Defining a symbolic function with an integral

Hi all,

I'm trying to define a symbolic function to represent the continuous time Fourier transform given by,

with the following code:

syms f;
CFTx(f) = int(x(t)*exp(-1i*2*pi*f*t), t, -Inf, Inf);

However, when I attempt to evaluate CFTx(s) at a some point s, say s = 10 the following is returned (directly from my command window):

CFTx(10)

ans =

int(exp(-pi*t*20i)*(5*cos(15*pi*t) + 2*sin(12*pi*t) + 5*exp(-(t - 25)^2/2)), t, -Inf, Inf)

I have tried evaluating this expression using the subs() function but so far have not had any luck.

Any advice is much appreciated.

Answer 1

[Star Strider]

Your approach is correct, Your implementation needs to substitute the limits of integration with symbolic variables you can then substitute later.

Example —

syms f t T 
x(t) = (5*cos(15*pi*t) + 2*sin(12*pi*t) + 5*exp(-(t - 25)^2/2));
CFTx(f) = int(x(t)*exp(-1i*2*pi*f*t), t, -T, T);
CFTx(f,T) = simplify(CFTx, 'Steps',20)

Then substitute for appropriate values of ‘f’ and ‘T’.

[David Beallor]

Unfortunately I'm still seeing the same results. Here's my code:

syms t;
x(t) = 5*exp(-(t-25)^2/2);
n(t) = 5*cos(15*pi*t) + 2*sin(12*pi*t);
y(t) = x(t) + n(t);

syms s T;
CFTy(s) = int(y(t)*exp(-1i*2*pi*s*t), t, -T, T);
CFTy(s,T) = simplify(CFTy, 'Steps', 20);

a = subs(CFTy(s,T), [s T], [10, Inf]);

disp(a)

and here's the output...

int(exp(-pi*t*20i)*(5*cos(15*pi*t) + 2*sin(12*pi*t) + 5*exp(-(t - 25)^2/2)), t, -Inf, Inf)

Can you think of what else might be causing this?

Thanks a lot for your help!

[Star Strider]

My pleasure!

Most likely, you cannot have infinite limits, especially with a time-domain function that is periodic and may have an arbitrary value at ±Inf. I would choose a finite value for ‘T’, and then choose a frequency for the evaluation. Note that ‘pure’ periodic functions ‘exist’ in the Fourier transform only at the individual frequencies at which they are defined. They do not exist elsewhere.

I would do this:

a = simplify(subs(CFTy(s,T), [s], [10]), 'Steps',20)

and then substitute a finite value of ‘T’.

If you want to get the Fourier transform with infinite limits, use the Symbolic Math Toolbox fourier function:

Y(w) = fourier(y, t, w);
Y(w) = simplify(Y, 'Steps',20)

Y(w) =
5*pi*(dirac(w - 15*pi) + dirac(w + 15*pi)) - pi*(dirac(w - 12*pi) - dirac(w + 12*pi))*2i + 5*2^(1/2)*pi^(1/2)*exp(- (w + 25i)^2/2 - 625/2)

The dirac functions are due to the periodic functions existing only at those frequencies.

Plot or evaluate ‘Y(w)’ at the values of ‘w’ (‘omega’ or 2*pi*f) you want.