Principles of Digital Data Transmission

Digital Communication Systems

• By the end of 1990s, digital format was dominating.
• Examples: CD, MP3, HDTV, DVD, Blu Ray, VoIP, 4k, 8k etc.
• This chapter is concerned with transmitting digital data over a channel.
• We begin with binary data, i.e., symbols 0 and 1.
• We assign a waveform or pulse to each of the two symbols.
• At the receiver, the received pulses are detected and converted back to binary data.

Digital Communication System Block Diagram

• The diagram includes components like the message signal, baseband modulation, digital carrier, multiplexer, channel, and regenerative repeater.

Source

• The input takes the form of a data set, digital audio (PCM, DM, LPC), digital video, telemetry data, etc.
• Discussion is confined to binary data (two symbols).
• M-ary communication (M symbols) is discussed in Section 7.7 in the textbook.

Line Coder

• Converts digital input into electrical pulses or waveforms.
• This process is called line coding or transmission coding.
• There are many ways to encode binary data.

Types of Line Codes

• Examples include on-off (RZ), polar (RZ), bipolar (RZ), on-off (NRZ), and polar (NRZ).

Multiplexer

• Channels typically have much wider bandwidths than individual sources.
• To utilize capacity effectively, multiplex several sources on one channel.
• Digital multiplexing can be time division or frequency division multiplexing.
• Code division multiplexing can also be used by using a set of orthogonal codes.

Regenerative Repeater

• Spaced at regular intervals on a transmission line to detect incoming signals and generate “clean” pulses for further transmission.
• This effectively eliminates noise.
• The bit rate $R_b$ needs to be estimated from the incoming signal.
• Return to zero (RZ) line codes allow for the extraction of $R_b$ (clock recovery) by rectifying the signal.
• A line code is transparent in which the bit pattern does not affect timing accuracy (such as RZ).

RZ Clock Recovery

• An on-off signal (a) is a sum of a random polar signal (b) and a clock frequency periodic signal (c).

Why Line Coding?

• Transmission bandwidth as small as possible.
• Power efficiency: For a given bandwidth, power must be as low as possible.
• Error detection and correction capability: Some line codes allow for error detection.
• PSD properties that fit channel: Some channels do not pass DC, and then DC-free codes are needed.

PAM Line Code

• A generic pulse $p(t)$ is considered, where $P(f)$ is its Fourier transform.
• The line code symbol at time $k$ is $a<em>k$ starting at time $t = kT</em>b$.
• The $k^{th}$ symbol is transmitted as $a<em>kp(t - kT</em>b)$.
• The baseband signal in a pulse train is of the form $y(t) = \sum a<em>kp(t - kT</em>b)$.
• The line coder determines the symbol set \left{ a_k \right}.
• The values of $a_k$ are random and depend on the line coder input.
• $y(t)$ is a pulse-amplitude-modulated (PAM) signal.
• $S<em>y(f) = |P(f)|^2 S</em>x(f)$

PSD of a Line Code

• On-off, polar, and bi-polar are special cases of PAM modulation, where $a_k$ takes values 1, 0, and -1 randomly.
• PSD of $y(t)$ depends on $a_k$ and $p(t)$, so if the pulse shape changes, the PSD changes.
• Can be overcome by having signal $x(t)$ with $T<em>b$ spaced impulses with area $a</em>k$ passed through a filter with impulse response $p(t)$ and transfer function $P(f)$: $S<em>y(f) = |P(f)|^2 S</em>x(f)$.

PSD of a Line Code (Continued)

• We need to determine the autocorrelation function $R_x(\tau)$ of the impulse train $x(t)$.
• This can be achieved by considering the impulses as limiting forms of the rectangular function.
• For a pulse width $\epsilon \rightarrow 0$, the pulse height is given by $h<em>k = \frac{a</em>k}{\epsilon} \rightarrow \infty$.
• The strength or area of the $k^{th}$ impulse is $a<em>k = \epsilon h</em>k$.
• The autocorrelation of the rectangular pulse train $\hat{x}(t)$ is given by $R<em>{\hat{x}}(\tau) = \lim</em>{T \rightarrow \infty} \frac{1}{T} \int_{-T/2}^{T/2} \hat{x}(t)\hat{x}(t - \tau) dt$.

