1
\$\begingroup\$

We know that squaring a signal doubles its bandwidth. Can you please justify this case in time domain for non sinusoid signals,say Rect(t/T)? On squaring the signal i don't see any changes in tine domain but convolution in frequency domain confirms double bandwidth. Please help on what I am missing here.

Reference: Effect on signal bandwidth due to multiplication?

\$\endgroup\$
6
  • \$\begingroup\$ squaring a sinusoidal signal doubles its frequency, which is not quite the same thing. \$\endgroup\$
    – Neil_UK
    Jul 5, 2017 at 5:40
  • \$\begingroup\$ Squaring a square wave gives DC. \$\endgroup\$
    – Chu
    Jul 5, 2017 at 6:08
  • \$\begingroup\$ How can you fit together what you say. On one hand you say squaring a square wave has no effect (and this can be true) but on the other hand spectrum convolving confirms double the bandwidth? You must be contradicting yourself somewhere? \$\endgroup\$
    – Andy aka
    Jul 5, 2017 at 11:25
  • 1
    \$\begingroup\$ "but convolution in frequency domain confirms double bandwidth" - ??? sinc(x) * sinc(x) = sinc(x) (* denotes convolution) \$\endgroup\$
    – Alfred Centauri
    Jul 5, 2017 at 11:41
  • 1
    \$\begingroup\$ @AlfredCentauri perfectly correct and well-said. Just in case anyone didn't understand that, a single pulse (i.e. a pulse in a square wave) has a sinc function in the f-domain and convolving in f-domain (multiplying in t-domain) produces exactly the same sinc function despite there being a time difference in the t-domain. \$\endgroup\$
    – Andy aka
    Jul 5, 2017 at 13:17

1 Answer 1

Reset to default
3
\$\begingroup\$

Here is your problem: "We know that squaring a signal doubles its bandwidth."

That is not true in general. It is true for sinusoidal signals.

Probably you assume it must be true also for linear combinations of sinusoidal signal (i.e. Fourier sums/integerals which can be used to approximate a rectangular signal), but this is not the case. This reasoning works only for linear operations (e.g. multiplying by a constant, taking the derivative, integrating) but squaring is not a linear operation.

(in other words: the linear combination of squares is not the same as the square of linear combinations)

Therefore the fact that your statement is true for single (pure) sinusoidal signals does not mean that it is also true for linear combination of sinusoidal signals (Fourier sum/integeral; e.g. rectangular signals).

\$\endgroup\$
2
  • \$\begingroup\$ The square of the sum of sinusoids will give a component at twice the highest sinusoid's frequency, so relative amplitudes have a significant impact. \$\endgroup\$
    – Chu
    Jul 5, 2017 at 8:52
  • \$\begingroup\$ @Chu: ...but what about amplitude and phase relation of those components at double frequency? It's important that they are not just the linear combination of the origianl amplitudes (including phases). In the case of a rectangular signal the amplidues and phases of those components of higer frequency obviously turn out to be exactly those of the original signal. \$\endgroup\$
    – Curd
    Jul 5, 2017 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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