2011
October Fall 18, 2011
EE113: Digital Signal Processing
Prof. Abeer Alwan
TA: Jianshu Chen, and Sen Li
Solution to HW 3
1. (Problem 5.1(part a,b))
Solution:
1
(a) The system y (n) = 3 y (n 1) + x(n 1) is relaxed, which implies that for input x(n) =
(n), the output y (n) = 0, n < 0.
1
y (0) = y (1) + (1) = 0
3
1
y (1) = y (0) + (0) = 1
3
1
1
y (2) = y (1) + (1) =
3
3
.
.
.
1
y (n) = y (n 1) + (n 1) =
3
1
3
n1
we can see that systems impulse response is
h(n) =
1 n1
,
3
n>0
n0
0,
1
(b) The system y (n) = 3 y (n 1) + x(n + 1), y (1) = 1
1
y (1) = 1 = 3 y (2) + (0) = y (2) = 0, then we can see that y (n) = 0, n < 1
1
y (0) = y (1) + (1) =
3
1
y (1) = y (0) + (2) =
3
1
3
1
3
1
y (2) = y (1) + (3) =
3
.
.
.
1
3
2
3
1
y (n) = y (n 1) + (n + 1) =
3
1
3
n+1
we can see that systems output is y (n) = ( 1 )n+1 , n 0. Thus, the impulse response of
3
the system is
n+1
1
, n0
3
h(n) = 1,
n = 1
0,
n < 1
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2. (Problem 5.8)
Solution:
MA(Moving Average): FIR LTI system is also called MA system.
AR(Auto-Regressive): systems are described by dierence equations of the form
M
y (n) =
ak y (n k ) + b0 x(n)
k=1
ARMA: systems are described by dierence equations of the form
M
y (n) =
N
ak y (n k ) +
k=1
bk x(n k )
k=1
(a) ARMA
(b) ARMA
(c) MA
(d) ARMA
1
Note: for (d), we can write h(n) in a recursive form y (n) = 2 y (n 1) + 1 x(n 2)
2
3. (Problem 5.9 (part a,b,c))
Solution:
(a) FIR
(b) IIR
(c) IIR
4. (Problem 5.14)
Solution:
We have a LTI system with impulse response:
1
15
1
h(n) = (n) + (n 1) (n 2)
2
8
2
(a) parallel cascade: Let
1
h1 (n) = (n)
2
15
1
h2 (n) = (n 1) (n 2)
8
2
to make sure h(n) = h1 (n) + h2 (n).
1
(b) series cascade: we know that h(n) have 3 nonzero value [ 2 , 15 , 1 ], and h(n) = h1 (n)
8
2
h2 (n), thus we assume that h1 (n) and h2 (n) have 2 nonzero value at position 0 and 1,
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ie, the nonzero values are h1 (0), h2 (0), h1 (1), h2 (1)
h(n) = h1 (n) h2 (n)
1
=
h1 (k )h2 (n k )
k=0
h(0) = h1 (0)h2 (0) =
1
2
(1)
h(1) = h1 (0)h2 (1) + h1 (1)h2 (0) =
h(2) = h1 (1)h2 (1) =
15
8
(2)
1
2
(3)
1
we can choose h1 (0) = 2, h1 (1) = 2 , h2 (0) = 1 , h2 (1) = 1 to satisfy the equations
4
above.
1
Thus, we have h1 (n) = 2 (n) 2 (n 1) and h2 (n) = 1 (n) + (n 1) to make sure
4
h(n) = h1 (n) h2 (n)
5. (Problem 5.15)
Solution:
We can rewrite the system to be
y (n) =
3n
x(3n 1) cos
0,
,
n 5
n < 5
(a) linear = True
Let y1 (n) = S{x1 (n)}, y2 (n) = S{x2 (n)} and x(n) = x1 (n) + x2 (n)
y (n) = [x1 (3n 1) + x2 (3n 1)] cos
= x1 (3n 1) cos
n
3
n + x2(3n 1) cos
n
3
3
= y1 (n) + y2 (n)
(b) causal = False
Let n = 1, we have y (1) = x(2) cos , which means the output of the system depends
3
on the future value of input. Thus, the system is not causal.
