Solutions to Additional Problems
6.26. A signal x(t) has Laplace transform X(s) as given below. Plot the poles and zeros in the s-plane
and determine the Fourier transform of x(t) without inverting X(s).
(a) X(s) = s2
...
Solutions to Additional Problems
6.26. A signal x(t) has Laplace transform X(s) as given below. Plot the poles and zeros in the s-plane
and determine the Fourier transform of x(t) without inverting X(s).
(a) X(s) = s2s+5 2+1 s+6
X(s) = (s + j)(s − j)
(s +3)(s +2)
zeros at: ±j
poles at: −3, −2
X(jω) = X(s)|s=jω
=
−ω2 + 1
−ω2 + 5jω + 6
Pole−Zero Map
Real Axis
Imag Axis
−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Figure P6.26. (a) Pole-Zero Plot of X(s)
(b) X(s) = s2s+2−s+1 1
X(s) = (s +1)(s − 1)
(s + 0.5 − j3 4)(s + 0.5 + j3 4)
zeros at: ±1
1poles at: −1 ± j√3
2
X(jω) = X(s)|s=jω
=
−ω2 − 1
−ω2 + jω + 1
Pole−Zero Map
Real Axis
Imag Axis
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Figure P6.26. (b) Pole-Zero Plot of X(s)
(c) X(s) = s−14 + s−22
X(s) = 3(s − 10 3 )
(s − 4)(s − 2)
zero at:
10
3
poles at: 4, 2
X(jω) = X(s)|s=jω
=
1
jω − 4
+
2
jω − 2
2Pole−Zero Map
Real Axis
Imag Axis
0 0.5 1 1.5 2 2.5 3 3.5 4
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Figure P6.26. (b) Pole-Zero Plot of X(s)
6.27. Determine the bilateral Laplace transform and ROC for the following signals:
(a) x(t) = e−tu(t +2)
X(s) = −∞ ∞ x(t)e−st dt
= −∞ ∞ e−tu(t +2)e−st dt
= −∞2 e−t(1+s) dt
=
e2(1+s)
1 + s
ROC: Re(s) > -1
(b) x(t) = u(−t +3)
X(s) = −∞ 3 e−st dt
=
−e−3s
s
ROC: Re(s) < 0
3(c) x(t) = δ(t +1)
X(s) = −∞ ∞ δ(t +1)e−st dt
= es
ROC: all s
(d) x(t) = sin(t)u(t)
X(s) = 0∞ 21j ejt − e−jt e−st dt
= 0∞ 21j et(j−s) dt − 0∞ 21j e−t(j+s) dt
=
12j
j−−1s − j +1 s
=
1
(1 + s2)
ROC: Re(s) > 0
6.28. Determine the unilateral Laplace transform of the following signals using the defining equation:
(a) x(t) = u(t − 2)
X(s) = 0−∞ x(t)e−st dt
= 0−∞ u(t − 2)e−st dt
= 2∞ e−st dt
=
e−2s
s
(b) x(t) = u(t +2)
X(s) = 0−∞ u(t +2)e−st dt
= 0−∞ e−st dt
=
1 s
(c) x(t) = e−2tu(t +1)
4X(s) = 0−∞ e−2tu(t +1)e−st dt
= 0−∞ e−t(s+2) dt
=
1
s + 2
(d) x(t) = e2tu(−t +2)
X(s) = 0−∞ e2tu(−t +2)e−st dt
= 0−2 et(2−s) dt
=
e2(2−s) − 1
2 − s
(e) x(t) = sin(ωot)
X(s) = 0−∞ 21j ejωot − e−jωot e−st dt
=
12j
0−∞ et(jωo−s) dt − 0−∞ e−t(jωo+s) dt
=
12j
jω−o 1− s − jωo1+ s
=
ωo
s2 + ω2
o
(f) x(t) = u(t) − u(t − 2)
X(s) = 0−2 e−st dt
=
1 − e−2s
s
(g) x(t) = sin(πt), 0 < t < 1
0, otherwise
X(s) = 0−1 21j ejπt − e−jπt e−st dt
