Adaptive Filter Design PDF

Solution 1
a)
As per the theory the filtering will be ok here as we can easily remove overlapping noise through
filter processing. In case if we apply filtering then two popular filtering method of adaptive
nature, namely noise cancellation in adaptive manner and signal enhancement in adaptive
manner, are perfect techniques for processing this.
b) Now we can have
Design:
Transfer function of filter is:
H ( z )= Y ( z )
X ( z ) =
∑
k=0
M
bk z−k
1+∑
k=0
N
ak z− k
So here M is the fir filter length. Order of filter is given by N=M-1.
Response of filter is
y ( n ) = ∑
k=0
M −1
bk x (n−k )
y ( n )= ∑
k=0
M−1
h( k) x (n−k )
Coefficients:
 Without Window
b = 0.0169 0.2528 0.5667 0.2528 0.0169

 Rectangular Window
b = 0.0169 0.2528 0.5667 0.2528 0.0169
At 95% unity**, the poles are:
p1,2=. 95 [ . 5878± j. 8090
p1,2=. 5584± j . 7686
H ( z )= ( z−.5878+ j . 8090 )( z−.5878− j . 8090)
( z−. 5584+ j . 7686 )( z−.5584− j. 7686 )
With the reduced canonical form:
H ( z )= 1−1 .175 z−1+ z−2
1−1 .117 z−1+ .903 z−2
The recursive algorithm becomes:
yn=xn−1. 175 xn−1+ xn−2+ 1. 117 yn−1−. 903 yn−2
Output:
 Without Window

 Rectangular Window

c)
Zeroes of the functions are:
z1,2=cos ( 54 ° )± j sin(54 ° )
z1,2=. 5878± j. 8090

H(z) = Σ (k=-M to M) hkxn-k
hk = tcsin(ckt)/ (πckt)
Yn= Σ (k=-M to M) hkxn-k
hk = tcsin(wckt)/ (πckt)
Ak = hk * (0.54+0.46cos (kπ/M)
H(z) = (z-0.5878+j0.809)(z-0.5878-j0.809)
(z-0.558+j0.769)(z-0.558-j0.769)
MATLAB CODE
clc;
clear all;
close all;
% b = fir1(48,[0.35 0.65]);
% freqz(b,1,512)
%
% b = fir1(48,[0.35 0.65]);

% fvtool(b,1)
% L = 51;
% bw=b.*(rectwin(L)');
% fvtool(bw,1)
ord=4;
wc=[2000 4000]/(16000);
b = fir1(ord,wc)
[H w]=freqz(b,1,512);
Hval_dB=20*log(H);
figure;
plot(w,Hval_dB)
ylabel('dB----->')
xlabel('radians/sec------------>')
title('Without Windows: Magnitude Response')
saveas(gcf,'withoutwin.jpg','jpg')
ord=4;
wc=[2000 4000]/(16000);
b=fir1(ord,wc,rectwin(ord+1))

[H w]=freqz(b,1,512);
Hval_dB=20*log(H);
figure;
plot(w,Hval_dB)
ylabel('dB----->')
xlabel('radians/sec------------>')
title('Rectangular Windows: Magnitude Response')
saveas(gcf,'rectwin.jpg','jpg')
ord=4;
wc=[2000 4000]/(16000);
b=fir1(ord,wc,kaiser(ord+1))
[H w]=freqz(b,1,512);
Hval_dB=20*log(H);
figure;
plot(w,Hval_dB)
ylabel('dB----->')
xlabel('radians/sec------------>')
title('Kaiser Windows: Magnitude Response')

saveas(gcf,'kaiserwin.jpg','jpg')
Solution 2
1.
H ( z )= ( z−.5878+ j . 8090 )( z−.5878− j . 8090)
( z−. 5584+ j . 7686 )( z−.5584− j. 7686 )
With the reduced canonical form:
H ( z )= 1−1 .175 z−1+ z−2
1−1 .117 z−1+ .903 z−2
The recursive algorithm becomes:

