, the system is NOT memoryless and NOT causal. (iii)
y (t ) = x(t ) * h(t ) =
∞
∞
−∞
−∞
∫ x(τ )h(t − τ )dτ = 2 ∫ e
−4( t −τ )
t
u (t − τ ) x(τ )dτ = 2 ∫ e−4(t −τ ) x(τ )dτ = 2e −4t −∞
t
∫e
4τ
x(τ )dτ .
−∞
From the input-output relationship, we observe that the output at time t depends on the values of the input from time (−∞, t). Therefore, the system is NOT memoryless. However, it is causal. (iv)
∞
y (t ) = x(t ) * h(t ) =
∫
−∞
x(τ )h(t − τ )dτ =
∞
∫
−∞
1 − e−4(t −τ ) u (t − τ ) x(τ )dτ =
t
∫ 1 − e
−∞
−4( t −τ )
x(τ )dτ .
From the input-output relationship, we observe that the output at time t depends on the values of the ▌ input from time (−∞, t). Therefore, the system is NOT memoryless. However, it is causal. Problem 3.14
(i)
System (i) is invertible with the impulse response h1i(t) of its inverse system given by h1i (t ) = 15 δ(t + 2) .
(ii)
System (ii) will be invertible if there exists an impulse response h 2i (t ) such that
114
Chapter 3 h 2(t ) ∗ h 2i (t ) = δ(t ) . Substituting the value of h2(t), we get
h 2i (t ) + h 2i (t + 2) = δ(t ) h 2i (t ) = δ(t − 2) − h 2i (t − 2 ) .
which simplifies to
Substituting the value of h 2i (t − 2) = δ(t − 4) − h 2i (t − 4 ) in the earlier expression gives
h 2i (t ) = δ(t − 2) − δ(t − 4) + h 2i (t − 4 ) . Iterating the above procedure yields, ∞
h 2 i (t ) =
∑ (−1) m +1 δ(t − 2m) .
m =1
Therefore, the system is invertible with the impulse response of the inverse system given above. (iii) System (iii) will be invertible if there exists an impulse response h 3i (t ) such that h3(t ) ∗ h3i (t ) = δ(t ) . Substituting the value of h3(t), we get
h3 i (t + 1) + h3i (t − 1) = δ(t ) h3i (t ) = δ(t − 1) − h3 i (t − 2 ) .
which simplifies to
Substituting the value of h3i (t − 2) = δ(t − 3) − h3 i (t − 4 ) in the earlier expression yields
h3i (t ) = δ(t − 1) − δ(t − 3) + h3 i (t − 4 ) . Iterating the above procedure yields, h3i (t ) =
∞
∑ (−1)m +1 δ(t + 1 − 2m) .
m =1
(iv)
System (iv) will be invertible if there exists an impulse response h 4i (t ) such that h 4(t ) ∗ h 4i (t ) = δ(t ) . Substituting the value of h4(t), we get ∞
∫ h4 i (τ)u (t − τ)dτ = δ(t )
−∞
t
which simplifies to
∫ h4 i (τ)dτ = δ(t ) .
−∞
Differentiating both sides of the above expression with respect to t, we obtain h 4i (t ) =
d dt
(δ(t ) ) .
In other words, system (iv) is an integrator. As expected, its inverse system is a differentiator.
Solutions
(v)
115
System (v) will be invertible if there exists an impulse response h5i (t ) such that h5(t ) ∗ h5i (t ) = δ(t ) . Substituting the value of h5(t), we obtain ∞
∫ h5 i (τ )rect( t −4τ )dτ = δ(t ) ,
−∞
t +4
∫ h5 i (τ)dτ = δ(t ) ,
which simplifies to
t −4 t +4
∫
which is expressed as
t −4
∫ h5 i (τ)dτ
h5 i (τ )dτ −
− ∞
− ∞
Substitute α = τ − 4
Substitute α = τ + 4
t
∫
or,
= δ(t ) ,
t
∫ h5 i (α − 4 )dα = δ(t )
h5 i (α + 4 )dα −
−∞
−∞
Taking the derivative of both sides of the equation with respect to t, we obtain h5 i (t + 4 ) − h5 i (t − 4 ) =
d dt
(δ(t ) ) .