=
π(1 + e−s)
s2 + π2
56.29. Use the basic Laplace transforms and the Laplace transform properties given in Tables D.1 and
D.2 to determine the unilateral Laplace transform of the following signals:
(a) x(t) = dt d {te−tu(t)}
a(t) = te−tu(t) ←−−−→ Lu A(s) = 1
(s +1)2
x(t) = d
dta(t)
Lu
←−−−→ X(s) = s
(s +1)2
(b) x(t) = tu(t) ∗ cos(2πt)u(t)
a(t) = tu(t)
Lu
←−−−→ A(s) = 1
s2
b(t) = cos(2πt)u(t)
Lu
←−−−→
s
s2 + 4π2
x(t) = a(t) ∗ b(t)
Lu
←−−−→ X(s) = A(s)B(s)
X(s) = 1
s2(s2 + 4π2)
(c) x(t) = t3u(t)
a(t) = tu(t)
Lu
←−−−→ A(s) = 1
s2
b(t) = −ta(t)
Lu
←−−−→ B(s) = d
dsA(s) = −s32
x(t) = −tb(t)
Lu
←−−−→ X(s) = d
dsB(s) = s64
(d) x(t) = u(t − 1) ∗ e−2tu(t − 1)
a(t) = u(t)
Lu
←−−−→ A(s) = 1
s
b(t) = a(t − 1)
Lu
←−−−→ B(s) = e−s
s
c(t) = e−2tu(t) ←−−−→ Lu C(s) = 1
s + 2
d(t) = e−2c(t − 1) ←−−−→ Lu D(s) = e−(s+2)
s + 2
x(t) = b(t) ∗ d(t)
Lu
←−−−→ X(s) = B(s)D(s)
X(s) = e−2(s+1)
s(s +2)
6(e) x(t) = 0t e−3τ cos(2τ)dτ
a(t) = e−3t cos(2t)u(t) ←−−−→ Lu A(s) = s + 3
(s +3)2 + 4
−∞ t a(τ)dτ ←−−−→ Lu 1s −∞ 0− a(τ)dτ + A(ss)
X(s) = s + 3
s((s +3)2 +4)
(f) x(t) = t dt d (e−t cos(t)u(t))
a(t) = e−t cos(t)u(t) ←−−−→ Lu A(s) = s + 1
(s +1)2 + 1
b(t) = d
dta(t)
Lu
←−−−→ B(s) = s(s +1)
(s +1)2 + 1
x(t) = tb(t)
Lu
←−−−→ X(s) = − d
dsB(s)
X(s) = −s2 − 4s − 2
(s2 + 2s +2)2
6.30. Use the basic Laplace transforms and the Laplace transform properties given in Tables D.1 and
D.2 to determine the time signals corresponding to the following unilateral Laplace transforms:
(a) X(s) = s+2 1 s+3 1
X(s) = 1
s + 2
+
−1
s + 3
x(t) = e−2t − e−3t u(t)
(b) X(s) = e−2s ds d (s+1) 1 2
A(s) = 1
(s +1)2
Lu
←−−−→ a(t) = te−tu(t)
B(s) = d
dsA(s)
Lu
←−−−→ b(t) = −ta(t) = −t2e−tu(t)
X(s) = e−2sB(s) ←−−−→ Lu x(t) = b(t − 2) = −(t − 2)2e−(t−2)u(t − 2)
(c) X(s) = (2s+1) 1 2+4
B(s
a
)
Lu
←−−−→ ab(at)
1
(s +1)2 + 4
Lu
←−−−→
12
e−t sin(2t)u(t)
x(t) = 1
4
e−0.5t sin(t)u(t)
7(d) X(s) = s ds d22 s21+9 + s+3 1
A(s) = 1
s2 + 9
Lu
←−−−→ a(t) = 1
3
sin(3t)u(t)
B(s) = d
dsA(s)
Lu
←−−−→ b(t) = −ta(t) = − t
3
sin(3t)u(t)
C(s) = d
dsB(s)
Lu
←−−−→ c(t) = −tb(t) = t2
3
sin(3t)u(t)
D(s) = sC(s)
Lu
←−−−→ d(t) = d
dtc(t) − c(0−) = 23t sin(3t)u(t) + t2 cos(3t)u(t)
E(s) = 1
s + 3
Lu
←−−−→ e(t) = e−3tu(t)
x(t) = e(t) + d(t) = e−3t + 23t sin(3t) + t2 cos(3t) u(t)
6.31. Given the transform pair cos(2t)u(t)
Lu
←−−−→ X(s), determine the time signals corresponding to
the following Laplace transforms:
(a) (s +1)X(s)
sX(s) + X(s)
Lu
←−−−→
d
