# Homework Help: Setting up a power spectral density problem

1. Apr 6, 2013

### Evo8

1. The problem statement, all variables and given/known data
Im having some trouble setting up this problem.

$x(t)$ is a polar signal with random binary data. The data is independent from bit to bit. $P(0)=P(1)=\frac{1}{2}$ Assume $a_n=+/-1$.

The fundamental pulse shape is attached.

I need to find the Power spectral density of x(t)

2. Relevant equations

$$S_y(f)=\frac{|P(f)^2|}{Tb}$$

3. The attempt at a solution

Ok so as i mentioned I am having trouble setting up this problem. From the fundamental pulse shape I can determine that my pulse width is $\frac{T_b}{3}$

Im not sure how to treat the $P(0)=P(1)=\frac{1}{2}$.

I think I need to find my $x(t)$ and calculate the fourier transform of it $x(f)$ or $p(f)$ in my formula above. Then square it and divide by the pulse width. If this is correct. What my $x(t)$? How do I go about finding it from the information given.

Any ideas?

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2. Apr 6, 2013

### I like Serena

P(0)=P(1)=1/2 would mean that the probability that the next bit is 0 or a 1 is fifty-fifty.

I'm not familiar with a "polar signal" or how a "fundamental pulse shape" ties in.
Perhaps you can provide a little more detail?

3. Apr 6, 2013

### Evo8

Well from my understanding its the way the 1's and 0's are transmitted. For polar a 1 is transmitted by the pulse $p(t)$ and 0 is transmitted by $-p(t)$.

There are also on/off signaling and bipolar signaling that this section speaks about.

For on/off 1 is transmitted by pulse p(t) and 0 is transmitted by no pulse. For bi-polar 0 is transmitted by no pulse and 1 is transmitted by either + or - p(t) pulse. I've attached a screenshot of the section that talks about polar signaling for further reference.

EDIT: This link on the second page shows a good graphical representation of the different types of signaling for line coding.

http://www.freewebs.com/angsuman/chapter4.pdf

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4. Apr 7, 2013

### I like Serena

Good!

I can see now that there is a mix-up for the symbol P.
The symbol P in your problem statement has 2 different meanings.
In the statement P(0)=P(1)=1/2 the symbol P means the probability on either a 0 or a 1.
In the power spectral density formula $S_y(f)= \dfrac{|P(f)|^2}{T_b}$, the function P(f) is the fourier transform of the pulse p(t).

Furthermore, in your last attachment, an example is given for polar signaling (7.13 and 7.14).
In this example they use a rectangular pulse of width $T_b/2$.

I believe you need to redo this example with a rectangular pulse of width $T_b/3$.

5. Apr 7, 2013

### Evo8

I was thinking the same thing but that makes a lot of the information given in the question just extra and unnecessary. Like the probability of the next 0 or 1 etc.

I'll give it a try. Thanks for the help ILS!

6. Apr 7, 2013

### I like Serena

The extra information identifies the problem as a polar NRZ signaling problem with white noise binary data.
In particular it's not unipolar, nor (alternating) bipolar, nor RZ.

That's why you can reapply the example, since it has the same classification.

7. Apr 7, 2013

### Evo8

I see. This makes sense.

One more question regarding this.

When finding the dc in p(t) it would be when f=o correct? So I would take my Fourier transform x(f) and find x(0)?

So $x(f)=\frac{T_b}{3}sinc^2(\frac{\pi f T_b}{3})$

and $x(0)=\frac{T_b}{3}$?

8. Apr 7, 2013

### I like Serena

If P(f) is the fourier transform of p(t), then yes the dc in p(t) is equal to P(f=0).

A function x(f) is not defined in your problem.
Do you perhaps mean X(f) representing the fourier transform of x(t)?
Either way, you're not supposed to do anything with x(t) or X(f).
The formula is only about P(f), the fourier transform of p(t).

I haven't tried to figure out the answer yet.
But how did you get this?
What is the function p(t) for starters?
And remember, you need P(f) and not x(f).

Certainly not, x(0) is your signal amplitude at t=0.
It has nothing to do with a pulse width.

9. Apr 7, 2013

### Evo8

I guess I made the assumption that $x(t)$ and $p(t)$ were interchangeable. The problem question says the signal $x(t)$ is the polar signal defined by.....

The example in the text uses $p(t)$ and never mentons $x(t)$. I thought it was just different notation since this question is not from the book.

This is what I did. Keep in mind i just used $x(t)$ instead of $p(t)$.

