how to plot a periodic function?
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Hey there,
I have to construct a periodic function of a signal according to the image shown in order to apply fourier series into it. I tried using piecewise command but I failed to get an appropriate output. What is the point that I'm missing? My code is shown below:
clc;
clear all;
a=3.2;
t1=5;
t2=7;
t3=12;
t4=17;
%constructing partial periodic function
syms a(x)
T=17;
i=-2;
interval=[0 -i*T];
pw=[];
while i<=diff(interval/(2*T))
a(x)=piecewise(i*T<=mod(x,17)<=t1+(i*T),3.2,t1+(i*T)<=mod(x,17)<=t2+(i*T),-1.6*(x-i*T)+11.2*4*i,t2+(i*T)<=mod(x,17)<=t3+(i*T),-0.64*(x-i*T)+4.48*4*i,t3+(i*T)<=mod(x,17)<=t4+(i*T),-3.2);
i=i+1;
pw=[pw a ];
end
pw
fplot(pw,interval)
0 件のコメント
採用された回答
Paul
2022 年 12 月 1 日
Hi Zafer,
To compute the Fourier series, you really only need to define one period of the function, because the defining integrals are only taken over one period. So you can use piecewise to define one period of the function.
a=3.2;
t1=5;
t2=7;
t3=12;
t4=17;
syms f(t)
f(t) = piecewise(t < t1,a, t1<=t<t2,-a/(t2-t1)*(t-t1)+a, t2<=t<t3, -a/(t3-t2)*(t-t2), t3<=t<t4,-a, 0);
fplot(f(t),[0 t4])
If you want to plot the periodic fucntion, then use mod as the argument into f, not in the definition of f
fplot(f(mod(t,t4)),[-50 50])
その他の回答 (2 件)
Walter Roberson
2022 年 12 月 1 日
編集済み: Walter Roberson
2022 年 12 月 1 日
You do not need any loop. You generate one cycle based upon t1 t2 t3 t4 and piecewise(). Then you substitute mod(Time,t4) to the piecewise, to end up with a piecewise that is cyclic every interval of t4.
Once you have the piecewise() then you can ask to rewrite(EXPRESSION, 'heaviside') to get an expression that, in theory, can be pass through fourier() . In practice the mod() is going to cause problems.
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