# Maximum Frequency of a Signal

Discussion in 'Homework Help' started by jegues, Feb 1, 2012.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
If I have a signal,

$f(t) = 2048 + 700cos(2\pi31.25t) - 1100sin(2\pi125t)$

How do I go about finding the maximum frequency?

What do they really mean by maximum frequency anyways?

(I haven't taken any signal processing courses yet so my understanding between these distinctions and how to solve for them is still quite minimal)

Thanks again!

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Is sampling the context? If so, the greater frequency component is 125Hz, and thus your sampler must have at least twice that.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I'm not quite sure, but I think it is within a sampling context.

My professor hinted that it should somehow relate to the maximum slope.

The question was stated as,

Nov 25, 2009
5,151
1,266
5. ### holnis Member

Nov 25, 2011
50
4
The frequencies present in the signal above are

F1=31.25
F2=125

Thus Fmax = 125 Hz and according to the sampling theorem:

Fs > 2Fmax = 250 hz

The Nyquist rate is Fn = 2Fmax. ==> Fn = 250 Hz.