Digital Signal Processing case study on Discrete Hilbert & Schur algorithm, and MATLAB implementation of IIR filter.

1. DSP CASE STUDY
Submitted by :-
Vaibhav Tayal
07411502816
ECE-II, 3rd
Year
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Q2. Summarize the two algorithms :-
(i) Discrete Hilbert transform
(ii) Schur algorithm
A2. (i) discrete Hilbert transform:-
Signals can be classified into two classes: 1. Analytical signals(for instance
x(t)=Asinwt), 2. Experimental signals (measured signals).An experimental
signal represents a signal observed during a limited interval of time. It is a
sample of the original signal, which characterized a physical process of interest.
The practical limitations of the systems used to analyze analogical sigals impose
that the experimental analogical signals had a limited frequency band.
If the original signal does not have a limited frequency band , then low pass
filtering needs to be applied.
In the case of Hilbert transform, it’s a known fact that the signal x(t) needs to be
causal (that is, x(t)=0, for t<0).
The sampled signal x[n] is in this case a non-periodic sequence, real and causal.
In such a case, we can talk of the discrete Hilbert transform applied to the
sequence x[n].
When x(t) is a periodic signal, x[n] is a periodic sequence and we cannot talk of
causality (the periodic term implies the sequence extension from -∞ to +∞). A
calculus algorithm for the discrete Hilbert transform in this case imposes the
condition that the discrete fourier transform of the complex analytical sequence
to be equal to zero in the interval of negative frequencies.

2. DISCRETE HILBERT TRANSFORM. (CALCULUS ALGORITHM).
Having a signal x(t) defined on the time interval [0,tN], using a sampling period
Te, we obtain the discrete signal x[n] :
x[n]=x(nTe), n∈0,N−1 where Te = tN/N.
the sampling frequency fe is chosen so that the frequency fe/2 is greater or equal
to the least significant frequency from the spectrum of x(t). we consider the
discrete frequency step , f0=fe/2 , o=2/N*fe respectiviely.
The discrete fourier transform is :
TFD{x[n]} = X{k} = ∑x[n]e-jnk N , k∈0,N−1
And the inverse discrete fourier transform DFT-1
is:
TFD-1
{X[k]}= x[n] ∑ 
The sample of the spectral density corresponding to frequency k0 is
determined with the relation:
X(jk0) = Te X[k]
On the other hand :
X[k]*
= X[N-k] = X[-k]. which shows that the sample X[N-k] = X[-k] has a
correspondent sample of the spectral density, with the negative frequency X(-
k0).
Similarly, the discrete Hilbert transform is defined :
H{x[n]}= x[n]=TFD-1
{X[k]} .

3. (ii). Shcur- cohn algorithm :-
To simplify the description of this algorithm, we first take up the analysis of the
stability domain of a 2nd-order transfer function. This particular case leads to a
simplification of the stability criteria imposed on the denominator of the transfer
function. Unfortunately, it cannot be applied to transfer functions of an order
greater than 2. We also present the Schur-Cohn stability algorithm based on the
transfer function of an all-pass filter, allowing us to establish equivalence
relation between the Schur coefficients and the reflection coefficients.
Let there be a second-order transfer function defined as follows:
H2(z) =
The poles of H2 (z) = are equal to:
P1 = √ and P1 = √
And its xeros are defined as follows :
P1 = √ and P1 = √
Depending on the values taken by 1a and 2a , the poles can be real or complex.
For example, when 2 21 4aa < , the poles are complex conjugates of each other.
Otherwise, they are real. To ensure stability, the poles of the transfer function
must be located within the unit circle in the z-plane, i.e.:
1 1 < p and 1 2 < p
this constrain implies that the following two inequalities are satisfied:
a2= p1p2<p1p2<1
and
a1<1+a2
These Relations make it possible to define a triangle in the ( 1a , 2a ) plane
where the filter is stable and which is called the stability triangle. This triangle

4. depicted in Figure F.1. is a simple tool for testing the stability as it is based on
the values of the filter’s coefficients.
therefore , the schur-cohn algorithm is written as follows:
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Q3. Design a digital type-1 chebyshev low pass filter operating at a
sampling rate of 100kHz with a passband edge frequency of 10kHz,
a passband ripple 0.4 db & minimum stopband attenuation of 50db at
50kHz using impulse invariance method and bilinear transformation
method. Determine the order of the analog filter prototype. Plot the
phase & magnitude response. Compare the performance of the filter.

5. MATLAB Implementation:
clc;
clear all;
close all;
rp = 0.4
rs = 50
wp = 10000
ws = 30000
fs = 100000
w1 = 2*wp/ws;
w2 = 2*ws/fs;
[n,wn] =
cheb1ord(w1,w2,rp,rs,'s');
n
[b,a] = cheby1(n,rp,wn,'low','s')
[z,p,k] = cheb1ap(n,rp)
[p,q,r,s] = zp2ss(z,p,k)
[bs,as] = ss2tf(p,q,r,s)
[hs,oms] = freqs(bs,as,fs)
subplot(3,2,1)
plot(oms,abs(hs))
title('magnitude response of
analog filter')
ylabel('Gain-->')
xlabel('frequency-->')
subplot(3,2,2)
plot(oms,angle(hs))
title('Phase response of analog
filter')
ylabel('Phase-->')
xlabel('frequency-->')
[bz,az] = impinvar(b,a,fs);
w = 0:pi/512:pi-(pi/512);
[h,om] = freqz(bz,az,512,fs)
m = (abs(h));
an = angle(h)
subplot(3,2,3)
plot(om,m)
title('using impulse invariance')
ylabel('Gain-->')
xlabel('frequency-->')
subplot(3,2,4)
plot(om,an)
title('using impulse invariance')
ylabel('Phase-->')
xlabel('frequency-->')
[bx,ax] = bilinear(b,a,fs);
[hx,omx] = freqz(bx,ax,512,fs)
mx = abs(h);
anx = angle(h)
subplot(3,2,5)
plot(omx,mx)
title('using bilinear
transformation')
ylabel('Gain-->')
xlabel('frequency-->')
subplot(3,2,6)
plot(omx,anx)
title('using bilinear
transformation')
ylabel('Phase-->')
xlabel('frequency-->')

6. The order of the analog filter is 17.
Comparision of Digital & Analog filter:
Response of digital filter using impulse invariance method &bilinear
transformation is better than the response of the analog filter prototype.
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