***Still need help in 2021 - not fully clear still on Jan 20th ***

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.

Using :

  • $T_s$ = Symbol Time Period
  • $R_s$ = Symbol Rate
  • $Sinc(t/T_s)$

Logic of Bandwidth of Sinc or Raised Cosine roll off = 0

  • Time-Domain Sinc : Every lobe is width $T_s$ but main lobe is $2T_s$ because its symmetrical on zero
  • Frequency-Domain Rectangle : Positive frequency Width of $\frac{1}{T_s}$ (But its Symmetrical on zero and so BW include negative frequency)
  • So...BW $= 2\frac{1}{T_s} =2 R_s$

Logic of Bandwidth Rectangle

  • Time-Domain Rectangle : Width $T_s$ (symmetrical on zero)
  • Frequency-Domain Sinc : Every lobe width is $\frac{1}{T_s}$ but main lobe is $\frac{2}{T_s}$ because its symmetrical on zero
  • So... BW $= \frac{2}{T_s} = 2R_s$

But I have read that's not right, the Zero ISI Filter BW = $R_s/2$ when the roll off factor is zero. Why has my logic above got this wrong?

***Still need help in 2021 - not fully clear still on Jan 20th ***


Suggest you take a look at a similar question I posted Sample rates, Samples per Symbol, and Digital Pulse Shaping

In general, the Positive BW for an RRC filter is $$BW_{pos} = (1+a)\frac{R_b}{2\log_2(M)} = (1+a)\frac{R_s}{2}$$ where $R_b$ is the bit rate, $a$ is the excess bw, $R_s$ is the symbol rate, and $M$ is the constellation size.

Since you're dealing with sinc pulses here, the minimum B.W. zero-ISI pulse, setting $a=0$ simplifies the RRC to the sinc pulse and gives $BW_{pos} = \frac{R_s}{2}$. The double-sided BW is then $BW=R_s$ and thus we can say the required complex sampling rate to satisfy nyquist is $f_s = R_s$

The double sided BW is $R_s$ and thus if we're using complex sampling we only need $f_s=R_s$ or in other words 1 complex sample per symbol.

If we want to increase the excess BW parameter $a$ then we also must increase our complex sampling rate according to $(1+a)R_s$ to avoid aliasing. In general we like to oversample however since otherwise the filtering of images in DACs & interpolation stages becomes difficult.

EDIT: Additional derivation on the boxcar in frequency:

The rect function is defined as (copied from wiki because I don't want to rewrite it) Rect

So if our boxcar in frequency has value $T_s$ over the interval $[-\frac{1}{2T_s},+\frac{1}{2T_s}]$ we let $B=\frac{1}{2T_s}$ and that gives us $$T_s rect(T_s f)$$

The fourier transform of this (again copied from wiki): sinc_in_time

and plugging in for $B=\frac{1}{2T_s}$ and multiplying by $T_s$ we get $$T_s 2 \frac{1}{2T_s} sinc(2 \frac{1}{2T_s} t) = sinc(\frac{t}{T_s})$$

So we can clearly see the 2-sided BW of the box-car is $\frac{1}{T_s}$ and the sinc will have zero crossings at multiples of $T_s$. If we imagine using these as pulses at a rate of $\frac{1}{T_s}$ it will look like the following sinc pulses

Note i've taken $T_s=1$ here. We place the sinc at multiples of $T_s$ (ive only plotted the first three) and we can clearly see that at each symbol center there is zero contribution from the adjacent pulses due to the zero crossing locations - this is what the zero-ISI property promised us.

  • $\begingroup$ Hi, Thanks for the reply. A Rectangular pulse (or boxcar) in time domain of width Ts centered on 0 produces a sinc in the frequency domain with x-domain crossings every 1/Ts, but the main lobe (centered on 0) spans 1/Ts in positive frequency and 1/Ts in negative frequency. So main lobe BW = 2/Ts = 2Rs ... How is that wrong? $\endgroup$ – Natalie Johnson Jan 19 at 19:51
  • $\begingroup$ I have looked at your other topic you linked but still cant understand why my logic in my bullet points leads to the incorrect rule for zero ISI bandwidth. $\endgroup$ – Natalie Johnson Jan 19 at 19:56
  • $\begingroup$ @NatalieJohnson I added a section going over the boxcar in freq to sinc in time. Does that help or is there something still confusing? My point is that the boxcar in freq has a total width of $\frac{1}{T_s}$ and that is thus its bandwidth. So im not sure why your bullet stated the bandwidth is twice that. $\endgroup$ – user67081 Jan 19 at 22:29
  • $\begingroup$ I am still not following this ... sinc(t/Ts) has crossings every Ts that are zero. Also, in the frequency domain this produces a cut off frequency as 1/Ts. This means the double sided spectrum must be 2/Ts since the cut off is symmetrical on zero. Many images of this on google search $\endgroup$ – Natalie Johnson Jan 20 at 22:25

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.


To more directly address the points in the question...

  • Bandwidth of pulses: the FT of a perfect rectangular pulse is a sinc function which has infinite bandwidth.

  • Bandwidth of zero-ISI filter: the FT of a sinc function is a perfect rectangular pulse. If the zeros in the sinc function are at $kT_s$, where $k \in (-\infty, \infty)$, then the bandwidth is $\frac{1}{T_s}=R_s$. In case the time domain signal is purely real, then the FT will be symmetric meaning that the negative frequencies give you no more information then the positive frequencies. This cuts the bandwidth in half to $\frac{R_s}{2}$.

enter image description here

  • $\begingroup$ 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
  • $\begingroup$ @NatalieJohnson I have again further edited my answer to address your comments below. $\endgroup$ – Engineer Jan 20 at 14:05

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