which can be expressed as ∞
∑ dtd (δ(t − 4 − 8m)) .
h5 i (t ) =
m =0
(vi)
System (vi) will be invertible if there exists an impulse response h 6i (t ) such that h6(t ) ∗ h6i (t ) = δ(t ) . Substituting the value of h6(t), we obtain ∞
∫ h6 i (τ )e
− 2(t − τ)
u (t − τ)dτ = δ(t ) ,
−∞
which simplifies to
e − 2t
t
∫ h6 i (τ )e
2τ
dτ = δ(t )
−∞ t
∫ h6 i (τ )e
or,
2τ
dτ = δ(t )e 2t .
−∞
Taking the derivative of both sides of the equation with respect to t, we obtain h6 i (t )e 2t = or,
d dt
(δ(t )e ) = e
h6 i (t ) =
2t
d dt
2t d dt
(δ(t )) + 2δ(t )e 2t
(δ(t )) + 2δ(t ) .
▌
116
Chapter 3
Problem 3.15
By inspection, v(t ) = x(t ) + h2 (t ) ∗ y (t ) , and y (t ) = v(t ) ∗ h1 (t ) . Therefore,
y (t ) = [ x (t ) + h2 (t ) ∗ y (t )] ∗ h1 (t ) = x (t ) ∗ h1 (t ) + h2 (t ) ∗ h1 (t ) ∗ y (t ) .
By rearranging the terms on both sides of the equation, we obtain
[δ(t ) − h2 (t ) ∗ h1 (t )] ∗ y (t ) = h1 (t ) ∗ x(t ) .
▌
Problem 3.16
The output of the system is given by
y (t ) = x (t ) ∗ h ( t ) = e
jω0t
∞
∗ h(t ) =
∫ h(τ )e
jω0 ( t −τ )
dτ = e
jω0t
−∞
∞
Defining H (ω) =
∫ h(t )e
jωt
∞
∫ h(τ )e
− jω0τ
dτ .
−∞
dt , the output is given by
−∞
y (t ) = e jωot H (ω) |ω=ωo .
▌
Problem 3.17
e jωot → e jωot H (ω ) |ω =ωo = e jωot H (ω0 ) .
In P3.16, it is shown that
Applying the linearity property, we obtain
Ae
j (ωo t +θ )
= ( Ae jθ ) e jωot → ( Ae jθ ) e jωot H (ωo ) = Ae
j (ωo t +θ )
H (ωo ) .
By expressing H(ω) in polar format, H (ω) = H (ω) e j < H ( ω) with |H(ω)| being the magnitude and
Ae
j (ωot +θ )
→ Ae
j (ωot +θ +< H ( ω0 ) )
H (ω0 ) .
Decomposing the above expression into its real and imaginary components A cos(ωo t + θ ) + jA sin (ωo t + θ) → A cos(ωo t + θ+ < H (ω0 ) ) H (ωo ) + jA sin (ωo t + θ+ < H (ω0 ) ) H (ωo ) Since the impulse response of the system is real-valued, the real part of the output arises due to the real part of the input, and the imaginary part of the output arises due to the imaginary part of the input. Therefore, by separating the real and imaginary components, we obtain: A cos(ωo t + θ ) → A H (ωo ) cos(ωo t + θ+ < H (ω0 ) ) and
A sin (ωo t + θ ) → A H (ωo ) sin (ωo t + θ+ < H (ω0 ) ) .
