How Do You Apply Fourier Transform to a Truncated Convolution Integral?

[lost87]

Hey, this is my first post, great forum! You've really helped me a lot of times.
I have a problem solving an integro-differential equation. It involves a term of the form: integration over [t, +infinity) of f(s)*exp(t-s)ds.
I have to solve the equation using Fourier transform, and most of the other terms are ok, but I have a problem with this one. I know that if the limits of integration where (-infinity, +infinity) it would be a convolution and it's Fourier transform would be simple, but I don't know what to do in this case. I tried to find the Fourier transform using the logic of the proof of the convolution theorem, but I end up with a result with t and s. Do you have any ideas on what to do? Any ideas and help would be really welcome!
Thanks!

 

[Hootenanny]

Hey, this is my first post, great forum! You've really helped me a lot of times.
I have a problem solving an integro-differential equation. It involves a term of the form: integration over [t, +infinity) of f(s)*exp(t-s)ds.
I have to solve the equation using Fourier transform, and most of the other terms are ok, but I have a problem with this one. I know that if the limits of integration where (-infinity, +infinity) it would be a convolution and it's Fourier transform would be simple, but I don't know what to do in this case. I tried to find the Fourier transform using the logic of the proof of the convolution theorem, but I end up with a result with t and s. Do you have any ideas on what to do? Any ideas and help would be really welcome!
Thanks!

Hi lost87 and welcome to Physics Forums.

Could you perhaps post the full integro-differential equation so that we can take a look at it?

 

[MisterX]

So, what you are describing is like convolution, only integrating over s>=t. I think there is way you could restate that integral as an equivalent integral from negative infinity to infinitiy by using a different function in place of f(s).

 

[lost87]

Hi lost87 and welcome to Physics Forums.

Could you perhaps post the full integro-differential equation so that we can take a look at it?

Hey, the equation has the form: f'=af(t)+b(integral of f(s)*exp(t-s) ds from t to +infinity) +cf''+e(integral of f(s)*exp(-(t-s)) ds from -infinty to t). I can find the Fourier transform for most of the terms, but I don't know what to do with the terms with the integral. I hope this helps.

 

[lost87]

So, what you are describing is like convolution, only integrating over s>=t. I think there is way you could restate that integral as an equivalent integral from negative infinity to infinitiy by using a different function in place of f(s).

Hey, thanks for your reply, but I am not sure that I understand what you are suggesting. Will that involve a change of the variables s, t?

 

[MisterX]

Hey, thanks for your reply, but I am not sure that I understand what you are suggesting. Will that involve a change of the variables s, t?

No, I was suggesting replacing f(s) with a different function of s, so that the bounds of the integral could be changed to from -infinity to infinity, and the integral would be equal to the integral you described.

Is this a homework problem?

 

[lost87]

No, I was suggesting replacing f(s) with a different function of s, so that the bounds of the integral could be changed to from -infinity to infinity, and the integral would be equal to the integral you described.

Is this a homework problem?

Hey, thanks again for your reply. No it's not a homework problem, it's just something I am working on in my dissertation

 

[MisterX]

u(s-t) evaluates to zero when s < t

\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds + \int^{\infty}_{t}f(s)u(s-t)e^{t-s} = \int^{\infty}_{t}f(s)e^{t-s}ds
as
\int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds = 0

Now, to get the Fourier transform of \int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds, we have to figure out what two functions [of t] are being convolved. I hadn't thought this all the way through so I may have been misleading when I suggested replacing f(s). Actually, we'll have to group the unit step with the exponential to use the convolution theorem.

Let
g(\tau) = u(-\tau)e^{\tau}
so that
g(t-s) = u(s-t)e^{t-s}

and
\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{\infty}_{-\infty}f(s)g(t-s)ds

so

\int^{\infty}_{-\infty}( \int^{\infty}_{t}f(s)e^{t-s}ds)e^{-j\omega t}dt = \mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \mathcal{F}\left\{ f(t) * g(t) \right\}

= \mathcal{F}\left\{ f(t) \right\}\mathcal{F}\left\{ g(t) \right\} = \mathcal{F}\left\{ f(t) \right\} \mathcal{F}\left\{ u(-t)e^{t} \right\}

\mathcal{F}\left\{ u(t)e^{-t} \right\} = \frac{1}{1 + j\omega}
so
\mathcal{F}\left\{ u(-t)e^{t} \right\} = \frac{1}{1 - j\omega}

finally

\mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \frac{\mathcal{F}\left\{ f(t) \right\} }{1 - j\omega}

 

[lost87]

u(s-t) evaluates to zero when s < t

\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds + \int^{\infty}_{t}f(s)u(s-t)e^{t-s} = \int^{\infty}_{t}f(s)e^{t-s}ds
as
\int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds = 0

Now, to get the Fourier transform of \int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds, we have to figure out what two functions [of t] are being convolved. I hadn't thought this all the way through so I may have been misleading when I suggested replacing f(s). Actually, we'll have to group the unit step with the exponential to use the convolution theorem.

Let
g(\tau) = u(-\tau)e^{\tau}
so that
g(t-s) = u(s-t)e^{t-s}

and
\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{\infty}_{-\infty}f(s)g(t-s)ds

so

\int^{\infty}_{-\infty}( \int^{\infty}_{t}f(s)e^{t-s}ds)e^{-j\omega t}dt = \mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \mathcal{F}\left\{ f(t) * g(t) \right\}

= \mathcal{F}\left\{ f(t) \right\}\mathcal{F}\left\{ g(t) \right\} = \mathcal{F}\left\{ f(t) \right\} \mathcal{F}\left\{ u(-t)e^{t} \right\}

\mathcal{F}\left\{ u(t)e^{-t} \right\} = \frac{1}{1 + j\omega}
so
\mathcal{F}\left\{ u(-t)e^{t} \right\} = \frac{1}{1 - j\omega}

finally

\mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \frac{\mathcal{F}\left\{ f(t) \right\} }{1 - j\omega}

This is great! I can't explain how helpful this has been! Thanks for everything!