# Confused on Pulse Filter Bandwidth, Raw Data Bandwidth and Symbol Rate Relation

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

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.