2 H (s ) = s (s + 4 )
Re{s} > 0
The system is unstable.
15
Chapter 9 9.33 A causal LTI system H (s ) =
x(t ) = e
t
Problem Solution
s +1 s 2 + 2s + 2
, ∞ < t < +∞
Determine and sketch the response y (t )
2 X (s ) = 2 s 1
1 < Re{s} < 1 Re{s} > 1
2(s + 1) (s + 1)(s 1) s 2 + 2s + 2
s +1 H (s ) = 2 s + 2s + 2 Y (s ) = X (s )H (s ) =
2 2 4 y (t ) = + e t u ( t ) + e t (cos t )u (t ) + e t (sin t )u (t ) 5 5 5
17
Chapter 9
Problem Solution
9.35 Consider a causal LTI system with the input x(t ) and output y (t ) . (b) Is this system stable?
2
②
1
③
1
④
2
σ
① Re{s} < 2 anticausal , unstable ② -2 < Re{s} < 1 noncausal , unstable
③ -1 < Re{s} < 1
noncausal , stable Causal , unstable
12
④ Re{s} > 1
Chapter 9 9.31 Consider a continuous-time LTI system (a) Determine H (s ) .
Determine the values of the constants αand β.
x(t ) = βe t u (t ) ← → x( t ) ← →
x(t ) + αx( t ) ← →
β
s +1
Re{s} > 1 Re{s} < 1
αβ
α = 1 1 β= 2
β
s +1
β
s +1 s +1
Re{s} < 1
1 1 h(t ) = + e t u ( t ) e 2t u ( t ) 3 3
13
Chapter 9 9.32 Consider a causal LTI system ,
1. x(t ) = e
2t
Problem Solution
dh(t ) 2. + 2h(t ) = e 4t u (t ) + bu (t ) dt
1 2t (-∞ < t < +∞ ) → y (t ) = e 6
(-∞ < t < +∞ )
b——unknown constant
Determine the system function H (s )and b. Solution sH (s ) + 2 H (s ) = 1 + b s+4 s
2
}ห้องสมุดไป่ตู้
Re{s} > 4
8
Chapter 9
Problem Solution
(i ) x(t ) = δ (t ) + u (t )
X (s ) = 1+ 1 s
Re{s} > 0
(j) x(t ) = δ (3t ) + u (3t )
1 x(t ) = δ (t ) + u (t ) 3
1 1 X (s ) = + 3 s
Homework: 9.2 9.5 9.7 9.8 9.9 9.13 9.21(a,b,i,j)
9.22(a,b,c,d) 9.28 9.31 9.32 9.33 9.35 9.45
1
Chapter 9 9.2 Consider the signal x(t ) = e 5t u (t 1) (a) Evaluate X (s )
14
Chapter 9
1 2t y (t ) = H (2)e = e 6
2t
Problem Solution
(-∞ < t < +∞ )
1 b(2 + 4) + 2 H (2) = = b =1 2(2 + 2)(2 + 4 ) 6
b (s + 4 ) + s H (s ) = s (s + 2 )(s + 4 )
(a )
(b )
1 1 2s + 4 + = s + 1 s + 3 (s + 1)(s + 3)
1 s +1 = 2 s 1 s 1
1 , 2
0, 1
(c)
s3 1 = s 1 2 s + s +1
1, 0
3
Chapter 9
Problem Solution
9.7 How many signals have a Laplace transform that may be s 1 expressed as
Problem Solution
(b) g (t ) = Ae 5t u ( t t0 ) G (s ) = X (s ) A, t0 , ROC = ?
L (t ) ←→ 1 e u
5t
s+5
s L e e 5(t 1)u (t 1) ←→
s+5
e (s +5 ) L e 5t u (t 1) ←→ Re{s} > 5 s+5 e (s +5 ) L e 5e 5(t 1)u ( t + 1) ←→ Re{s} < 5 s+5 g (t ) = e 5t u ( t + 1)
2 4 + X (s ) = Re{s} > 3 s+3 s+4
x(t ) = 2e 3t + 4e 4t u (t )
(
)
6
Chapter 9
Problem Solution
t 9.13 Let g (t ) = x(t ) + αx( t ) , where x(t ) = βe u (t ) s and G (s ) = 2 -1 < Re{s} < 1 s 1
+
(β αβ )s β αβ s = G (s ) = 2 s2 1 s 1
7
Chapter 9 9.21 Determine the Laplace transform.
Problem Solution
(a ) x(t ) = e 2t u (t ) + e 3t u (t )
X (s ) = 1 1 2s + 5 + = Re{s} > 2 s + 2 s + 3 (s + 2 )(s + 3)
(s + 2)(s + 3)(s 2 + s + 1)
in its region of convergence?
jω
Poles :
1 3 + j 2 2
s1 = 2, s1 = 3, s3, 4
1 3 = ± j 2 2
3 2
1 3 j 2 2
σ
There are four signals.
4
Chapter 9
Problem Solution
9.8 Let x(t ) be a signal that has a rational Laplace transform with exactly two poles,located at s = -1 and s = -3. If g (t ) = e 2t x(t ) and G ( jω ) convergence,determine whether x(t ) is left sided, right sided,or two sided.
b (s + 4 ) + s H (s ) = s (s + 2)(s + 4)
Causal Re{s} > 0
x(t ) = e 2t (-∞ < t < +∞ ) is the eigenfunction of system.
s = 2 Re{s} > 0 H (s ) s=2 is finite
A = 1 t 0 = 1
2
Chapter 9
Problem Solution
9.5 For each of the following algebraic expressions for the Laplace transform of a signal ,determine the number of zeros located in the finite s-plane and the number of zeros located at infinity:
s (cos 3t )u ( t ) ← → 2 s +9 Re{s} < 0