# Not able to plot a proper graph for the equation.

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Kartikeya 2024 年 6 月 7 日
コメント済み: Kartikeya 2024 年 6 月 8 日
y = (1*1*((25)-(x.^2)))/((x.^4)-((x.^2)*((1*(1.78))+(25)+1))+(1*(25)));
fplot(y);
i used this code but am only getting a straight line instead of the 2 peak FRF that I want.
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Kartikeya 2024 年 6 月 8 日
@Torsten@Sam Chak Is it possible to get a graph like the following one for my case?
Kartikeya 2024 年 6 月 8 日
@Torsten@Sam Chak Thanks for the answers guys.

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### 採用された回答

John D'Errico 2024 年 6 月 8 日

syms x
y = (1*1*((25)-(x.^2)))/((x.^4)-((x.^2)*((1*(1.78))+(25)+1))+(1*(25)))
y =
First, where are the poles? A pole lives where the denominator is zero.
[N,D] = numden(y)
N =
D =
droot = solve(D)
droot =
double(droot)
ans = 4x1
0.9650 5.1816 -0.9650 -5.1816
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So we need to look between -6 and 6.
fplot(y,[-6,6])
NO. It does not look the way you want it to look. But it is your function. You set the problem. If it does not look like what you want, then maybe you are starting with the wrong function.
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John D'Errico 2024 年 6 月 8 日

That could be. And you could have offered that as an answer. But my mind reading abilities are sometimes lacking, only answering the question posed, instead of answering a different question. If I always tried that, I'd end up chasing into infinitely many rabbit holes.
Kartikeya 2024 年 6 月 8 日
@Sam Chak @Torsten @John D'Errico Guys I got the whole thing. Just had to use the absolute values of the y variable that I got. Thanks for your help, all of you.
x = 0:0.1:10;
y = (1*1*((25)-(x.^2)))./((x.^4)-((x.^2)*((1*(1.78))+(25)+1))+(1*(25)));
plot(y);
xlim([0 100]);
ylim([0 inf]);
xlabel('Natural Frequency Ratio');
ylabel('Amplitude x_0');
z = abs(y);
disp(z);
plot(z);

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