false

Textbook Notes
(369,018)

United States
(206,160)

University of Maryland
(1,931)

ENEE 322
(3)

Steven Marcus
(3)

Chapter

Unlock Document

Electrical & Computer Engineering

ENEE 322

Steven Marcus

Spring

Description

Homework #3
2.3
Solution 1:
1 n▯2
Since the output is the convolution between the inp2t xu[n▯2]( )
and the impulse response h[n] = u[n + 2].
y[n] = x[n] ▯ h[n]
= ( )n▯2u[n ▯ 2] ▯ u[n + 2]
2
X1
1 k▯2
= (2) u[k ▯ 2]u[n ▯ k + 2]
k=▯1
By replacing k with m + 2 in the summation, we obtain
1
X 1 m
y[n] = ( ) u[m]u[n ▯ m]
m=▯1 2
Note that u[m] and u[n ▯ m] implies that m ▯ 0 and m ▯ n respectively.
n
X 1m
y[n] = ( )
m=0 2
= 2[1 ▯ ( )+1]u[n]
2
Solution 2:
Note that
1 n
x[n] = (2) u[n] ▯ ▯[n ▯ 2]
h[n] = u[n] ▯ ▯[n + 2]
Thus, the output y[n] is
y[n] = x[n] ▯ h[n]
1
= ( ) u[n] ▯ ▯[n ▯ 2] ▯ u[n] ▯ ▯[n + 2]
2
1 n
= ( ) u[n] ▯ ▯[n ▯ 2] ▯ ▯[n + 2] ▯ u[n]
2
1 n
= ( ) u[n] ▯ u[n]
2
Xn 1
= ( )k
k=0 2
1 n+1
= 2[1 ▯ 2) ]u[n]
1 8
x[n]*h[n]
6
4
2
0
0 5 10 15 20 25
Figure 1: Output y[n] (Problem 2.4)
2.4
Solution 1
▯
1; 3 ▯ n ▯ 8
x[n] = 0; otherwise
= u[n ▯ 3] ▯ u[n ▯ 9]
▯
1; 4 ▯ n ▯ 15
h[n] =
0; otherwise
= u[n ▯ 4] ▯ n[n ▯ 16]
Thus, the output is
y[n] = x[n] ▯ h[n]
= (u[n ▯ 3] ▯ u[n ▯ 9]) ▯ (u[n ▯ 4] ▯ u[n ▯ 16])
= u[n ▯ 3] ▯ u[n ▯ 4] ▯ u[n ▯ 3] ▯ u[n ▯ 16]
▯u[n ▯ 9] ▯ u[n ▯ 4] + u[n ▯ 9] ▯ u[n ▯ 16]
= u[n] ▯ u[n] ▯ (▯[n ▯ 7] ▯ ▯[n ▯ 13] ▯ ▯[n ▯ 19] + ▯[n ▯ 25])
Xn Xn Xn Xn
= 1 ▯ 1 ▯ 1 + 1
k=7 k=13 k=19 k=25
8
< n ▯ 6; 7 ▯ n < 13
6; 13 ▯ n < 19
=
: 24 ▯ n;19 ▯ n < 25
0; otherwise
The output y[n] is shown in Figure 1.
Solution 2:
We can directly calculate the convolution of x[n] and h[n], which is shown
in Figure 2.
2 1. Filp x[m], then we obtain x[▯m].
2. Shift x[▯m] by n, then we obtain x[n▯m]. When n increases to 7, x[n▯m]
and h[n] overlap. And the overlapping area continues to increase when n
increases.
3. When n 2 [12;18], the overlapping area does not change, which means the
convolution value is constant at [12;18]:
4. When n increases to 19, the overlapping area begins to decrease.
5. There is no overlap when n > 25.
Therefore, the output
8
> n ▯ 6; 7 ▯ n < 13
< 6; 13 ▯ n < 19
y[n] =
: 24 ▯ n;19 ▯ n < 25
0; otherwise
2.7
(a)
Given that
x[n] = ▯[n ▯ 1]
We see that
X
y[n]= x[k]g[n ▯ 2k]
k=▯1
X
= ▯[k ▯ 1]g[n ▯ 2k]
k=▯1
= g[n ▯ 2]
= u[n ▯ 2] ▯ u[n ▯ 6]
(b)
Given that
x[n] = ▯[n ▯ 2]
We see that
1
X
y[n]= x[k]g[n ▯ 2k]
k=▯1
1
X
= ▯[k ▯ 2]g[n ▯ 2k]
k=▯1
= g[n ▯ 4]
= u[n ▯ 4] ▯ u[n ▯ 8]
3 2 2
x[m] x[7m]
1.8 1.8
h[m] h[m]
1.6 1.6
1.4 1.4
1.2
1.2
1 1
0.8 0.8
0.6 0.6
0.4 0.4
0.2 0.2
0 0
10 5 0 5 10 15 20 10 5 0 5 10 15 20
Step 1 Step 2
2 2
x[13m] 1.8 x[19m]
1.8 h[m] h[m]
1.6 1.6
1.4 1.4
1.2
1.2
1 1
0.8 0.8
0.6
0.6
0.4 0.4
0.2 0.2
0 0
10 5 0 5 10 15 20 10 5 0 5 10 15 20
Step 3 Step 4
2
1.8 x[24m]
h[m]
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
10 5 0 5 10 15 20
Step 5
Figure 2: 2.8
4 (c)
The input to the system in part (b) is the same as the input in part (a)
shifted by 1 to the right. If S is time invariant then the system output obtained
in part (b) has to be the same as the output in part (a) shifted by 1 to the right.
Clearly, this is not the case. Therefore, the system is not LTI.
(d)
If x[n] = u[n], then
1
X
y[n] = x[k]g[n ▯ 2k]
k=▯1
1
X
= g[n ▯ 2k]

More
Less
Related notes for ENEE 322

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.