$x(t)=rect(\frac{t}{(\frac{T_b}{3})})=rect(\frac{3t}{T_b})$

I used a table of fourier transforms to find $X(f)$ Ive shown the pair below. This is the same form the text uses from what I can see.

$$∏(\frac{t}{\tau}) \ \ → \tau sinc(\pi ft\ \tau)$$

so I end up with $X(f)=\frac{T_b}{3}Sinc(\frac{\pi f T_b}{3})$

Then to calculate $S_y(f)=\frac{|X(f)|^2}{T_b}$ I end up with $S_y(f)=\frac{T_b}{9}Sinc^2(\frac{\pi f T_b}{3})$

And to find the DC. I took my $X(f)$ and found $X(0)$ by plugging in the 0 and solving. Leaving only the $\frac{T_b}{3}$ term.

10. Apr 7, 2013

### I like Serena

No. They are not interchangeable.
The function x(t) is the signal that contains many 0's and 1's.
Each 0 and 1 is encoded by the pulse function -p(t) respectively p(t).

As far as I can tell all your texts use x(t) consistently.
That you don't see it in the attachment is only because it would be on the previous page that was not included.

Yes. Very confusing. But more or less correct for p(t).

One small issue: the pulse in your attachment is negative.
So you would have $p(t)=-\text{rect}(\frac{3t}{T_b})$.

That should be $\Pi(\frac{t}{\tau}) \ \ \to \ \ \tau \text{sinc}(\pi f \tau)$, that is, without the t.

Yes. That is correct, although the result for P(f) should be negative.
The result for $S_y(f)$ is still the same.

Hmm, the question does not ask for the DC.
Any reason you are calculating it?

Second, you calculated the DC of a single pulse, which would yield $P(0)=-\frac{T_b}{3}$.
This is also what you can deduce from the given pulse shape.

But if you want the DC component of the actual signal x(t), then this would be zero.
It is zero because you would have on average the same number of positive pulses as negative pulses.

11. Apr 8, 2013

### Evo8

ivsdcsdc
Hmm. ok well im asked to find the PDF of $x(t)$ not $p(t)$ so im confused as how to go about that not that I have the PSD of $p(t)$

Ok, This makes sense.

Yes this was an error on my part. The t should be omitted.

This also makes sense. Thanks for the check on this.

The second part of the problem asks for the DC. I didnt add it as I thought I would be ok once I got the first part. Hmm ok I follow this.

So im still confused on the x(t) and p(t) PSD calculation. I looked back in my text and didnt see x(t) referenced anywhere...

12. Apr 8, 2013

### I like Serena

The PSD you have is of $x(t)$ and to find it, you need $P(f)$.

On pages 8 and 9 in your pdf file they explain how to get from x(t) to a general formula for the PSD of x(t).
They do use slightly different symbols.
Instead of $S_y(f)$ they use $P_x(f)$, signifying the PSD for the signal x(t).
Instead of $p(t)$ they use $f(t)$ for the pulse.
And instead of $P(f)$ they use $F(f)$ for the fourier transform of the pulse.

In your scan with the example they pick up this general formula and apply it to a polar signal in equations (7.10a) up to (7.12), the last being your PSD for a polar signal x(t) containing binary white noise.

On page 8 of your pdf they give $X_T(f)$ in formula (2) which is the fourier transform of the signal x(t) in a finite interval T.
The DC component would be $\displaystyle\lim_{T \to \infty} X_T(0)$.
There is little sense in trying to calculate this though. The result will be zero since you can expect as many positive pulses as negative pulses.

13. Apr 8, 2013

### Evo8

Woops! Its actually asking to fine the DC power of x(t). This basically changes everything...

14. Apr 9, 2013

### Evo8

I see in my text theres a section that talks about the PSD and power.

$x(t)$ is a power signal and $X_T(t)$ is an engergy signal. So the PSD is $S_X(f)=lim_{T→ \infty} \frac{|X_T(f)|^2}{T}$

and the power is $P_X=\int_{-\infty}^{\infty} S_X(f)df=2 \int_0^{\infty} S_X(f)df$

I know $S_y(f)=|P(f)|^2 S_X(f)$ I already have $S_y(f)$ can I just plug it in and solve for $S_X(f)$?

Now im confused between the diference of $S_y(f) \ and \ S_X(f)$

Thanks again for the help

15. Apr 9, 2013

### I like Serena

Good!

That does not look right.
How did you get by it?

As you can see on page 8 of your pdf, the signal $x(t)$ is a convolution of the bit values $a_n$ and the pulse shape.
In the frequency domain a convolution becomes a multiplication.
$$X_T(f) = P(f)\cdot\sum_n a_n e^{-j2\pi f n T_b}$$
The second part is the fourier transform of the bit values $a_n$.

In other words, your $S_X(f)$ already contains $P(f)$, so it makes no sense to multiply $S_X(f)$ yet again with $P(f)$.
Your $S_X(f)$ is the same as $S_y(f)$. Yet again just another choice of symbols.

You may want to make a list of concepts with their corresponding symbols as they vary from text to text.

The DC power of the signal would be $S_y(0)$.

16. Apr 10, 2013

### Evo8

I agree. Its doesnt make sense thats where I was getting confused.

I understand. Thanks again ILS!