Laboratory Exercise 4

LINEAR, TIME-INVARIANT DISCRETE-TIME SYSTEMS: FREQUENCY-DOMAIN REPRESENTATIONS

4.1 TRANSFER FUNCTION AND FREQUENCY RESPONSE

Project 4.1 Transfer Function Analysis

Answers:

Q4.1 The modified Program P3_1 to compute and plot the magnitude and phase spectra of

a moving average filter of Eq. (2.13) for 0 ≤ ω ≤ 2π is shown below:% Program P3_1% Evaluation of the DTFT clf;% Compute the frequency samples of the DTFTM = input('Desired length of the filter = ');w = 0:pi/(M^2):2*pi;num = (1/M).*ones(1,M);h = freqz(num,1, w);% Plot the DTFTsubplot(2,1,1)plot(w/pi,abs(h));gridtitle('Magnitude Spectrum |H(e^{j\omega})|')xlabel('\omega /\pi');ylabel('Amplitude');subplot(2,1,2)plot(w/pi,angle(h));gridtitle('Phase Spectrum arg[H(e^{j\omega})]')xlabel('\omega /\pi');ylabel('Phase in radians');This program was run for the following three different values of M and the plots of the corresponding frequency responses are shown below:

M=4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1Magnitude Spectrum |H(e jω)|

ω /π

Am

plit

ude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-4

-2

0

2

4Phase Spectrum arg[H(e jω)]

ω /π

Phase in radia

ns

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1Magnitude Spectrum |H(e jω)|

ω /π

Am

plitu

de

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-4

-2

0

2

4Phase Spectrum arg[H(e jω)]

ω /π

Phase in radia

ns

M=6

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1Magnitude Spectrum |H(e jω)|

ω /π

Am

plitu

de

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-4

-2

0

2

4Phase Spectrum arg[H(e jω)]

ω /π

Phase in radians

M=3

The types of symmetries exhibited by the magnitude and phase spectra are even and odd , respectively.

The type of filter represented by the moving average filter is low pass filter.

The results of Question Q2.1 can now be explained as follows the moving average filter is the low pass filter so the signal s2 is suppressed because s2 is the high frequency signal.

Q4.2 The plot of the frequency response of the causal LTI discrete-time system of Question Q4.2 obtained using the modified program is given below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1Magnitude Spectrum |H(e jω)|

ω /π

Amplitude

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2

-1

0

1

2Phase Spectrum arg[H(e jω)]

ω /π

Phase in radians

The type of filter represented by this transfer function is band pass filter.

Q4.3 The plot of the frequency response of the causal LTI discrete-time system of Question Q4.3 obtained using the modified program is given below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1Magnitude Spectrum |H(e jω)|

ω /π

Amplitude

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4

-2

0

2

4Phase Spectrum arg[H(e jω)]

ω /π

Phase in radians

The type of filter represented by this transfer function is band pass filter.

The difference between the two filters of Questions 4.2 and 4.3 is that the attenuation of the phase spectrum of the filter of the Question 4.3 is faster than the filter of the Question

4.2.

I shall choose the filter of Question Q4.2 for the following reason. The attenuation of the phase spectrum of the filter of the Question 4.3 is faster so the amount of the signal distortion is more significant.

Q4.4 The group delay of the filter specified in Question Q4.4 and obtained using the function grpdelay is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

3

4

5

6

7

8Groupdelay of |H(e jω)|

ω /π

Seconds

From this plot we make the following observations: the difference phase is the largest at the frequency of 0.3 that means the amount of the signal distortion is the largest at the frequency of 0.3.

Q4.5 The plots of the first 100 samples of the impulse responses of the two filters of Questions 4.2 and 4.3 obtained using the program developed in Question Q3.50 are shown below:

0 10 20 30 40 50 60 70 80 90 100-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15the inverse of a rational z-transform

0 10 20 30 40 50 60 70 80 90 100-1.5

-1

-0.5

0

0.5

1

1.5

2x 10

7 the inverse of a rational z-transform

From these plots we make the following observations: that the filter of the question 4.2 is stable and the one of the question 4.3 is not stable.

The transfer function in the question 4_2

Q4.6 The pole-zero plots of the two filters of Questions 4.2 and 4.3 developed using zplane are shown below:

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Part

Imag

inar

y P

art

-1.5 -1 -0.5 0 0.5 1 1.5

-1

-0.5

0

0.5

1

Real Part

Imag

inar

y P

art

From these plots we make the following observations: that the filter of the question 4.2 is casual and stable and the filter of the question 4.3 is not stable..

