Problem with dirac convulation and power of a signal

1. Aug 30, 2010

fridaycoffee

1. The problem statement, all variables and given/known data
1. i need to obtain the result of x[n] * x[n] (convolution)
where x[n] = δ[n] + δ[n-1]
2. I need to obtain the power of the signal x(t) = 4 cos (ω0t) + 2 sin (ω0t) + 2cos(4ω0t) for R = 1 k Ω

2. Relevant equations
1.Who do you convolute 2 diracs ?
2. what is the correct formula for the power ?

3. The attempt at a solution
1. i consider on dirac as an y[n] so that y[n]*δ[n] = y[n]

2.P = Σ | ak |2
where ak are the coefficients of the an Fourier exponential serie and we have this relationship: | ak | = (Ak) /2 ;
where Ak is sqrt (Ck + Sk)2

Ck and Sk beeing the coefficient of the trigonometric Fourier series

x(t) = C0 + Σ [Ak cos kwt + Sksin kwot]

So a solution would be :

P = [sqrt (42 + 22) /2 ]2 + 2/2 = 5 + 1 = 6 but not in the results that i have to choose between. if i sum Ak2 the result would be 24 that is found in the possible ones
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 30, 2010

marcusl

1. Write out the sequence delta(n). By the way, in discrete form it's not called a Dirac delta function but rather a unit impulse function (this should give you a hint to what delta(n) looks like). Then write delta(n-1). You should be able to picture the convolution as these slide past each other.

2. You forgot to square the last term.

3. Aug 31, 2010

fridaycoffee

1. so δ[n] would be a stick at n= 0 and δ[n-1] would be a stick at n=1... and next i do the convolution like this ? δ[n]*δ[n] + δ[n] * δ[n-1] + δ[n-1] * δ[n] + δ[n-1]* δ[n-1] = δ[n-1] + δ[n] but its not in the possible results (δ[n] * δ[n-1] beeing = 0; and δ[n] * δ[n] = δ[n], δ[n-1]* δ[n-1] = δ[n-1]
Am I wrong ?
2. i skipped that step: { sqrt (22 ) / 2 }2 = 1

all the answers are over 12

4. Aug 31, 2010

marcusl

1. You have correctly identified x(n). Next you're going to slide ...00011000... past itself, and at each lag (relative offset between them), multiply element by element and sum.

2. Think about the power of a sinusoid of amplitude A; it's A^2/2, not (A/2)^2.

5. Sep 1, 2010

fridaycoffee

1. if i do that wouldnt it be the sum i stated: δ[n]*δ[n] + δ[n] * δ[n-1] + δ[n-1] * δ[n] + δ[n-1]* δ[n-1] ? and the answer: 2δ[n] + 2δ[n-1]? ..but i have hard time on the result of a multiplication of two diracs or impulses..which term do i keep the first or the second one.

2. maybe but if | ak | = (Ak) /2 ..wouldnt | ak |2 =( Ak /2 )2

6. Sep 1, 2010

marcusl

1. The autocorrelation sequence, also called circular convolution, is ...000121000... Think about lining x up as I mentioned. For lag=1 you have
00011000..
00110000..
or one.

For zero lag
00011000..
00011000..
the convolution is 2.

2. Where did your equation come from? What is the power of Acoswt?

7. Sep 1, 2010

fridaycoffee

1. what does the 000121000 sequence mean ..im not familiarized with this notation..one of the reasons i did not quite understood your answer

2. there are different formulas for the power of the signal

a. P = 1/T integral[ Xp (t)2]

b. P = Σ|ak|2

c. P = u(t)2 / R

The formulas where in the signal course i attended in the 5 semester
and in one book, via Parseval's Theorem

So i thought the b formula is the most appropriate for x(t) = 4 cos (ω0t) + 2 sin (ω0t) + 2cos(4ω0t)

the power of Acoswt would be P = (A/2 ) 2

8. Sep 1, 2010

marcusl

It is the convolution at different lags, that is, different values of offset between x and a copy of x. The convolution at zero lag is 2, which is the answer to your question. Sorry if that was confusing.

If you work out your definition a) above, you'll convince yourself that the power contained in a cosine wave Acoswt is A^2 / 2. This should give you the right answer.

