Download Solutions of Chapter 5 Part 2/2

Solutions of Chapter 5
Part 2/2
Problem 5.3-18 (a) Find the zero-state response of an LTID system with transfer function
H[z] =
z
(z + 0.2)(z − 0.8)
and the input x[n] = (e)(n+1) u[n].
(b) Write the difference equation relating the output y[n] to input x[n].
Solution:
(a)
x[n] = (e)(n+1) u[n] = e(en u[n]) ⇔ X[z] = e
Y [z] = H[z]X[z] =
Therefore
z
z−e
ez2
(z + 0.2)(z − 0.8)(z − e)
Y [z]
ez
1.32
0.186
1.13
=
=
−
−
z
(z + 0.2)(z − 0.8)(z − e) z − e z + 0.2 z − 0.8
z
z
z
Y [z] = 1.32
− 0.186
− 1.13
z−e
z + 0.2
z − 0.8
y[n] = [1.32(e)n − 0.186(−0.2)n − 1.13(0.8)n ] u[n]
(b) From
H[z] =
Y [z]
z
=
X[z] (z + 0.2)(z − 0.8)
we have
(z2 − 0.6z − 0.16)Y [z] = zX[z]
Hence, the corresponding difference equation of the system is
y[n + 2] − 0.6y[n + 1] − 0.16y[n] = x[n + 1]
or
y[n] − 0.6y[n − 1] − 0.16y[n − 2] = x[n − 1]
Problem 5.3-23 Find h[n], the unit impulse response of the systems described by the following equations:
(a) y[n] + 3y[n − 1] + 2y[n − 2] = x[n] + 3x[n − 1] + 3x[n − 2]
(b) y[n + 2] + 2y[n + 1] + y[n] = 2x[n + 2] − x[n + 1]
(c) y[n] − y[n − 1] + 0.5y[n − 2] = x[n] + 2x[n − 1]
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Solution:
(a) Write the system in advanced form
y[n + 2] + 3y[n + 1] + 2y[n] = x[n + 2] + 3x[n + 1] + 3x[n]
or
(E 2 + 3E + 2)y[n] = (E 2 + 3E + 3)x[n]
Thus,
H[z] =
z2 + 3z + 3
Y [z] z2 + 3z + 3
= 2
=
X[z] z + 3z + 2 (z + 1)(z + 2)
H[z]
z2 + 3z + 3
3/2
1
1/2
=
=
−
+
z
z(z + 1)(z + 2)
z
z+1 z+2
3
z
1 z
H[z] = −
+
2 z+1 2 z+2
[
]
3
1
n
n
h[n] = δ [n] − (−1) + (−2) u[n]
2
2
(b)
z2Y [z] + 2zY [z] +Y [z] = 2z2 X[z] − zX[z]
H[z] =
Therefore
2z2 − z
z(2z − 1)
Y [z]
= 2
=
X[z] z + 2z + 1
(z + 1)2
H[z]
2z − 1
2
3
=
=
−
2
z
(z + 1)
z + 1 (z + 1)2
z
z
H[z] = 2
−3
z+1
(z + 1)2
h[n] = [2(−1)n − 3n(−1)n ] u[n] = (2 − 3n)(−1)n u[n]
(c) Performing z-transform of the system yields
1
0.5
2
Y [z] − Y [z] + 2 Y [z] = X[z] + X[z]
z
z
z
Thus,
H[z] =
1 + 2z
Y [z]
z(z + 2)
= 2
=
1
0.5
X[z] 1 − z + z2
z − z + 0.5
H[z]
(z + 2)
= 2
z
z − z + 0.5
Using the pair 12c in Table 1, with
1
A = 1, B = 2, a = −0.5, |γ |2 = 0.5, |γ | = √
2
√
−2.5
π
r = 5.099, β = cos−1 (0.5 5) = , θ = tan−1 (
) = −1.373
4
0.5
we have
π
1
h[n] = 5.099( √ )n cos( n − 1.373) u[n]
4
2
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Problem 3.4-3 (used in Problem 5.5-3) A moving average is used to detect a trend of a rapidly fluctuating
variable such as the stock market average. A variable may fluctuate (up and down) daily, masking its longterm (secular) trend. We can discern the long-term trend by smoothing or averaging the past N values of the
variable. For the stock market average, we may consider a 5-day moving average y[n] to be the mean of the
past 5 days’ market closing values x[n], x[n − 1], · · · , x[n − 4]. (a) Write the difference equation relating y[n]
to the input x[n].
Solution:
(a) y[n] = 15 (x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4])
Problem 5.5-3 Find the frequency response for the moving average system in Prob.3.4-3. The input-output
equation of this system is given by
1 4
y[n] = ∑ x[n − k]
5 k=0
Solution: The advance operator form of the equation is
1
E 4 y[n] = (E 4 + E 3 + E 2 + E + 1)x[n]
5
Thus, the transfer function of the system is
[
]
Y [z] 1 z4 + z3 + z2 + z + 1
H[z] =
=
X[z] 5
z4
The frequency response is
jΩ
H[e ] =
=
=
[
]
1 e j4Ω + e j3Ω + e j2Ω + e jΩ + 1
5
e j4Ω
1 − j2Ω j2Ω
e
[e + e jΩ + 1 + e− jΩ + e− j2Ω ]
5
1 − j2Ω
e
[1 + 2 cos Ω + 2 cos 2Ω]
5
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Problem 5.5-5 For an LTID system specified by the equation
y[n + 1] − 0.5y[n] = x[n + 1] + 0.8x[n]
(a) Find the amplitude and the phase response.
(b) Find the system response y[n] for the input x[n] = cos(0.5n − π3 ).
Solution: For the given system, we have the transfer function
H[z] =
z + 0.8
z − 0.5
(a) The frequency response is
H[e jΩ ] =
e jΩ + 0.8 (cos Ω + 0.8) + j sin Ω
=
e jΩ − 0.5 (cos Ω − 0.5) + j sin Ω
Amplitude response: From
|H[e jΩ ]|2 = |H[e jΩ ]|H ∗ [e jΩ ] = |H[e jΩ ]|H[e− jΩ ] =
we have
(
|H[e ]| =
jΩ
(e jΩ + 0.8)(e− jΩ + 0.8) 1.64 + 1.6 cos Ω
=
(e jΩ − 0.5)(e− jΩ − 0.5)
1.25 − cos Ω
1.64 + 1.6 cos Ω
1.25 − cos Ω
) 12
The phase response is
sin Ω
sin Ω
) − tan−1 (
)
cos Ω + 0.8
cos Ω − 0.5
∠H[e jΩ ] = tan−1 (
(b) For the given input, Ω = 0.5. The amplitude response is
|H[e j0.5 ]|2 =
Thus,
1.64 + 1.6 cos(0.5)
= 8.174
1.25 − cos(0.5)
1
H[e j0.5 ] = (8.174) 2 = 2.86
The phase response is
∠H[e j0.5 ] = 0.2784 − 0.9037 = −0.6253 rad
Therefore, the system response is
y[n] = 2.86 cos(0.5n −
π
− 0.6253) = 2.86 cos(0.5n − 1.6725)
3
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