0
$\begingroup$

Let us consider $x_1(t)$, $x_2(t)$, $x_3(t)$, all the same within some some duration 0 to $T$ but all different outside this interval. Now let us multiply each of these signals with $w(t)$, a window function - nonzero from 0 to $T$ but 0 outside this interval. So the multiplication of this $w(t)$ with each of $x_i(t)$ will give the same signal. This should be same as Inverse Fourier Transform of $W(\omega)$ convolved with $X_i(\omega)$. Does this mean convolution of the same signal with different signals can give the same result? Any comments? Is there a pitfall in my interpretation?

Also I see mathematically equations neatly entered in questions and answers on this site? Where should I start to learn on how to write the equations?

$\endgroup$
  • $\begingroup$ To enter equations, use latex notation between dollar signs. See the edits I made to your question to get started. $\endgroup$ – MBaz Apr 30 '15 at 23:52
  • $\begingroup$ The answer to the question in the title is Yes. The signals $x_i(t) = \operatorname{sinc}(t/T_i)$ are the impulse responses of different ideal lowpass filters of different bandwidths. Assume that $T_1 > T_2 > T_3$. Then, $$x_1(t)\star x_2(t) = x_1(t)\star x_3(t) = x_1(t).$$ Don't try to verify this statement in the time domain via convolution: just use $$X_1(f)X_2(f) = X_1(f)X_3(f) = X_1(f)$$ based on the properties of ideal LPFs. (Drawing a sketch of the LPF transfer functions might help....) $\endgroup$ – Dilip Sarwate May 1 '15 at 1:15
  • $\begingroup$ Thanks a lot. I do not know Latex. How do I get onboard Latex quickly. Is there a quick reference so I can use it directly on this site? $\endgroup$ – Seetha Rama Raju Sanapala May 1 '15 at 4:20
0
$\begingroup$

Convolution is filtering.

Consider creating a filter with a rectangular response in the frequency domain. Then any signal with the same content within the filter's passband, no matter what the signal content outside the passband of the filter, should result in only the content within the passband after convolution with the impulse response of the filter. Thus, multiple inputs (signals with different stuff outside of the passband) to a convolution with this filter should result in the same output.

(Your opening statement merely swaps the frequency and time domains for the rectangular function, making this relationship harder to recognize.)

This should work even for a non-rectangular and finite length FIR filter with at least one zero in it's transform (maybe at a nice integer submultiple of the sample rate). Add any magnitude of sinusoidal input at the frequency of the zero, and the convolution result or filter output remains unchanged (except perhaps for numerical issues/clipping/quantization/etc.).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.