PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

DTFT, DFT, ZT, and FFT 495

fC(3) = 1(3) + 0(2) + 2(1) + 4(2) + 2(5)

fC(3) = 4 + 0 + 2 + 8 + 10 = 24

fC(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)

fC(4) = 1(5) + 0(3) + 2(2) + 4(1) + 2(2)

fC(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 fC(n) = f 1 (n) (^) ° f 2 (n), which results in fC(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
Fk f n e
n
N
()
() jnkN




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 T he 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