Linear system, input/output signal

[fluidistic]

Homework Statement

I'm stuck on the following problem:
A linear system has an output of $G(\omega ) e^{-i \omega t}$ to an input signal of $e^{-i\omega t}$ where omega is arbitrary.
If the input signal has the form $f(t) = \begin{cases} 0 \\ e^{-\lambda t} \end{cases}$ where lambda is a constant, then the output signal is $F(t)= \begin{cases} 0 \\ (1-e^{-\alpha t })e^{-\lambda t} \end{cases}$ where alpha is another constant.
1)Calculate $G(\omega )$.
2)Calculate the output signal of the input signal $f(t)=A \delta (t)$.

Homework Equations

Not really sure.

The Attempt at a Solution

Let $\lambda t =i\omega t \Rightarrow \lambda = i \omega$. Since alpha and lambda are constants, I can write $\frac{\alpha}{\lambda}=c$. Therefore $G(\omega ) =1-e^{-c\lambda t}=1-e^{-\frac{i\alpha \omega t}{\lambda}}$.
This would be my answer for part 1). Does this look correct?
2)I've no idea how to solve this. I guess I must use my expression for $G (\omega )$ but I really don't see how this help.
Any tip is welcome.

 

[I like Serena]

Hi fluidistic,

I suspect λ is supposed to be real.
In that case your answer to (1) is not correct.

If you have a linear system, the typical way to deal with it, is to make Fourier transforms.
If the system gets an input i(t) and has an output o(t), then the response function H(ω) is given by:
$$H(ω) = \frac {\mathscr{F}_t[o(t)](ω)} {\mathscr{F}_t[i(t)](ω)}$$
Do you know how to calculate that in your respective cases?

 

[fluidistic]

Hi fluidistic,

I suspect λ is supposed to be real.
In that case your answer to (1) is not correct.

If you have a linear system, the typical way to deal with it, is to make Fourier transforms.
If the system gets an input i(t) and has an output o(t), then the response function H(ω) is given by:
$$H(ω) = \frac {\mathscr{F}_t[o(t)](ω)} {\mathscr{F}_t[i(t)](ω)}$$
Do you know how to calculate that in your respective cases?

Hi and thanks for the lesson, I had no idea about this!
But I'm totally confused. Is your $H (\omega )$ what they call $G( \omega )$?

 

[I like Serena]

Hi and thanks for the lesson, I had no idea about this!
But I'm totally confused. Is your $H (\omega )$ what they call $G( \omega )$?

Yes.
TBH, I expected that you were studying exactly this when you came up with this problem.
So I'm a bit surprised that you aren't.

I avoided the use of your G(ω) since its introduction in your problem is unusual to me.
However, if you do the math you'll see that G(ω) and H(ω) are the same.

Btw, as an alternative to the Fourier transform you could also use the laplace transform.
The theory is the same. Both transforms are identical with just a different choice of the arguments.
Seeing that you have another thread on the laplace transform, did you perhaps study that?

 

[fluidistic]

Yes.
TBH, I expected that you were studying exactly this when you came up with this problem.
So I'm a bit surprised that you aren't.

I avoided the use of your G(ω) since its introduction in your problem is unusual to me.
However, if you do the math you'll see that G(ω) and H(ω) are the same.

Btw, as an alternative to the Fourier transform you could also use the laplace transform.
The theory is the same. Both transforms are identical with just a different choice of the arguments.
Seeing that you have another thread on the laplace transform, did you perhaps study that?

Ok thanks. Basically I'm self studying a mathematical course (in view of taking a final exam in December); I'm solving all the problems I can from a list of problems on a sheet given by a professor. That's the only exercise talking about "signals", I had never seen this before.
Ok so yeah I've showed that H(\omega ) = G (\omega ). I in fact reached $H(\omega ) =G (\omega ) \frac{\int _{-\infty }^\infty e^{-2i \omega t dt }}{\int _{-\infty }^\infty e^{-2i \omega t dt }}=G(\omega )$.
For part 1), I reached that $G(\omega ) = \frac{\alpha}{\alpha + \lambda + i \omega}$. (I hope it's OK, if not I'll post my work).
I'll try part 2) now, thanks so far.

Edit: I notice I forgot to mention in my OP that f(t) and F(t) are worth 0 for t<0 and are different from 0 for t>0!

Edit 2: For part 2) I get that o(t) is worth $\frac{A\alpha }{2\pi}\int _{-\infty}^{\infty } \frac{e^{i\omega t }}{\alpha + \lambda + i \omega }d\omega$. This does not make any sense, I've made some errors somewhere.

 

[I like Serena]

Part 1 is correct.
Good!

For part 2 you should look up the inverse Fourier transform instead of trying to calculate it yourself.
Didn't you look it up for part 1 as well? I did.

You can find it for instance here on wiki:
http://en.wikipedia.org/wiki/Fourier_transform#Square-integrable_functions

 

[fluidistic]

Part 1 is correct.
Good!

For part 2 you should look up the inverse Fourier transform instead of trying to calculate it yourself.
Didn't you look it up for part 1 as well? I did.

You can find it for instance here on wiki:
http://en.wikipedia.org/wiki/Fourier_transform#Square-integrable_functions

Ok thanks!
Well no, I have not looked into any table for part 1). So I looked at the wikipedia's article for part 2) and the answer is ... $o(t)=A\alpha \Omega (t)e^{-(\lambda + \alpha ) t}$ where the omega function is the Heaviside step "function". I hope this result is right.
Unfortunately we cannot look at any integral/DE/integral transforms tables during the final exam (worth 100% of the grade). So I guess I'll have to solve more problems than what they give us.
Thank you very much for all your help I like Serena!

 

[I like Serena]

I'm kind of curious.
In this problem a number of the symbols deviate from the conventions I'm used to.

Just now, you used $\Omega(t)$ for the Heaviside step function.
But I only know $H(t)$, $u(t)$, or $\Theta(t)$ to be used for this function.
Where does the $\Omega(t)$ come from?

As for solving the integral, it helps when you realize that in part 1 you already did the same transform in the opposite direction.

 

[fluidistic]

I'm kind of curious.
In this problem a number of the symbols deviate from the conventions I'm used to.

Just now, you used $\Omega(t)$ for the Heaviside step function.
But I only know $H(t)$, $u(t)$, or $\Theta(t)$ to be used for this function.
Where does the $\Omega(t)$ come from?

As for solving the integral, it helps when you realize that in part 1 you already did the same transform in the opposite direction.

Sorry for the "omega" rather than the "theta" convention for the Heaviside step function, I made a mistake in calling it omega while it was a capital theta, my fault.
And yes you are right, I did not notice I did the same transform but in the opposite direction. So my result looks good (maybe I made a sign mistake in the last result though. I'll have to recheck this up).

 

[I like Serena]

Ah, okay.
Btw, your result for part 2 is also correct.

 

[fluidistic]

Ah, okay.
Btw, your result for part 2 is also correct.

Ok thanks for all.