# Fourier Transform and Parseval's Theorem

1. Dec 10, 2015

### roam

1. The problem statement, all variables and given/known data
Using Parseval's theorem,

$$\int^\infty_{-\infty} h(\tau) r(\tau) d\tau = \int^\infty_{-\infty} H(s)R(-s) ds$$

and the properties of the Fourier transform, show that the Fourier transform of $f(t)g(t)$ is

$$\int^\infty_{-\infty} F(s)G(\nu-s)ds$$

2. Relevant equations
Fourier transform for $f(t)g(t)$ is defined as:

$$\int^\infty_{-\infty} f(t)g(t) e^{-2 \pi \nu t} dt$$

3. The attempt at a solution
So starting from the definition of Fourier transform:

$$\int^\infty_{-\infty} f(t)g(t) e^{-2 \pi \nu t} dt$$

So, do we need to ignore the exponential term here? If we ignore it, we can apply Parseval's theorem to get the frequency domain:

$$\int^\infty_{-\infty} f(t)g(t) dt = \int^\infty_{-\infty} F(s) G(- s) d s$$

Now, what property of the Fourier transform can I use to get $G(-s) \implies G(\nu-s)$?

I don't understand what the $(\nu-s)$ part means, does it indicates some sort of shifting or delay in the input?

Any help is greatly appreciated.

2. Dec 10, 2015

### Samy_A

I think you forgot an $i$ in the exponent in the definition of Fourier transform.

You cannot just ignore the exponential term in $\int^\infty_{-\infty} f(t)g(t) e^{-2 \pi iv t} dt$

Start with $G(v-s)$ for a fixed $v$.
Write out the formula for $G(v-s)$, and you will note that $G(v-s)=\mathcal F (g_v(-s))$ for some function $g_v$ ($\mathcal F$ denotes the Fourier transform).

Using that knowledge about $G(v-s)$, apply Parseval's theorem to $\int^\infty_{-\infty} F(s)G(v-s)ds$ and see what you get ...

It is clearer when you write something like:
the Fourier transform of $f(t)g(t)$ in $\nu$ is $$\int^\infty_{-\infty} F(s)G(\nu-s)ds$$

Last edited: Dec 10, 2015
3. Dec 11, 2015

### roam

Thank you for your input. In $g_v$ what does the subscript $v$ mean? Should it not be $g_t$ since $G(s)$ is the Fourier transform of $g(t)$? (In my course a function denoted by a capital letter is the Fourier transform of the function denoted by the corresponding small letter)

So from the definition, the formula for the $G(v-s)$ part is:

$$G(v-s) = \int^\infty_{-\infty} g(t) e^{-i 2 \pi st} dt$$

Here I used the duality property of the Fourier transform i.e. $F(t) \iff f(-s)$

Is this the right formula?

4. Dec 11, 2015

### Samy_A

There is no $v$ in your right hand side expression for $G(v-s)$, so that can't be correct.

I really meant $g_v$. Just pick one $v$ as a constant and let's work with that for a while.

In general, the definition of the Fourier transform is:
$$G(u) = \int^\infty_{-\infty} g(t) e^{-2 \pi iut} dt$$

I deliberately used another name for the variable here.
Now set $u=v-s$ in that definition. You get:
$$G(v-s) = \int^\infty_{-\infty} g(t) e^{-2 \pi i(v-s)t} dt=\int^\infty_{-\infty} \Big(g(t)e^{- 2 \pi ivt}\Big)e^{-2 \pi i(-s)t} dt$$

Compare that last expression with the definition of the Fourier transform. What we have there is the Fourier transform of the function between the big brackets at the point $-s$. So now name the function between the big brackets $g_v$, and what we have is that:
$$g_v(t)=g(t)e^{-2 \pi ivt}$$
$$G(v-s)=\mathcal F(g_v(-s))$$
Or, if you prefer the convention of your course:
$$G(v-s)=G_v(-s)$$
Now apply this to evaluate $\int^\infty_{-\infty} F(s)G(v-s)ds$, by using Parseval's theorem.

Last edited: Dec 11, 2015
5. Dec 12, 2015

### Samy_A

Never mind. I thought something was wrong, but I was wrong about that.

My sincere apologies for my confusion.

Last edited: Dec 12, 2015
6. Dec 12, 2015

### roam

I see. Thank you very much for the clear explanation.

$$\int^\infty_{-\infty} F(s) G(\nu -s) ds = \int^\infty_{-\infty} F(s) G_\nu (-s) ds$$

using Parseval's theorem this becomes:

$$\int^\infty_{-\infty} f(t) g_\nu (t) dt = \int^\infty_{-\infty} f(t) g(t) e^{-2 \pi i \nu t} dt$$

Is this then sufficient to 'show' that $\int^\infty_{-\infty} F(s) G(\nu -s)ds$ is the Fourier transform of $f(t)g(t)$?

7. Dec 12, 2015

### Samy_A

Yes, because taking the two equations together, you now proved that for any $\nu$: $$\int^\infty_{-\infty} f(t) g(t) e^{-2 \pi i \nu t}=\int^\infty_{-\infty} F(s) G(\nu -s) ds$$
The left hand side is by definition the Fourier transform of the function $f.g$ at point $\nu$.

As an aside, $\int^\infty_{-\infty} F(s) G(\nu -s) ds$ is called the convolution of F and G at $\nu$. (More here and here)

Last edited: Dec 12, 2015
8. Dec 12, 2015