dtx(t) + x(t)
= [−2 sin(2t) +cos(2t)] u(t)
(b) X(3s)
X(s
a
)
Lu
←−−−→ ax(at)
x(t) = 1
3
cos(2
3
t)u(t)
(c) X(s +2)
X(s +2)
Lu
←−−−→ e−2tx(t)
x(t) = e−2t cos(2t)u(t)
(d) s−2X(s)
B(s) = 1
s
X(s)
Lu
←−−−→ −∞ t x(τ)dτ
Lu
←−−−→ −∞ t cos(2τ)u(τ)dτ
Lu
←−−−→ 0t cos(2τ)dτ
8B(s)
Lu
←−−−→
12
sin(2t)
1 s
B(s)
Lu
←−−−→ 0t 1 2 sin(2τ)dτ
Lu
←−−−→
1 − cos(2t)
4
u(t)
(e) ds d e−3sX(s)
A(s) = e−3sX(s) ←−−−→ Lu a(t) = x(t − 3) = cos(2(t − 3))u(t − 3)
B(s) = d
dsA(s)
Lu
←−−−→ b(t) = −ta(t) = −t cos(2(t − 3))u(t − 3)
6.32. Given the transform pair x(t)
Lu
←−−−→ 2s
s2+2, where x(t) = 0 for t < 0, determine the Laplace
transform of the following time signals:
(a) x(3t)
x(3t)
Lu
←−−−→
13
X(s
3
)
X(s) = 23s
(3s)2 + 2
=
6s
s2 +18
(b) x(t − 2)
x(t − 2)
Lu
←−−−→ e−2sX(s) = e−2s 2s
s2 + 2
(c) x(t) ∗ dt d x(t)
b(t) = d
dtx(t)
Lu
←−−−→ B(s) = sX(s)
y(t) = x(t) ∗ b(t)
Lu
←−−−→ Y (s) = B(s)X(s) = s[X(s)]2
Y (s) = s s22+ 2 s 2
(d) e−tx(t)
e−tx(t) ←−−−→ Lu X(s +1) = 2(s +1)
(s +1)2 + 2
9(e) 2tx(t)
2tx(t)
Lu
←−−−→ −2
d
ds X(s) = (4s2s2+2) − 82
(f) 0t x(3τ)dτ
0t x(3τ)dτ ←−−−→ Lu Y (s) = X3(s3s)
Y (s) = 2
s2 +18
6.33. Use the s-domain shift property and the transform pair e−atu(t) ←−−−→ Lu s+1a to derive the
unilateral Laplace transform of x(t) = e−at cos(ω1t)u(t).
e−atu(t) ←−−−→ Lu 1
s + a
x(t) = e−at cos(ω1t)u(t)
=
12
e−at ejω1t + e−jω1t u(t)
Using the s-domain shift property:
X(s) = 1
2 (s − jω11) + a + (s + jω11) + a
=
12
2(s + a)
(s + a)2 + ω12
=
(s + a)
(s + a)2 + ω12
6.34. Prove the following properties of the unilateral Laplace transform:
(a) Linearity
z(t) = ax(t) + by(t)
Z(s) = 0∞ z(t)e−st dt
= 0∞ (ax(t) + by(t)) e−st dt
= 0∞ ax(t)e−st dt + 0∞ by(t)e−st dt
= a 0∞ x(t)e−st dt + b 0∞ y(t)e−st dt
= aX(s) + bY (s)
(b) Scaling
z(t) = x(at)
10Z(s) = 0∞ x(at)e−st dt
=
1 a
0∞ x(τ)e− as τ dτ
=
1 a
X(s
a
)
(c) Time shift
z(t) = x(t − τ)
Z(s) = 0∞ x(t − τ)e−st dt
Let m = t − τ
Z(s) = −∞τ x(m)e−s(m+τ) dm
If x(t − τ)u(t) = x(t − τ)u(t − τ)
Z(s) = 0∞ x(m)e−sme−sτ dm
= e−sτX(s)
(d) s-domain shift
z(t) = esotx(t)
Z(s) = 0∞ esotx(t)e−st dt
= 0∞ x(t)e−(so−s)t dt
= X(s − so)
(e) Convolution
z(t) = x(t) ∗ y(t)
= 0∞ x(τ)y(t − τ) dτ; x(t), y(t) causal
Z(s) = 0∞ 0∞ x(τ)y(t − τ) dτ e−st dt
= 0∞ 0∞ x(τ)y(m) dτ e−sme−sτ dm
= 0∞ x(τ)e−sτ dτ 0∞ y(m)e−sm dm
= X(s)Y (s)
11(f) Differentiation in the s-domain
z(t) = −tx(t)
Z(s) = 0∞ −tx(t)e−st dt
= 0∞ x(t)ds d (e−st) dt
= 0∞ ds d x(t)e−st dt
Assume: 0∞(.) dt and ds d (.) are interchangeable.
