how to do implement difference equation in matlab

Hi i am stuck with this question

Write a MATLAB program to simulate the following difference equation 8y[n] - 2y[n-1] - y[n-2] = x[n] + x[n-1] for an input, x[n] = 2n u[n] and initial conditions: y[-1] = 0 and y[0] = 1

(a) Find values of x[n], the input signal and y[n], the output signal and plot these signals over the range, -1 = n = 10.

The book has told to user filter command or filtic

my code is down kindly guide me about initial conditions

[moonman]

I have done in this way

n=0:10;
a = [8 -2 -1];                     % left hand side of difference equation
b = [1 1 0];                % right hand side of difference equation

x=2.^n;
k=filter(b,a,x)
plot(n,k)

Is it correct How to cater for initial conditions y[-1]=0 and y[0]=1

[moonman]

I have to do

Find values of x[n], the input signal and y[n], the output signal and plot these signals over the range, -1 = n = 10.

[moonman]

can somebody guide me

[Fangjun Jiang]

My advice:

2. Use proper punctuation mark, e.g. comma, period, question mark.

3. Read your own question again after posting, e.g. what is x[n] = 2n u[n]?

4. Update your question with comments, not answers.

[Ahmed ElTahan]

Here is a function I have built to calculate it with added example. https://www.mathworks.com/matlabcentral/fileexchange/56142-output-estimation-difference-equation

Answer 1

[Honglei Chen]

I think your b is incorrect, it should be [1 1] instead. To accommodate for the initial value of y and x, you need to translate them into the corresponding filter state. filter command is an implementation of direct form II transpose, so you can use filtic to convert y and x to the state.

Here is an example, where n runs from 1 to 10. Based on you example, x[0] is 1.

n = 1:10;
a = [8 -2 -1];
b = [1 1];
yi = [1 0];
xi = 1;
zi = filtic(b,a,yi,xi)
y = filter(b,a,2.^n,zi)

BTW, I doubt if the input is really 2.^n as this becomes unbounded very quickly. Are you sure it's not 2.^(-n)?

HTH

[moonman]

Thanks Honglei this is the question of book. I hope it will be correct.U can read it again

*Write a MATLAB program to simulate the following difference equation: 8y[n] - 2y[n-1] - y[n-2] = x[n] + x[n-1] for an input, x[n] = 2n u[n] and initial conditions: y[-1] = 0 and y[0] = 1. (Hint : Try the commands ‘filter’ or ‘filtic’). a. Find the values of x[n], the input signal and y[n], the output signal and plot these signals over the range, -1 = n = 10. (All plots must be properly labeled). * So can u help me for this

[moonman]

Is this solution Ok

n = -1:10;
a = [8 -2 -1];   % left hand side of difference equation
b = [1 1 0];  % right hand side of difference equation
yi = [1 0];
xi = 0;
x=2*n;
zi = filtic(b,a,yi,xi)
y = filter(b,a,2*n,zi)
subplot(2,1,1)
plot(n,y)
xlabel('--------n----------->')
ylabel('-------Values------->')
title(' Plot of output signal y over the range from n=-1 to n=10')

subplot(2,1,2)
plot(n,x)
xlabel('--------n----------->')
ylabel('-------Values------->')
title(' Plot of input signal x over the range from n=-1 to n=10')

[Honglei Chen]

You are on the right track in general, but there are two things I want to point out:

1. If your x is 2*n, then x(0) is 0.
2. You already have y[0], y[-1], x[0], x[-1], so you don't need to compute them again.

Basically you should update your code to use

n = 1:10

and

xi = 0

This way, the y you get is y(1) through y(10). You can then concatenate y(0) and y(-1) to y to form the total vector. Similar things can be done for x too.

HTH

[moonman]

Any more help will be appreciated Can anybody else confirm me that my code is correct

[Walter Roberson]

You are working on a homework question. You do the best you can and if *you* cannot find an errors in your work, then you submit your answer and take your chances.

Have you considered working the results out manually and comparing them to the computed results?

[Honglei Chen]

I think what Walter suggests is a great way to proceed. Or you can do a program, as Fangjun suggested to output the result using a program and see if the two approaches agree. Debug and verification is just as important as programming.

BTW I think I mention in my post that you have issue with your coefficient and that is not fixed yet.

