DSP Project 01

8/4/2019 DSP Project 01

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PROJECT REPORT-1

DIGITAL SIGNAL PROCESSING

ELEN 5316/4304

Instructor: Dr.Gleb Tcheslavski

10/13/2010

Submitted by:

Md. Khairuzzaman-L20258945

Adhitya Govardhanam-L20238253



1. Implement one clock-cycle of the Fundamental form of the Signal Flow Graph as a

stand alone Matlab module/function. Your module should be able to implement an

arbitrary SOS: compute the output and the next states for the given input,

states, and filter’s coefficients.

Solution:

For the Fundamental Direct form SFG (for SOS):

[] = +

Output

The MATLAB coding for Fundamental Direct form SFG:

1. function [y,Sn] = SFG(x,a,b,S1,S2,S3,S4)

% Function for a Fundamental Direct form SFG for SOS

% inputs:

% x = input signal % a = [a0,a1,a2] b = [b0,b1,b2] % S1, S2, S3, S4 = initial states of the system

% outputs: % y = output signal % Sn = Matrix for next states of the system

2. a = a/a(1); % normalization of the coefficients b = b/a(1);

3. if nargin < 7, S4 = 0; end % set S1, S2, S3, S4 if not designated if nargin < 6, S3 = 0; end if nargin < 5, S2 = 0; end if nargin < 4, S1 = 0; end

4. N = length(x); % length of input vector

5. y = zeros(N,1); % pre-allocating y



6. Sn = [S1; S2; S3; S4]; % Set initial value of states

7. % State-Space Description

A = [0 0 0 0; 1 0 0 0; b(2) b(3) -a(2) -a(3); 0 0 1 0]; B = [1; 0; b(1); 0]; Ct = [b(2) b(3) -a(2) -a(3)]; D = b(1);

8. for n = 1:N

y(n) = Ct*Sn + D*x(n); % calculates the nth element of y

Sn = A*Sn + B*x(n); % calculates the next states end

Description of above steps:

In step -1, SFG is a MATLAB function whose inputs are x, a, b, S1, S2, S3, S4

and the outputs are y, Sn (here Sn is a matrix contains next states of the system)

respectively. x contains values of input signal, whereas S1, S2, S3 & S4 are the

initial states of the system. a and b are three element vectors with elements

a0,a1,a2 and b0, b1, b2. The output y contains values of output signal and Sn

matrix contains next states of the system.

In step-2, the function does normalization of the coefficients. In step-3, the function sets the values of S1, S2, S3 and S4 to 0, if the are not

given by the user when the function is called.

In step-4, the length of the input vector x, is measured by the function and saved

in the variable N. This allows to know how many iterations the program must take

and the size the vector made to pre-allocate memory for the output needs to be.

In step-5, the function pre allocates the vector y as zeroes.

In step-6, the function sets initial values of S1, S2, S3, S4 into output Sn.

In step-7, the function calculates 4 vectors for state-space description (A, B, Ct,

D). In step-8, the function calculates the value of y and Sn.



2. Test your module to verify its correct functionality. Hint : remember that t he SFG is

an implementation of LCCDE.

Solution:

In this part, we load coefficient from P1_coeff.mat to obtain the values of a and b. Now,we randomly select three values for x as [1 17 36] and we calculate the output using the

function SFG( ) assuming S1, S2, S3, S4 are zero. So now, we recalculate the values

manually using the formulas listed below:

MATLAB Calculations:

>> load P1_coeff.mat

>> x = [1 17 36];

>> [y, Sn] = SFG(x, a, b)

y =

0.1497

2.7849

9.7209

>> Sn

Sn =

36.0000

17.0000

9.7209

2.7849

>> a

a = 1.0000 -1.2944 0.6400



>> b

b = 0.1497 0.0462 0.0374

Formulas :

Here, a0 = 1; a1 = -1.2944; a2 = 0.64; b0 = 0.1479; b1 = 0.0462; b2 = 0.0374

Clock-cycle 1

x1 = 1; S1,1 = 0; S2,1 = 0; S3,1 = 0; S4,1 = 0;

y1 = 0.0462*0 + 0.0374*0 – (-1.2944)*0 – 0.64*0 + 0.1479*1 = 0.1479

S1,2 = x1 = 1

S2,2 = S1,1 = 0

S3,2 = 0.0462*0 + 0.0374*0 – (-1.2944)*0 – 0.64*0 + 0.1479*1 = 0.1479

S4,2 = S3,1 = 0

Clock-cycle 2

x2 = 17; S1,2 = 1; S2,2 = 0; S3,2 = 0.1479; S4,2 = 0;

y2 = 0.0462*1 + 0.0374*0 – (-1.2944)*0.1479 – 0.64*0 + 0.1479*17 = 2.7520

S1,3 = x2 = 17

S2,3 = S1,2 = 1

S3,3 = 0.0462*1 + 0.0374*0 – (-1.2944)*0.1479 – 0.64*0 + 0.1479*17 = 2.7520



S4,3 = S3,2 = 0.1479

Clock-cycle 3

x3 = 36; S1,3 = 17; S2,3 = 1; S3,3 = 2.7520; S4,3 = 0.1479;

