-3
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Let suppose

x(t)=$\sum\limits_{k=-∞}^∞ R(t-kT)$

$R(t) = \begin{cases}1 &[0,2T] \\ 0 & \text{otherwise} \end{cases}$

x(t) is the input to an ideal bandpass filter with $\text{BandWidth} = \dfrac{1}{(2T)}$
and $\text{Center Frequency} = \dfrac{L}{(T)}$

How can i find the output y(t). any help will be appreciated.

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  • $\begingroup$ looks like $$x(t)=2 \quad \forall t$$. is that what you want? $\endgroup$ – robert bristow-johnson Feb 7 '15 at 2:17
  • $\begingroup$ well, okay, maybe $$ x(t)=3 $$ for $t=kT \quad k\in\mathbb{Z}$ but those are infinitely thin points and don't amount to anything when integrated. $\endgroup$ – robert bristow-johnson Feb 7 '15 at 2:19
  • $\begingroup$ How x(t) is 2 or 3? isn't it a periodic rectangular signal? i want the response y(t). $\endgroup$ – dfeast Feb 7 '15 at 3:40
  • $\begingroup$ $x(t)$ is periodic with period $T$. at integer values of the period (when $t=kT$ and $k$ is an integer), then 3 of those $R(t-kT)$ rectangular functions add up. otherwise it's 2. you've said nothing about $y(t)$ until now. how is $y(t)$ defined? $\endgroup$ – robert bristow-johnson Feb 7 '15 at 12:46
  • $\begingroup$ y(t)=x(t)*h(t) or Y(f)=X(f).H(f) $\endgroup$ – dfeast Feb 7 '15 at 23:54
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Because the off sets between translated copies of the top hat and the translation is half the top had width your function is a constant taking the value 2 except possibly at integer values of time when it takes the value 3, but this may be because you have defined the function incorrectly, but that does not matter since these points constitute a set of measure 0. So $x(t)=1\ a.e.$, hence the only non-zero frequency component in its spectrum that at $\omega=0$ (or $f=0$ depending how you chose to define the FT). If the DC component is not in the pass band you get nothing out. If it is in the passband the output is $2$ (you will have removed the glitches at integer times).

... I can't help but suspect you have errors in the question as posted, mainly because of the mysterious appearance of $L$, and the answer changes its nature if the width of the top hat ceases to be twice the translation between copies.

If I assume you have posted the question with errors the basic idea if you do not have a constant is that the output of the band-pass filter is the sum of the frequency components of the Fourier series (since the signal is periodic) corresponding to the signal that fall in the pass band.

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  • $\begingroup$ let say L is 4. $\endgroup$ – dfeast Feb 8 '15 at 3:13
  • $\begingroup$ If $L=4$ the bottom of the pass band is $f_c-B/2 = 4/T - 1/(4T)>0$ so it does not include DC and the output is zero. $\endgroup$ – Conrad Turner Feb 8 '15 at 4:33

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