1. Homework 3
1. Suppose we have two four-point sequences x[n] and h[n] as follows:
(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.

2. Homework 3 Solution: The formula of four-point DFT .
(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] = = 2.
(b) Substitute h[n] to the formula, then
H[0] = = 15,
H[1] = 1+(-2j)+(-4)+8j = -3+6j,
H[2] = 1+(-2)+4+(-8) = -5,
H[3] = 1+2j+(-4)+(-8j) = -3-6j.

3. Homework 3 Solution (cont.): (c) y[n] = x[n] 4 h[n] = .
(d) Y[k] = X[k]·H[k].
Y[0] = 0, Y[1] = -6+12j, Y[2] = 0, Y[3] = -6-12j.
The formula of of four-point IDFT
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.

4. Homework 3
2. Suppose that a computer program is available for computing the DFT
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

5. 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
Therefore, The IDFT of X[k] can be obtained

6. 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.
H1(ejw) 2 2
x[n]
y1[n]
H2(ejw)
x[n]
y2[n]

7. Homework 3 X(e2jw) X(e2jw) H1(ejw) H1(ejw) 2 2 X(ejw) Y1(ejw)
Solution:
We can analyze the system in the frequency domain:
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))}.
X(e2jw)
X(e2jw) H1(ejw)
H1(ejw) 2 2
X(ejw)
Y1(ejw)