y3 = 0.0462*17 + 0.0374*1 – (-1.2944)*2.7520 – 0.64*0.1479 + 0.1479*36 = 9.6147

S1,4 = x3 = 36

S2,4 = S1,3 = 17

S3,4 = 0.0462*17 + 0.0374*1 – (-1.2944)*2.7520 – 0.64*0.1479 + 0.1479*36 = 9.6147

S4,4 = S3,3 = 2.7520

3. Load the f ilter coefficients from the file P1_coe ff .mat by the command:

“load P1_coe ff .mat ”. Inspect the given coefficients. What wil l yo u

conclude about the resulting filter’s stability in BIBO sense? Verify by

appropriate tests.

Solution

Coefficient Stability Test:

The coefficients loaded from "P1_coeff.mat" are

a = [1, -1.2944, 0.64] = [a0, a1, a2], and

b = [0.1497, 0.0462, 0.0374] = [b0, b1, b2].

The tests for BIBO stability of a SOS are

|a1| < 1 + a2, and

|a2| < 1.

We tested for BIBO stability using the above equations and acquired

|a1| = |-1.2944| = 1.2944 < 1 + 0.64 = 1.64, and



|a2| = 0.64 < 1.

Since the system meets these requirements the system is BIBO stable.

MATLAB Stability Test code


>> x = rand (1000, 1);

>> [y, Sn] = SFG(x, a, b);

>> plot(y)



Conclusions:

Also we tested the stability by inputting a vector of 1000 randomly generated numbers

and graphed the output of the function in the above graph. Then we checked the graph of

the output to see if the function remains stable. Since the plot of the output does not

shout up towards infinity at any point on the graph we conclude that the system is stable.

4. Load the P1.wav file to your workspace by the command “wavread” . The

command “whos” will show you which variables you have in your workspace.

You can play the loaded sound by the command “ sound ” . Next, filter the signal

you loaded from the wav- file by the SOS of the system you have implemented

with the coeff icients given in vectors a and b. Hint : SFG represent one clock cycle…Play the filtered signal by the command sound . What are your observations?

Hint : you may save your filtered signal in another wav-f ile and play it later on

another computer by any multimed ia player. You would need to use the command

“wavwr it e” in this case. Note: depending on a CPU, t he filter ing may take

several minutes!

Solution:


>> x1 = wavread('P1.wav');

>> whos

Name Size Bytes Class Attributes

a 1x3 24 double

b 1x3 24 double

x1 91781x1 734248 double

>> sound(x1)

>> [y1, Sn] = SFG(x1, a, b);

>> sound(y1)



Steps:

We first loaded the system coefficients from the "P1_coeff.mat" file.

Next we have loaded the "P1.wav" file using the "wavread" command and

saved it under the variable x1.

Next we played the original sound from x1 using the sound command. Then we filtered x1 and played the output, y1, using the sound command.

The difference from the original sound to the filtered sound was that the

noises have mostly been filtered out and the volume i.e. amplitude

attenuated also.

Conclusions:

This filter eliminates noises and attenuates the amplitude of the input signal.



5. Simulate a cascade of two identical SOS with the same coefficients and repeat

the filtering in 4. What are your observations?

Solution:

>> [y2, Sn] = SFG(y1,a,b);>> sound(y2)

Procedure:

We ran the output signal from question 4, y1, though the function a second time

using the same a, b, Sn parameters as we used in question 4.

Our observation is that the difference from the original sound to the filtered

sound was that the noises had been mostly filtered out.

6. (Graduate Students only) finally, determine whether any state and/or coefficient

scaling is needed to implement your module in hardware. Modify your SOS to implement

necessary scaling. Illustrate by an appropriate block diagram. Note: State and/or

coefficient scaling is not reall y required for a Matlab model: it ’ s important for the

hardware implementations. There f ore, the assignment DOES NOT ask you to perform

scaling; just make your module ready f or it

Solution: To verify the requirement of the state scaling, we need to check the condition

∑

And

Here,



MATLAB CODING:

function [h3_sum, hn] = sum_h(a, b, cs, n)

ct = [0 0 1 0]; A = [0 0 0 0; 1 0 0 0; cs*b(2) cs*b(1) -a(2) -a(3); 0 0 1 0]; b = [0; 0; cs*b(1); 0];

h3_sum = 0;

for i = 1:n hn = ct * A^(i-1) * b;

h3_sum = h3_sum + hn;end

Output:

>> [h3_sum, hn] = sum_h(a, b, 1, 10000)

h3_sum =

0.4332

hn =

0

Since the above condition is satisfied we do not need state scaling.

To verify the requirement of the coefficient scaling, we need to check the condition

From the stability triangle: || ||

And the coefficient value from P1_coeff.mat file is



a1 = -1.29 and a2 = 0.64

Here, || || , so we don’t need coefficient scaling too.

Conclusion:

Since, both the conditions are satisfied we do not need any type of scaling andhence no modification is required.

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DSP Project 01