(c) time-invariant = False
Suppose input is x1 (n) = x(n n0 ), then y1 (n) = x(3(n n0 ) 1) cos
(d) stable = True
Assume x(n) Bx , and we know that | cos
3n
3n
= y (n n0 ).
| 1, then we have
|y (n)| = |x(3n 1)|| cos
n ||u(n)|
3
|x(3n 1)|
Bx
6. (Problem 5.19)
Solution:
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(a) We can express (n) = u(n) u(n 1), and the system is LTI, then we know that
h(n) = S{ (n)}
= S{u(n) u(n 1)}
= S{u(n)} S{u(n 1)}
= y (n) y (n 1)
1
2
=
n
n1
1
2
u(n)
u(n 1)
from above, we can see that h(n) = 0, n < 0, which indicates the system is CAUSAL.
The length of h(n) is innite, which means the system is IIR. To determine the stability
of the system, we do the following calculation:
1
2
h(n) = (n)
n
u(n 1), n 0
2
|h(n)| = 1 +
n=
n=1
1
4
=1+
n=1
n=0
n
|2
n
n
1
4
=
=
1
2
|
4
<
3
Then we conclude that the system is BIBO STABLE.
1
(b) We can rewrite x(n) = 1 (n + 1) + (n) + 2 (n 1), then we can directly get the
3
output to be
1
1
y (n) = h(n + 1) + h(n) + h(n 1)
3
2
n+1
1
1n
1
=
u(n + 1)
u(n) +
3
2
2
n1
1
2
1
3
1
2
1
2
1
2
1
2
n
1
2
n1
n1
u(n)
1
2
u(n 1)
n2
4
3
n+1
=
1
2
n
n2
u(n 1)
+
1
2
1
2
u(n + 1) +
u(n 2)
u(n)
1
2
u(n 1)
u(n 2)
7. (Problem 6.1) Solution:
To solve the problems here, we need to call upon the following properties of convolution.
h(n) (n n0 ) = h(n n0 )
x(n) [h(n) w(n)] = [x(n) h(n)] w(n)
(4)
x(n) h(n) = h(n) x(n)
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Now, let us solve problem (a)(d).
(a)
x(n) h(n 1) = x(n) [h(n) (n 1)]
= [x(n) h(n)] (n 1)
= y (n) (n 1)
= y (n 1)
(b)
x(n 1) h(n) = [x(n) (n 1)] h(n)
= [ (n 1) x(n)] h(n)
= (n 1)[ x(n) h(n)]
= (n 1) y (n)
= y (n) (n 1)
= y (n 1)
(c)
x(n 1) h(n 2) = x(n 1) [h(n) (n 2)]
= [x(n 1) h(n)] (n 2)
= y (n 1) (n 2)
= y (n 3)
(d)
x(n) h(n 3) = x(n) [h(n) (n 3)]
= [x(n) h(n)] (n 3)
= y (n) (n 3)
= y (n 3)
8. (Problem 6.6) Solution:
(a) Analytical Method
Let x(n) and y (n) denote
n+1
x(n) =
1
4
n1
y (n) =
1
2
u(n 2)
u(n + 1)
We are asked to evaluate the convolution of x(n) with y (n), which can be expressed as
z (n) = x(n) y (n)
+
=
x(k )y (n k )
k=
+
=
k=2
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1
4
k+1
1
2
nk1
u(n k + 1)
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Figure 1: Graphical method to evaluate convolution in Problem 9 (Problem 6.6 in textbook).
