m(t)*m(t) in time domain is equivalent to the convolution of m(w) and m(w) in frequency domain.Thus if the bandwidth of m(t) is a known quantity then how is the bandwidth of m(t)*m(t) is determined?

  • 2
    \$\begingroup\$ Hint. What is the highest frequency component of sin(t) * sin(t)? \$\endgroup\$
    – user16324
    Mar 29, 2014 at 16:46
  • \$\begingroup\$ Do you understand what "convolution" means? Saying that you know the "bandwidth" of a signal means that you know some general facts about \$m(\omega)\$, even if you don't know its precise definition. \$\endgroup\$
    – Dave Tweed
    Mar 29, 2014 at 17:04

1 Answer 1


Your bandwidth will double. You can derive this using trig identities:

enter image description here

DC and 2*theta components will form unless you filter them out. I ran a quick simulation in MATLAB to confirm. F(m(t)) on the top-left, F(m(t)*m(t)) on top-right. Bottom-left is the the top-left zoomed in, and vice verse.

enter image description here enter image description here conclusions:

  1. m(t) has a passband from 4kHz to 6kHz

  2. m(t)*m(t) has twice the bandwidth, and massive DC componets

Side note, not related to your question:

This property is actually exploited in QPSK demodulation. Here the double frequency is eliminated almost perfectly by mathematically performing something called "integrate-and-dump". (Or you can always just use a low pass filter too). Any good communications textbook will describe this in detail.


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