Signals and Systems - Electrical Engineering

7.2 Uniform Sampling 435

0 1 2 3 4 5 6 78

− 1

0

1

t (sec)

0 0.5 1 1.5 2 2.5 3 3.5

−1.5

− 1

−0.5

0

0.5

1

1.5

0

0.0050.010.0150.020.0250.030.0350.040.0450.05

− 1

−0.5

0

0.5

1

t (sec)

x(

t),

y(

t)

− 10 − 8 − 6 − 4 − 2 0 2 4 6 8

0

0.2

0.4

|X

(Ω

)|

−0.4−0.3 −0.2−0.1 0 0.1 0.2 0.3

0

0.2

0.4

f (KHz)

|Y

(Ω

)|

× 10 −^3

× 10 −^3

× 10 −^3

x(

t),

xr

(t

)

(a) (b)

Analog signal

Sampled signal

FIGURE 7.8
Aliasing: sampling simulation ofx(t)=cos( 2000 πt)usingfs= 800 samples/sec. (a) Plots display the original
signalx(t)and the sampled signaly(t)(it looks like a lower-frequency signal being sampled). The sprectra ofx(t)
andy(t)are shown below (|Y()|is periodic and displays a lower frequency than|X()|). (b) Sinc interpolation
for three samples and the whole signal. The reconstructed signalxr(t)is a sinusoid of period0.5× 10 −^2 or
200 -Hz frequency due to aliasing.

The following function implements the sampling and computes the Fourier transform of the analog
signal and of the sampled signal using the fast Fourier transform. It gives the range of frequencies for
each of the spectra.

function [y,y1,X,fx,Y,fy] = sampling(x,L,fs)

%

% Sampling

% x analog signal

% L length of simulated x

% fs sampling rate

% y sampled signal

% X,Y magnitude spectra of x,y

% fx,fy frequency ranges for X,Y

%

fsim = 20000; % analog signal sampling frequency

% sampling with rate fsim/fs

delta = fsim/fs;

y1 = zeros(1,L);