Can someone explain in simple words /graphs the necessary and sufficient Nyquist criterion for zero ISI in frequency domain, namely:

The constant folded Fourier Transform.

  • 3
    $\begingroup$ Do you have a reference for where you are seeing this term: "The constant folded Fourier Transform"? $\endgroup$
    – Robert L.
    Nov 29, 2018 at 21:46
  • 2
    $\begingroup$ I'd suggest trying to explain in more detail what it is that you don't understand. Any general explanations and figures we can present will be very similar to those already present on countless websites and textbooks. $\endgroup$
    – MBaz
    Nov 29, 2018 at 21:57
  • $\begingroup$ "Digital Communication Receivers, Synchronization, Channel Estimation, and Signal Processing" by Heinrich Meyr, page 65-66 $\endgroup$ Nov 29, 2018 at 21:57
  • 1
    $\begingroup$ What @MBaz said: not many of us will have that book, and you'll need to explain what you don't understand, because otherwise our explanation would be similar to the book, and hence, likely a waste of time. $\endgroup$ Nov 29, 2018 at 22:42
  • $\begingroup$ How does a flat folded spectrum look like vs an example of a non-flat folded spectrum? Is there a plot that compares both? $\endgroup$ Nov 29, 2018 at 23:53

3 Answers 3


The Wikipedia entry for the Nyquist ISI criterion has a good explanation.

Suppose that the symbols to be transmitted are regarded as an impulse train (one impulse every $T$ seconds). At the transmitter, the impulse train is applied to the transmitter pulse-shaping filter producing the transmitted signal, a pulse train, which is then transmitted over the channel. At the receiver, the pulse train (shaped by whatever the channel does to it) is applied to the receiver filter (possibly a matched filter) and the filter output is sampled every $T$ seconds. We regard the "channel" as a linear filter comprising the entire cascade of transmitter filter, channel, and receiver filter, whose input is an impulse train and (normalized) output is a continuous-time signal $h(t)$ that is sampled every $T$ seconds. There is no ISI if $h(t)$ is such that $$h(nT) = \begin{cases} 1, & n=0,\\0, & n\neq 0.\end{cases}\tag{1}$$ Note that there may or may not be ISI in between the sampling instants, bute we don't care; there is no constraint on what $h(t)$ might be between the sampling instants $nT$.

Let's denote by $h_0(t)$ the normalized channel response with the smallest bandwidth that satisfies $(1)$. Then, $h_0(t) =\operatorname{sinc}\left(\frac tT\right)$ whose Fourier transform $H_0(f)$ is $T \operatorname{rect}(fT)$. For reasons that will become obvious soon, we set $\operatorname{rect}\left(\pm\frac 12\right)=\frac 12$ instead of leaving it undefined as is the case in the usual definition of $\operatorname{rect}$. Note that $H_0(f) = 0$ for $|f| > \frac{1}{2T}$, while for $|f| \leq \frac{1}{2T}$, the graph of $H_0(f)$ is very nearly a perfect rectangle with base $\left[-\frac{1}{2T},\frac{1}{2T}\right]$ and height $T$ above the $f$ axis, with the minor glitch that $H_0\left(\pm\frac{1}{2T}\right) = \frac T2$.

Now, while $\operatorname{sinc}\left(\frac tT\right)$ is the smallest bandwidth impulse response such that $(1)$ holds, there are infinitely many channel impulse responses $h(t)$ of bandwidth larger than $\frac{1}{2T}$ that satisfy $(1)$. This is because the only constraint that $(1)$ imposes on $h(t)$ is that it must be such that $H(f)$ satisfies $$\sum_{k=-\infty}^{\infty}H\left(f - \frac kT\right) = T ~ \text{for all} ~f, -\infty < f < \infty.\tag{2}$$ Let's take this fancy notation apart a little. For $k=0$, the LHS of $(2)$ contains the term $H(f)$. For $k=1$, we have $H\left(f - \frac 1T\right)$ which is just $H(f)$ slid to the right by $\frac 1T$ Hz and so on. Thus the LHS of $(2)$ is just the sum of replicas of $H(f)$ repeated at intervals of $\frac 1T$ Hz along the frequency axis, and the sum of these replicas has value $T$ at all points on the frequency axis. Eq. $\mathbf{(2)}$ is the frequency-domain criterion for no ISI that the OP was asking about.

Does $(2)$ hold for $H_0(f)$? Sure it does; as described above, $H_0(f)$ is very nearly a perfect rectangle of height $T$ and base $\left[-\frac{1}{2T},\frac{1}{2T}\right]$ while $H_0\left(f - \frac 1T\right)$ is very nearly a perfect rectangle of height $T$ with base $\left[\frac{1}{2T},\frac{3}{2T}\right]$. At the glitch point $f = \frac{1}{2T}$, \begin{align} H_0(f)\big\vert_{f=\frac{1}{2T}} &= \frac T2,\\ H_0\left(f - \frac 1T\right)\big\vert_{f=\frac{1}{2T}} &= \frac T2,\\ H_0\left(f - \frac kT\right)\big\vert_{f=\frac{1}{2T}} &= 0, & k \neq 0,1 \end{align} and so the LHS of $(2)$ has value $T$ at $f = \frac{1}{2T}$. The incredulous reader should carry out this exercise at other glitch points too to verify that it works.