PSD of a Line Code (Continued)

$R<em>{\hat{x}}(\tau) = \lim</em>{T \rightarrow \infty} \frac{1}{T} \int_{-T/2}^{T/2} \hat{x}(t)\hat{x}(t - \tau) dt$ (1)

• $R_x(\tau)$ is an even function of $\tau$, as such we only consider \tau > 0.
• To start, consider the case of \tau < \epsilon, where (1) is the area under the signal $\hat{x}(t)$, multiplied by $\hat{x}(t)$ delayed by $\tau$ for \tau < \epsilon.
• As can be seen, the area associated with the $k^{th}$ pulse is $(\epsilon - \tau)h<em>k^2$, where R{\hat{x}}(\tau < \epsilon) = \lim{T \rightarrow \infty} \frac{1}{T} \sumk (\epsilon - \tau)hk^2 = \lim{T \rightarrow \infty} \frac{1}{T} \sumk ak^2 \frac{(\epsilon - \tau)}{\epsilon^2} = \frac{R0}{\epsilon Tb} (1 - \frac{\tau}{\epsilon}),
  and $R<em>0 = \lim</em>{T \rightarrow \infty} \frac{T<em>b}{T} \sum</em>k a_k^2$.

Line Code PSD Signals

• Illustrations showing impulse trains and pulse shapes for autocorrelation calculations.

PSD of a Line Code (Continued)

• Recall $R<em>0 = \lim</em>{T \rightarrow \infty} \frac{T<em>b}{T} \sum</em>k a_k^2$.
• During the averaging interval $T$ where $T \rightarrow \infty$, there are $N$ pulses where $N \rightarrow \infty$, where $N = \frac{T}{T_b}$.
• Hence $R<em>0 = \lim</em>{N \rightarrow \infty} \frac{1}{N} \sum<em>k a</em>k^2 = \langle a<em>k^2 \rangle$, where the summation is over $N$ pulses; as such, $R</em>0$ is the time average of the squared pulse amplitude.
• Since $R<em>x(\tau)$ is an even function, we have $R</em>{\hat{x}}(\tau) = \frac{R<em>0}{\epsilon T</em>b} (1 - \frac{|\tau|}{\epsilon})$ , |\tau| < \epsilon

PSD of a Line Code (Continued)

• What happens for \tau > \epsilon?
• The next overlap will happen at the $(k + 1)^{th}$ pulse and $\tau$ approaches $T_b$.
• Repeating previous argument, we have $R<em>1 = \lim</em>{T \rightarrow \infty} \frac{T<em>b}{T} \sum</em>k a<em>ka</em>{k+1} = \lim<em>{N \rightarrow \infty} \frac{1}{N} \sum</em>k a<em>ka</em>{k+1} = \langle a<em>ka</em>{k+1} \rangle$.
• For multiples of $\pm nT<em>b$, we have that the height of the pulses centered $\pm nT</em>b$ is $\frac{R<em>n}{(\epsilon T</em>b)}$, where
  $R<em>n = \lim</em>{T \rightarrow \infty} \frac{T<em>b}{T} \sum</em>k a<em>ka</em>{k+n} = \lim<em>{N \rightarrow \infty} \frac{1}{N} \sum</em>k a<em>ka</em>{k+n} = \langle a<em>ka</em>{k+n} \rangle$

Line Code PSD Signals

• Illustrations detailing pulse amplitude and autocorrelation for calculating PSD.

PSD of a Line Code (Continued)

• To find $R_x(\tau)$, let $\epsilon \rightarrow 0$, and each triangular pulse becomes an impulse function with the same area as before.

• The $n^{th}$ pulse has height $\frac{R<em>n}{(\epsilon T</em>b)}$ and area $\frac{R<em>n}{T</em>b}$.

