1B Paper 6: CommunicationsHandout 4: Digital Baseband Modulation

Ramji Venkataramanan

Signal Processing and Communications LabDepartment of Engineeringramji.v@eng.cam.ac.uk

Lent Term 2016

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Data TransmissionWe have seen how analogue sources can be digitised. E.g., AnMPEG or QuickTime file is a stream of bits

←→ . . . 10110010001101010 . . .

Now we have to transport those bits across a channel:

(Digitised source)

110 001 100 111

Transmitter Channelwaveform

inputReceiver

waveform

output

110 000 100 111

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110001

Channel Encoder

Modulator

010110100

Channelwaveform

inputDemodulator

waveform

output

Channel Decoder

010010110

110001

Tx Rx

The transmitter (Tx) does two things:

1. Encoding: Adding redundancy to the source bits to protectagainst noise

2. Modulation: Transforming the coded bits into waveforms.

The receiver (Rx) does:• Demodulation: noisy output waveform −→ output bits• Decoding: Try to correct errors in the output bits and recover

the source bits3 / 36

Modulation/DemodulationWe’ll first consider the modulation and demodulation blocksassuming that the channel encoder/decoder are fixed, and look atthe design of the channel encoder and decoder later.

(Encoded bits)10110100

Modulator +x(t)

Demodulatory(t)

10100110

n(t)

We now study a digital baseband modulation technique calledPulse Amplitude Modulation (PAM) & analyse its performanceover an Additive White Gaussian Noise (AWGN) channel

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The Symbol ConstellationThe digital modulation scheme has two basic components.

1. The first is a mapping from bits to real/complex numbers, e.g.

0→ −A, 1→ A (binary symbols)

00→ −3A, 01→ −A, 10→ A, 11→ 3A (4-ary symbols)

The set of values the bits are mapped to is called the constellation,e.g., the 4-ary constellation above is {−3A, A, A, 3A}.Once we fix a constellation, a sequence of bits can be uniquelymapped to constellation symbols. E.g., with constellation {−A, A}0 1 0 1 1 1 0 0 1 0 −→ −A, A,−A, A, A, A,−A,−A, A,−A

With constellation {−3A,−A, A, 3A}, the same sequence of bits ismapped as

01 01 11 00 10 −→ − A, −A, 3A, −3A, A

In a constellation with M symbols, each symbol represents log2 Mbits

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The Pulse Shape

2. The second component of Pulse Amplitude Modulation is aunit-energy baseband waveform denoted p(t), called the pulseshape. E.g., a sinc pulse or a rect pulse:

p(t) =1√T

sinc(πt

T

)or p(t) =

{1√T

for t ∈ (0, T ]

0 otherwise

T is called the symbol time of the pulse

A sequence of constellation symbols X0, X1, X2, . . . is used togenerate a baseband signal as follows

xb(t) =∑

k

Xkp(t − kT )

Thus we have the following important steps to associate bits witha baseband signal xb(t):

. . . 0 1 0 1 1 1 0 0 1 0 . . . −→ X0, X1, X2, . . . −→∑

k

Xkp(t−kT )

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Rate of TransmissionThe modulated baseband signal is xb(t) =

∑k Xkp(t − kT ).

With the rect pulse shape

p(t) =

{1√T

for t ∈ (0, T ]

0 otherwise

and Xk ∈ {+A,−A}, xb(t) looks like

t

A/√T

−A/√T

T 2T 3T 4T

Every T seconds, a new symbol is introduced by shifting the pulseand modulating its amplitude with the symbol.

The transmission rate is 1T symbols/sec or log2 M

T bits/second

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Desirable Properties of the Pulse Shape p(t)

p(t) is chosen to satisfy the following important objectives:

1. We want p(t) to decay quickly in time, i.e., the effect ofsymbol Xk should not start much before t = kT or last muchbeyond t = (k + 1)T

2. We want p(t) to be approximately band-limited.For a fixed sequence of symbols {Xk}, the spectrum of xb(t) is

Xb(f ) = F[∑

k

Xkp(t − kT )

]= P(f )

∑

k

Xk e−j2πfkT

Hence the bandwidth of xb(t) is the same as that of the pulse p(t)

3. The retrieval of the information sequence from the noisyreceived waveform xb(t) + n(t) should be simple and relativelyreliable. In the absence of noise, the symbols {Xk}k∈Z shouldbe recovered perfectly at the receiver.

