DTFT, DFT, ZT, and FFT
495
f C (3) = 1(3) + 0(2) + 2(1) + 4(2) + 2(5)
f C (3) = 4 + 0 + 2 + 8 + 10 = 24
f C (4) = f 1 (0)f 2 (4) + f 1 (1)f 2 (3) + f 1 (2)f 2 (2) + f 1 (3)f 2 (1) + f 1 (4)f 2 (0)
f C (4) = 1(5) + 0(3) + 2(2) + 4(1) + 2(2)
f C (4) = 5 + 0 + 4 + 4 + 4 = 17
Then the circular convolution of the sequence f 1 (n) = [1 0 2 4 2] with f 2 (n) = [2 1 2 3 5]
is given by f C (n) = f 1 (n) ° f 2 (n), which results in f C (n) = [1 0 2 4 2] ° [2 1 2 3 5] =
[18 27 32 24 17].
R.5.110 Let us now illustrate the process of evaluating DFT by hand for the sequence
f(n) = {1 0 2 4 2}.
ANALYTICAL Solution
Recall that
F k
f n e
n
N
j nk N
( )
( )
ϭ
ϭ
Ϫ
Ϫ
ր
0
1
2
∑
then
F(k) = f(0) + f(1)e
−j2k/5
+ f(2)e
−j22k/5
+ f(3)e
−j23k/5
+ f(4)e
−j24k/5
F(k) = 1 + 2e
−j4k/5
+ 4e
−j23k/5
+ 2e
−j24k/5
F(0) = 1 + 2 + 4 + 2 = 9
F(1) = 1 + 2e
−j42/5
+ 4e
−j6/5
+ 2e
−j8/5
= −3.2361 + 3.0777i
F(2) = 1 + 2e
−j4(2)/5
+ 4e
−j6(2)/5
+ 2e
−j8(2)/5
= 1.2361 + 0.7265i
F(3) = 1 + 2e
−j4(3)/5
+ 4e
−j6(3)/5
+ 2e
−j8(3)/5
= 1.2361 − 0.7265i
F(4) = 1 + 2e
−j4(4)/5
+ 4e
−j6(4)/5
+ 2e
−j8(4)/5
= −3.2361 + 3.0777i
R.5.111 The coeffi cients of the DFT of the sequence f(n) = [1 0 2 4 2] just evaluated in R.5.110
are verifi ed by reevaluating the DFT equations by using MATLAB as well as fft
command, indicated as follows:
MATLAB Solution
>> fn = [1 0 2 4 2];
% time sequence
>> DFT _ Fk = (fft(fn))’
% DFT using fft
DFT _ Fk =
9.0000
-3.2361 - 3.0777i
1.2361 + 0.7265i
1.2361 - 0.7265i
-3.2361 + 3.0777i
CRC_47760_CH005.indd 495
CRC_47760_CH005.indd 495
7/23/2008 1:47:27 PM
7/23/2008 1:47:27 PM