plotting a function using a function

Question

jvfa 2023-9-9

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编辑： jvfa 2023-9-14

Hello everyone, I like to ask how do i plot a function that integrates a certain function

Cn = integral(function of the sawtooth * exp(-1i*2*pi*n*f*t), 0, T);

so far this is my working code which only plots the sawtooth wave function. I have made multiple changes to it trying to figure out how to plot Cn. I'm still fairly new to matlab. maybe I'm missing something.

close all; figure; t=linspace(0,3,1500); m=linspace(0,3,1500); T=1; f=1/T; amp=1; sawf=amp/2; k=plot(NaN,NaN); 1i; for n=1:1:500 sawf = sawf - (amp/pi)*(1/n)*(sin(2*pi*n*f*t)); set(k, 'XData',t,'YData',sawf); pause(0.01); end

%cn1= integral(sawf*((cos(2*pi*1*f*t))-(-(1i)*sin(2*pi*1*f*t))),0,T); %cn1= sawf*(sin(2*pi*n*f*t)/(2*pi*n*f))+((i)cos(2*pi*n*f*t)/(2*pi*n*f)); %cn = (1./T)*cn1; %set(k, 'XData',m,'YData',cn);

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Dyuman Joshi 2023-9-9

编辑：Dyuman Joshi 2023-9-9

在 MATLAB Online 中打开

integral requires a function handle as an integrand. You have not supplied any function, just a numerical value.

Also, I don't understand how this -

%1
cn1= integral(sawf*((cos(2*pi*1*f*t))-(-(1i)*sin(2*pi*1*f*t))),0,T);

results in this -

%2
cn1= sawf*(sin(2*pi*n*f*t)/(2*pi*n*f))+((i)cos(2*pi*n*f*t)/(2*pi*n*f));

or how did you arrive at line 2.

or was it supposed to be n instead of 1 (next to pi inside the trignometric terms) in line 1?

jvfa 2023-9-9

Hi, can you elaborate where i where i got wrong on what you said about integral? because it's really a big part of understanding the matlab function of integration that I lack.

those comments don't worry about them they were just my attempts at understanding the function. %1 was originally n then i tried substituting 1 for n to see what the value is sort of doing trial and error. for %2 i manually integrated the thing by hand, given that i have the value for sawf so i can move it out of the integral sign then just integrated exp(-2i*pi*n*f*t)dt instead

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Answer 1

Bruno Luong 2023-9-9

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https://ww2.mathworks.cn/matlabcentral/answers/2018806-plotting-a-function-using-a-function#answer_1306026

编辑：Bruno Luong 2023-9-9

在 MATLAB Online 中打开

The coefficients Cn = 1i/(2*pi*n).

I'm not very good in symbolic calculation, I never own the tollbox license.

syms t

syms n integer

sawtooth = t;

Cn = simplify(int(sawtooth*exp(-1i*2*pi*n*t), t, 0, 1))

Cn =

C0 = simplify(int(sawtooth, t, 0, 1)) % funny the above general formula is wrong for n=0

C0 =

Cnfun = matlabFunction(Cn);

n = -10:10;

Cnval = Cnfun(n);

Cnval(n==0) = subs(C0); % no idea why I need to do this

figure

hold on

plot(n, real(Cnval))

plot(n, imag(Cnval))

legend('real','imag')

xlabel('n')

ylabel('C(n)')

t = linspace(0,3,1000)';

f = @(t) mod(t,1)

f = function_handle with value:

@(t)mod(t,1)

fFourier = sum(Cnval.*exp(1i*2*pi*n.*t),2);

figure

plot(t, f(t), 'r', t, fFourier, 'b')

Warning: Imaginary parts of complex X and/or Y arguments ignored.

xlabel('t')

legend('f', 'Fourier approximation')

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Paul 2023-9-9

编辑：Paul 2023-9-9

在 MATLAB Online 中打开

Hi Bruno,

Regarding the concern about the general formula being wrong ...

syms t
syms n integer
sawtooth = t;

Don't use simplify to see what's happening (always safer to use sym(pi) )

Cn = int(sawtooth*exp(-1i*2*sym(pi)*n*t), t, 0, 1)

Cn =

int comes up with a general solution that doesn't take into account the special case of n = 0. However, note that it's sort of the right answer for n = 0 in the sense that

limit(Cn,n,0)

ans =

This situation comes up all of the time with Fourier integrals. I'm not sure if it's reasonable to expect that int identify all of the special cases for all parameters in the integrand and then provide a result in terms of piecewise that captures each special case. I've often wondered about this issue.

I don't know the inner workings of simplify, but I'll note that we can do this

[num,den] = numden(Cn)

num =

den =

num = simplify(num)

num =

And then

num/den

ans =

which is standard for the Symbolic toolbox, i.e., n/n will always be replaced with 1, even if n = 0 is a possibility.

C0 = simplify(int(sawtooth, t, 0, 1)) % funny the above general formula is wrong for n=0

C0 =

What I do is identify the roots of the denominator of the output of int, before doing any simplification, and then do a piecewise and limit. The roots can be found using solve, for example, but here it's easy to do by hand

Cn = piecewise(n == 0, limit(Cn,n,0), Cn)

Cn =

Then simplify at the end

Cn(n) = simplify(Cn)

Cn(n) =

The drawback to this approach is that piecewise is not supported by matlabFunction

try
    Cnfun = matlabFunction(Cn);
catch ME
    ME.message
end
ans = 'Unable to generate code for piecewise for use in anonymous functions. Use option 'File' to write code for piecewise to a file instead.'

So I just evaluate Cn symbolically and convert to double (if numeric is preferred)

n = -10:10;

%Cnval = Cnfun(n);

Cnval = double(Cn(n));

% subs wasn't needed here

% Cnval(n==0) = subs(C0); % no idea why I need to do this

figure

hold on

stem(n, real(Cnval))

stem(n, imag(Cnval))

legend('real','imag')

xlabel('n')

ylabel('C(n)')

t = linspace(0,3,1000)';

f = @(t) mod(t,1);

fFourier = sum(Cnval.*exp(1i*2*pi*n.*t),2);

figure

plot(t, f(t), 'r', t, fFourier, 'b');

Warning: Imaginary parts of complex X and/or Y arguments ignored.

xlabel('t')

legend('f', 'Fourier approximation')

Bruno Luong 2023-9-9

@Paul thanks

jvfa 2023-9-10

hi bruno, after studying your comment and code i finally understand. thank you, alsp for paul for the correction

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plotting a function using a function

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