computing assignment 1(pramod kumar surana)

7/30/2019 Computing Assignment 1(Pramod Kumar Surana)

1/39

Computing Assignment 1 (EE 679)

Name- Pramod Kumar Surana

Roll No.-123070005

1. Given the following specification for a single-formant resonator, obtain the transfer

function of the filter H(z). Plot its magnitude response (dB magnitude versus frequency)

and impulse response.

F1 (formant) = 1 kHz

B1(bandwidth) = 200 Hz

Fs (sampling freq) = 16 kHz

Ans.- The general transfer function of a single formant resonator is given by

()

( )( )

Where

r is related to bandwidth B by B 2(1-r) when r 1. is resonance frequency. B and

should be in radians.

Here Fs= 16 kHz so =1000*2*pi/16000 =pi/8;

B=200*2*pi/16000 =pi/40;

r=1-pi/80 = (80-pi)/80;

A can be chosen depending on gain requirements, I have chosen A as 1/13;

Program for calculating frequency and impulse response of the above system is:

r=(80-pi)/80; %%Assign r

w0=pi/8; %%Assign w0

%%Assign two conjugate poles

p1=r*exp(1i*w0);


2/39

p2=r*exp(-1i*w0);

ww=0:5:8000;

H=filt(1/13,[1 -(p1+p2) p1*p2],1/16000); %%realize filter trans. func.

t=0:1/16000:2/100;

h=impulse(H,t); %%calculate impulse response over 0 to 20 ms

plot(t.*1000,h); %%plot impulse response

xlabel('time in ms');

ylabel('amplitude');

Hw=freqresp(H,exp(1i*ww.*(pi/8000))); %%calculate frequency response

n=size(Hw,3);

HH=reshape(Hw,n,1); %%convert freq. response to 2D matrix

figure;

semilogx(ww,mag2db(abs(HH))); %%plot frequency response

xlabel('frequency in Hz');

ylabel('gain in dB');

grid on;

Fig 1- Impulse response of the single formant resonator for Ques. 1

0 2 4 6 8 10 12 14 16 18 20-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

time in ms

amplitude


3/39

Fig 2- Frequency response of the single formant resonator for Ques. 1

2. Excite the above resonator (filter) with a source given by an impulse train of F0 = 100

Hz. Compute the output of the source-filter system over the duration of 0.5 second. Plot the

time domain waveform. Also play it out and comment on the sound quality.

Ans. - The program for generating the impulse train of 100 Hz for .5 second duration and passing

it through the filter designed in question 1 is as follows- (h is the impulse response of the filter

designed in question 1, it is necessary to execute the code in program 1 before this code in orderto pass the pulse train through filter)

fs=16000; %%set sampling frequency

rep=1:1:50; %%no. of periods

D=rep.*1e-2; %%set time period of .1 s (F0=100Hz)

te=0:1/fs:.5; %%create sample points

100

101

102

103

104

-35

-30

-25

-20

-15

-10

-5

0

5

10

frequency in Hz

gainindB


4/39

w=1e-3; %%width of pulse

yp=1/4*(pulstran(te,D,@rectpuls,w)); %%generate pulse train

plot(te,yp); %%plot pulse train

axis([0 .5 -.5 .5]);

xlabel('time in seconds');


ww=conv(h,yp); %%pass the pulse train through filter

te1=0:1/fs:.52;

figure;

plot(te1,ww); %%plot the filter output



wavwrite(yp,fs,'pk.wav'); %%covert the output to sound file

Fig 3- Pulse train of frequency 100 Hz and duration .5 seconds

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

time in seconds

amplitu

de


5/39

Fig 4- Output of filter designed in question 1 when excited by pulse train shown in figure 3.

Fig 5- Enlarged version of filter output waveform

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

time in seconds

amplitude


6/39

Comment-

Output waveform is periodic due to periodicity of the pulse train and like the repetition of

the impulse response. Sound obtained after passing the pulse train through the filter is a bit harsh

and not like natural sounds.

3. Vary the parameters as indicated below and comment on the differences in waveform

and sound quality for the different parameter combinations.

