# Filter transfer function for impulse response

12 次查看（过去 30 天）
ACICY2020-7-30

Hello, guys
I need help
How to get filter transfer function from this scheme. Then i need make impulse response from filter transfer function and calculate impulse response the first 20 readings.
Thanks for help
I tried i get transfer function H(z)=z^-1*0.25/1+z^-1*0.25 <--- is this right?

### 回答（1 个）

Hi ACICY,
I think you just wanted to confirm if the transfer function placed is apt or not.
You made a good try, but seems that the direct path in the numerator is missed.
After looking at it, i feel the transfer function is 1.
Here is how it is,
Consider the adder in the top path as a node from left. It takes in two paths, one directly from X(z) and other from the path of delay, and multiplier. Implies the equation at that node is
X(z) + (z^-1)*0.25*X(z) % Let this be S(z)
Now lets see at the adder at the second half of the circuit, one path is the resultant sum of first path, the other path is the Y(z) , delay, multipler. Impies the equation at that node is
% Inward flow to the adder
S(z) + (z^-1)*Y(z)*(-0.25) => S(z) - 0.25*(z^-1)*Y(z)
% The outward flow of the adder is Y(z)
% Equating the both
S(z) - 0.25*(z^-1)*Y(z) = Y(z)
=> S(z) = Y(z)*(1+(0.25*(z^-1)))
=> X(z) + (z^-1)*0.25*X(z) = Y(z)*(1+(0.25*(z^-1)))
=> X(z)*(1+(0.25*(z^-1))) = Y(z)*(1+(0.25*(z^-1)))
=> X(z) = Y(z)
=> Y(z)/X(z) = 1
=> H(z) = 1
Hope this helps.
Regards,
Sriram
##### 2 个评论显示隐藏 1更早的评论
Robert U 2020-8-14
Hi ACICY,
if transfer function is one it means that the output will reproduce the input without change.
Reproduction of first 20 time instants:
[y,t] = step(tf(1,1,1),20);
Visualization:
step(tf(1,1,1),20)
Kind regards,
Robert

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