# Finding sampling frequency for analog to digital conversion

1. Apr 13, 2013

### Jd303

The input signal to an analog to digital converter is x(t) = 5.4 cos (134.5πt + 0.1π).
The output from the converter is y(n) = 5.4 cos (0.47πn - 0.1π).
Compute the sampling frequency (samples per second) of the analog to digital converter.
Hint: the continuous-time input signal is under-sampled in this case.

I have been doing questions like this recently, without too much trouble. However this one has me a little confused. I don't understand the theory behind how the value 0.1*pi changes to -0.1*pi in the output signal.

-If the output was 5.4cos(0.47*pi*n + 0.1*pi) I would do the following:
- fs = 2*134.5/0.47 = 572.34

-My attempt at this question is not much more than a guess, I have done:
3*fo - fs = 3*134.5 - 572.34 = 168.84

I would greatly appreciate anyone who is able to shed some light on this problem. Thanks!

2. Apr 13, 2013

### rude man

Where do you get this from? Why "3fo"? Wy work with the 3rd harmonoc of fo?

What is the basic function the a/d performs? Your input signal is a sine wave of one and only one frequency, and the output is again a sinusoid of one and only one frequency. How are the three fundamental frequencies related to each other?

So focus on the fundamental frequencies of all three signals: input, sampling and output.

Hint No. 1: there are two and only two possible sampling frequencies involved. How does your superhet radio work? Its local oscillator performs the same function as the sampling a/d converter.

In order to resolve which of the two possible frequencies is the sampler, you then need to look at the phases of the input and output signals.

Hint No. 2:
sin(x)cos(y) = 1/2 sin(x-y) + 1/2 sin(x+y) or
cos(x)sin(y) = -1/2 sin(x-y) + 1/2 sin(x+y)

Hint no. 3: the a/d output is low-pass filtered so any signals well above the output frequency are attenuated to essentially zero. This should have been specified in the problem statement.

Last edited: Apr 13, 2013