• Hence, from Fig 7.5e, we have $R<em>x(\tau) = \frac{1}{T</em>b} \sum<em>{n=-\infty}^{\infty} R</em>n \delta(\tau - nT_b)$.

• Hence, the PSD $S<em>x(f)$ is the Fourier transform of $R</em>x(\tau)$, i.e.,

  $S<em>x(f) = \frac{1}{T</em>b} \sum<em>{n=-\infty}^{\infty} R</em>n e^{-jn2\pi fT_b}$.

• Recognising that $R<em>{-n} = R</em>n$, because $R_x(\tau)$ is an even function of $\tau$, we have

  $S<em>x(f) = \frac{1}{T</em>b} \left[ R<em>0 + 2 \sum</em>{n=1}^{\infty} R<em>n \cos(n2\pi fT</em>b) \right]$ (2)

PSD of a Line Code (Continued)

Finally! We can now determine the PSD of the line code as

$S<em>y(f) = |P(f)|^2 S</em>x(f) = \frac{|P(f)|^2}{T<em>b} \left[ \sum</em>{n=-\infty}^{\infty} R<em>n e^{-jn2\pi fT</em>b} \right] = \frac{|P(f)|^2}{T<em>b} \left[ R</em>0 + \sum<em>{n=1}^{\infty} R</em>n \cos(n2\pi fT<em>b) \right]$ Thus, the PSD of the line code is fully characterised by its $R</em>n$ and the pulse shaping transfer function $P(f)$.

Polar Signalling PSD

• For a polar line code, $R<em>0 = 1$ and $R</em>n = 0$ for $n \ge 1$ (see textbook for derivation). As such, $S<em>y(f) = \frac{|P(f)|^2}{T</em>b} R<em>0 = \frac{|P(f)|^2}{T</em>b}$.

• Assume rectangular pulses of width $T<em>b/2$, i.e. $p(t) = \Pi \left( \frac{t}{T</em>b/2} \right) = \Pi \left( \frac{2t}{T<em>b} \right)$, and $P(f) = \frac{T</em>b}{2} \text{sinc} \left( \frac{\pi fT_b}{2} \right)$.

• Therefore, $S<em>y(f) = \frac{T</em>b}{4} \text{sinc}^2 \left( \frac{\pi fT_b}{2} \right)$.

Polar Signalling PSD (Continued)

• Clear that most power density situated at DC.

• Essential bandwidth - first non-DC null $(2R_b)$ Hz for polar line code.

• This is 4x the theoretical bandwidth (Nyquist BW) required to transmit $R_b$ pulses per second.

• For a full-width pulse, the essential BW is still twice the theoretical BW.

Polar Signalling PSD (Continued)

• Power density at DC is a problem for repeaters.
• Polar signaling does not allow for error correction.
• Polar signaling is the most efficient in terms of power (lowest detection error probability amongst all line codes ).
• Polar signaling is transparent - except when using full-width pulses.

Constructing DC Null Using Pulse Shaping

• Since $S<em>y(f) = |P(f)|^2 S</em>x(f)$, we can select $p(t)$ such that $P(f)$ has a DC null.
• Because $P(f) = \int<em>{-\infty}^{\infty} p(t)e^{-j2\pi ft} dt$, we have $P(0) = \int</em>{-\infty}^{\infty} p(t) dt$.
• Hence, if a pulse $p(t)$ integrates to zero, $P(0) = 0$.
• Consider a Manchester code aka split-phase code aka twinned binary code.

DC-Free Manchester Code

• Illustrations showing the basic pulse $p(t)$ and the transmitted waveform for binary data sequence using Manchester signaling.

On-Off Signalling PSD

• 1 - pulse transmitted, 0 - no pulse transmitted.

• $a_k$ assumed equally probable to be a 0 or 1.

• Hence, $R<em>0 = \lim</em>{N \rightarrow \infty} \frac{1}{N} \left[ \frac{N}{2}(1)^2 + \frac{N}{2}(0)^2 \right] = \frac{1}{2}$.

