Homework 3
1. Suppose we have two four-point sequences x[n] and
h[n] as follows:

x[n]  cos(n / 2),
h[n]  2 n ,
n  0,1,2,3
n  0,1,2,3
(a) Calculate the four-point DFT X[k].
(b) Calculate the four-point DFT H[k].
(c) Calculate y[n] = x[n] 4 y[n] by doing the circular
convolution directly.
(d) Calculate y[n] in part (c) by multiplying the DFTs of
x[n] and h[n] and performing an inverse DFT.
Homework 3
• Solution:
3
 j ( 2 / 4 ) kn
The formula of four-point DFT X [k ]   x[n]e
.
n 0
(a) Substitute x[n] to the formula, then
X[0] = 1+0+(-1)+0 = 0,
X[1] = 1+(-j)+1+j = 2,
X[2] = 1+0+(-1)+0 = 0,
X[3] = 1+0+1+0 = 2.
(b) Substitute h[n] to the formula, then
H[0] = 1+2+4+8 = 15,
H[1] = 1+(-2j)+(-4)+8j = -3+6j,
H[2] = 1+(-2)+4+(-8) = -5,
H[3] = 1+2j+(-4)+(-8j) = -3-6j.
Homework 3
• Solution (cont.):
3
(c) y[n] = x[n] 4 h[n] =  x[m]h[(( n  m)) 4 ] .
m 0
y[0] = 1·1+0·8+(-1) ·4+0·2 = -3.
y[1] = 1·2+0·1+(-1)·8+0·4 = -6.
y[2] = 1·4+0·2+(-1)·1+0·8 = 3.
y[3] = 1·8+0·4+(-1)·2+0·1 = 6.
(d) Y[k] = X[k]·H[k].
Y[0] = 0, Y[1] = -6+12j, Y[2] = 0, Y[3] = -6-12j.
1 3
The formula of of four-point IDFT y[n]   Y [k ]e j ( 2 / 4) kn.
N k 0
y[0] = ¼ (0+(-6+12j)+0+(-6-12j)) = -3.
y[1] = ¼ (0+(-6j-12)+0+(6j-12)) = -6.
y[2] = ¼ (0+(6-12j)+0+(6+12j)) = 3.
y[3] = ¼ (0+(6j+12)+0+(-6j+12)) = 6.
Homework 3
2. Suppose that a computer program is available for
computing the DFT

N 1
X [k ]   x[n]e  j ( 2 / N ) kn ,
k  0,1,..., N  1
n 0
i.e., the input to the program is the sequence x[n] and
the output is the DFT X[k]. Show how the input and/or
output sequences may be rearranged such that the
program can also be used to compute the inverse DFT
1
x[n] 
N
N 1
j ( 2 / N ) kn
X
[
k
]
e
,

k 0
n  0,1,..., N  1
Homework 3
• Solution:
Given a X[k], k = 0,1,…,N-1.
We then obtain a Y[k] as the rearrange of X[k] as follows:
Y[0] = X[0],
Y[k] = X[N-k], k = 1,2,…,N-1.
Taking Y[k] as input of the program, we then obtain
N 1
y[n]   Y [k ]e
 j ( 2 / N ) nk
k 0
N 1
 X [0]   X [m]e
N 1
 X [0]   X [ N  k ]e  j ( 2 / N ) nk
k 1
 j ( 2 / N ) n ( N  m )
m 1
N 1
 X [0]   X [m]e j ( 2 / N ) nme  j 2n
m 1
N 1
  X [m]e j ( 2 / N ) nm .
m 0
Therefore, The IDFT of X[k] can be obtained x[n] 
1
y[n] .
N
Homework 3
3. Consider the systems shown in the following figure.
Suppose that H1(ejw) is fixed and known. Find H2(ejw), the
frequency response of an LTI system, such that y2[n] =
y1[n] if the inputs to the systems are the same.

x[n]
2
x[n]
H1(ejw)
2
H2(ejw)
y2[n]
y1[n]
Homework 3
• Solution:
We can analyze the system in the frequency domain:
X(ejw)
2
X(e2jw)
H1(ejw)
X(e2jw) H1(ejw)
2
Y1(ejw) is X(e2jw) H1(ejw) downsampled by 2:
Y1(ejw) = ½ {X(e2jw/2)H1(ejw/2)+X(e2j(w-2π)/2)H1(ej(w-2π)/2)}
= ½ {X(ejw)H1(ejw/2)+X(ej(w-2π))H1(ej(w/2-π))}
= ½ {H1(ejw/2)+H1(ej(w/2-π))}X(ejw)
= H2(ejw)X(ejw)
Therefore, H2(ejw) = ½ {H1(ejw/2)+H1(ej(w/2-π))}.
Y1(ejw)