Need help, How do filter function work?

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Miguel Faria
Miguel Faria 2018-1-20
评论： William Rose 2018-1-22
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.
  2 个评论
Rik
Rik 2018-1-20
Have a read here and here. It will greatly improve your chances of getting an answer.
William Rose
William Rose 2018-1-20
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.

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Miguel Faria
Miguel Faria 2018-1-21
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.
  1 个评论
William Rose
William Rose 2018-1-22
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

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