Fourier transform using Convolution

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Nurhan Aydinalp 2020-12-23

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/701022-fourier-transform-using-convolution

编辑： Paul 2020-12-24

I have two signals x(t) = sin(2.*pi.*t)/(pi.*t) and y(t) = x(t) I want to calculate z(t) = x(t)*y(t) and z(JW).I should plot x(t), x(JW), y(t), y(JW) and z(t), z(JW) using subplot. z(JW)=(1/(2pi))*(convolution(x(t),y(t))), I have the following code:w = [-6.*pi 6.*pi];

syms x(t)
x(t) = sin(2.*pi.*t)./(pi.*t);
 
subplot(3,2,1)
fplot(t,x(t));
title('x(t) vs t');
xlabel('time');
ylabel('x(t)')
 
X_J_W = fourier(x(t));
subplot(3,2,2)
fplot(X_J_W,w);
title('X(JW) vs w')
ylabel('X(JW)')
xlabel('W')
 
syms y(t)
y(t) = sin(2.*pi.*t)./(pi.*t);
subplot(3,2,3)
fplot(t,y(t));
title('y(t) vs t');
xlabel('time');
ylabel('y(t)')
 
Y_J_W = fourier(y(t));
subplot(3,2,4)
fplot(Y_J_W,w);
title('Y(JW) vs w')
ylabel('Y(JW)')
xlabel('W')
 
syms z(t)
z(t) = x(t).*y(t);
subplot(3,2,5)
fplot(z(t));
C_X_Y = conv(X_J_W,Y_J_W,'full');
 
Z_J_W = (1./(2.*pi)*(C_X_Y));
subplot(3,2,6)
fplot(Z_J_W,w)

in the convolution part I get

Error using conv2
Invalid data type. First and second arguments must be numeric or logical.
Error in conv (line 43)
c = conv2(a(:),b(:),shape);

and I do not know how to fix it.

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回答（1 个）

Matt J 2020-12-23

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/701022-fourier-transform-using-convolution#answer_582852

You must use int to implement a symbolic convolution integral. conv is for numeric convolution.

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Paul 2020-12-23

在 MATLAB Online 中打开

Code as posted results in Z_J_W being a function of both t and tao, at least when I run it in 2019A. Hence fplot doesn't know which variable to plot against or what value to assume for the other.

I think that the expression for C_X_Y is not correct. I think it's integrating from tao to t wrt to t, when it should be integration from 0 to t wrt tao (also, see comment below). I think this would be the correct syntax for the integration and to make C_X_Y a function:

C_X_Y(t) = int(x(tao).*y(t-tao),tao,0,t);

However, when I run this I don't get a nice closed form expression for C_X_Y.

Similarly, you probably want

X_J_W(w) = fourier(x(t));

and the same for Y_J_W.

z(t) is coded as the product of x(t) and y(t). Then Z_J_W is coded as some scaled value of the convolution of x(t) and y(t), which doesn't really follow assuming that Z_J_W is supposed to be the Fourier transform of z(t).

Looking back at your original question, it's not clear what you're trying to do.

Is z(t) the convolution of x(t) and y(t)? In which case Z_J_W = X_J_W * Y_J_W

Or is z(t) the the product of x(t) and y(t)? in which case Z_J_w = (1/2/pi)*convolution(X_J_W(w),Y_J_W(w))

The expression for the convolution of x(t) and y(t) implies that both are zero for t < 0. But x(t) and y(t) are not defined that way, so the call to fourier just assumes that x(t) and y(t) are nonzero for t < 0.

Matt J 2020-12-24

在 MATLAB Online 中打开

Truncating the convolution seems to help:

syms x(t) y(t) z(t) c(t) X(w) Y(w) tau

x(t) = sin(2.*pi.*t)./(pi.*t);

y(t) = sin(2.*pi.*t)./(pi.*t);

z(t)=x(t).*y(t);

X(w) = fourier(x(t));

Y(w) = fourier(y(t));

c(t)=int(x(tau).*y(t-tau),tau,-100,+100);

fplot(c(t))

Paul 2020-12-24

编辑：Paul 2020-12-24

在 MATLAB Online 中打开

Nurhan,

Why compute the convolution of x(t) and y(t)? I thought the problem at hand is related to the product of x(t) and y(t).

If z(t) = x(t)y(t), then

Z(w) = conv(X(w),Y(w))/2/pi:

>> syms u
>> Z(w)=int(X(u)*Y(w-u),u,-inf,inf)/2/pi;
>> Z(w)
 
ans =
 
-((heaviside(- w - 4*pi)*(w + 4*pi))/2 - w*heaviside(-w) + (heaviside(4*pi - w)*(w - 4*pi))/2)/pi
 
>> fplot(Z(w),[-20 20])

The result can be confirmed by numerically computing the Fourier transform of z(t):

>> fun=matlabFunction(z(t)*exp(-1j*w*t));
>> wr=-20:.1:20;
>> for ii=1:numel(wr),q(ii)=integral(@(t)fun(t,wr(ii)),-20,20);end
>> hold on
>> plot(wr,real(q),'ro'),grid