DC removal flter
in [Matlab]
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From: david on 4 Jul 2010 15:54 Hello, i am not very good at matlab but i need to implement this in matlab, Y(z) = 1 - z exp -1 X(z) = 1 - 0.95*z exp -1 H(z) = Y/Z This is a DC removal filter.
From: Faraz Afzal on 4 Jul 2010 17:40 "david " <int78005(a)stud.uni-stuttgart.de> wrote in message <i0qost$fvu$1(a)fred.mathworks.com>... > Hello, i am not very good at matlab but i need to implement this in matlab, > > > Y(z) = 1 - z exp -1 > X(z) = 1 - 0.95*z exp -1 > > H(z) = Y/Z > > This is a DC removal filter. Hi David, Use either of the given two in simulink to serve the purpose of DC removal Discrete FIR filter, or Discrete filter... Regards, Faraz
From: david on 4 Jul 2010 18:08 hi, i need to programm it not to use simulink
From: Faraz Afzal on 4 Jul 2010 18:53 "david " <int78005(a)stud.uni-stuttgart.de> wrote in message <i0qost$fvu$1(a)fred.mathworks.com>... > Hello, i am not very good at matlab but i need to implement this in matlab, > > > Y(z) = 1 - z exp -1 > X(z) = 1 - 0.95*z exp -1 > > H(z) = Y/Z > > This is a DC removal filter. y = [1 -1] x = [1 -0.95] [z ,p ,k] = tf2zp(y,x) Regards, Muhammad Faraz
From: Wayne King on 4 Jul 2010 19:46 "david " <int78005(a)stud.uni-stuttgart.de> wrote in message <i0qost$fvu$1(a)fred.mathworks.com>... > Hello, i am not very good at matlab but i need to implement this in matlab, > > > Y(z) = 1 - z exp -1 > X(z) = 1 - 0.95*z exp -1 > > H(z) = Y/Z > > This is a DC removal filter. If this is the filter you want to implement, I think you want the pole just inside the unit circle at z=1 and the zero on the unit circle at z=1 so, A = [1 -0.95]; % this gives you a pole at z=0.95+j0 B = [1 -1]; % this gives you a zero at z=1+j0 zplane(B,A); % see the pole-zero plot fvtool(B,A); % see the magnitude response I think the constant-coefficient difference equation you are going for is: y(n)-0.95*y(n-1)=x(n)-x(n-1) Wayne
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