Need help, How do filter function work?

y[n] − 0.4y[n − 1] = x[n]; y[−1] = 1.

Need response to input signal, x[n] = ((0.4)^n)u[n] to n = −5, ..., 30.

[Rik]

Have a read here and here. It will greatly improve your chances of getting an answer.

[William Rose]

Mr. Faria, You state that y is known at time n=-1, but you also say you want to see the system response to a signal which starts at n=-5. That leaves me unsure about whether you want to know y starting at time n=-5 or starting at time n=0. The Z transform of the system is Y(z)=H(z)*X(z), where H(z)=B(z)/A(z), where B and A are the numerator and denominator polynomials (polynomials in z^-1). In this case, the numerator polynomial is B=1, and the denominator polynomial is A=1-0.4*z^-1. Therefore the vector of numerator coefficients is b=1 and the vector of denominator coefficients is a=[1,-0.4]. If we ignore the bit about initial conditions on y (which means we assume y=0 at times in the past, then we can compute y at times from -5 to +30 using the filter function, as follows:

 n=[-5:30];
 ustep=(n>=0);
 x=ustep.*0.4.^n;
 a=[1 -.4];b=1;
 y=filter(b,a,x);
 plot(n,x,'ro-',n,y,'bx-');
 xlabel('Time'); legend('X','Y');

You could get the same result for y using a for loop. I will call it y2 this time:

y2(1)=0;
for i=2:36,y2(i)=.4*y2(i-1)+x(i); end

The solution above started at n=-5 and did not obey the required initial condition y(-1)=1. If we want to apply the initial condition at n=-1, then we will use Matlab's filter to compute the output starting at time n=0. No time for more now. Maybe later.

Answer 1

[Miguel Faria]

My problem is: This is what I have : y(n) - 0.4y(n-2) = x(n) , y(-1) = 1 , x(n) = (0.4)^n * u(n)

I started to solve: a = [1, -0.4] b = 1

n = -5:1:30.

y= filter(b , a , x , filtic(b,a,1))

And my problem is: How do I define this? --> x(n) = (0.4)^n * u(n) I already tried it: (0.4)^n * (ones(1,30)) but it does not.

[William Rose]

I assume that by u(n) you mean the unit step function: u=0 for n<0, u=1 for n>=0. I did give you x(n) in the code I posted earlier. I wrote:

n=[-5:30];   %vector of times
ustep=(n>=30); %boolean vector with 5 false values
               %followed by 31 true values
x=ustep.*0.4.^n;  %element-wise multiply and
                  %element-wise power