# Finding the sampled signal - Fourier series and integration problems.

1. Mar 5, 2013

### Evo8

1. The problem statement, all variables and given/known data
The signal g(t) is band limited to B Hz and is sampled by a periodic pulse train $PT_{s}(t)$ made up of a rectangular pulse of width $1/8B$ second (centered at the origin) repeating at the nyquist rate (2B pulses per second). Show that the sampled signal $\bar{g}(t)$ is given by

$\bar{g}(t)=\frac{1}{4} g(t)+\sum_{n=1}^\infty sin(\frac{n \pi}{4}) g(t) cos(4nBt)$

2. Relevant equations
$$a_0=\frac{1}{T_0} \int_{T_0} g(t) dt$$
$$a_n=\frac{2}{T_0} \int_{T_0} g(t) \cos(n\omega_0 t) dt$$
$$b_n=\frac{2}{T_0} \int_{T_0} g(t) \sin(n\omega_0 t) dt$$

3. The attempt at a solution
Ok so my fourier series and integration skills are weak at best. Ontop of that im rusty. I think i know how to do this but i get stuck on some simple but paramount steps.

In a similar problem from a previous edition text they give the following pulse train. $PT_s(t)=C_0+\sum_{n=1}^\infty \cos(n\omega_s t)$ I see how this helps as it is in the same form as the sampled signal im trying to find $\bar{g}(t)$.

My book had the following to work with
$$\bar{g}(t)=g(t)\delta T_s(t)=\sum_{n} g(nT_s) \delta(t-nT_s)$$
How do they get the pulse train from this? I just dont see it.

For the first coefficient $a_0$ I get something like this:$$a_0=\frac{1}{T_0} \int_{\frac{-1}{16B}}^\frac{1}{16B} g(t) dt$$
$$2B\int_{\frac{-1}{16B}}^\frac{1}{16B} g(t) dt$$
$$=2B[\frac{1}{16B}+\frac{1}{16B}]$$
$$=\frac{1}{4}$$
Which is the first term in $\bar{g}(t)$
Now i got the limits for the integral from the pulse width of $\frac{1}{8B}$ being centered at the origin. So the equation would be 1 between $\frac{-1}{16B} and \frac{1}{16B}$

Is this correct? Am I on the right track? I have my doubts....

Thanks for any help!

2. Mar 5, 2013

### rude man

Can you find out if the pulses forming the sampled signal $\bar{g}(t)$ have flat tops or do they follow the contour of g(t) for the duration of a pulse? If using finite-width sampling pulses this distinction is important.

Never mind, I think I know.

This is just multiplying your g(t) by the Fourier series of the described pulse train.

The pulse train has unit height, is centered at t = 0, has width w = 1/8B and period T = 1/2B. From this, can you write the Fourier series of the pulse train?

Last edited: Mar 5, 2013
3. Mar 6, 2013

### Evo8

Thanks for your response rude man.

Im not sure I understand how to find the pulse train equation from the description. I understand the characteristics that are described but am still unsure of how to get the $PT_s$ equation. How do i determine the form from the given information?

Thanks again

4. Mar 6, 2013

### rude man

OK, so let's look at the sampling pulse train by itself (no g(t) here):

One form of the Fourier series for any periodic function is
f(t) = 1/2 A_0 + ∑(A_n cos nwt + B_n sin nwt), n = 1 to ∞.

If your function is even, which is to say f(-t) = f(t), then you only need the A_n terms. This is the case here because the problem specifically states that the pulse is " ... centered at the origin". This is one of the important things you need to review about Fourier series.

So now you need to solve for the A_n coefficients. Dig up the integral for the coefficients, perform the integration over one period, and you have represented the sampling pulse train by its Fourier series. Multiply the series by g(t) and you're done.

5. Mar 6, 2013

### Evo8

Why do only th $A_n$ terms apply when my function is even?

If using my $A_n$ from above I get something like this.
$$A_n=\frac{2}{T_0} \int_{T_0} g(t) cos(n\omega_0 t)dt$$
$$=4B \int_{T_0} g(t) cos(4 \pi nB)$$
$$=4B cos(4 \pi nB)$$

Where does the sin term come from in my $\bar{g}(t)$? I would have assumed from the $A_b$ equation...

6. Mar 6, 2013

### rude man

Because if the function is even you can't have any sine terms because the sine function is odd. Basic Fourier series theory which you need to review.

I said not to include g(t) in this ... you are supposed to be deriving the series for the pulse train which has nothing to do with g(t).

Your expression for the A_n would apply if you were given g(t) & trying to express g(t) in a Fourier series.
You are not even given g(t) so how could you do this even if you wanted to?

7. Mar 6, 2013

### Evo8

Opps. Yes your correct you did mention not to include g(t). So what would my expression for A_n be?

Just $A_n=\frac{2}{T_0} \int_{T_0} dt$? In this case wouldnt I get the same result as I posted before? It seems I didnt include the g(t) there i just included it in the "math" which is weird i know.

I would still see$A_n=4B \cos(4 \pi n B)$

8. Mar 6, 2013

### rude man

No.

Look up the integral expression in your textbook for the A_n coefficients. It's got cosines in it.

9. Mar 6, 2013

### Evo8

Ok. My text doesnt really have fourier series covered but ive been reffering to this site. The Fourier Transform - Fourier series specifically this page of the site. Equation 7 is $A_n=\frac{2}{T} \int_0^{T} cos(\frac{2 \pi n t}{T})dt$

this has a cosine function in it...