4.2 TYPES OF TRANSFER FUNCTIONS

Project 4.2 Filters

% Program P4_1% Impulse Response of Truncated Ideal Lowpass Filterclf;fc = 0.25;n = [-6.5:1:6.5];y = 2*fc*sinc(2*fc*n);k = n+6.5;stem(k,y);title('N = 13');axis([0 13 -0.2 0.6]);xlabel('Time index n');ylabel('Amplitude');grid;

Answers:

Q4.7 The plot of the impulse response of the approximation to the ideal lowpass filter obtained using Program P4_1 is shown below:

0 2 4 6 8 10 12-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6N = 13

Time index n

Amplitude

The length of the FIR lowpass filter is 13

The pole – zero plot of the filter of question 4_2

The pole – zero plot of the filter of question 4_3

The statement in Program P4_1 determining the filter length is n = [-6.5:1:6.5];

The parameter controlling the cutoff frequency is fc.

Q4.8 The required modifications to Program P4_1 to compute and plot the impulse response of the FIR lowpass filter of Project 4.2 with a length of 20 and a cutoff

frequency of ω c = 0.45 are as indicated below:% Program P4_1% Impulse Response of Truncated Ideal Lowpass Filterclf;fc = 0.45/(2*pi);n = [-10:1:10];y = 2*fc*sinc(2*fc*n);k = n+10;stem(k,y);title('N = 21');axis([0 20 -0.2 1]);xlabel('Time index n');ylabel('Amplitude');grid;

The plot generated by running the modified program is given below

0 2 4 6 8 10 12 14 16 18 20-0.2

0

0.2

0.4

0.6

0.8

N = 21

Time index n

Amplitude

Q4.9 The required modifications to Program P4_1 to compute and plot the impulse response of the FIR lowpass filter of Project 4.2 with a length of 15 and a cutoff frequency of

ω c = 0.65 are as indicated below:% Program P4_1% Impulse Response of Truncated Ideal Lowpass Filterclf;fc = 0.65/(2*pi);n = [-7.5:1:7.5];y = 2*fc*sinc(2*fc*n);k = n+7.5;stem(k,y);title('N = 21');axis([0 20 -0.2 1]);xlabel('Time index n');ylabel('Amplitude');grid;

The plot generated by running the modified program is given below:

0 2 4 6 8 10 12 14 16 18 20-0.2

0

0.2

0.4

0.6

0.8

N = 21

Time index n

Amplitude

Q4.10 The MATLAB program to compute and plot the amplitude response of the FIR lowpass filter of Project 4.2 is given below:

% Program P4_1% Impulse Response of Truncated Ideal Lowpass Filterclf;fc = 0.25;n = [-6.5:13/511:6.5];y = 2*fc*sinc(2*fc*n);Y=fft(y);Plot(abs(Y));Xlable(‘w/pi’);Ylable(‘amplitude’);Plots of the amplitude response of the lowpass filter for several values of N are shown below:

0 100 200 300 400 500 6000

5

10

15

20

25

30

35

40

45

From these plots we can make the following observations that the transition of the filter is approximately immediate.

A copy of Program P4_2 is given below:

% Program P4_2% Gain Response of a Moving Average Lowpass Filterclf;M = 2;num = ones(1,M)/M;[g,w] = gain(num,1);plot(w/pi,g);gridaxis([0 1 -45 0])xlabel('\omega /\pi');ylabel('Gain in dB');title(['M = ', num2str(M)])

Answers:

Q4.11 A plot of the gain response of a length-2 moving average filter obtained using Program P4_2 is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-45

-40

-35

-30

-25

-20

-15

-10

-5

0

ω /π

Gain in dB

M = 2

From the plot it can be seen that the 3-dB cutoff frequency is at 0.5pi.

Q4.12 The required modifications to Program P4_2 to compute and plot the gain response of a cascade of K length-2 moving average filters are given below:

% Gain Response of a cascade of K length -2 moving Average %Lowpass Filterclf;k=input('the number of sections of a cascade system:');M = 2;num = ones(1,M)/M;w = 0:pi/255:pi;h = freqz(num,1,w);ha=h.^k;g = 20*log10(abs(ha));plot(w/pi,g);gridaxis([0 1 -45 0])xlabel('\omega /\pi');ylabel('Gain in dB');title(['M = ', num2str(M)])The plot of the gain response for a cascade of 3 sections obtained using the modified program is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-45

-40

-35

-30

-25

-20

-15

-10

-5

0

ω /π

Gain in dB

M = 2

From the plot it can be seen that the 3-dB cutoff frequency of the cascade is at 0.3pi.

