Is the result of convolution of a heaviside function with a sinc low pass with cut off at fc, say 25khz, band limited ((sin(2 x Pi x fc x i))/(Pi x i))?

I've been working it out in frequency domain (taking ft of the signals and multiplying them). The ft of the sinc is a rectangular function with cut off at +/-fc. If I pass a continuous 5khz sine, I get a 5khz sine at output, if I pass a continuous 30khz sine, I get a 0 at output, and if I pass an impulse I get a band limited impulse with components upto 25khz at my output. All these results are band limited.

What about heaviside function? I am unable to get the result to look Bandlimited. The Fourier transform of heaviside step function looks very complicated. What am I doing wrong.

I am unable to embed what I've tried but I am linking the equivalent:

  1. https://math.stackexchange.com/questions/736749/fourier-transform-of-sinc-function . This results in a rectangular function from -fc to +fc.
  2. http://www.thefouriertransform.com/pairs/sinusoids.php . The result is energy concentrated at frequency of the sinusoid. If this frequency is inside the rectangular function, after multiplication, I Will get the result as same as before multiplication. If this frequency is outside the rectangular function, I will get 0 as result.
  3. I am unable to come at a proper derivation for the heaviside function. It feels too complicated. Looking at resources. http://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/Heaviside_step_function.html
  • $\begingroup$ What do you mean by "work it out in the frequency domain"? You should know the answer already if you did that properly. $\endgroup$ – TimWescott Oct 1 '20 at 21:02
  • $\begingroup$ Instead of convolving in time domain, I multiplied their Fourier transforms. $\endgroup$ – Manuel Jenkin Oct 1 '20 at 21:09
  • $\begingroup$ Can you show your work? How can a finite number times zero be anything but zero? $\endgroup$ – TimWescott Oct 1 '20 at 21:26
  • $\begingroup$ Did you multiply their actual Fourier transforms analytically, or were you doing this numerically with FFTs? That's two different things. $\endgroup$ – TimWescott Oct 1 '20 at 21:27
  • $\begingroup$ Thank you. I have updated what I've done so far. $\endgroup$ – Manuel Jenkin Oct 1 '20 at 21:49

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