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Orthonormality of pulse shifts

Consider the third objective, namely, simple and reliable detection.

To achieve this, the pulse is chosen to have the following“orthonormal shifts” property:

∫ ∞

−∞p(t − kT )p(t −mT ) dt =

{1 if k = m0 if k ̸= m

(1)

We’ll see how this property makes signal detection at the Rx simple

• This property is satisfied by the rect pulse shape

p(t) =

{1√T

for t ∈ (0, T ]

0 otherwise

• The sinc pulse p(t) = 1√T

sinc(

πtT

)also has orthonormal

shifts! (You will show this in Examples Paper 9, Q.2)

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Time Decay vs. Bandwidth Trade-offThe first two objectives say that we want p(t) to:

1. Decay quickly in time

2. Be approximately band-limited

But . . . faster decay in time ⇔ larger bandwidth

Consider the pulse p(t) = 1√T

sinc(

πtT

)

p(t)

0

P(f)

The sinc is perfectly band-limited to W = 12T

But decays slowly in time |p(t)| ∼ 1|t|

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Next consider the rectangular pulse

p(t) =

{1√T

for t ∈ (0, T ]

0 otherwise

1p(t) |P(f)|

This pulse is perfectly time-limited to the symbol interval [0, T ).But . . .

• Decays slowly in freq. |P(f )| ∼ 1|f |

• Main-lobe bandwidth = 1T

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In practice, the pulse shape is often chosen to have a raised cosinespectrum

0 12T

−12T

P (f)

Bandwidth slightly larger than 12T ; decay in time |p(t)| ∼ 1

|t|3

A happy compromise!

• More on raised cosine pulses in 3F4

• For intuition, it often helps to envision xb(t) with a rect pulse,though it is never used in practice

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PAM DemodulationNow, assume that we have picked a constellation and a pulse shapesatisfying the objectives, and we transmit the baseband waveform

xb(t) =∑

k

Xkp(t − kT )

over a baseband channel y(t) = xb(t) + n(t)

xb(t) y(t)

n(t)

How does the receiver recover the information symbols{X0,X1, X2, . . .} from y(t)?

• This process is called demodulation

• We will see that the orthonormal shift property of p(t) leadsto a simple and elegant demodulator

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Matched Filter DemodulatorLet us first understand the operation assuming no noise, i.e.,

y(t) = xb(t) =∑

k

Xkp(t − kT )

h(t) = p(−t)Filter

y(t)r(t)

t = mT

r(mT )

y(t) is passed through a filter with impulse response h(t) = p(−t)

This is called a matched filter. The filter output is

r(t) = y(t) ⋆ h(t) = xb(t) ⋆ h(t) (assuming no noise)

=

∫ ∞

−∞xb(τ)h(t − τ)dτ

=

∫ ∞

−∞xb(τ)p(τ − t)dτ

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The Output of the Matched Filter

h(t) = p(−t)Filter

y(t)r(t)

t = mT

r(mT )

r(t) =

∫ ∞

−∞xb(τ)p(τ − t)dτ =

∑

k

Xk

∫ ∞

−∞p(τ − kT )p(τ − t)dτ

By sampling the filter output at time t = mT , m = 0, 1, 2, . . ., youget

r(mT ) =∑

k

Xk

∫ ∞

−∞p(τ − kT )p(τ −mT )dτ = Xm

because of the orthonormal shifts property of p(t)∫ ∞

−∞p(τ − kT )p(τ −mT )dτ =

{1 if k = m0 if k ̸= m

The orthonormal shifts property is crucial for this demodulator towork!

