Good morning
a continuous time signal x(t) is obtained in the output of an ideal low-pass filter with wc=1000 pi.
If a sampling with a pulse train is done on x(t) which of these sampling period is the most appropriate to recuperate x(t) from the sampled signal.
a) T=0.5 10^(-3) s
b) T=2 10^(-3) s
c) T= 10^(-4) s
My answer was both a and c.
Is my answer correct?
Thank you for your help.

 

I'll assume "wc" is really \(\omega_c\), the cutoff frequency of a brick-wall filter. If ω=1000π, then f=500Hz
Sampling periods are 500μs, 2ms and 100μs, sampling rates of 2kHz, 500Hz and 10kHz respectively.
Whilst both a and c are appropriate, semantically, only one of them can be the most appropriate, so it would have to be c, because it would make the design of the reconstruction filter a little bit easier than a.

 

I'll assume "wc" is really \(\omega_c\), the cutoff frequency of a brick-wall filter. If ω=1000π, then f=500Hz
Sampling periods are 500μs, 2ms and 100μs, sampling rates of 2kHz, 500Hz and 10kHz respectively.
Whilst both a and c are appropriate, semantically, only one of them can be the most appropriate, so it would have to be c, because it would make the design of the reconstruction filter a little bit easier than a.

Thank you
I would like to know why the design with a period of 100μs is more simple.

 

Thank you
I would like to know why the design with a period of 100μs is more simple.

Because you need the most attenuation possible at half the sampling frequency. If your signal is bandwidth-limited to 500Hz, then it is easier to design the filter. With either 10kHz or 2kHz sampling rate your filter passband is 500Hz, so you need a steeper slope to achieve the desired attenuation by 2kHz than by 10kHz.

 

Because you need the most attenuation possible at half the sampling frequency. If your signal is bandwidth-limited to 500Hz, then it is easier to design the filter. With either 10kHz or 2kHz sampling rate your filter passband is 500Hz, so you need a steeper slope to achieve the desired attenuation by 2kHz than by 10kHz.

Tahnk you

 

I do think that the question is rather too vague.