# Signal Bandwidth

Discussion in 'Homework Help' started by Peytonator, Nov 11, 2010.

1. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
Hi there,

Say I have a Signal

X(s) = 1 / [ (s/4) + 1 ] (s = jω)

then the bandwidth is obviously √4 = 2.

And if I have

X(s) = 1 / [ (s^2/4) + s + 1 ]

It's also √4 = 2.

But what if I have

X(s) = 1 / [ (s^3/4) + s^2 + s + 1 ]

Is it still √4 = 2?

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I can't nail the theoretical background right now, but I don't think the two last TFs have a knee frequency of 2rads/sec.

Check this useful application to see what I mean: http://www.williamsonic.com/BodePlot/

The BW is at the frequency where amplitude drops at -3dB.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The cubic can be resolved into

s^3+4s^2+4s+4=(s+3.1304)(s^2+0.8696s+1.2778)

So there are poles at -3.1304, -0.4348+j1.0434 & -0.4348-j1.0434

One would expect the -3dB point to be less than √(1.2778) rad/sec - the 2nd order part ωo value.

The underdamped second order component will also probably produce a peak in the spectrum.

As Georacer points out the -3dB points for each case will differ. His suggestion of using Bode plot is a good one.