Z(s) = d
ds 0∞ x(t)e−st dt
Z(s) = d
ds X(s)
6.35. Determine the initial value x(0+) given the following Laplace transforms X(s):
(a) X(s) = s2+51s−2
x(0+) = lim
s→∞
sX(s) = s
s2 + 5s − 2 = 0
(b) X(s) = s2+2 s+2 s−3
x(0+) = lim
s→∞
sX(s) = s2 + 2s
s2 + 2s − 3 = 1
(c) X(s) = e−2s s26+2 s2+s−s2
x(0+) = lim
s→∞
sX(s) = e−2s 6s3 + s2
s2 + 2s − 2 = 0
6.36. Determine the final value x(∞) given the following Laplace transforms X(s):
(a) X(s) = s22+5 s2+3 s+1
x(∞) = lim
s→0
sX(s) = 2s3 + 3s
s2 + 5s + 1 = 0
(b) X(s) = s3+2 s+2 s2+s
x(∞) = lim
s→0
sX(s) = s + 2
s2 + 2s + 1 = 2
12(c) X(s) = e−3s s2(ss+2) 2+12
x(∞) = lim
s→0
sX(s) = e−3s 2s2 + 1
(s +2)2 =
14
6.37. Use the method of partial fractions to find the time signals corresponding to the following unilateral Laplace transforms:
(a) X(s) = s2+3 s+3 s+2
X(s) = s + 3
s2 + 3s + 2 =
A
s + 1
+
B
s + 2
1 = A + B
3 = 2A + B
X(s) = 2
s + 1
+
−1
s + 2
x(t) = 2e−t − e−2t u(t)
(b) X(s) = 2ss22+10 +5ss+6 +11
X(s) = 2s2 +10s +11
s2 + 5s + 6 = 2 −
1
(s +2)(s +3)
1
(s +2)(s +3) =
A
s + 2
+
B
s + 3
0 = A + B
1 = 3A + 2B
X(s) = 2 − 1
s + 2
+
1
s + 3
x(t) = 2δ(t) + e−3t − e−2t u(t)
(c) X(s) = s22+2 s−s1+1
X(s) = 2s − 1
s2 + 2s + 1 =
A
s + 1
+
B
(s +1)2
2 = A
−1 = A + B
x(t) = 2e−t − 3te−t u(t)
(d) X(s) = s3+3 5ss+4 2+2s
X(s) = 5s + 4
s3 + 3s2 + 2s =
A s
+
B
s + 2
+
C
s + 1
130 = A + B + C
5 = 3A + B + 2C
4 = 2A
X(s) = 2
s
+
−3
s + 2
+
1
s + 1
x(t) = 2 − 3e−2t + e−t u(t)
(e) X(s) = (s+2)( ss22−+2 3 s+1)
X(s) = s2 − 3
(s +2)(s2 + 2s +1) =
A
s + 2
+
B
s + 1
+
C
(s +1)2
1 = A + B
0 = 2A + 3B + C
−3 = A + 2B + 2C
X(s) = 1
s + 2
+
−2
(s +1)2
x(t) = e−2t − 2te−t u(t)
(f) X(s) = s2+2 3s+2 s+10
X(s) = 3s + 2
s2 + 2s +10 =
3(s +1) − 1
(s +1)2 + 32
x(t) = 3e−t cos(3t) − 13e−t sin(3t) u(t)
(g) X(s) = (s+2)( 4s2+8 s2+2 s+10 s+5)
X(s) = 2
s + 2
+
2(s +1)
(s +1)2 + 22 +
−2
(s +1)2 + 22
x(t) = 2e−2t + 2e−t cos(2t) − e−t sin(2t) u(t)
(h) X(s) = (s+2)( 3s2+10 s2+6 s+10 s+10)
X(s) = 3s2 +10s +10
(s +2)(s2 + 6s +10) =
A
s + 2
+
Bs + C
