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Nyquist zero-ISI criteria; one pulse shaping method

Short answer

Nyquist criterion: overall pulse p(nTb)=1 for n=0 and 0 otherwise, achieved if folded spectrum is constant. The raised cosine pulse satisfies this with roll-off factor α, bandwidth B_T = (Rb/2)(1+α), and provides fast tail decay, reducing ISI even with timing jitter.

State Nyquist pulse shaping criteria for Zero ISI. Discuss any one pulse shaping method of ISI reduction.

Nyquist criterion for zero ISI

For an overall pulse p(t)p(t) sampled every TbT_b seconds, there is no ISI if

p(nTb)={1,n=00,n≠0⟺∑n=−∞∞P(f−nTb)=Tbp(nT_b) = \begin{cases} 1, & n = 0 \\ 0, & n \ne 0 \end{cases} \quad\Longleftrightarrow\quad \sum_{n=-\infty}^{\infty} P\left(f - \frac{n}{T_b}\right) = T_b

The ideal solution is the rectangular spectrum of bandwidth B0=Rb/2B_0 = R_b/2, p(t)=sinc(t/Tb)p(t) = \text{sinc}(t/T_b), but it is not realizable and its slowly decaying tails make it very sensitive to timing errors.

Raised cosine pulse shaping

The raised cosine spectrum keeps the zero-ISI property but rolls off gradually:

P(f)={12B0,0≤∣f∣<f114B0[1+cos⁡π(∣f∣−f1)2B0−2f1],f1≤∣f∣<2B0−f10,∣f∣≥2B0−f1P(f) = \begin{cases} \dfrac{1}{2B_0}, & 0 \le |f| < f_1 \\[2mm] \dfrac{1}{4B_0}\left[1 + \cos\dfrac{\pi(|f| - f_1)}{2B_0 - 2f_1}\right], & f_1 \le |f| < 2B_0 - f_1 \\[2mm] 0, & |f| \ge 2B_0 - f_1 \end{cases}

with B0=Rb2B_0 = \frac{R_b}{2}, roll-off factor α=1−f1B0\alpha = 1 - \frac{f_1}{B_0} (0≤α≤10 \le \alpha \le 1).

Transmission bandwidth:

BT=B0(1+α)=Rb2(1+α)B_T = B_0(1 + \alpha) = \frac{R_b}{2}(1+\alpha)

Pulse:

p(t)=sinc(2B0t) cos⁡(2παB0t)1−16α2B02t2p(t) = \text{sinc}(2B_0t)\,\frac{\cos(2\pi\alpha B_0 t)}{1 - 16\alpha^2B_0^2t^2}

The sinc factor keeps zeros at t=nTbt = nT_b (zero ISI); the second factor makes the tails decay as 1/∣t∣31/|t|^3.

 P(f)
   |---------.                alpha = 0
   |----.     '.  .           alpha = 0.5
   |      '.    '.  '.        alpha = 1
   |        '.    '.  '.
   +----------+-----+----+--> f
   0       Rb/2  3Rb/4  Rb
α\alphaBandwidthRemarks
0Rb/2R_b/2Ideal Nyquist, unrealizable
0.50.75Rb0.75R_bCommon practical choice
1RbR_bFull cosine roll-off, easy filters, tails decay fastest

Example: Rb=10R_b = 10 kbps, α=0.5\alpha = 0.5 gives BT=5000×1.5=7.5B_T = 5000 \times 1.5 = 7.5 kHz.

Advantages: realizable filters, small ISI under timing jitter, wide eye opening. Cost: extra bandwidth αB0\alpha B_0. The filter is often split equally as root raised cosine filters at transmitter and receiver.

✓ Answer checked by Claude Opus 5.5 (high effort)

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