• To calculate $R<em>n$, we consider the product $a</em>ka<em>{k+n}$, where both $a</em>k$ and $a_{k+n}$ are equally probable to be a 0 or 1.

• Therefore, the product $a<em>ka</em>{k+n}$ will be 1, one in four times and zero three in four times, i.e.

  $R<em>n = \lim</em>{N \rightarrow \infty} \frac{1}{N} \left[ \frac{N}{4}(1)^2 + \frac{3N}{4}(0)^2 \right] = \frac{1}{4}$, $n \ge 1$.

• Using the calculated values for $R<em>n$, $S</em>x(f)$ can be obtained:

  $S<em>x(f) = \frac{1}{4T</em>b} + \frac{1}{4T<em>b^2} \sum</em>{n=-\infty}^{\infty} e^{-jn2\pi fT<em>b} = \frac{1}{4T</em>b} + \frac{1}{4T<em>b^2} \sum</em>{n=-\infty}^{\infty} \delta \left( f - \frac{n}{T_b} \right)$.

On-Off Signalling PSD (2)

• The spectrum of the on-off waveform is given by

  $S<em>y(f) = \frac{|P(f)|^2}{4T</em>b} \left[ 1 + \frac{1}{T<em>b} \sum</em>{n=-\infty}^{\infty} \delta \left( f - \frac{n}{T_b} \right) \right]$.

• NOTE: the spectrum contains a continuous part due to $P(f)$ and a discrete part owing to the pulse train

• The discrete part can be nullified if the pulse shape $p(t)$ is chosen such that $P \left( \frac{n}{T_b} \right) = 0$, for $n = \pm 1, \pm 2, . . .$.

• In the case of a half-width rectangular pulse, we have

  $S<em>y(f) = \frac{T</em>b}{16} \text{sinc}^2 \left( \frac{\pi fT<em>b}{2} \right) \left[ 1 + \frac{1}{T</em>b} \sum<em>{n=-\infty}^{\infty} \delta \left( f - \frac{n}{T</em>b} \right) \right]$.

On-Off Signalling PSD

• A graph illustrating the power spectral density (PSD) of an on-off signal.

On-Off Signalling - Advantages/Disadvantages

• It is less immune to noise interference than polar signaling
• Noise immunity depends on amplitude difference 0 to 1 (on-off) vs. -1 to 1 (polar)
• Could consider 0 to 2, but that would consume more power
• Average power for on-off signalling is $P<em>{\text{on-off}} = \frac{2E}{T</em>b}$, where $E$ is the pulse energy. $P_{\text{on-off}}$ is twice the value required for the polar signal
• A long string of zeros causes a lack of signal, which leads to synchronization loss.

Bipolar Signalling - Alternate Mark Inversion (AMI)

• 0 - no pulse, 1 - pulse $p(t) = -p(t - T_b)$

• Bi-polar signalling actually uses three symbols: $\left[ -p(t), 0, p(t) \right]$

• Half $a<em>k$s are 0, the other half are either 1 or -1, with $a</em>k^2 = 1$, hence

  $R<em>0 = \lim</em>{N \rightarrow \infty} \frac{1}{N} \sum<em>k a</em>k^2 = \lim_{N \rightarrow \infty} \frac{1}{N} \left[ \frac{N}{2} (\pm 1)^2 + \frac{N}{2} (0)^2 \right] = \frac{1}{2}$.

Bipolar Signalling - Alternate Mark Inversion (AMI) (2)

• Now consider $a<em>ka</em>{k+1}$ for $R<em>1$. There are four possibilities: 11, 10, 01, 00. The product $a</em>ka<em>{k+1}$ will be zero for the last three cases. As such, 3N/4 combinations have $a</em>ka<em>{k+1} = 0$ and N/4 combinations have $a</em>ka<em>{k+1} = 1$. Since 11 always alternate, the product $a</em>ka_{k+1} = -1$ when 11 occurs. Hence

  $R<em>1 = \lim</em>{N \rightarrow \infty} \frac{1}{N} \left[ \frac{N}{4} (-1) + \frac{3N}{4} (0) \right] = -\frac{1}{4}$.

