How to see the freq response of a transfer function
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I am having Transfer Function
H(z)= .5z(2)+1/ [z(2)+1]
i want to plot its freq response using omega axis
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Wayne King
2011 年 11 月 13 日
Hi, I'm going to assume that by z(2) you mean z^{-2}.
You can do a couple things.
[H,W] = freqz([1 0 0.5],[1 0 1]);
plot(W,20*log10(abs(H)));
set(gca,'xlim',[0 pi]); grid on;
xlabel('Radians/sample');
Or
fvtool([1 0 0.5],[1 0 1]);
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その他の回答 (6 件)
moonman
2011 年 11 月 13 日
1 件のコメント
Wayne King
2011 年 11 月 13 日
Hi, I don't think that is correct. If you look at the pole-zero plot for that
zplane([0.5 0 1],[1 1]);
That frequency response peaks near pi radians/sample or 1 in omega/pi.
You have a lowpass filter in your figure.
moonman
2011 年 11 月 13 日
1 件のコメント
Wayne King
2011 年 11 月 13 日
I know which one is correct, but I don't want to just give you the answer. You have to keep in mind that
[H,W] = freqz(....) is giving you zero frequency first. Your book is plotting zero frequency in the middle. That may be confusing you. What is the value of your transfer function at zero frequency?
Look at H(1) for you possibilities that is the transfer function at 0 frequency.
Wayne King
2011 年 11 月 13 日
Also, you see that your plotting frequency response has zero as 0-1j and 0+1j. That means there are zeros on the unit circle at those frequencies (angles).
Use
zplane(B,A)
to look for those zeros.
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moonman
2011 年 11 月 13 日
2 件のコメント
Wayne King
2011 年 11 月 13 日
You're right.
They are just plotting the magnitude, so at zero frequency they have 8.
Look at:
[H,W] = freqz([1 0 1],[1 -0.75]);
H(1)
moonman
2011 年 11 月 13 日
1 件のコメント
Wayne King
2011 年 11 月 13 日
Because I saw that the magnitude was zero at -0.5 and 0.5, which corresponds to angles of -pi/2 and pi/2
exp(1j*pi/2) = 0+1j
exp(-1j*pi/2) = 0-1j
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