Finding sampling frequency for analog to digital conversion

[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!

 

[rude man]

The input signal to an analog to digital converter is x(t) = 5.4 cos (134.5πt + 0.1π).
-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!

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.