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我正在尝试为 p=7 绘制一个简单多项式 (1+z)^(2p) 的极点/零图。我的代码如下：

p = 7;
rCoeffs = [1 1];
for ii=1:2*p-1
    rCoeffs = conv(rCoeffs, [1 1]);
end
zplane(real(rCoeffs),1);

该图显示以下内容：

在此处输入图像描述

我不明白为什么零是复数。我认为所有的零都应该位于 z=-1 但是这个图显示了一个圆圈。当 p 很小时不会发生这种情况，但是我再次在网上看到了一些显然是由 zplane 生成的图，它们在特定点上显示了大量的零。

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2 回答 2

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Basically you're looking for 14 solutions to the equation given the setup. Unfortunately the general solution to polynomial equations of order 5 or greater doesn't exist and must be found numerically. The solution provided is correct only in that is approximately what the algorithm thinks you're looking for.

I would add that Nathan's method works as intended, and if you change it slightly, you will see all the solutions to the above equation.

z = tf('z',1)
H = (1+z)^(2*7);
[p,z1] = pzmap(H)
z1 % solution to H = 0
(1+z1)^(2*7)
于 2013-02-24T00:52:46.920 回答
0

更简单的方法：

p = 7

z = zpk('z',0.1);
H = (1+z)^(2*p);

pzmap(H)
于 2013-02-23T23:30:30.047 回答