s2 + 6s +10
3 = A + B
10 = 6A + 2B + C
10 = 10A + 2C
X(s) = 1
s + 2
+
2(s +3)
(s +3)2 + 1 −
6
(s +3)2 + 1
x(t) = e−2t + 2e−3t cos(t) − 6e−3t sin(t) u(t)
14(i) X(s) = 2s2+11 s2+5 s+16+ s+6e−2s
X(s) = 2s2 +11s +16 + e−2s
s2 + 5s + 6 = 2 +
s + 4
s2 + 5s + 6 +
e−2s
s2 + 5s + 6
X(s) = 2 + −1
s + 3
+
2
s + 2
− e−2s 1
s + 3
+ e−2s 1
s + 2
x(t) = 2δ(t) + 2e−2t − e−t u(t) + e−2(t−2) − e−3(t−2) u(t − 2)
6.38. Determine the forced and natural responses for the LTI systems described by the following differential equations with the specified input and initial conditions:
(a) dt d y(t) + 10y(t) = 10x(t), y(0−) = 1, x(t) = u(t)
X(s) = 1
s
Y (s)(s + 10) = 10X(s) + y(0−)
Y f(s) = 10X(s)
s +10
=
10
s(s +10)
=
1 s
+
−1
s +10
yf(t) = 1 − e−10t u(t)
Y n(s) = y(0−)
s +10
yn(t) = e−10tu(t)
(b) dt d22 y(t) + 5 dt d y(t) + 6y(t) = −4x(t) − 3 dt d x(t), y(0−) = −1, dt d y(t) t=0− = 5, x(t) = e−tu(t)
Y (s)(s2 + 5s +6) − 5 + s + 5 = (−4 − 3s) 1
s + 1
Y (s) = −1
(s +1)(s +2)(s +3) +
s
(s +2)(s +3)
= Y f(s) + Y n(s)
Y f(s) = −0.5
s + 1
+
−2
s + 2
+
2.5
s + 3
yf(t) = −0.5e−t − 2e−2t + 2.5e−3t u(t)
Y n(s) = −2
s + 2
+
3
s + 3
yn(t) = −2e−2t + 3e−3t u(t)
(c) dt d22 y(t) + y(t) = 8x(t), y(0−) = 0, dt d y(t) t=0− = 2, x(t) = e−tu(t)
Y (s)(s2 + 1) = 8X(s) + 2
15Y f(s) = 4
s + 1
+
4
s2 + 1 −
4s
s2 + 1
yf(t) = 4 e−t +sin(t) − cos(t) u(t)
Y n(s) = 2
s2 + 1
yn(t) = 2 sin(t)u(t)
(d) dt d22 y(t) + 2 dt d y(t) + 5y(t) = dt d x(t), y(0−) = 2, dt d y(t) t=0− = 0, x(t) = u(t)
Y (s)(s2 + 2s +5) = sX(s) + sy(0−) + 2y(0−)
Y f(s) = 1
(s +1)2 + 22
yf(t) = 1
2
e−t sin(2t)u(t)
Y n(s) = 2(s +2)
(s +2)2 + 1
yn(t) = 2e−t cos(t)u(t)
6.39. Use Laplace transform circuit models to determine the current y(t) in the circuit of Fig. P6.39
assuming normalized values R = 1Ω and L = 1 2 H for the specified inputs. The current through the
inductor at time t = 0− is 2 A.
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 2e Peter Olver, Chehrzad Shakiban.png)
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