PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

538 Practical MATLAB® Applications for Engineers

20 30

time index n

100 40 50 60 70

0

0.5

1

1.5

− 10

− 10 100 20 30 40 50 60 70

− 4

− 2

0

Amplitude

Amplitude

2

4

6

× 10 −^15

error2 = abs[abs[corr[f1(n), f2(n)]]-abs[conv[f1(-n), f2(n)]]]

error1 = corr[f1(n), f2(n)]-conv[f1(n), f2(-n)]

FIGURE 5.51
Error plots of part e of Example 5.11.

Example 5.12

The objective of this example is to explore the relation between the FT of a continuous

time function and its discrete version.

Let f(t) = 3e−5tu(t).

Create the script fi le FT_DFT that returns the following plots:

a. f(t) versus t, over the range 0 ≤ t ≤ 1 (Figure 5.52)

b. Magnitude and phase of [DFT[f(n)]] versus w (Figure 5.53)

c. Magnitude and phase of [FT[f(t)]] versus w (Figure 5.54)

d. error_mag = magnitude of [FT[f(t)]] − magnitude of [DFT[f(n)]] versus w and

error_phase = phase of [FT[f(t)]] − phase of [DFT[f(n)]] versus w (Figure 5.55)

MATLAB Solution

% Script file: FT _ DFT

N = 526;t = linspace(0,5,N);

fn =3*exp(-5*t); fs = 8000;Ts = 1/fs;

ws = 2*pi/Ts;

F = fft(fn);

Fdc = fftshift(F)*Ts;

W = ws*(-N/2:(N/2)-1)/N;

figure(1)

plot (t(1:100),fn(1:100));title (‘[f(t)=3*exp(-5*t)] vs t’)

ylabel (‘Amplitude’); xlabel (‘time’);axis([0 1 0 3.5]);

figure(2) % discrete frequency analysis

subplot (2,1,1)

plot (W,abs(Fdc));title(‘abs[DFT[f(n)]] vs w’);ylabel(‘Magnitude’);