But, what about the other infinitely many $h(t)$'s promised just a couple of paragraphs ago? Well, let $\alpha \leq 1$ and consider $$h_{\alpha}(t)= \frac{\cos\left(\alpha \frac{\pi t}{T}\right)}{1-\left(2\alpha \frac tT\right)^2}\cdot\operatorname{sinc}\left(\frac tT\right)$$ and $$H_{\alpha}(f) = \begin{cases}T, & 0 \leq |f| \leq \frac{1-\alpha}{2T},\\ \dfrac T2 \left[1 - \sin\left(\frac{\pi T\left(|f| - \frac{1}{2T}\right)}{\alpha}\right)\right], & \frac{1-\alpha}{2T} \leq |f| \leq \frac{1+\alpha}{2T},\\ 0, & |f| > \frac{1+\alpha}{2T},\end{cases}\tag{3}$$ which is called a raised cosine spectrum with rolloff factor $\alpha$. Note that the bandwidth $\frac{1+\alpha}{2T}$ is $\alpha\%$ larger than the minimum bandwidth $\frac{1}{2T}$ achieved by $H_0(f)$. The reader is invited to verify that $H_\alpha(f)$ does satisfy $(2)$.

Finally, what's all this "constant folded Fourier transform" that the OP is asking about? It is merely a graphical (or paper-folding) way of thinking about $(2)$. The essential idea is that is we "fold" the part of the spectrum of $H_{\alpha}(f)$ that sticks out beyond $f = \pm \frac{1}{2T}$ back into the interval $\left[-\frac{1}{2T},\frac{1}{2T}\right]$, then the part that gets folded back exactly compensates for the roll-off between $\frac{1-\alpha}{2T}$ and $\frac{1}{2T}$, making the value of this folded spectrum exactly $T$ from $f = -\frac{1}{2T}$ to $f = +\frac{1}{2T}$, that is, the folded Fourier transform has constant value $T$.

  • $\begingroup$ Dilip, it certainly is not your fault, but I wish that filter frequency responses were expressed as dimensionless, which means that impulse responses have dimension of 1/time. All of your (continuous-time) impulse responses and the corresponding frequency responses should be multiplied by $\frac{1}{T}$. $\endgroup$ Sep 17, 2021 at 4:09
  • $\begingroup$ @robertbristow-johnson I don't understand your comment at all. Are you saying that, for example, $h_0(t)$ should have been $\frac 1T \operatorname{sinc}\left(\frac tT\right)$instead of $\operatorname{sinc}\left(\frac tT\right)$ the way I wrote it, and $H_0(f)$ should have been $\operatorname{rect}(fT)$ instead of $T\operatorname{rect}(fT)$ as I wrote? $\endgroup$ Sep 17, 2021 at 13:35
  • $\begingroup$ It's a dimensional analysis issue. If $H(f)$,is dimensionless (as it should be), then $h(t)$ has dimension of time${}^{-1}$. The $T$ factor multiplying $H(f)$ should be moved to a $T^{-1}$ multiplying $h(t)$. $\endgroup$ Sep 17, 2021 at 17:33
  • $\begingroup$ This is related to the thing I had been bitching about regarding the sampling and reconstruction theorem as presented in most EE texts that deal with it. $\endgroup$ Sep 17, 2021 at 18:00

The short answer is you will see that when you follow that symmetry, the resulting transform (the impulse response in time) will all have nulls spaced at T where T is the symbol duration which by definition is zero-ISI. Work through those transforms as well as ones without such symmetry and I believe it will become very clear to you what is going on. The simplest case to start is obvious: a rectangular function in one domain is a Sinc function in the other. If it is a rectangular function with width 1/T in frequency, it is a SInc in time with nulls at integer multiples of T.

Another very simple example is to then convolve the two rectangular fucntions in frequency to get a triangular fucntion which will have this "folded frequency" relationship. This is simply multiplying in time, so although we changed the Sinc response, we have not changed the location of the nulls.


Just saw this and thought it would be fun to add something to what has already been said.

The Nyquist No-ISI criteria is a rule for avoiding interference between symbols in digital communications. In order for a symbol pulse to meet the criteria, that symbol's pulse amplitude must be zero at the start of any other symbol period. The criteria itself has a formal mathematical definition that can most likely be found in your textbook and any other textbook on digital communications, so I will skip that here. There are three cases to consider where pulse shapes may meet this criteria.

  1. When the length of the pulse shape is shorter than the symbol period. One example of this is the RZ pulse shape. The orthonormal RZ pulse shape is defined as $p(t) = \sqrt{2/T_{s}}$ for $0 \le t \lt T_{s}/2$ and an example is shown in the graph below.

enter image description here


You can see the symbols do not interfere and there is actually extra space between each symbol. This is why it is called "Return to Zero."

  1. When the length of the pulse shape is equal to the symbol period. One example of this is the NRZ pulse shape. The orthonormal NRZ pulse shape is defined as $p(t) = \sqrt{1/T_{s}}$ for $0 \le t \lt T_{s}$ and an example is shown in the graph below.

enter image description here


You can see the symbols do not interfere, but there is also no extra space between symbols. This is why it is called "Non-Return to Zero."

  1. When the pulse shape period is greater than the symbol period. The most common example of this is the SRRC pulse shape. A pure SRRC pulse is bandlimited, so it is infinite in the time domain. This is desirable because it takes less bandwidth to send an SRRC pulse with the same energy as an NRZ pulse. I'm not going to attempt to write out the entire function, but a graph of it is shown below.

enter image description here


You can see that the function is zero at the start of each other symbol period and therefore satisfies the Nyquist No-ISI criteria even though the signal itself is infinite in the time domain.

It's also worth mentioning that the Nyquist No-ISI criteria only holds for an ideal channel. Most real-world channels apply a filtering effect to the transmitted waveform that stretches out the symbols and causes some ISI.


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