0
$\begingroup$

I understand the Nyquist criterion in the frequency domain, which specifies that for zero inter-symbol interference (ISI), the signal's bandwidth should be limited to $\frac{R_{b}}{2}$, where ${R_{b}=\frac 1T}$ is the symbol rate or

$\sum_{k=-\infty}^{\infty}H\left(f - \frac kT\right) = T ~ \text{for all} ~f, -\infty < f < \infty$

However, if the roll-off factor is greater than 1, leading to a bandwidth as an example of $\frac{3}{2T}$ will there be any ISI introduced? How does exceeding affect the system, and is there a trade-off in terms of bandwidth and ISI when ? Or is it just a question of reduced spectral efficiency?

Improve this question
$\endgroup$
5
  • $\begingroup$ Your statement that the bandwidth should be limited to $R_b/2$ is wrong. It needs to be at least $2\cdot R_b$, not at most $R_b/2$. (for example, use a very short rectangle as time domain pulse shape. It has infinite bandwidth, yet definitely no ISI. $\endgroup$
    – Marcus Müller
    Commented Aug 24 at 18:50
  • 1
    $\begingroup$ @MarcusMüller I believe the min bandwidth is $R_b$. Consider a 10101 pattern as the highest possible data rate- we can still send that with a sine wave at frequency $R_b/2$, and thus two-sided bandwidth of $R_b$. Generally with good pulse shaping the RF bandwidth approaches the symbol rate. Am I mixing anything up? $\endgroup$
    – Dan Boschen
    Commented Aug 25 at 2:40
  • $\begingroup$ @DanBoschen no, it's probably me, and I got the complex/real bandwidth mixed up in my head. $\endgroup$
    – Marcus Müller
    Commented Aug 25 at 11:12
  • $\begingroup$ @DanBoschen Yet I think the remark that the requirement is regarding min bandwidth is not maximum is correct, is it not? Otherwise it makes no sense. $\endgroup$
    – Yair M
    Commented Aug 25 at 17:42
  • $\begingroup$ @YairM If you are saying there is a requirement on BW (not the maximum it can be but the minimum int can be), then yes I agree. $\endgroup$
    – Dan Boschen
    Commented Aug 25 at 20:29

1 Answer 1

Reset to default
2
$\begingroup$

The Nyquist Zero ISI criterion does not limit the signal's maximum bandwidth, but it does result in a lower limit for the double-sided bandwidth to be greater than $1/T_s$ Hz where $T_s$ is the symbol duration in seconds.

The formula that the OP gives is correct:

$$\sum_{k=-\infty}^{\infty}H\left(f - \frac kT_s\right) = T_s ~ \text{for all} ~f, -\infty < f < \infty$$

Where for clarity I replaced $T$ with $T_s$, as the symbol duration. $h(t)$ can be interpreted as the pulse shape itself, and therefore can be convolved with a series of modulation symbols each represented as weighted impulses. $H(f)$ is therefore simply the Fourier Transform of the pulse shape. With equiprobable randomly distributed symbols, the power spectral density of a pulse-shaped modulated waveform will take on the power spectral density of the pulse (this then makes it very easy to evaluate general conditions from the pulse shape alone).

The Nyquist ISI criterion states that frequency-shifted copies of $H(f)$ must add up to a constant value. This will occur when the bandwidth is greater than or equal to $1/T_s$ (and to be very clear here I am referring to the two-sided bandwidth, which is the same bandwidth we would see in the passband waveform at RF), and the spectrum is even symmetric about it's center and odd symmetric at $\pm \frac{1}{2T_s}$ as I demonstrate graphically below. If we meet these conditions then zero ISI will result!

Nyquist Zero ISI in freq domain

A spectral shape with those conditions will achieve the Nyquist Zero ISI criterion, but those are not the only cases. Consider the extreme of rectangular pulses in time: If a pulse for each symbol only lasts for the symbol duration $T_s$, clearly there will not be any ISI. The frequency response for a rectangular pulse of duration $T_s$ is a Sinc function in frequency with the first nulls occurring at $\pm 1/T_s$. There is no odd symmetry at $\pm 1/(2T_s)$ in this case yet shifted versions of a Sinc function by $k/T_s$ with nulls spaced at $k/T_s$ for $k$ being all integers except for $0$ will still all add up to a constant value! That specifically is the Nyquist Zero ISI criterion.

For further details and demonstration of "Zero ISI", please see my answer posted for DSP.SE #40094.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.