(a) F1 = 300 Hz, B1 = 100 Hz; F1=1200 Hz, B1 = 200 Hz

(b) F0 = 80 Hz; F0 = 150 Hz

Ans- (a) first case F1=300 Hz, B1=100 Hz, Fs=16 kHz, F0=100 Hz

So here we get B1= 100*2*pi/16000= pi/80 radians;

So r=1- pi/(80*2)= (160-pi)/160;

W0=300*2*pi/16000= 3*pi/80 radians;

Code for generating above filter is-

r=(160-pi)/160; %%set value of r

w0=3*pi/80; %%set value of w0

p1=r*exp(1i*w0); %%set poles

p2=r*exp(-1i*w0);

ww=0:5:8000;

H=filt(1/13,[1 -(p1+p2) p1*p2],1/16000); %%realize filter

t=0:1/16000:2/100;

h=impulse(H,t); %%compute impulse response




Hw=freqresp(H,exp(1i*ww.*(pi/8000))); %%compute freq. response

n=size(Hw,3);

HH=reshape(Hw,1,n);

figure;


7/39

semilogx(ww,mag2db(abs(HH))); %%plot freq. response



grid on;

Fig 6- Impulse response of filter with F1=300 Hz and B1=100 Hz

0 2 4 6 8 10 12 14 16 18 20-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

time in ms

am

plitude


8/39

Fig 7- Frequency response of filter with F1=300 Hz and B1=100 Hz

100

101

102

103

104

-40

-30

-20

-10

0

10

20

30

frequency in Hz

gainindB

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-1.5

-1

-0.5

0

0.5

1

1.5

2

time in seconds

amplitude


9/39

Fig 8-Output of filter with F1=300 Hz and B1=100 Hz when excited with 100 Hz pulse train

Comment-

Waveform of output has damping sinusoids with less frequency than the question 1 filter

as the resonance frequency of this filter is only 300 Hz. Sound appears to be somewhat softer

than the sound output of F1= 1 kHz filter.

Second case F1=1200 Hz, B1=200 Hz, Fs=16 kHz, F0=100 Hz

So here we get B1= 200*2*pi/16000= pi/40 radians;

So r=1- pi/(40*2)= (80-pi)/80;

W0=1200*2*pi/16000= 12*pi/80 radians;

Code for generating above filter is-

r=(80-pi)/80; %%set value of r

w0=12*pi/80; %%set value of w0

p1=r*exp(1i*w0); %%set poles

p2=r*exp(-1i*w0);

ww=0:5:8000;

H=filt(1/13,[1 -(p1+p2) p1*p2],1/16000); %%realize filter

t=0:1/16000:2/100;h=impulse(H,t); %%compute impulse response




Hw=freqresp(H,exp(1i*ww.*(pi/8000))); %%compute freq. response

n=size(Hw,3);

HH=reshape(Hw,1,n);

figure;

semilogx(ww,mag2db(abs(HH))); %%plot freq. responsexlabel('frequency in Hz');


grid on;


10/39



0 2 4 6 8 10 12 14 16 18 20-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

time in ms

amplitude

100

101

102

103

104

-40

-35

-30

-25

-20

-15

-10

-5

0

5

10

frequency in Hz

gainindB


11/39

Fig 11- Output of filter with F1=1200 Hz and B1=200 Hz when excited by 100 Hz pulse train

Code for exciting the above filter with 100 Hz pulse train is-fs=16000; %%set sampling frequency


D=rep.*1e-2; %%set time period of .1 s (F0=100Hz)





axis([0 .5 -.5 .5]);

xlabel('time in seconds');ylabel('amplitude');

ww=conv(h,yp); %%pass the pulse train of .5 second

duration through filter

te1=0:1/fs:.05;

figure;

plot(te1,ww(1:801)); %%plot the filter output

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


12/39




Comment-

Output has more variations due to high resonance frequency. Slightly more sharper sound

than 1 kHz filter, difficult to distinguish perceptually from the 1 kHz filters sound.

(b) F0 = 80 Hz; F0 = 150 Hz

First case F0 = 80 Hz

The code for generating 80 Hz impulse train and passing through the filter in question 1 is as

follows (code for questions 1 filter should be executed first)-



D=rep.*1.25e-2; %%set time period of .1 s (F0=100Hz)





axis([0 .5 -.5 .5]);




te1=0:1/fs:.05;

figure;

plot(te1,ww(1:801)); %%plot the filter output





13/39

Fig 12- Pulse train with frequency of 80 Hz

Fig 13- Output of filter realized in question 1 when excited with 80 Hz pulse train

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

time in seconds

amplitude

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


14/39

Fig 14- Enlarged view of output

Comment-

Output is periodic but period is more as compared to the output due to 100 Hz pulse train.

Sound has less continuity when heard (something like trrrrrrr).