[Divya K M]

output of difference equation should be discrete but the output waveform is coming continuous

[Aditya Kumar Singh]

use stem instead of plot... you should get the correct waveform after that

Answer 2

[Mohazam Awan]

%%DSP LAb Task 4
% Difference equation implementation in matlab
%
clc
clear all
close all

% using filter function
n=-5:1:10;
index=find(n==0);
x=zeros(1,length(n));
x(index)=1;
subplot(2,2,1)
stem(n,x)
grid on 
axis tight

b=[1 0];
a=[1 -2];
y=filter(b,a,x);
subplot(2,2,2)
stem(n,y,'filled','r')
grid on 
axis tight

% Now without filter function
y1=zeros(1,length(n));
for i=1:length(n)
    if(n(i)<0)
        y1(i)=0;
    end
    if (n(i)>=0)
        y1(i)=2*y1(i-1)+x(i);
    end
end

subplot(2,2,3)
stem(n,y1,'filled','k')
grid on 
axis tight

[Divya K M]

Is this code for given question if so means can I explain

[Daniel]

Why do you use negative samples in the n vector? I am trying to use only positve samples and it doesnt work

Answer 3

[Fangjun Jiang]

It might be a filter. But I thought all the assignment was asking you to do is to write a for-loop to generate the y series data based on the equation and the initial conditions.

[moonman]

no book has told to use filter is my code correct?? How to cater for initial conditions y[-1]=0 and y[0]=1

Answer 4

[BHOOMIKA MS]

% Define the symbolic variable syms z n

% Define the Z-transform of the right-hand side 2^n rhs_z = 1 / (1 - 2*z^(-1));

% Define the left-hand side: Y(z) - 2z^(-1)Y(z) + z^(-2)Y(z) lhs_z = (1 - 2*z^(-1) + z^(-2)) * sym('Y(z)');

% Set up the equation for Y(z) eq = lhs_z == rhs_z;

% Solve for Y(z) Y_z = solve(eq, 'Y(z)');

% Simplify the expression for Y(z) Y_z_simplified = simplify(Y_z);

% Perform partial fraction decomposition on Y(z) Y_z_decomp = partfrac(Y_z_simplified, z);

% Display the decomposed Y(z) disp('Decomposed Y(z):'); disp(Y_z_decomp);

% Now take the inverse Z-transform for each term y_n = iztrans(Y_z_decomp);

% Display the time-domain solution y(n) disp('The time-domain solution y(n) is:'); disp(y_n);

% Create a numerical sequence for plotting % Define the range for n (e.g., n from 0 to 10) n_values = 0:10;

% Evaluate y(n) for each n using subs (substitute n into the expression) y_values = double(subs(y_n, n, n_values));

% Plot the solution figure;

% Plot y(n) stem(n_values, y_values, 'filled', 'LineWidth', 2); title('Time-domain solution y(n)'); xlabel('n'); ylabel('y(n)'); grid on;

% Add labels to the graph for clarity text(0, y_values(1), ['y(0) = ', num2str(y_values(1))], 'VerticalAlignment', 'bottom', 'HorizontalAlignment', 'right');

Answer 5

[BHOOMIKA MS]

% Define the symbolic variable syms z n

% Define the Z-transform of the right-hand side 2^n rhs_z = 1 / (1 - 2*z^(-1));

% Define the left-hand side: Y(z) - 2z^(-1)Y(z) + z^(-2)Y(z) lhs_z = (1 - 2*z^(-1) + z^(-2)) * sym('Y(z)');

% Set up the equation for Y(z) eq = lhs_z == rhs_z;

% Solve for Y(z) Y_z = solve(eq, 'Y(z)');

% Simplify the expression for Y(z) Y_z_simplified = simplify(Y_z);

% Perform partial fraction decomposition on Y(z) Y_z_decomp = partfrac(Y_z_simplified, z);

% Display the decomposed Y(z) disp('Decomposed Y(z):'); disp(Y_z_decomp);

% Now take the inverse Z-transform for each term y_n = iztrans(Y_z_decomp);

% Display the time-domain solution y(n) disp('The time-domain solution y(n) is:'); disp(y_n);

% Create a numerical sequence for plotting % Define the range for n (e.g., n from 0 to 10) n_values = 0:10;

% Evaluate y(n) for each n using subs (substitute n into the expression) y_values = double(subs(y_n, n, n_values));

% Plot the solution figure;

% Plot y(n) stem(n_values, y_values, 'filled', 'LineWidth', 2); title('Time-domain solution y(n)'); xlabel('n'); ylabel('y(n)'); grid on;

% Add labels to the graph for clarity text(0, y_values(1), ['y(0) = ', num2str(y_values(1))], 'VerticalAlignment', 'bottom', 'HorizontalAlignment', 'right');