We notice that u(n k + 1) = 1 for k n + 1 and u(n k + 1) = 0 for k n + 2. Hence,
we need to compute z (n) in two separate cases:
i. If n + 1 < 2, i.e., n < 1 or equivalently, n 0, we have z (n) = 0.
ii. If n + 1 2, i.e., n 1, then, we have
n+1
k+1
1
2
nk1
1
2
n+1 n+1
1
2
k
1
2
n+3 n+1
1
2
k2
1
2
k
z (n) =
k=2
=
=
=
=
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1
2
1
2
1
4
k=2
k=2
n+3 n1
k=0
n+2
1
1
2
n
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In summary, can we write the convolution result as
z (n) =
1
2
n+2
n
1
2
1
u(n 1)
(b) Graphical Method
We sketch out sequences x(k ), y (k ), and y (n k ) in Figure 1. We notice that there is
overlap between x(k ) and y (n k ) only when n 1. Therefore, we have, for n 1,
n+1
k+1
1
2
nk1
1
2
n+1 n+1
1
2
k
1
2
n+3 n+1
1
2
k2
1
2
k
z (n) =
k=2
=
=
=
=
1
2
1
2
1
4
k=2
k=2
n+3 n1
k=0
n+2
1
1
2
n
For n 0, the output z (n) = 0.
(c) MATLAB Method
Example of MATLAB code is shown below and the result is given in Figure 2.
% Evaluate convolution by MATLAB
clear all
close all
n = -2:10;
x = (1/4).^(n+1).*(n>=2);
y = (1/2).^(n-1).*(n >=-1);
z = conv(x,y,same);
%------------subplot(3,1,1);
stem(n,x,k,Linewidth,2,Markersize,8);
xlabel($n$,Interpreter,Latex,Fontsize,16)
ylabel($x(n)$,Interpreter,Latex,Fontsize,16)
axis([-2 15 0 max(x) ])
set(gca,Fontsize,12)
%------------subplot(3,1,2);
stem(n,y,k,Linewidth,2,Markersize,8);
xlabel($n$,Interpreter,Latex,Fontsize,16)
ylabel($y(n)$,Interpreter,Latex,Fontsize,16)
axis([-2 15 0 max(y)])
set(gca,Fontsize,12)
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x(n )
0.01
0
2
0
2
4
6
0
2
4
6
0
2
4
6
y (n )
4.00
8
10
12
14
8
10
12
14
8
10
12
14
2.00
0
2
z (n )
n
n
0.04
0.02
0
2
n
Figure 2: MATLAB method to evaluate convolution in Problem 9 (Problem 6.6 in textbook).
%------------subplot(3,1,3);
stem(z,k,Linewidth,2,Markersize,8);
xlabel(n)
ylabel(z(n))
xlabel($n$,Interpreter,Latex,Fontsize,16)
ylabel($z(n)$,Interpreter,Latex,Fontsize,16)
set(gca,Fontsize,12)
axis([-2 15 0 max(z)])
9. (Problem 6.13) The result is shown in Figure 3 in the last page.
10. (Problem 6.15) Solution:
Since x(n) is nonzero for 3 n 4, we can write the expression of z (n) as
z (n) = x(n) y (n) =
4
x(k )y (n k ) =
k=
x(k )y (n k )
(5)
k=3
Now we analyze (a)(d) one by one.
(a) False. For n < 3, we have for any k that satises 3 k 4,
n k < 3 k 3 (3) = 6
Since y (n) is nonzero only for 7 n 9, we thus have
y (n k ) = 0,
for n < 3 and 3 k 4
Then, for any n < 3, all the terms in the sum of (5) are zero. Hence, the claim that
z (n) = 0 for n < 3 is true. However, the other part of the claim that z (n) = 0 for n > 12
is false. This is because for n > 12 and 3 k 4, we have
n k > 12 k 12 4 = 8
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In other words, n k 9, and y (n k ) can be nonzero for n k = 9. To see this, let
us consider y (13), which can be expressed as
4
z (13) =
x(k )y (13 k )
k=3
= x(3)y (16) + x(2)y (15) + x(1)y (14) + + x(3)y (10) + x(4)y (9)
= x(4)y (9)
where both x(4) and y (9) can be nonzero. Then, z (13) may not be zero.