Second case F0 = 150 Hz

The code for generating 150 Hz impulse train and passing through the filter in question 1 is as

follows (code for questions 1 filter should be executed first)-


rep=1:1:75; %%no. of periodsD=rep.*1e-2/1.5; %%set time period of .1 s

(F0=100Hz)





0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

time in seconds

amplitude


15/39

axis([0 .5 -.5 .5]);




te1=0:1/fs:.52;

figure;

plot(te1,ww); %%plot the filter output




Fig 15- Pulse train with frequency of 150 Hz

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

time in seconds

amplitude


16/39

Fig 16- Output of filter realized in question 1 when excited with 150 Hz pulse train

Fig 17- Enlarged view of output waveform in figure 16.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

time in seconds

amplitude


17/39

Comment-

Sound appears to have less trrrr like effect. Sound is sharper.

4. In place of the simple single-resonance signal, synthesize the following more realistic

vowel sounds at two distinct pitches (F0 = 80 Hz, F0 = 220 Hz). Keep the bandwidths

constant at 100 Hz for all formants. Duration of sound: 0.5 sec

Vowel F1, F2, F3

/a/ 730, 1090, 2440

/i/ 270, 2290, 3010

/u/ 300, 870, 2240

Ans- Vowel /a/

Filter for /a/-

The filter can be realized as some of three filters with single resonances equal to formant

frequencies.

Here r = 1- pi/160 = (160-pi)/pi;

And w01 = 73*pi/80; w02 = 109*pi/80; w03= 244*pi/80;

()

( )()

( )()

( )()

Where pi = r*exp(j*w0i) for i= 1,2,3

Code for realizing this filter is

r=(160-pi)/160;

w01=73*pi/800; %%first formant

w02=109*pi/800; %%second formant

w03=244*pi/800; %%third formant

%%poles for first formant

p11=r*exp(1i*w01);

p12=r*exp(-1i*w01);


18/39

%%poles for second formant

p21=r*exp(1i*w02);

p22=r*exp(-1i*w02);

%%poles for third formant

p31=r*exp(1i*w03);

p32=r*exp(-1i*w03);

ww=0:10:8000;

H1=filt(1/52,[1 -(p11+p12) p11*p12],1/16000); %%tf of first filter

H2=filt(1/33,[1 -(p21+p22) p21*p22],1/16000); %%tf of second filter

H3=filt(1/16,[1 -(p31+p32) p31*p32],1/16000); %%tf of third filter

Ha=H1+H2+H3; %%tf of composite filter with

three formants

t=0:1/16000:3/100;

ha=impulse(Ha,t); %%compute impulse response of

filter

plot(t.*1000,ha); %%plot impulse response



Haw=freqresp(Ha,exp(1i*ww.*(pi/8000))); %%compute frequency response

n=size(Haw,3);

HaH=reshape(Haw,n,1);

figure;

semilogx(ww,mag2db(abs(HaH))); %%plot frequency response



grid on;


19/39

Fig 18- Impulse response of the filter for generating vowel /a/

Fig 19- Frequency response of the filter for generating vowel /a/

0 5 10 15 20 25 30-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

time in ms

amplitude

101

102

103

104

-40

-35

-30

-25

-20

-15

-10

-5

0

5

10

frequency in Hz

gainindB


20/39

Code for generating /a/ with pitches 80 and 220 Hz is as follows-

fs=16000;

rep=1:1:66;

D=rep.*1.25e-2;

te=0:1/fs:.5;

w=1e-3;

yp=1/4*(pulstran(te,D,@rectpuls,w));

ww=conv(ha,yp);

te1=0:1/fs:.05;

plot(te1,ww(1:801));



wavwrite(yp,fs,'pka80.wav');

fs=16000;

rep=1:1:110;

D=rep.*1e-2/2.2;

w=1e-3;

yp1=1/4*(pulstran(te,D,@rectpuls,w));

ww1=conv(ha,yp1);

figure;

plot(te1,ww1(1:801));



wavwrite(yp1,fs,'pk220.wav');


21/39

Fig 20- Waveform of vowel /a/ generated by pulse train of 80 Hz

Fig 21- Enlarged view of fig 20

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


22/39

Fig 22- Waveform of vowel /a/ generated by pulse train of 220 Hz


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


23/39

Comments-

Sound generated with 80 Hz pitch is not so good in quality (have more trrrrr like

effect), sound generated with 220 Hz is like some musical instrument sound.

Vowel /i/

Filter for /i/-


frequencies.

