# Power spectral density of signal

Discussion in 'Electronics Homework Help' started by evol_w10lv, Nov 13, 2013.

1. ### evol_w10lv

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Feb 19, 2013
First of all, I have to calculate power spectral density S(jw) of this signal:

It looks something like that:

t>=0
But it seems to me, that I can't use this standart formula:

So.. I see that there is signal multiplication- exponent and sinus, but still not clear how to get spectral density S(jw).
Any ideas?

2. ### Harald KappModeratorModerator

10,071
2,152
Nov 17, 2011
Why not? Of course it is the equation to use.
Since s(t)=0 for t<0 the integral ranges only from 0...infinity and since exp(-jwt) -> 0 for t->infinity you can solve the integral using the standard rules of analysis.

3. ### evol_w10lv

73
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Feb 19, 2013
Also I noticed, that we can't use the same "w" in exponent and sinus.
So, when I use this formula, then I get:

But what about range? Yes, it's from 0 to infinity, but I still don't understand some maths there. I guess, that when limit is 0, than all equation is 0. But what about infinity?
-U*exp(-t*(a+jw)) = -U [when t=inf]
What abot cos(t*w0) and sin(t*w0), when t=inf?

4. ### Harald KappModeratorModerator

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Nov 17, 2011
what is exp(-a*t) for t=infinity?

5. ### evol_w10lv

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Feb 19, 2013
It's 0, but then all equation is 0, when sin(w*t) = 0 and cos(w*t)=0.
Now I'm trying to find solution in other way, using this property:

I think that I can try in this way, because s(t) is multiplication of two signals.
S1(t) = U*e^(-a*t) and S2(t) = sin(w0*t)

6. ### Harald KappModeratorModerator

10,071
2,152
Nov 17, 2011
If
integral f(t)dt = F(t)
then
integral from 0 to infinity (f(t) dt = F(infinity)-F(0)

Even if F(infinity)=0, the first term, F(0) is not.

7. ### evol_w10lv

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Feb 19, 2013
Yes, of course.. again I made akward mistake in my notes.
Then result shoud be:

But it don't give same signal as it was at the beggining, when I use inverse formula:

What could be wrong there?

8. ### Harald KappModeratorModerator

10,071
2,152
Nov 17, 2011
I'm sorry, I'll be off for the rest of the day.

My math handbook tells me:
integral (exp(ax)*sin(bx)dx = exp(ax)/(a^2+b^2)*(a*sin(bx)-b*(cos(bx))
and also exp(ax)*exp(bx) = exp((a+b)x)
These two equations should be sufficient to solve the task.

9. ### evol_w10lv

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Feb 19, 2013
I found out, that we can't just use standart formula, because a sine wave spectral density has only a delta function at the carrier frequency since the signal contains just
one spectral component namely the carrier frequency.
Here is something simular, but there's used cosinus.

For now I haven't got ideas, how to continue this task. Of course I'll try to think something.

10. ### Harald KappModeratorModerator

10,071
2,152
Nov 17, 2011
Sorry, that's only partly correct.
Right, a pure sine in the time domain evaluates to a delta function in the frequency domain.
But you don't have a pure sine. Yours is modulated by an exponential term. This leads to a spread of the energy in the frequency domain. Look up modulation for a more detailed explanation.
If you use the equations i put here yesterday, use the second one to combine the two exp() terms under the integral into one exp term. Then you are left with an equation like the first I gave you. This can be integrated. Insert the limits (0, inf) and evaluate to get the solution.

11. ### evol_w10lv

73
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Feb 19, 2013
I integrated with your equations and here is result:

It's the same as before:

If I'm right with all limits, then it should be the correct answer, but, when I use this formula:

(inverse formula of spectral density or inverse Fourier trasnsform formula)
I should get back s(t) as it was at the beginng, but it's not. Maybe you can see my mistake somewhere in calculations? Or maybe I can't use inverse formula in this case?

12. ### Harald KappModeratorModerator

10,071
2,152
Nov 17, 2011
What do you get instead? Where's the difference to f(t)?

Depending on whether you transform to F(f) or F(jw) different factors come into play which have to be taken into account when transforming back from F(f) or F(jw) to f(t).

13. ### evol_w10lv

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Feb 19, 2013
Just noticed, that I have to integrate back by frequency dw.
Then formula should be:

But I don't know, how to integrate this and also w0 befrore depended by time also.
It seems to me, that you obviously know about this stuff.. so I can believe for you, that after integration I should get back s(t) as it was at the beginng.
I had doubts about sinus and delta function, but actually you explained it in post #10.
If we are speaking about requirement for the integral to converge, I wasn't convinced about this method, because I'm not sure wheather this 'infinity' criterie is acceptable or not when we got sinus.

As I understand, you are sure, that it is correct answer?

14. ### Harald KappModeratorModerator

10,071
2,152
Nov 17, 2011
That doesn't show in your original equation. It will complicate matters as you will have to put w0=w(t) into the equation.
For the start, set w0=constant.
The denominator under the integral is a constant and can be placed outside the integral. That leaves you with a function exp(iwt)/(a^2+2iwa-w^2+w0^2). Lok up possible solutions in your math handbook. You may need to apply some of the laws of integration.
Also note that exp(iwt)=cos(wt)+i*sin(wt) which may come n handy when the result of your integration contains exp-functions that you want to convert to a sine.

Well, I haven't checked in detail (no time), but it looks reasonable. You may wan to look at Wolframalpha which will take some of the mechanical calculations from you. You will not find the full solution there, however.

Good luck, I'll be off now for the weekend.

Last edited: Nov 15, 2013

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Feb 19, 2013