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Demodulation with Noisy y(t)Now consider the noisy case. The receiver gets y(t) = x(t) + n(t)

h(t) = p(−t)Filter

y(t)r(t)

t = mT

r(mT )

The matched filter output is

r(t) = y(t) ⋆ h(t) = xb(t) ⋆ h(t) + n(t) ⋆ h(t)

=∑

k

Xk

∫ ∞

−∞p(τ − kT )p(τ − t)dτ +

∫ ∞

∞n(τ)p(τ − t)dτ

Sampling at t = mT , m = 0, 1, 2, . . ., we now get

r(mT ) = Xm + Nm

where Nm is noise part of the filter output at time mT :

Nm =

∫ ∞

∞n(τ)p(τ −mT )dτ

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Properties of the NoiseLet us denote r(mT ), the sampled output at time mT , by Ym.

Ym = Xm + Nm, m = 0, 1, 2, . . .

Note that this is a discrete-time channel. We have converted thecontinuous-time problem into a discrete-time one of detecting thesymbols Xm from the noisy outputs Ym.

• To do this, we first need to understand the properties of thenoise Nm. Recall that

Nm =

∫ ∞

∞n(τ)p(τ −mT )dτ

• Nm is a random variable whose distribution depends on thestatistics of the random process n(t).

You will learn about random processes and their characterisation in 3F1

& 3F4, but this is outside the scope of this course. For now, we will

directly specify the distribution of Nm and analyse the detection problem.17 / 36

Ym = Xm + Nm, m = 0, 1, 2, . . .

Modelling n(t) as a Gaussian process leads to the followingimportant characterisation of Nm:

• For each m, Nm is a Gaussian random variable with zeromean, and variance σ2 that can be estimated empirically

• Further N1,N2, . . . are independent

• Thus the sequence of random variables {Nm}, m = 0, 1, . . .are independent and identically distributed as N (0, σ2).

Detection• At the Rx, how do we detect the information symbol Xm from

Ym for m = 0, 1, . . .?

• Remember that each Xm belongs to the symbol constellation

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Detection for Binary PAMLet’s start with a simple binary constellation, then generalise.

Consider a constellation where each Xm ∈ {−A, A}. This is calledbinary PAM or BPSK (‘Binary Phase Shift Keying’)

Y = X + N

The detection problem is now:Given Y , how to decide whether X = A or X = −A?

Observe that:Y = A + N if X = A and Y = −A + N if X = −A

• N is distributed as N (0, σ2)• Therefore the pdf f (Y |X = A) is Gaussian with mean A and

variance σ2

• Similarly the pdf f (Y |X = −A) is Gaussian with mean −Aand variance σ2

Note: Adding a constant to a random variable just shifts themean, does not change the shape of the distribution

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−A A

f (Y |X = −A) f (Y |X = A)

Y

Let X̂ denote the decoded symbol. When the symbols A and −Aare a priori equally likely, the optimal detection rule is:

X̂ = A if f (Y | X = A) ≥ f (Y | X = −A)

X̂ = −A if f (Y | X = −A) > f (Y | X = A)

“Choose the symbol from which Y is most likely to have occurred”

• This decoder is called the maximum-likelihood decoder• This decoder is intuitive and seems sensible, and is in fact, the

optimal detection rule when all the constellation symbols areequally likely (we will not prove this here)

• It is then a special case of the Maximum a Posteriori (MAP)detection rule, which minimises the probability of detectionerror (module 4F5)

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−A A

f (Y |X = −A) f (Y |X = A)

Y

The detection rule can be compactly written as

X̂ = arg maxx∈{A,−A}

f (Y |X = x)

X̂ = arg maxx∈{A,−A}

f (Y |X = x)

= arg maxx∈{A,−A}

1√2πσ2

e−(Y −x)2/2σ2= arg min

x∈{A,−A}(Y − x)2

Thus the detection rule is just: X̂ = A if Y ≥ 0, X̂ = −A if Y < 0“Choose the constellation symbol closest to the output Y ”

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Decision Regions

The detection rule partitions the space of Y (the real line) intodecision regions.For binary PAM, we just derived the following decision regions:

−A A

X̂ = AX̂ = −A

Y

Q: When does the detector make an error?

