Sampling a continuous-time signal, aliasing/Nyquist

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The discussion centers on sampling a continuous-time signal and the implications of aliasing based on the Nyquist theorem. When sampling the analog signal x(t) = cos(2πft) at 1 kHz, aliasing occurs for frequencies above 0.5 kHz and below 1 kHz, while it does not occur for frequencies below 0.5 kHz. Specifically, for f = 1.01 kHz and f = 0.99 kHz, aliasing is present, resulting in output frequencies of 0.01 kHz and -0.01 kHz, respectively. In contrast, for f = 0.49 kHz, there is no aliasing, and the output frequency remains at 0.49 kHz. The discussion also touches on the cutoff frequency of the low-pass filter and its relationship to the input frequency.
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Homework Statement


The analog signal x(t) = cos(2pi f t) is sampled at a rate of 1 kHz, using ideal
impulse sampling, to obtain the sampled signal x^(s)(t). The sampled signal is then sent through an ideal
lowpass filter with transfer function H(2pi f ) = 0.001 rect (0.001 f ).
(a) If f =1.01kHz, what is the output frequency from the filter? Is there aliasing or not?
(b) If f = 0.99kHz, what is the output frequency from the filter? Is there aliasing or not?
(c) If f = 0.49kHz, what is the output frequency from the filter? Is there aliasing or not?

Homework Equations


To perfectly reconstruct a signal, we need: sampling rate > 2*signal frequency

The Attempt at a Solution


I think these are correct, not 100% sure though:
(a)
aliasing occurs, as 1.00 kHz !> 2 × 1.01 kHz
fout = 1.01 kHz – n × 1.00 kHz = 1.01 kHz – 1 × 1.00 kHz = 0.01 kHz

(b)
aliasing occurs, as 1.00 kHz !> 2 × 0.99 kHz
fout = 0.99 kHz – n × 1.00 kHz = 0.99 kHz – 1.00 kHz = - 0.01 kHz

(c)
aliasing does not occur, as 1.00 kHz > 2 × 0.49 kHz
fout = 0.49 kHz
 
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alc95 said:

Homework Statement


The analog signal x(t) = cos(2pi f t) is sampled at a rate of 1 kHz, using ideal
impulse sampling, to obtain the sampled signal x^(s)(t). The sampled signal is then sent through an ideal
lowpass filter with transfer function H(2pi f ) = 0.001 rect (0.001 f ).
Not sure I understand this terminology. What really is the ideal low-pass cutoff frequency? Is it really a function of input frequency f? Do you have to divide by 2π? Etc. ?

I don't know your method, and as I say I'm not sure what your cutoff filter really looks like, but it basically seems to work.

Just FYI I first determine the cutoff frequency of the low-pass filter = f0, the I take the sampling frequency fs and my input frequency f and find spectral components:

f, |fs - f|, fs + f, |2fs - f|, 2fs + f, ...

and then compare each component against the cutoff frequency f0.
The only unaliased signal is at f so anything below f0 that is a mix of f and fs is an aliased component.
 

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