• Now consider $a<em>ka</em>{k+2}$ for $R_2$. Similar reasoning can be used, but now there are eight possible input bit combinations, in that case

  $R<em>2 = \lim</em>{N \rightarrow \infty} \frac{1}{N} \left[ \frac{N}{8} (1) + \frac{N}{8} (-1) + \frac{6N}{8} (0) \right] = 0$. NOTE: There is an error in the textbook.

Bipolar Signalling - Alternate Mark Inversion (AMI) (3)

• In general $R<em>n = \lim</em>{N \rightarrow \infty} \frac{1}{N} \sum<em>k a</em>ka_{k+n}$.

• Now consider $a<em>ka</em>{k+2}$ for $R_2$. Similar reasoning can be used, but now there are eight possible input bit combinations, in that case :

  $R<em>2 = \lim</em>{N \rightarrow \infty} \frac{1}{N} \left[ \frac{N}{8} (1) + \frac{N}{8} (-1) + \frac{6N}{8} (0) \right] = 0$.

• For n > 2, the product $a<em>ka</em>{k+n}$ can be 1, -1 or 0. Moreover an equal number of combinations have values 1 and -1, hence $R_n = 0$

• Therefore, $R_n = 0$, for n > 1.

Bipolar Signalling - Alternate Mark Inversion (AMI) PSD

• Hence, from eq. (2), the PSD of a bipolar signal is

  $S<em>y(f) = \frac{|P(f)|^2}{2T</em>b} \left[ 1 - \cos(2\pi fT<em>b) \right] = \frac{|P(f)|^2}{2T</em>b} \sin^2(\pi fT_b)$.

• A DC null will be present because of the squared sine factor

• In the case of a half-width rectangular pulse, we have $S<em>y(f) = \frac{T</em>b}{4} \text{sinc}^2 \left( \frac{\pi fT<em>b}{2} \right) \sin^2(\pi fT</em>b)$.

Bipolar Signalling - Alternate Mark Inversion (AMI) PSD (2)

• Essential bandwidth is $R<em>b = 1/T</em>b$ - half that of polar signalling using same half-width pulse.

Bipolar Signalling - Advantages/Disadvantages

• Spectrum has a DC null
• It is bandwidth-efficient
• It has single-error detection capability (alternating pulse rule)
• Requires twice as much power as a polar signal
• Bipolar signalling is not transparent
• Two versions of bipolar signalling that are transparent:
  • High-density bipolar (HDB) signalling
  • Binary with N zero substitution (BNZS) signalling.

Pulse Shaping

• The PSD $S_y(f)$ can be controlled by the choice of line code or pulse shape
• Previously, we considered controlling $S_y(f)$ through the choice of line code
• Now, $p(t)$ will be manipulated to influence $S<em>y(f)$ through $S</em>y(f) = |P(f)|^2 S_x(f)$.

Intersymbol Interference

• In the last section, we considered the case where $p(t)$ was a half-width rectangular pulse

• Strictly speaking, a half-width rectangular pulse has infinite bandwidth, and therefore also $S_y(f)$ (essential BW is finite though)

• PSD power in range f > Rb is suppressed for transmission over the channel with BW $R</em>b$

• Suppression of spectrum in f > Rb limits BW and causes time spreading beyond the allotted time interval $T</em>b$

• This causes interference with neighboring pulses - known as intersymbol interference (ISI)

• ISI is not noise, but distortion

Apparent Impasse

• Bandlimited signals require infinite time duration
• Time-limited signals require infinite bandwidth
• How can we transmit finite pulses over a bandlimited channel without introducing ISI??
• We can relax the ISI constraint - that is, no ISI only at the sample instances
• Nyquist proposed three criteria for pulse shaping, where pulses are allowed to overlap
• We will consider the first criterion

Nyquist Criterion for Zero-ISI - First Criterion

• Choose a pulse shape that has nonzero amplitude at its center and zero amplitude at $t = \pm nT<em>b, (n = 1, 2, 3, . . .)$, where $T</em>b$ is the separation between transmitted pulses, i.e.