The above result implies that an LTIC system only changes the magnitude and phase of the sinusoidal input. The output is still sinusoidal with the same fundamental frequency as that of the input signal. ▌ Problem 3.18
Express
− 3 sin (2πt + π / 4 ) → 5 cos(2πt )
Solutions
117
3sin ( 2π t + 5π / 4 ) → 5sin ( 2π t + π / 2 ) = 3 × (5 / 3) sin ( 2π t + 5π / 4 − 3π / 4 ) .
as
A sin (ωo t + θ ) → A H (ωo ) sin (ωo t + θ+ < H (ω0 ) ) ,
Comparing with
H (ωo ) = 5 / 3 and < H (ω0 ) = − 3π / 4 .
we note that
The transfer function of the system at ω = 2π is therefore given by H (ω) ω = 2 π =
5 3
e − j 3π / 4 .
▌
Problem 3.19
y (t ) + 4 y (t ) + 8 y (t ) = x (t ) + x (t ) with
x (t ) = e −4 t u(t ), y (0) = 0, and y (0) = 0.
The Matlab code is included below with both the analytical and computational plots included in Fig. S3.19.1. Problem 3.19(i) Analytical Solution
0.1 0.05 0 -0.05 -0.1
0
2
4
6
8
10 time (t)
12
14
16
18
20
0
2
4
6
8
10 time (t)
12
14
16
18
20
0.1 Computed Solution
(i)
0.05 0 -0.05 -0.1
Fig. S3.19.1: Analytical (top) and computational (bottom) plots for Problem 3.2 part (i) % MATLAB Code for Problem 3.19(i) tspan = [0:0.02:20]; %Analytical Solution from Problem 3.2 t = tspan; yanalytical = 3/8*(exp(-2*t).*cos(2*t)-exp(-2*t).*sin(2*t)-exp(-4*t)); subplot(211); plot(t,yanalytical); title('Problem 3.19(i)'); xlabel('time (t)'); ylabel('Analytical Solution'); grid on %Computational Solution y0 = [0; 0] [t2,y] = ode23('myfunc4problem3_19a',tspan,y0); subplot(212); plot(t2,y(:,2)); xlabel('time (t)'); ylabel('Computed Solution'); grid on % Include the following function in a separate file < myfunc4problem3_19a.m>
118
Chapter 3
function [ydot] = myfunc4problem3_19a(t,y) ydot(1,1) = -4*y(1) - 8*y(2) - 3*exp(-4*t)*(t >= 0); ydot(2,1) = y(1); %---end of the function----------------------
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.
The Matlab code is included below with both the analytical and computational plots included in Fig. S3.19.2. Problem 3.19(ii) Analytical Solution
2
1
0
-1
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2 Computed Solution
(ii)
1
0
-1
Fig. S3.19.2: Analytical (top) and computational (bottom) plots for Problem 3.2 part (ii) % MATLAB Code for Problem 3.19(ii) tspan = [0:0.02:20]; %Analytical Solution from Problem 3.2 t = tspan; yanalytical = -0.1962*exp(-5.2361*t)+2.1169*exp(0.7639*t)+0.0793*cos(6*t); yanalytical = yanalytical+0.0983*sin(6*t); subplot(211); plot(t,yanalytical); title('Problem 3.19(ii)'); xlabel('time (t)'); ylabel('Analytical Solution'); myaxis = axis; grid on %Computational Solution y0 = [0; 2] [t2,y] = ode23('myfunc4problem3_19b',tspan,y0); subplot(212); plot(t2,y(:,2)); xlabel('time (t)'); ylabel('Computed Solution'); axis(myaxis); grid on % Include the following function in a separate file < myfunc4problem3_19b.m> function [ydot] = myfunc4problem3_19b(t,y) ydot(1,1) = -6*y(1) - 4*y(2) - 6*sin(6*t) + cos(6*t); ydot(2,1) = y(1);
Solutions
(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.