Here r = 1- pi/160 = (160-pi)/pi;

And w01 = 27*pi/80; w02 = 229*pi/80; w03= 301*pi/80;

()

( )()

( )()

( )()



r=(160-pi)/160;

w01=27*pi/800; %%first formantw02=229*pi/800; %%second formant



p11=r*exp(1i*w01);

p12=r*exp(-1i*w01);


p21=r*exp(1i*w02);

p22=r*exp(-1i*w02);

%%poles for third formantp31=r*exp(1i*w03);

p32=r*exp(-1i*w03);

ww=0:10:8000;


24/39




Hi=H1+H2+H3; %%tf of composite filter with

three formants

t=0:1/16000:3/100;

ha=impulse(Ha,t); %%compute impulse response of

filter

plot(t.*1000,ha); %%plot impulse response



Hiw=freqresp(Ha,exp(1i*ww.*(pi/8000))); %%compute frequency response

n=size(Hiw,3);HiH=reshape(Hiw,n,1);

figure;

semilogx(ww,mag2db(abs(HiH))); %%plot frequency response



grid on;

Fig 24Impulse response of the filter for generating vowel /i/

0 5 10 15 20 25 30-0.1

-0.05

0

0.05

0.1

0.15

time in ms

amplitude


25/39

Fig 25Frequency response of the filter for generating vowel /i/

Code for generating vowel /i/ with 80 and 220 Hz pitch-

fs=16000;

rep=1:1:66;

D=rep.*1.25e-2;

te=0:1/fs:.5;

w=1e-3;


ww=conv(hi,yp);

te1=0:1/fs:.53;

te2=0:1/fs:.05;

plot(te1,ww);



figure;




wavwrite(yp,fs,'pki80.wav');

101 102 103 104-40

-35

-30

-25

-20

-15

-10

-5

0

5

10

frequency in Hz

gainindB


26/39

fs=16000;

rep=1:1:110;

D=rep.*1e-2/2.2;

w=1e-3;yp1=1/4*(pulstran(te,D,@rectpuls,w));

ww1=conv(hi,yp1);

figure;

plot(te1,ww1);



figure;

plot(te2,ww1(1:801));

xlabel('time in seconds');ylabel('amplitude');

wavwrite(yp1,fs,'pki220.wav');

Fig 26- Vowel /i/ generated with 80 Hz pulse train

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


27/39


Fig 28- Vowel /i/ generated with 220 Hz pulse train

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


28/39


Comments-

Sound generated with 80 Hz pitch is not so good in quality ( have more trrrrr effect),

sound generated with 220 Hz is like some musical instrument sound. Identifying the vowel is

more difficult at 80 Hz pitch.

Vowel /u/

Filter for /u/-


frequencies.

Here r = 1- pi/160 = (160-pi)/pi;

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


29/39

And w01 = 30*pi/80; w02 = 87*pi/80; w03= 224*pi/80;

()

( )()

( )()

( )()



r=(160-pi)/160;

w01=30*pi/800; %%first formant

w02=87*pi/800; %%second formant



p11=r*exp(1i*w01);p12=r*exp(-1i*w01);


p21=r*exp(1i*w02);

p22=r*exp(-1i*w02);

%%poles for third formant

p31=r*exp(1i*w03);

p32=r*exp(-1i*w03);

ww=0:10:8000;




Hu=H1+H2+H3; %%tf of composite filter with

three formants

t=0:1/16000:3/100;

hu=impulse(Ha,t); %%compute impulse response of

filter

plot(t.*1000,hu); %%plot impulse response



Hiw=freqresp(Hu,exp(1i*ww.*(pi/8000))); %%compute frequency response


30/39

n=size(Hiw,3);

HuH=reshape(Huw,n,1);

figure;

semilogx(ww,mag2db(abs(HuH))); %%plot frequency response



grid on;

Fig 30Impulse response of the filter for generating vowel /u/

0 5 10 15 20 25 30-0.1

-0.05

0

0.05

0.1

0.15

time in ms

amplitude


31/39

Fig 31Impulse response of the filter for generating vowel /u/

Code for generating vowel /u/ with 80 and 220 Hz pitches is-

fs=16000;

rep=1:1:66;

D=rep.*1.25e-2;

te=0:1/fs:.5;

w=1e-3;


ww=conv(hu,yp);

te1=0:1/fs:.53;

te2=0:1/fs:.05;

plot(te1,ww);



figure;




wavwrite(yp,fs,'pku80.wav');