(b) False. z (0) must be zero. To see this, let us write z (0) as
4
4
z (0) =
x(k )y (0 k ) =
k=3
x(k )y (k )
(6)
k=3
We already know that y (n) is nonzero only for 7 n 9, thus for 3 k 4, we have
4 k 3, which implies y (k ) = 0. Then, z (0) is zero.
(c) False. Let us consider the example that x(n) = 1 for all 3 n 4, and y (n) = (1)n
for 7 n 9. Then, z (5) can be expressed as
4
z (5) =
x(k )y (5 k )
k=3
= x(3)y (8) + x(2)y (7) + + x(4)y (1)
= x(3)y (8) + x(2)y (7)
= 1 + (1) = 0
(d) True. Since (a)(c) are false, this claim is true.
11. (Problem 6.32) Solution:
Let us rst evaluate the impulse response of the system. The system has to be relaxed to
ensure it is LTI, i.e., y (n) should be zero before the input is added. If x(n) = (n), then
y (1) = 0. The impulse response is the solution to the following dierence equation:
1
h(1) = y (1) = 0
h(n) = h(n 1) (n 2),
2
We can solve h(n) as
1
h(0) = h(1) (2) = 0
2
1
h(1) = h(0) (1) = 0
2
1
h(2) = h(1) (0) = 1
2
1
1
h(3) = h(2) (1) = (1)
2
2
1
12
h(4) = h(3) (2) =
(1)
2
2
.
.
.
h(n) =
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1
2
n2
u(n 2)
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Next, we evaluate the response to the input in Figure 6.19 of the textbook. The input can
be written in the following form:
x(n) = (n + 2) + 2 (n + 1) 2 (n 1) + 2 (n 2)
Then, the output becomes:
y (n) = h(n) x(n)
= h(n) [ (n + 2) + 2 (n + 1) 2 (n 1) + 2 (n 2)]
= h(n) (n + 2) + 2h(n) (n + 1) 2h(n) (n 1) + 2h(n) (n 2)
= h(n + 2) + 2h(n + 1) 2h(n 1) + 2h(n 2)
=
1
2
n
u(n) 2
1
2
n1
u(n 1) + 2
1
2
n3
u(n 3) 2
1
2
n4
u(n 4)
12. (Problem 6.34) Solution:
We rst solve the implies response of the system. The system has to be relaxed to ensure
it is LTI, i.e., y (1) = 0. The impulse response is the solution to the following dierence
equation:
1
h(n) = h(n 1) + (n 1),
2
h(1) = y (1) = 0
Then, we can have
1
h(0) = h(1) + (1) = 0
2
1
h(1) = h(0) + (0) = 1
2
1
1
h(2) = h(1) + (1) =
2
2
12
1
h(3) = h(2) + (2) =
2
2
.
.
.
Hence, we conclude that the impulse response is
h(n) =
1
2
n1
u(n 1)
Next, we evaluate the response to the two inputs respectively.
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(a) The response to the unit step function is
y (n) = u(n) h(n)
+
=
u(k )h(n k )
k=
+
=
h(n k )
k=0
+
=
k=0
n1
=
1
2
k=0
0
1
=2
0
=
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1
2
nk1
u(n k 1)
1
2
nk1
if n 1
if n < 1
1
2
n+1
if n 1
if n < 1
1
2
n+1
u(n 1)
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(b) The response to the one-side exponential sequence is
y (n) =
n
1
4
u(n) h(n)
+
=
k=
+
=
k=0
+
=
k=0
+
=
k=0
+
=
k=0
n1
=
k=0
0
1
=
2
0
=
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1
2
k
1
4
u(k )h(n k )
1
4
k
1
4
k
1
4
k
1
2
n+k1
h(n k )
1
2
nk1
1
2
nk1
u(n k 1)
u(n k 1)
u(n k 1)
n+k1
1
2
if n 1
if n < 1
n2
1
4
n1
if n 1
if n < 1
n2
1
4
n1
u(n 1)
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Figure 3: Graphical method to evaluate convolution in Problem 10 (Problem 6.13 in textbook).
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