Q4.13 The required modifications to Program P4_2 to compute and plot the gain response of the highpass filter of Eq. (4.42) are given below:

% Program P4_2% Gain Response of a Moving Average Lowpass Filterclf;M=input('length of the highpass filter:');num = zeros(1,M);for k=0:M-1,

num(k+1)=num(k+1)+(-1)^k;

endnum=num/M;[g,w] = gain(num,1);plot(w/pi,g);gridaxis([0 1 -14 0])xlabel('\omega /\pi');ylabel('Gain in dB');title(['M = ', num2str(M)])

The plot of the gain response for M = 5 obtained using the modified program is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-14

-12

-10

-8

-6

-4

-2

0

ω /π

Gain in dB

M = 5

From the plot we can see that the 3-dB cutoff frequency is at 0.82pi

Q4.14 From Eq. (4.16) for a 3-dB cutoff frequency ω c at 0.45π we obtain α = 0.08.

Substituting this value of α in Eqs. (4.15) and (4.17) we arrive at the transfer function of the first-order IIR lowpass and highpass filters, respectively, given by

HLP(z) =1

1

08.01

)1(46.0−

−

−+z

z

HHP(z) = 1

1

08.01

)1(54.0−

−

−−z

z

The plots of their gain responses obtained using MATLAB are shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-14

-12

-10

-8

-6

-4

-2

0

ω /π

Gain in dB

lowpass filter wc=0.45pi

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-14

-12

-10

-8

-6

-4

-2

0

ω /π

Gain in dB

highpass filter wc=0.45pi

From these plots we observe that the designed filters do meet the specifications.

A plot of the magnitude response of the sum HLP(z) + HHP(z) obtained using MATLAB

is given below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-14

-12

-10

-8

-6

-4

-2

0

ω /π

Gain in dB

lowpass filter wc=0.45pi

From this plot we observe that the two filters are allpass-complementary.

A plot of the sum of the square-magnitude responses of HLP(z) and HHP(z) obtained

using MATLAB is given below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω /π

square amplitude

square-amplitude response of highpass filter wc=0.45pi

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω /π

square amplitude

square-amplitude response of highpass filter wc=0.45pi

From this plot we observe that the two filters are power-complementary

Q4.15 From Eq. (4.24), we get substituting K = 10, B = 1.866.

Substituting this value of B and ω c = 0.3π in Eq. (4.23) we obtain α = -0.311.

Using this value of α in Eq. (4.22) we arrive at the transfer function of the cascade of 10 IIR lowpass filters as

HLP,10 (z) =1 – α2

⋅1+ z−1

1 – αz−1

10

=10

1

1

311.01

)1(6555.0

+

+−

−

z

z

Substituting ω c = 0.3π in Eq. (4.16) we obtain α = 0.325.

Using this value of α in Eq. (4.15) we arrive at the transfer function of a first-order IIR lowpass filter

HLP,1(z) =1 – α2

⋅1 + z−1

1 –αz−1 =1

1

325.01

)1(3375.0−

−

−+z

z

The gain responses of HLP,10(z) and HLP,1(z) plotted using MATLAB are shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

ω /π

square amplitude

gain response of lowpass filter wc=0.3pi

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

ω /π

square amplitude

gain response of 10th-orderlowpass filter wc=0.3pi

From these plots we make the following observation the transition of the 10th-order lowpass filter is more immediate than the first-order lowpass filter.

Q4.16 Substituting ω o = 0.61π in Eq. (4.19) we get β = cos(0.61π) = -0.339.

Substituting ∆ω3dB = 0.15π in Eq. (4.20) we get (1 + α2)cos(0.15π) − 2α = 0,

whose solution yields α =0.6128 and α = 1.6319 .

Substituting the value of β and the first value of α in Eq. (4.18) we arrive at the transfer function of the IIR bandpass transfer function

HBP,1(z) = 21

2

6128.0547.01

)1(1936.0−−

−

++−

zz

z

Substituting the value of β and the second value of α in Eq. (4.18) we arrive at the transfer function of the IIR bandpass transfer function

HBP,2(z) = 21

2

6319.1892.01

)1(316.0−−

−

++−−

zz

z

Next using the zplane command we find the pole locations of HBP,1(z) and

HBP,2(z) from which we conclude that the stable transfer function HBP(z) is given

by

HBP,1(z) = 21

2

6128.0547.01

)1(1936.0−−

−

++−

zz

z

The plot of the gain response of the stable transfer function HBP(z) obtained using

MATLAB is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-25

-20

-15

-10

-5

0

ω /π

Gain in dB

bandpass filter BW=0.15pi

Using the same value of α and β in Eq. (4.21) we next obtain the transfer function of a stable IIR bandstop filter as

HBS(z) =21

21

6128.0547.01

)678.01(8064.0−−

−−

++++zz

zz

The plot of the gain response of the transfer function HBS(z) obtained using

MATLAB is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-35

-30

-25

-20

-15

-10

-5

0

ω /π

Gain in dB

bandstop filter BW=0.15pi

From these plots we observe that the designed filters do meet the specifications.