  $p(t) = \begin{cases} 1 &amp; t = 0 \ 0 &amp; t = \pm nT<em>b, \text{where } T</em>b = 1/R_b \end{cases}$ (3)

  *Do we know of such a pulse?

Zero-ISI Pulse Example

$p(t) = \text{sinc}(\pi R<em>b t), \quad P(f) = \frac{1}{R</em>b} \Pi \left( \frac{f}{R_b} \right)$

• Has BW of $R<em>b/2$ - can transmit $R</em>b$ pulses per second over BW $R_b/2$
• This is the theoretical maximum transmission rate

Practical Challenges of Using a Sinc Pulse

• The sinc pulse is of infinite duration
• A truncation would increase its bandwidth
• Signal decays too slowly, at a rate of $1/t$
• What would happen in the case of timing deviations/jitter?

Nyquist’s First Criterion for Zero ISI

• Consider a pulse $p(t) \Longleftrightarrow P(f)$ where the bandwidth of $P(f)$ is on the interval $(R<em>b/2, R</em>b)$

• We sample the pulse $p(t)$ every $T<em>b$ seconds by multiplying $p(t)$ by $\delta</em>{T<em>b}(t)$, where $\delta</em>{T<em>b}(t)$ is a $T</em>b$ spaced impulse train

• Since $p(t)$ satisfies (3), the sampled signal $\bar{p}(t)$ is given by

  $\bar{p}(t) = p(t) \delta<em>{T</em>b}(t) = \delta(t)$

• Take the Fourier transform on both sides of the second equality, we obtain

  $\frac{1}{T<em>b} \sum</em>{n=-\infty}^{\infty} P(f - nR<em>b) = 1$, where $R</em>b = 1/T_b$, or

  $\sum<em>{n=-\infty}^{\infty} P(f - nR</em>b) = T_b$

Nyquist's First Criterion for Zero ISI - Example

• Over the range 0 < f < Rb: P(f) + P(f - Rb) = Tb \quad 0 < f < Rb

• Letting $x = f - R_b/2$, we have

  $P \left( x + \frac{R<em>b}{2} \right) + P \left( x - \frac{R</em>b}{2} \right) = T<em>b \quad |x| < 0.5R</em>b$

• Using the conjugate symmetry property, we have

  $P \left( \frac{R<em>b}{2} + x \right) + P^* \left( \frac{R</em>b}{2} - x \right) = T<em>b \quad |x| < 0.5R</em>b$

• Let us choose $P(f)$ to be real-valued and positive, then only $|P(f)|$ needs to satisfy the above equation

  $|P \left( \frac{R<em>b}{2} + x \right)| + |P^* \left( \frac{R</em>b}{2} - x \right)| = T<em>b \quad |x| < 0.5R</em>b$

$|P(f)|$ Properties for Zero ISI

• Note that $p(0.5R_b) = 0.5P(0)$

Bandwidth of Zero ISI $|P(f)|$ - Vestigial Spectrum

• The bandwidth in Hz of $P(f)$ is $0.5R<em>b + f</em>x$.
• Let $r$ be the ratio of the excess bandwidth $f<em>x$ to the theoretical minimum bandwidth $R</em>b/2$, i.e.
  r= excess bandwidth/ theoretical minimum bandwidth = fx/0.5Rb = 2fxTb.
• Therefore, the bandwidth of $P(f)$ is $B<em>T = \frac{R</em>b}{2} + r\frac{R<em>b}{2} = \frac{(1 + r)R</em>b}{2}$
• $r$ is known as the roll-off factor - also specified as a percentage (for example, for a bandwidth of 1.5$R_b$, $r = 0.5$ or 50 %)

Nyquist Zero ISI Filter

One family of spectra satisfies Nyquist’s first criterion:

$P(f) = \begin{cases} 1, &amp; |f| < R<em>b/2 - f</em>x \ \frac{1}{2} \left[ 1 - \sin \pi \left( \frac{f - R<em>b/2}{2f</em>x} \right) \right] & |f - R<em>b/2| < f</em>x \ 0, & |f| > R<em>b/2 + f</em>x \end{cases}$

Raised Cosine Filter

In the maximum bandwidth case, where $f<em>x = R</em>b/2$ (i.e., $r = 1$), we have
P(f) = 1/2 (1+cos(πfTb)Πf /2Rb (4)
= cos2(πfTb/2) Π (fTb/2) (5)

• This is known as the raised cosine or full-cosine roll-off filter/characteristic.

