- #1
Evo8
- 169
- 0
Homework Statement
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)##
Homework Equations
$$a_0=\frac{1}{T_0} \int_{T_0} g(t) dt$$
$$a_n=\frac{2}{T_0} \int_{T_0} g(t) \cos(n\omega_0 t) dt$$
$$b_n=\frac{2}{T_0} \int_{T_0} g(t) \sin(n\omega_0 t) dt$$
The Attempt at a Solution
Ok so my Fourier series and integration skills are weak at best. Ontop of that I am 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 I am 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 don't 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!