I have found that the convolution of two rect(f) functions (f=frequency) gives a triangular pulse of width 2 centred at 0 Hz. Can anyone tell me what the bandwidth of this pulse is? Is it F or 2F?

My professor wrote the bandwidth of a rect pulse (centred between -B and B) as B. Should it also be 2B?

It would be very helpful if you solved my doubt. (I have just started my analog and digital communications course).

What will the bandwidth be if the figure is right shifted such that the whole spectrum lies to the right of the origin? (Will it be 2 in this case?)

Triangular pulse

  • \$\begingroup\$ This sounds like a homework problem. We don't give answers to homework problems, but if you show how you tried to solve it or your reasoning and where you got stuck or have doubts, you will likely get some hints. \$\endgroup\$
    – John D
    Commented Aug 28, 2021 at 17:45
  • \$\begingroup\$ The question was to find the bandwidth of the signal x(t) = sinc^2(t). I took its fourier transform and then convoluted the two rect pulses (graphical convolution) to get a value of (1+f) for 0<f<-1 and a value of (1-f) for 0<f<1. This is how I got the triangular pulse. \$\endgroup\$ Commented Aug 28, 2021 at 17:49
  • \$\begingroup\$ Prof is correct and ought to be 99% of the time \$\endgroup\$ Commented Aug 28, 2021 at 17:56
  • \$\begingroup\$ So, is the bandwidth of the figure I have shown 1 instead of 2Hz? \$\endgroup\$ Commented Aug 28, 2021 at 17:57
  • 1
    \$\begingroup\$ No but 0+/-B is just BW=B, yet for 1000+/-B, BW=2B the difference between signals inside +B and -B is due to phase difference \$\endgroup\$ Commented Aug 28, 2021 at 18:03

2 Answers 2


When a real time domain signal is analysed, there is always a half power mirror image of the half power positive frequency components in the origin of the frequency domain (SNR is unaffected because noise is also half power); well the real frequency components of the real time domain signal are an exact mirror image, but the imaginary frequency components of the real time domain signal are conjugated, so it's a conjugate reflection.

The point is, in a real signal, the negative side of the frequency domain is a mathematical figment derivable from the positive side, and therefore the bandwidth only concerns the positive frequency domain, and therefore the baseband bandwidth is B. If it were modulated up to a higher frequency, the passband bandwidth would be 2B, so long as it weren't modulated with SSB, where it would be B.

Only real signals can be transmitted – complex signals are transmitted as real signals using in phase and in quadrature complements. The complex baseband bandwidth of the signal is 2B.

  • \$\begingroup\$ Thanks a lot for spending your valuable time in explaining the concept to me. \$\endgroup\$ Commented Aug 29, 2021 at 4:07

2B or not 2B ... that is the?

Regardless of the shape of a 2B symmetrical +/-B spectrum on some carrier f centre , if you shift it down to 0="Baseband" the resulting BW is Only B, not 2B.(mirror image)

  • consider it redundant spectrum -B=+B which is why SSB was used for many analog channels like TV (but complex SSB notch filter on IF/ so not practical for other phase-distortion reasons to apply to all forms of communication)
  • \$\begingroup\$ Thanks a lot for your answer. \$\endgroup\$ Commented Aug 29, 2021 at 4:07

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