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Still need help in 2021

I am confused about the relation between Sinc and Rectangle transform pair and how that relates the Bandwidth of Pulses, Bandwidth of Zero-ISI Filter and the Symbol Rate.

My thoughts... Using $T_s$ = Symbol Time Period, $R_s$ = Symbol Rate

Bandwidth of Pulses (Raw data):

  • Time-Domain Rectangle : Width $T_s$ (symmetrical on zero)
  • Frequency-Domain Sinc : Every lobe width is $\frac{1}{T_s}$ but main lobe is $\frac{2}{T_s}$.
  • So... Useful BW $= \frac{2}{T_s} = 2R_s$
  • So if you wanted to sample this properly it would need to be twice this under nyqust theorem, so minimum BW $= 4 R_s$ for nyquist criteria

Bandwidth of Zero-ISI Filter:

  • Time-Domain Sinc : Every lobe is width $T_s$ but main lobe is $2T_s$
  • Frequency-Domain Rectangle : Width of $\frac{1}{T_s}$ (Symmetrical on zero)
  • So...BW $= \frac{1}{T_s} = R_s$
  • So if you wanted to sample this properly it would need to be twice this under nyqust theorem, so minimum BW $= 2R_s$ for nyquist criteria

But I have read that's not right, the bandwidth of the Zero ISI Filter is $R_s/2$ when the roll off factor is zero

Still need help in 2021

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Nyquist says that you can send up to twice as many pulses (symbols) per second as the channel bandwidth $B$ with zero ISI, so you need $R_s \leq 2B$. That is all there is to it. The sinc pulse has zero excess bandwidth so the bandwidth of the signal is equal to the symbol rate, $B_s = R_s$.

About sampling, Nyquist says that you need to sample at least twice as fast as the bandwidth of the signal, so you need $F_s \geq 2B_s$, where $F_s$ is the sampling rate.

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  • $\begingroup$ Okay but the last part, should be Fs>2B ... why have you put Rs? $\endgroup$ – Natalie Johnson Jan 12 at 20:45
  • $\begingroup$ @NatalieJohnson Because you are sampling the signal, not the channel. $\endgroup$ – Engineer Jan 12 at 20:52
  • $\begingroup$ I think we are mixed up in terminology. I am using B as the bandwidth of the signal. Fs>=2B and I dont understand how you are saying the bandwidth of the signal equals the symbol rate $\endgroup$ – Natalie Johnson Jan 12 at 21:36
  • $\begingroup$ @NatalieJohnson ok I see what you mean. For the sinc pulse the bandwidth is equal to the symbol rate. I tried to simplify my answer/notation by adding that clarification. Thank you for bringing that up $\endgroup$ – Engineer Jan 13 at 12:19

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