MHB Fourier Transform of Periodic Functions

[Joppy]

A tad embarrassed to ask, but I've been going in circles for a while! Maybe i'll rubber duck myself out of it.

If $$f(t) = f(t+T)$$ then we can find the Fourier transform of $$f(t)$$ through a sequence of delta functions located at the harmonics of the fundamental frequency modulated by the Fourier Transform of the restricted function $$F_r(\omega).$$

$$f(t) \leftrightarrow F_r(\omega) \frac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta(\omega - \frac{2n\pi}{T})$$

Find the Fourier Transform of the following function,

$$f(t) = sin(t) + cos(t)$$

In order for this to be transformable, let's define our restrictive function $$f_r(t)$$ to be,

$$f_r(t) =

\left\{ \begin{array}{rl}
sin (t) + cos (t) &\mbox{ $$-\frac{1}{2}T \le t < \frac{1}{2}T$$} \\
0&\mbox{ otherwise}
\end{array} \right.

$$

This function is periodic in $$2\pi (T =2\pi)$$, and is neither odd or even.

Since $$f_r(t)$$ is absolutely integrable, we can find its Fourier Transform.

$$F_r(\omega) = \int_{-\frac{1}{2}T}^{\frac{1}{2}T} \,cos (t)e^{-i\omega t} + sin (t)e^{-i\omega t} dt$$

Now I've tried a few things from here to try make the process faster, but i can't seem to find any simplifications, especially since the function is neither odd or even, there can't be any cancellation of terms.

Of course I've tried brute forcing,$$F_r(\omega) = \int_{-\pi}^{\pi} \,cos (t)[cos(\omega t)) + isin(\omega t)] + sin (t)[cos(\omega t)) + isin(\omega t)]dt$$

and so on... But i yield nonsense results. If however this is the right, or at least one way of going about it, please let me know as I've probably just messed up the integration!

Furthermore, what do we do about $$\omega$$? Doesn't $$\omega = 2\pi f = \frac{2\pi}{T} = 1$$ in this case?

I can't seem to find many resources outside of my notes that deal with F.T of periodic functions in this way.. Hence, my needing to consult you guys! :).

Thanks for your time.

EDIT: I'm guessing i should take advantage of the linearity property and find transforms for sin(t) and cos(t) separately (restricted of course).

 

[Joppy]

Ok so i got $$F_r(\omega) = \pi (1+i)$$ by brute force.. Which gives,

$$f(t) \leftrightarrow \pi (1+i) \sum_{n=-\infty}^{\infty} \delta (\omega - n)$$..

But the answer is,

$$\pi(1-i) \delta(\omega - 1) + \pi(1+i)\delta(\omega + 1)$$

 

[I like Serena]

Hey Joppy! ;)

A periodic function has a Fourier series.
That is, it can be written as the sum of sines and cosines at discrete frequencies, which is exactly what you're doing here.

In your example $f(t)=\cos t + \sin t$, we have $T=2\pi$.
Btw, we shouldn't restrict $f$ - it's supposed to be periodic!

Obviously, this $f(t)$ is already the sum of sines and cosines.
We can write it as:
$$f(t) = \frac{a_0}2 + \sum a_n\cos nt + \sum b_n\cos nt$$
where $a_0=0, a_1=b_1=1$, and all other coefficients are zero.

From the wiki page, we can see that we can also write:
$$f(t) = \sum_{n=-\infty}^\infty c_n e^{-int}$$
where $c_0=0,\quad c_1=\frac 12{a_1-ib_1}=\frac 12(1-i),\quad c_{-1}=c_1^*=\frac 12(1+i)$, and all other coefficients are zero.

That is, we can write:
$$f(t) = \frac 12(1-i)e^{it} + \frac 12(1+i)e^{-it}$$

In other words, your $F_r(\omega)$ is:
$$F_r(\omega)= \begin{cases}
\frac 12(1-i) &\text{if } \omega=1 \\
\frac 12(1+i) &\text{if } \omega=-1 \\
0 &\text{otherwise}
\end{cases}$$
Oh, and you may have some normalization constant that we still need to multiply by or divide by, depending on which version of the Fourier transformation you're using.

Note that these are not delta functions.

 

[Joppy]

Hey Joppy! ;)

A periodic function has a Fourier series.
That is, it can be written as the sum of sines and cosines at discrete frequencies, which is exactly what you're doing here.

In your example $f(t)=\cos t + \sin t$, we have $T=2\pi$.
Btw, we shouldn't restrict $f$ - it's supposed to be periodic!

I'm aware of this. And we should restrict f if we want to take the transform of it (using the definition i posted).

Alternatively we can do what you have done, and just find the F.S of our function, and take the transform of it. But i think using

$$f(t) \leftrightarrow F_r(\omega) \frac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta(\omega - \frac{2n\pi}{T})$$

we can sneakily avoid having to take the Fourier series and just re-define our function for one period, and find the transform of that (since after this, the function will be totally integrable and will have a F.T.)?

Note that these are not delta functions.

But we can express them as delta functions?

 

[I like Serena]

I'm aware of this. And we should restrict f if we want to take the transform of it (using the definition i posted).

Alternatively we can do what you have done, and just find the F.S of our function, and take the transform of it. But i think using

$$f(t) \leftrightarrow F_r(\omega) \frac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta(\omega - \frac{2n\pi}{T})$$

we can sneakily avoid having to take the Fourier series and just re-define our function for one period, and find the transform of that (since after this, the function will be totally integrable and will have a F.T.)?

But we can express them as delta functions?

We should only restrict ourselves to the period over which we integrate.
As I said, we're getting a transform for a Fourier series, which is based on the concept of having a periodic signal.
That's actually how mister Fourier originally set it up.
If we restrict the function to be zero outside of one period, that's not what we're doing.

With your definition of $F_r(\omega)$, which is not the actual transform, since it's multiplied by a sum of delta functions, they cannot be expressed as delta functions.
The actual transform does have delta functions.
Do you want the actual transform? Or the $F_r(\omega)$ as you've defined it?

 

[Joppy]

Using the following Fourier Transform pairs we can easily find $F(\omega)$;

$cos(\Omega t) \rightleftharpoons \pi (\delta(\omega - \Omega) + \delta(\omega + \Omega))$

$sin(\Omega t) \rightleftharpoons - i \pi (\delta(\omega - \Omega) - \delta(\omega + \Omega))$.