After performing the integral I would end up with
$A_n=4B \int_0^{\frac{1}{2B}} cos(\frac{2 \pi nt}{\frac{1}{2B}})dt$
$=4B \int_0^{\frac{1}{2B}} cos(4B \pi nt)dt$

$=4B sin(2 \pi n)$

Am I on a better track here?

Thanks again for your help on this. I do appreciate it.

10. Mar 6, 2013

### rude man

Actually, eq. 7 has f(t) in it, does it not? However, you lucked out because in your case f(t) = 1 from t = -1/16B to 1/16B, repeating every 1/2B seconds. At all other times, f(t) = 0.

You therefore now have the right integral for computing the A_n coefficients of the Fourier series, but you are not using the correct limits.

First: fact: you can integrate from any point on the waveform to any other point as long as the total periodic interval is T = 1/2B. The limits do not have to be 0 to T. So, to make life easier, run the integration from t = 1/16B to t = 1/2B + 1/16B.

Look at your pulse train picture which you should have in front of you. For what values of t is your f(t) = 1? = 0? Then, choose your limits of integration accordingly.

(BTW I can't fathom why you are doing this problem without access to a Fourier series text. But, you did pick an excellent Website. I have it bookmarked myself.)

11. Mar 6, 2013

### Evo8

Ok,

Im a little confused by what to select for the limits. You first mention to try $\frac{1}{16B} to (\frac{1}{16B}+\frac{1}{2B})$ so the limits i evaluate my integral at are $\frac{1}{16B} to \frac{9}{16B}$
This yields $A_0=4B[sin(\frac{9 \pi n}{4}) - sin(\frac{\pi n}{4})]$ if I did it correctly.

Why not choose the limits of $+/-\frac{1}{16B}$?

Evaluating with these limits I get $8B sin(\frac{ \pi n}{4})$ this is of course if I did it correctly. I just re learned how to do basic integrals or "remembered" so I appologize for any blatant mistakes..

You also mention that I would solve for the coefficients of $A_n$ I was under the impression that the $A_n$ IS the coefficient? No?

I agree this is crazy without a fourier series text. I may stop by a book store this evening and see what I can find. However for now I'm "limited" to what i can find online.

12. Mar 6, 2013

### rude man

I meant the span of integration is from 1/16B to 1/2B + 1/16B. That does not mean the limits of your integral become those end-points. You are to integrate over those regions of t where your f(t) = 1 only.
Absolutely fine! I thought you might be confused by integrating from negative time but it's actually the easiest thing to do.
Sorry, that integration is incorrect. Way off. Need to check your math.

There is a lot of good stuff on-line but sometimes it's a bit heavy on the math. I would try to get an electrical engineering text where they apply the Fourier series to practical applications.

You also need to remember that ∫cos(ax)dx = (1/a)sin(ax)!
Another tricky part is when you solve for A_0. That coefficient will be of the form sin(x)/x with x = 0. Hint: = 1.

You have one final step: combining with g(t). Remember I said g_bar(t) = g(t) * sampling function.

13. Mar 7, 2013

### Evo8

Success!! Sort of....

Thanks for the integration help as well. After rereadeing a few intro to integral sections and the form you posted I understand now. Hopefully i wont make such mistakes in the future.

So now for my $A_n$ I end up with something like this. $A_n=\frac{2}{n \pi} Sin(\frac{\pi n}{4})$

For the $A_0$ I am a little confused as to what you mean by
are you saying that $\frac{sin(x)}{x} \ with\ x=0\ \ gives\ \frac{sin(0)}{0}\ which\ is\ \neq1$?

My integral goes something like this $$a_0=2B\int_{\frac{-1}{16B}}^{\frac{1}{16B}}1dt=\frac{1}{4}$$

However in a previous post
The 1/2 before the A_0 term makes me feel like something is wrong? I would end up with 1/8 not 1/4 in the pulse train..

I would end up with something like this $$\bar{g}(t)=\frac{1}{4}g(t)+\sum_{n=1}^\infty \frac{2}{n \pi} sin(\frac{n \pi}{4}) cos(4 \pi nBt)g(t)$$

So im left with two problems.... I "should" get 1/4 for that first term according to my texts exercise, and I have a $\pi$ in the cosine term where the text does not.. The quoted equation below is the expected "solution" that i am meant to show how its derived.

Where did the $\pi$ go?

14. Mar 7, 2013

### rude man

That is exactly right.
No. sin(x)/x = 1 if x = 0. cf. below.

If you evaluate the first term the way you did, it's the entire term, not the half. So you get 1/4 for the n = 0 coefficient as you should.

Never mind about the sin(x)/x for now. I'll just say that you can use the general expression for A_n to include n = 0 if you handle it right.

As to the missing π, you may tell your teach he made a typo or otherwise got it wrong. The π belongs!

Good going there in the stretch!

15. Mar 7, 2013

### Evo8

Thanks Rude man!

It is the text book that has the incorrect solution. Oddly enough I have an older edition of the same text with the "same" problem and the $\bar{g}(t)$ has the $\pi$.

I did some quick research on the sin(x)/x and I see what you mean. I dont fully understand/remember quite yet but i can at least buy it. Im still reading....

Thanks again for your help! I'm sure i'll have more questions in the near future...