• The inverse Fourier transform of eq (4) gives us the pulse $p(t)$, i.e.

  $p(t) = \frac{R<em>b \cos \pi R</em>b t}{1 - 4R<em>b^2 t^2} \text{sinc}(\pi R</em>b t)$

Raised Cosine Filter Observations

• Zero at multiples of $T<em>b$ AND $0.5T</em>b$
• Decays rapidly, at a rate of $1/t^3$
• Raised cosine-based pulse insensitive to timing inaccuracies
• Pulse generating filter is closely realisable

Controlled ISI - Partial Response Signalling

• Nyquist first criterion requires bandwidth larger than the theoretical minimum

• Possible to further reduce bandwidth by widening the pulse $p(t)$

• However - this WILL introduce ISI

• Since transmission is binary, with only two possible symbols, it will be possible to remove the interference - limited number of interference patterns

• Consider a pulse $p(t)$ which satisfies the following:

  $p(nT_b) = \begin{cases} 1, &amp; n = 0, 1 \ 0, &amp; \text{for all other } n \end{cases}$ (6)

• This leads to controlled ISI from the $k^{th}$ to the next pulse only

Possibilities in Controlled ISI Signalling

• Decision rule at the receiver:

  • If the sample value is positive $(present bit, previous bit) = (1,1)$
  • If the sample value is negative $(present bit, previous bit) = (0,0)$
  • If the sample value is zero, then present bit = NOT previous bit - knowledge of the previous bit allows us to know the present bit

• This is known as partial response signalling

• A pulse satisfying eq (6) is called a duobinary pulse

Example: Duobinary Pulse

• Question: What pulse would meet the requirements of eq (57)?

• Solution: From Problem 7.3-9 - only the following pulse $p(t)$ meets the requirements:

  $p(t) = \frac{\sin(\pi R<em>b t)}{\pi R</em>b t (1 - R_b t)}$

• The Fourier transform $P(f)$ of the pulse $p(t)$ is given by

  $P(f) = \frac{2}{R<em>b} \cos \left( \frac{\pi f}{R</em>b} \right) \Pi \left( \frac{f}{R<em>b} \right) e^{-j\pi f/R</em>b}$

Example: Duobinary Pulse (Plots)

• Illustrations of the minimum bandwidth pulse that satisfies the duobinary pulse criterion and its spectrum.

Duobinary Pulse - Properties

• Pulse not ideally realisable – noncausal with infinite duration
• However, it decays rapidly with $t$ at a rate $1/t^2$

Relationship Between Zero-ISI, Duobinary, and Modified Duobinary

• Let $p_a(t)$ satisfy the first Nyquist criterion.
• Let $p_b(t)$ be a duobinary pulse.
• It is clear that $p<em>a(kT</em>b)$ and $p<em>b(kT</em>b)$ only differ at $k = 1$
• One can construct $p<em>b(t)$ from $p</em>a(t)$ via $p<em>b(t) = p</em>a(t) + p<em>a(t - T</em>b)$.
• Taking the Fourier transform allows us to inspect duobinary signalling in terms of spectral bandwidth:
  Pb(f) = Pa(f)[1 + e^{-j2πfTb} ]
  |Pb(f)| = 2|Pa(f)|√2(1 + cos(2πfTb)) · | cos(πfTb)|
  *Spectral null at $2\pi f T<em>b = \pi$, or equivalently at $f = 0.5/T</em>b$ shows how duobinary signalling reshapes the PSD to reduce the essential BW to the minimum theoretical BW

Modified Duobinary