A plot of the magnitude response of the sum HBP(z) + HBS(z) obtained using MATLAB

is given below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

ω /π

amplitude

From this plot we observe that the two filters are allpass-complementary.

A plot of the sum of the square-magnitude responses of HBP(z) and HBS(z) obtained

using MATLAB is given below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω /π

square amplitude

bandpass filter BW=0.15 π

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω /π

square amplitude

bandstop filter BW=0.15 π

From this plot we observe that the two filters are power-complementary.

Q4.17 The transfer function of a comb filter derived from the prototype FIR lowpass filter of Eq. (4.38) is given by

G(z) = H0(zL) = 0.5(1+z-L)

Plots of the magnitude response of the above comb filter for the following values of L are shown below:

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω/π

amplitude

magnitude response of the above comb filter L=4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

ω/π

amplitude

magnitude response of the above comb filter L=6

From these plots we observe that the comb filter has mutilpe notches at ω k=

(2k+1)pi/L and L peaks at ω k = 2k*pi/L, where k = 0, 1, . . ., L-1.

Q4.18 The transfer function of a comb filter derived from the prototype FIR highpass filter of Eq. (4.42) with M = 2 is given by

G(z) = H1(zL) =0.5(1-z-L)

Plots of the magnitude response of the above comb filter for the following values of L are shown below -

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω/π

amplitude

magnitude response of the above comb filter L=4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω/π

amplitude

magnitude response of the above comb filter L=6

From these plots we observe that the comb filter has multiple notches at ω k =

k2pi/L

and multiple peaks at ω k = pi(2k+1)/L.

Q4.19 A copy of Program P4_3 is given below:

% Program P4_3% Zero Locations of Linear Phase FIR Filtersclf;b = [1 -8.5 30.5 -63];num1 = [b 81 fliplr(b)];num2 = [b 81 81 fliplr(b)];num3 = [b 0 -fliplr(b)];num4 = [b 81 -81 -fliplr(b)];n1 = 0:length(num1)-1;n2 = 0:length(num2)-1;subplot(2,2,1); stem(n1,num1);xlabel('Time index n');ylabel('Amplitude'); grid;title('Type 1 FIR Filter');subplot(2,2,2); stem(n2,num2);xlabel('Time index n');ylabel('Amplitude'); grid;title('Type 2 FIR Filter');subplot(2,2,3); stem(n1,num3);xlabel('Time index n');ylabel('Amplitude'); grid;title('Type 3 FIR Filter');subplot(2,2,4); stem(n2,num4);xlabel('Time index n');ylabel('Amplitude'); grid;title('Type 4 FIR Filter');pausesubplot(2,2,1); zplane(num1,1);title('Type 1 FIR Filter');subplot(2,2,2); zplane(num2,1);title('Type 2 FIR Filter');subplot(2,2,3); zplane(num3,1);title('Type 3 FIR Filter');subplot(2,2,4); zplane(num4,1);title('Type 4 FIR Filter');disp('Zeros of Type 1 FIR Filter are');disp(roots(num1));disp('Zeros of Type 2 FIR Filter are');disp(roots(num2));disp('Zeros of Type 3 FIR Filter are');disp(roots(num3));

disp('Zeros of Type 4 FIR Filter are');disp(roots(num4));

The plots of the impulse responses of the four FIR filters generated by running Program P4_3 are given below:

0 2 4 6 8-100

-50

0

50

100

Time index n

Amplitude Type 1 FIR Filter

0 5 10-100

-50

0

50

100

Time index n

Amplitude Type 2 FIR Filter

0 2 4 6 8-100

-50

0

50

100

Time index n

Amplitude Type 3 FIR Filter

0 5 10-100

-50

0

50

100

Time index n

Amplitude Type 4 FIR Filter

-1 0 1 2 3

-1

0

1

8

Real Part

Imaginary Part

Type 1 FIR Filter

-2 0 2 4

-2

-1

0

1

2

9

Real Part

Imaginary Part

Type 2 FIR Filter

-2 0 2 4 6

-2

0

2

8

Real Part

Imaginary Part

Type 3 FIR Filter

-1 0 1 2 3

-1

0

1

9

Real Part

Imaginary Part

Type 4 FIR Filter

From the plots we make the following observations:

Filter #1 is of length 9 with a symmetry impulse response and is therefore a Type 1st

linear-phase FIR filter.