The Matlab code is included below with both the analytical and computational plots included in Fig. S3.19.3. Problem 3.19(iii) Analytical Solution
4 2 0 -2
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Computed Solution
4 2
0
-2
Fig. S3.19.3: Analytical (top) and computational (bottom) plots for Problem 3.2 part (iii) % MATLAB Code for Problem 3.19(iii) tspan = [0:0.02:20]; %Analytical Solution from Problem 3.2 t = tspan; yanalytical = 2.36*exp(-t)+2.9*t.*exp(-t); yanalytical = yanalytical - 0.5*sin(t)+0.64*cos(2*t)+0.48*sin(2*t); subplot(211); plot(t,yanalytical); title('Problem 3.19(iii)'); xlabel('time (t)'); ylabel('Analytical Solution'); myaxis = axis; grid on %Computational Solution y0 = [1; 3] [t2,y] = ode23('myfunc4problem3_19c',tspan,y0); subplot(212); plot(t2,y(:,2)); xlabel('time (t)'); ylabel('Computed Solution'); axis(myaxis); grid on % Include the following function in a separate file < myfunc4problem3_19c.m> function [ydot] = myfunc4problem3_19c(t,y) ydot(1,1) = -2*y(1) - y(2) - cos(t) - 4*sin(2*t); ydot(2,1) = y(1);
(iv)
y (t ) + 4 y (t ) = 5 x (t ) with
x (t ) = 4te− t u(t ), y (0) = −2, and y (0) = 0.
119
120
Chapter 3 The Matlab code is included below with both the analytical and computational plots included in Fig. S3.19.4. Problem 3.19(iv) Analytical Solution
10 5 0 -5
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Computed Solution
10 5
0
-5
Fig. S3.19.4: Analytical (top) and computational (bottom) plots for Problem 3.2 part (iv) % MATLAB Code for Problem 3.19(iv) tspan = [0:0.02:20]; %Analytical Solution from Problem 3.2 t = tspan; yanalytical = -3.6*cos(2*t)-1.2*sin(2*t)+1.6*exp(-t)+4*t.*exp(-t); subplot(211); plot(t,yanalytical); title('Problem 3.19(iv)'); xlabel('time (t)'); ylabel('Analytical Solution'); myaxis = axis; grid on %Computational Solution y0 = [0; -2] [t2,y] = ode23('myfunc4problem3_19d',tspan,y0); subplot(212); plot(t2,y(:,2)); xlabel('time (t)'); ylabel('Computed Solution'); axis(myaxis); grid on % Include the following function in a separate file < myfunc4problem3_19d.m> function [ydot] = myfunc4problem3_19d(t,y) ydot(1,1) = - 4*y(2) +5*4*t.*exp(-t); ydot(2,1) = y(1);
(v)
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.
The Matlab code is included below with both the analytical and computational plots included in Fig. S3.19.5.
Solutions
121
Problem 3.19(v) Analytical Solution
40 20
0 -20
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Computed Solution
40
20
0
-20
Fig. S3.19.5: Analytical (top) and computational (bottom) plots for Problem 3.2 part (v) % MATLAB Code for Problem 3.19(iii) tspan = [0:0.02:20]; %Analytical Solution from Problem 3.2 t = tspan; %yanalytical = -0.25*exp(t)-0.75*exp(-t)-cos(t)+0.5*sin(t)+2; yanalytical = 1.50*sin(t)-2*cos(t)-t.*sin(t)-0.5*t.*cos(t) +2; subplot(211); plot(t,yanalytical); title('Problem 3.19(v)'); xlabel('time (t)'); ylabel('Analytical Solution'); myaxis = axis; grid on %Computational Solution y0 = [0; 0; 1; 0]; [t2,y] = ode23('myfunc4problem3_19e',tspan,y0); subplot(212); plot(t2,y(:,4)); xlabel('time (t)'); ylabel('Computed Solution'); axis(myaxis); grid on % Include the following function in a separate file < myfunc4problem3_19e.m> function [ydot] = myfunc4problem3_19e(t,y) ydot(1,1) = -2*y(2) - y(4) + 2; ydot(2,1) = y(1); ydot(3,1) = y(2); ydot(4,1) = y(3);
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