Evo8 said:
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.
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.
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})##
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})##.
I used a table of Fourier transforms to find ##X(f)## I've shown the pair below. This is the same form the text uses from what I can see.
$$∏(\frac{t}{\tau}) \ \ → \tau sinc(\pi ft\ \tau)$$
That should be ##\Pi(\frac{t}{\tau}) \ \ \to \ \ \tau \text{sinc}(\pi f \tau)##, that is, without the t.
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})##
Yes. That is correct, although the result for P(f) should be negative.
The result for ##S_y(f)## is still the same.
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.
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.