Filter #2 is of length 10 with a symmetry impulse response and is therefore a Type

2nd linear-phase FIR filter.

Filter #3 is of length 9 with an anti-symmetric impulse response and is therefore a

Type 3rd linear-phase FIR filter.

Filter #4 is of length 10 with an anti-symmetric impulse response and is therefore a

Type 4th linear-phase FIR filter.

From the zeros of these filters generated by Program P4_3 we observe that:

Filter #1 has no restriction of zeros.

Filter #2 has zeros at z = -1 .

Filter #3 has zeros at z = 1 and -1 .

Filter #4 has zeros at z = 1.

Plots of the phase response of each of these filters obtained using MATLAB are shown below:

0 0.5 1-30

-20

-10

0

ω/2π

radians

Type 1 FIR Filter

0 0.5 1-30

-20

-10

0

ω/2π

radians

Type 2 FIR Filter

0 0.5 1-30

-20

-10

0

ω/2π

radians

Type 3 FIR Filter

0 0.5 1-30

-20

-10

0

ω/2π

radians

Type 4 FIR Filter

From these plots we conclude that each of these filters have linear phase.

The group delay of Filter # 1 is 4

The group delay of Filter # 2 is 4.5

The group delay of Filter # 3 is 4The group delay of Filter # 4 is 4.5

Answers:

Q4.21 A plot of the magnitude response of H1(z) obtained using MATLAB is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

ω/π

amplitude

magnitude response of H1(Z)

From this plot we observe that the magnitude response has a maximum at ω =0.7020 pi with a value = 0.3289.

Using zplane we observe that the poles of H1(z) are outside the unit circle and

hence the transfer function is not stable.

Since the maximum value of the magnitude response of H1(z) is =0.3289, we scale

H1(z) by 0.3289 and arrive at a bounded-real transfer function

H2(z) = )651(0.3289

)1(5.121

1

−−

−

+++

zz

z

Q4.22 A plot of the magnitude response of G1(z) obtained using MATLAB is shown below:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

ω/π

amplitude

magnitude response of transfer function G1(Z)

From this plot we observe that the magnitude response has a maximum at ω = 0.64pi with a value = 1.7843

Using zplane we observe that the poles of G1(z) are inside the unit circle and

hence the transfer function is stable.

Since the maximum value of the magnitude response of G1(z) is = 1.7843 , we scale

G1(z) by 1.7843 and arrive at a bounded-real transfer function

G2(z) = )2(7843.1

121

1

−−

−

++−

zz

z

4.3 STABILITY TESTclf;dens =input('Denominator coefficients = ','s');denn=sscanf(dens,'%f');ki = poly2rc(denn');disp('Stability test parameters are');disp(ki);Answers:Q4.23 The pole-zero plots of H1(z) and H2(z) obtained using zplane are shown

below:

-3 -2 -1 0 1 2 3-1

-0.5

0

0.5

1

2

Real Part

Imaginary

Part

pole-zero plot of H1(z)

-3 -2 -1 0 1 2 3-1

-0.5

0

0.5

1

2

Real Part

Imaginary P

art

pole-zero plot of H2(z)

From the above pole-zero plots we observe that all poles of two transfer functions are inside of the unit circle.

Q4.24 Using Program P4_4 we tested the stability of H1(z) and arrive at the following

stability test parameters {ki}: -0.9989, 0.8500.

From these parameters we conclude that H1(z) is stable .

Using Program P4_4 we tested the stability of H2(z) and arrive at the following

stability test parameters {ki}:-1.0005, 0.8500.

From these parameters we conclude that H2(z) is stable .

Q4.25 Using Program P4_4 we tested the root locations of D(z) and arrive at the following stability test parameters {ki}: 0.7604, 0.9672, 2.6057, -0.5195, 0.3125.

From these parameters we conclude that all roots of D(z) are not inside the unit

circle.

Q4.26 Using Program P4_4 we tested the root locations of D(z) and arrive at the following stability test parameters {ki}: -0.6087, 0.7958, 0.6742, 0.5938, 0.6000

From these parameters we conclude that all roots of D(z) are inside the unit circle.

Ngày: 10/10/2009