For the past hour, I've been trying to understand what the book is trying to say here: enter image description here

This is Communication Systems, 5th Edition by Simon Haykin.

We know that the Nyquist criterion is enter image description here I'm unable to see how if the condition specified in the first image is true then that one in the second image becomes also true. I tried plugging W as $R_{b}/2$ and $R_{b}$ and then grouping terms to reach the condition but that didn't work and then after thinking for a bit it felt that the condition in the first image does not make sense. If $P(f)$ has bandwidth $W$ then the two other terms added to it will have no effect on it (each is $2W$ wide as well.)


2 Answers 2


The point that you are missing is that $P(f)$ as used in Eq.$(4.59)$ is no longer the minimum-bandwidth function that is nonzero only for $f \in [-W,W]$ but something that is nonzero for $f \in [-W(1+\alpha),W(1+\alpha)]$ where $0 < \alpha \leq 1$. The Nyquist criterion for no intersymbol interference is not that $P(f)$ has constant value for $f \in [-W,W]$ but that if we add up frequency-shifted versions of $P(f)$ as in Eq.$(4.53)$, the sum is a constant for all $f$. Now, regardless of what $P(f)$ actually is (we don't care diddlysquat whether $P(f)$ is band-limited or not)

$$S(f) = \sum_{n=-\infty}^\infty P(f-nR_b) = \sum_{n=-\infty}^\infty P(f-n(2W))\tag{1}$$

is a periodic function of $f$ with period (not necessarily least period) $2W$. Disbelievers should carefully compute what $S(f+2W)$ is, and after murmuring a few shibboleths such as $-\infty+1 = -\infty$ and $\infty+1 = \infty$, satisfy themselves that $S(f+2W)$ is indeed equal to $S(f)$. That being said, Eq,$(4.53)$ is saying that $S(f)$ has constant value for all $f$, that is, $P(f)$ must be such that when we add up frequency-shifted copies of $P(f)$, the sum is a constant for all $f$, and in particular for all $f\in [W,W]$ which is one period of $S(f)$.

So, let's go back to Eq.$(4.59)$ where $P(f)$ has support $[-W(1+\alpha),W(1+\alpha)]$ with $0 < \alpha \leq 1$. Then, for $n \geq 2$, $P(f\pm n(2W))$ has support that is disjoint from $[-W,W]$. Thus, if we are looking only at $f\in [-W,W]$, we can cheerfully discard $P(f\pm n(2W))$ as being irrelevant in the sum $(1)$ since we are restricting attention to $f\in [-W,W]$. Eq.(4.59) is thus saying that in view of our restricted gaze, what we need is for $P(f)$ should be such that $P(f) + P(f+2W) + P(f-2W)$ equals $\dfrac{1}{2W}$ for $f\in [-W,W]$. By periodicity of $S(f)$ in $(1)$, the sum in $(1)$ equals $\dfrac{1}{2W}$ for all $f$ too.


Look at the two in the time and frequency domain. In the time domain as we limit bandwidth it extends the significant portions of the impulse response in time (leading to ISI). What is accomplished here with a Nyquist Pulse is that we are allowed to constrain the bandwidth in such a way that still allows for a relatively long impulse response, but notably the impulse response goes through zero at each multiple of the symbol rate and therefore at those locations specifically there will be zero ISI.

For more specific details please see this related post: https://dsp.stackexchange.com/a/40098/21048


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