# Bandwidth of square wave

Discussion in 'Homework Help' started by EngIntoHW, Sep 30, 2010.

1. ### EngIntoHW Thread Starter Member

Apr 24, 2010
128
0
In communcation, data signals carry square waves which represent symbols.

The fourier transform of a square wave is sinc.

1. Does it mean that the fourier transform of a data signal (which is a sequence of time-shifted square waves) is a sum of phase-shifted sincs?

2. How can I find the spectrum (bandwidth) of a 1Mbps rated data signal?

Thank you

2. ### Nik Well-Known Member

May 20, 2006
55
3
Well, if you have really square ('fast') edges, your pulse train will smear across a wide spectrum. I suppose you could deconstruct it into a *very* long series of harmonics...

3. ### Ghar Active Member

Mar 8, 2010
655
73
For 1) I guess that has to be correct but I've never thought of that way.
If you repeat a waveform in the time domain you are sampling it in the frequency domain.

So if the spectrum of a square pulse is a sinc, the spectrum of a square wave (wave implies repetition) is a set of defined harmonics following a sinc envelope.

The bandwidth of a square wave is technically infinite so you need to come up with some definition to compute a value.

Finite rise and fall times add an extra decaying coefficient to the sinc function making upper frequency harmonics fall off faster. The bandwidth is still technically infinite... sometimes you'll see it defined as the corner frequency where this rise/fall time coefficient begins to work.
For rise time = fall time = tr, that's given by:
$f = \frac{1}{\pi t_r}$

Anyway, this infinite bandwidth problem is why you don't actually use square pulses in wireless digital communication.
http://en.wikipedia.org/wiki/Pulse-shaping
http://en.wikipedia.org/wiki/Raised_cosine