101

102

103

104-40

-35

-30

-25

-20

-15

-10

-5

0

5

10

frequency in Hz

gainindB


32/39

fs=16000;

rep=1:1:110;

D=rep.*1e-2/2.2;

w=1e-3;

yp1=1/4*(pulstran(te,D,@rectpuls,w));

ww1=conv(hu,yp1);

figure;

plot(te1,ww1);



figure;

plot(te2,ww1(1:801));



wavwrite(yp1,fs,'pku220.wav');

Fig 32- Vowel /u/ generated with 80 Hz pulse train

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


33/39

Fig 33- Enlarged view of figure 32

Fig 34- Vowel /u/ generated with 220 Hz pulse train

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


34/39

Fig 35- Enlarged view of figure 34

Comments-

Sound generated with 80 Hz pitch is not so good in quality (have trrrrr like effect),

sound generated with 220 Hz is like some musical instrument sound.

5. Signal Analysis:

Compute the DTFT magnitude spectrum of any 2 of the vowel sounds you have

synthesized. Use rectangular and Hamming windows of lengths: 40 ms, 100 ms. (i)

Comment on the similarities and differences between the different spectra. (ii) Estimate the

signal parameters from each of the spectra and compare with the ground-truth.

Ans. -The sound signals of vowel /u/ with 80 Hz pitch and /i/ with 220 Hz pitch are analyzed.

The code for computing spectrum of both signals with rectangular window of 40 ms and

hamming window of 100 ms is as follows (1024 point fft is used for rectangular windowed

signal and 2048 point fft is used for hamming windowed signal)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

time in seconds

amplitude


35/39

fs=16000; %%select sampling frequency

u80=wavread('pku80.wav'); %%read sound files

i220=wavread('pki220.wav');

u80=5.*(u80.');i220=5.*(i220.');

wr=rectwin(.04*fs); %%create rectangular window of 40 ms

wh=hamming(.1*fs); %%create hamming window of 100ms

wr=wr.';

wh=wh.';

%%apply both windows on u80 one by one

urw80=u80(16:(15+.04*fs)).*wr;

uhw80=u80(16:(15+.1*fs)).*wh;

%%apply both windos of i220 one by one

urw220=u80(16:(15+(.04*fs))).*wr;

uhw220=u80(16:(15+(.1*fs))).*wh;

fr = fs/2*linspace(0,1,513);

fh = fs/2*linspace(0,1,1025);

%%compute fourier transform

Uur=fft(urw80,1024);

Uuh=fft(uhw80,2048);

Uir=fft(urw220,1024);

Uih=fft(uhw220,2048);

plot(fr,mag2db(abs(Uur(1:513))));


ylabel('magnitude');figure;

plot(fh,mag2db(abs(Uuh(1:1025))));


ylabel('magnitude');

figure;

plot(fr,mag2db(abs(Uir(1:513))));


36/39



figure;

plot(fh,mag2db(abs(Uih(1:1025))));



Fig 36 -Magnitude spectrum of /u/ with 80 Hz pitch using rectangular window of size 40ms and

using 1024 point FFT.

0 1000 2000 3000 4000 5000 6000 7000 8000-40

-30

-20

-10

0

10

20

30

40

frequency in Hz

magnitude


37/39

Fig 37 -Magnitude spectrum of /u/ with 80 Hz pitch using hamming window of size 100ms and


0 1000 2000 3000 4000 5000 6000 7000 8000-100

-80

-60

-40

-20

0

20

40

frequency in Hz

magnitude


38/39

Fig 38 -Magnitude spectrum of /i/ with 220 Hz pitch using rectangular window of size 40ms and


0 1000 2000 3000 4000 5000 6000 7000 8000-40

-30

-20

-10

0

10

20

30

40

frequency in Hz

magnitude


39/39

Fig 39 -Magnitude spectrum of /i/ with 220 Hz pitch using hamming window of size 100ms and


(i) The spectrum in figure 37 and 39 has more spreading than in figure 36 and 38. Hence

hamming windowed spectrum has more spreading than rectangular windowed spectrum.

(ii)

The distance between two impulses in figure 36 and 37 is coming out to be 80 Hz that is equal to

the actual pitch period.

The distance between two impulses in figure 38 and 39 is coming out to be 220 Hz that is equal

to the actual pitch period.

The formants are not so apparent and are not measurable from the spectrums.

0 1000 2000 3000 4000 5000 6000 7000 8000-100

-80

-60

-40

-20

0

20

40

frequency in Hz

magnitude

computing assignment 1(pramod kumar surana)

Documents