Maximum Frequency of a Signal

Thread Starter

jegues

Joined Sep 13, 2010
733
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!
 

Georacer

Joined Nov 25, 2009
5,181
Is sampling the context? If so, the greater frequency component is 125Hz, and thus your sampler must have at least twice that.
 

Thread Starter

jegues

Joined Sep 13, 2010
733
Is sampling the context? If so, the greater frequency component is 125Hz, and thus your sampler must have at least twice that.
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,


What is the maximum frequency component of the function? What is the Nyquist Sampling​
Rate?
 

holnis

Joined Nov 25, 2011
49
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.
 
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