DSP Lab Experiments - Digital Signal Processing: An Introduction with MATLAB and Applications - page 330

Task 4

Now listen to a single-tone test audio signal saved as ''stone'' (e.g., 440 Hz sound
from http://www.mediacollege.com/audio/tone/download/ ). Find its sampling
frequency fs (it is 44.1k Hz). Read the signal as x. Add Gaussian noise n (of
power =-20 dB) to the signal and write the result as s = x+n in your directory
as an audio signal using wavwrite(s,fs,'stonen.wav') . Listen to the
corrupted signal and compare with the original one. Plot the time signals x and s
with their spectra versus the normalized frequency ranges fn = f/fs and
fN = f/(fs/2) . Now design a digital filter (LPF Butterworth) of order 10 and
normalized cut-off frequency wcr=wc/(fs/2) , where wcr is the cut-off fre-
quency which should be chosen carefully to reduce noise. Plot the frequency
response of the digital filter using freqz. Then filter the discrete signal s to get the
signal y . Plot y and its spectrum Y , then compare with the original and noisy
signals.

Experiment # 5: A Sinusoidal Digital Oscillator

Introduction

A digital oscillator is a system that generates an output waveform (like a sinusoid)
without a need for an input signal, except for a D.C. supply and perhaps a trigger
(like a delta function) at the starting time. The theory of operation is based on the
fact that a digital system with poles on the circumference of the unit circle in the z-
plane is neither stable nor divergent, but oscillatory (or, marginally stable).
To design a digital sinusoidal oscillator, we need a transfer function in the z-
domain whose impulse response is a sinusoid. Using Tables we can reach at the
following z-transform pair:

sin ð b Þ z 2
z 2 2cos ð b Þ z þ 1

h ð n Þ¼ sin ½ð n þ 1 Þ b Z H ð z Þ¼ ð 1 Þ

Since n represents the time count, b would represent the normalized radian
frequency X o ¼ x o T s ,

hence the frequency of oscillation is f o ¼ x o = 2p ¼

ð b = T s Þ= 2p ¼ð b = 2p Þ f s Hz ; and we should have j b j \p.
Note that the two poles of this system are the roots of z 2 - 2cos (b)z +1= 0,

which are given by:
p 1 ; 2 ¼ cos ð b Þ

p

¼ cos ð b Þ jsin ð b Þ¼ e jb

cos 2 ð b Þ 1 ð 2 Þ

Hence, the poles are exactly on the circumference of the unit circle, and the system
is neither stable nor unstable (oscillatory). Figure D.6 shows the implementation