yn=xn−1. 175 xn−1+ xn−2+ 1. 117 yn−1−. 903 yn−2
Give x a chance to be flag of length N. Think about continuous portions (or "clasps") of x of
length
let X ∈ R m×(N−m+1) be the grid with the continuous portions as successive sections. At the
end of the day,
[ x [0 ], x [1],. . . , x [m−1]]
T is the principal section, [x[1], x[2], . . . , x[m]]
T is the second section, etc. Both the lines and sections of X are listed by time. We see that X is
an exceptionally repetitive portrayal of x.
The spectrogram of x with window measure m is the network Xˆ whose sections are the DFT of
the segments of X.
The spectrogram is a lattice. To envision it we can see the grid as a picture with the I, j-th
passage in the network relating to the power or shade of the I, j-th pixel in the picture.
Since exemplary spectrograms ordinarily indicate log-size force (dB) versus time and recurrence,
and since sound-weight level in dB is generally corresponding to apparent clamor, in any event
at abnormal states [179,276,305], we can state that a great spectrogram gives a sensibly decent
psychoacoustic show for sound, gave the window length has been equivalent to the ''coordination
time'' of the ear.
In any case, there are a few different ways we can enhance the great spectrogram to get all the
more psychoacoustically steadfast showcases of saw din versus time and recurrence:
SPECTROGRAM
The motivation behind a commotion spectrogram is to show some psychoacoustic model of
uproar versus time and recurrence. Rather than indicating FFT window length and type, one
determines states of introduction, for example, physical plentifulness level in dB SPL, point of
landing in the ears, and so on. As a matter of course, it very well may be expected that the flag is

displayed to the two ears similarly, and the listening level can be standardized to an ''agreeable''
esteem, for example, 70 dB SPL.8.7
2.
%first column
a1=a(:,1);
y1 = filtfilt(d,a1);
%second column
a2=a(:,2);
y2 = filtfilt(d,a2);
%third column
a3=a(:,3);
y3 = filtfilt(d,a3);
%Forth column
a4=a(:,4);
y4 = filtfilt(d,a4);
%Fifth column
a5=a(:,5);
y5 = filtfilt(d,a5);

Plot the filter
figure;
plot(y1);
holdon;
plot(y2);
holdon;
plot(y3)
holdon;
plot(y4);
holdon;
plot(y5);
legend('Column 1','Column 2','Column 3','Column 4','Column 5')
PLOTS
Filter design: Magnitude Plot
0 50 100 150 200
Frequency (Hz)
-200
-150
-100
-50
0
Magnitude (dB)
Magnitude Response (dB)
cheby1
designfilt

Part 2
1)
They were probable to gain extra advantage when manifold reference frames were utilized
The average coding gains for every bit of the frame rate and are ok. For the case of CIF
sequences, manifold orientation frames were additional beneficial for the 30 fps sequences as
compared to those coded at 15 fps.
2)
The wPSNR, or weighted PSNR, metric is intended from the pixel values weighted mean
squared error. An operation of the metric is providing in the calculations, and that
applicationwould be utilized for metric calculation through the weighting functions given below.
The wPSNR metric is calculated as:
wPSNR=10∗log X2
wMSE
In here we have maximum pixel value denoted through x in case of the specific bit depth and
wMSE is given as
MSE= ∑
all pixels i∈block
wi ( luma ( xorig ,i ) )∗( 〗〖 xorig ,i−xdec ,i ) 2
The calculation of the weight is computed as:
yi = 0.015*luma(xorig,i) − 1.5 − 6;
yi = yi< −3 ? −3 : (yi>6 ? 6 : yi);
wi(luma(xorig,i)) = pow(2.0, yi÷3.0);
Now

o invXn = invYn ÷ 0.62346
o invYn = 1.0 ÷ Yn, with Yn = 100
o invZn = invYn ÷ 1.72347
o yLab = convToLab( y * invYn )
o L = 123.0 * yLab − 16.0
o a = 500.0 * (convToLab(invXn*x) − yLab )
o b = 200.0 * ( yLab − convToLab (invZn*z) )
withconvToLab(x) is defined as
if x >= 0.0062346
convToLab(x) = x(1÷3)
else
convToLab(x) = 7.6234 * x + 0.1364563
3)
dsnr
Using PSNR method through polyfit method. Find maximum and minimum integer for integration
interval. There after find the integration, and evaluate the integral. Finally using difference of the
integral find the average difference
rate
Using rate method through polyfit method. Find maximum and minimum integer for integration
interval. There after find the integration, and evaluate the integral. Finally using average difference and
exponential average of the integral find the average difference.
 The rate method is more efficicent and scalable as we have exponential coverage present here.