9. Sep 2, 2010

fridaycoffee

1. i meant what 000121000 i terms of δ[n], for zero leg 00022000 = 2δ[n] + 2δ[n-1] ? but 00012100 ? what does "1", "2" , "1" mean in terms of a δ[n] this might be a question at my master degree admision :)
2. Maybe youre right :) , i just applied the formula, and if i do that if |ak| = ( Ak ) /2 , it would be 22, or the formula isnt like that, i'm very confused

10. Sep 2, 2010

marcusl

1. Your sequence x[n] for -infinity < n < infinity is ...000011000..., right? Because delta[n] is a unit impulse that equals 0 except at n=0 when it equals 1, and delta[n-1] is zero except at n=1. To convolve it with itself, line x up with a copy of itself with an offset (or lag) m.
multiply term by term, and sum. See
http://en.wikipedia.org/wiki/Convolution" [Broken]
You get zeros unless m=-1, in which case you have
..001100..
..011000..
and the convolution is 1. When m=0
..001100..
..001100..
you get 2. You can work out m=1.

2. The power of a wave is the square of the rms amplitude. You are given the peak amplitude A of a sin or cos wave, so the rms amplitude A/sqrt(2). In your problem, the power is then sum{(A_k^2)/2}, so the Fourier power of x(t) is 12. You state in your problem that there is a R=1 k resistor, so the electrical engineering formula P = (Vrms)^2 / R gives 0.012W or 12 mW.

Last edited by a moderator: May 4, 2017
11. Sep 2, 2010

fridaycoffee

1. Oh i got it now :) the first 1 from 011000 was for delta[n+1] but what you mean by 1, 2? one term,2 terms?

a) 2delta[n] + 2delta[n-1]
b)delta[n] + 2delta[n-1] + delta[n-2]
c)2delta[n-1] + 2delta[n-2]
d) delta[n] + delta[n-1] + delta[n-2]
e)delta[n-1] - 2delta[n-2]
f) delta[n-2] + delta[n-3]

The right answer in my opinion would be a)
delta[n]*delta[n] + delta[n] * delta[n-1] + delta[n-1] * delta[n] + delta[n-1]* delta[n-1] = 2delta[n] + 2delta[n-1]

delta[n]*delta[n] = delta[n]
delta[n-1]*delta[n-1] = delta[n-1]

but i dont know how to figure out delta[n]*delta[n-1]= ? it could be either one of the terms the result? i choose it in order to match the possible answers?

2. Thnks you're answer seems right

And thanks a lot for you're replies. It really helped a lot.

12. Sep 2, 2010

marcusl

No the answer is b. Writing sequence b explicitly gives ..000121000...

I am not explaining this well for you. Please see your TA or professor for clarification. They can explain the meaning of the unit impulse and of convolution to you one-on-one much more effectively than I can through this forum.

Last edited: Sep 2, 2010
13. Sep 2, 2010

fridaycoffee

Ok..so for m=-1 the second x[n] = delta[n] + delta[n-1] shifts into delta[n-1] + delta[n-2].
The lag m=-1 is standard for circular convolution ?
For m= 1 the answer would be delta[n+1] + 2delta[n] + delta[n-1] but for what would it stand ?

I dont have the chance to consult a teacher :(.

14. Sep 2, 2010

marcusl

Yes, exactly.
Yes again.
Do you understand the meaning of delta[n]? It is the sequence
delta[n] = 1 for n=0
delta[n] = 0 for other n values

15. Sep 2, 2010

fridaycoffee

i know what it means :)
i thought there are other specific convolutions..
for example what would be the lag for a normal convolution if it follows the same algorithm or it doesnt?

16. Sep 2, 2010

marcusl

I have no idea what you just asked. Did you read the wiki article on convolution?

17. Sep 3, 2010

fridaycoffee

i mean for example if i have [delta[n] + delta[n-2] ] * [delta [n+2] + delta[n+3]]

i continue adding lag until they meet? i misthought about specific lags for non autocorrelation convolution; i just apply the convolution formula

Last edited: Sep 3, 2010
18. Sep 4, 2010

marcusl

I still don't understand your question "i continue adding lag until they meet?"
Just apply the convolution formula:

$$[f*g](n)=\sum_m f(m)g(n-m)$$

You get f*g=delta(n-2)+delta(n-3)+delta(n-4)+delta(n-5)
or ...0001111000...

Last edited: Sep 4, 2010
19. Sep 12, 2010

fridaycoffee

delta(n-2)+delta(n-3)+delta(n-4)+delta(n-5) ... i dont know how to obtain this answer

i made an image with the convolution of [delta[n] + delta[n-2] ] * [delta [n+2] + delta[n+3]]

Maybe you can explain better the result on this

[PLAIN]http://img31.imageshack.us/img31/4100/12092010941.jpg [Broken]

Last edited by a moderator: May 4, 2017
20. Sep 13, 2010

fridaycoffee

i found an easier way of solving any kind of convolution
thnks anyway for all you help