A: When X = A and Y < 0, or When X = −A and Y > 0

We will calculate the probability of error shortly, but let’s first findthe detection rule for general PAM constellations

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Detection for General PAM Constellations

The detection rule can easily be extended to a generalconstellation C• E.g., C may be the 3-ary constellation {−2A, 0, 2A} or a 4-ary

constellation {−3A,−A, A, 3A}• The maximum-likelihood principle is the same: “Choose the

constellation symbol from which y is most likely to haveoccurred ”

X̂ = arg maxx∈ C

f (Y |X = x)

= arg maxx∈ C

1√2πσ2

e−(Y −x)2/2σ2= arg min

x∈ C(Y − x)2

Thus, the detection rule for any PAM constellation boils down to:“Choose the constellation symbol closest to the output Y ”

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Example: 3-ary PAM

Y = X + N, N ∼ N (0, σ2)

What is the optimal detection rule and the associated decisionregions if X belongs to the 3-ary constellation {−2A, 0, 2A}?The “nearest symbol to Y ” decoding rule yields

X̂ =

−2A if Y < −A0 if − A ≤ Y < A2A if Y > A

−2A 2A

X̂ = 2AX̂ = −2A

−A A

X̂ = 0

Y0

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Probability of Detection Error

Y = X + N

Consider binary PAM with X ∈ {A− A}. The decision regions are:

−A A

X̂ = AX̂ = −A

Y

The detector makes an error when X = A and Y < 0, or whenX = −A and Y > 0

The probability of detection error is

Pe = P(X̂ ̸= X )

= P(X = −A)P(X̂ = A|X = −A) + P(X = A)P(X̂ = −A|X = A)

= 12P(X̂ = A | X = −A) + 1

2P(X̂ = −A | X = A)

(The symbols are equally likely ⇒ P(X = A) = P(X = −A) = 12

)25 / 36

Let us first examine P(X̂ = A | X = −A)

P(X̂ = A | X = −A) = P(Y > 0 | X = −A)

= P(−A + N > 0 | X = −A)

= P(N > A | X = −A)(a)= P(N > A)

(a) is true because the noise random variable N is independent ofthe transmitted symbol X . Similarly,

P(X̂ = −A | X = A) = P(Y < 0 | X = A)

= P(A + N < 0 | X = A)

= P(N < −A | X = A) = P(N < −A)

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The probability of detection error is therefore

Pe = 12P(X̂ = A | X = −A) + 1

2P(X̂ = −A | X = A)

= 12P(N > A) + 1

2P(N < −A)

(b)= P(N > A)

(c)= P

(N

σ>

A

σ

)

• (b) holds due to the symmetry of the Gaussian pdf N (0, σ2):

−A A0

P (N > A)P (N < −A)

• In (c), we have expressed the probability in terms of astandard Gaussian random variable with distribution N (0, 1)

• Recall from 1B Paper 7 (Probability) that if N is distributedas N (0, σ2) then N

σ is distributed as N (0, 1)

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The Q-function

The error probability is usually expressed in terms of theQ-function, which is defined as:

Q(x) =

∫ ∞

x

1√2π

e−u2/2du

x0

Q(x)N (0, 1) pdf

• Q(x) is the probability that a standard Gaussian N (0, 1)random variable takes value greater than x

• Also note that Q(x) = 1− Φ(x), where Φ(.) is the cdf of aN (0, 1) random variable

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Pe in terms of the signal-to-noise ratioThe probability of detection error is therefore

Pe = P(N > A) = P

(N

σ>

A

σ

)= Q

(A

σ

)= Q

(√Es

σ2

)

where Es is the average energy per symbol of the constellation:

Es =1

2(A2 + (−A)2) = A2

• For a binary constellation, each symbol corresponds to 1 bit.⇒ the average energy per bit Eb is also equal to A2 in thiscase

• For an M-point constellation, the average energy per symbolEs = Eb log2 M

• Ebσ2 is called the signal-to-noise ratio (snr) of the transmissionscheme

• Pe can be plotted as a function of the snr Ebσ2 . . .

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Pe vs snr for binary PAM

−2 0 2 4 6 8 10 12 14 16 1810

−16

10−14

10−12

10−10

10−8

10−6

10−4

10−2

100

snr Eb/σ

2 (dB)

Pe

To get Pe of 10−3, we need snr Eb/σ2 ≈ 9 dB30 / 36

Error Probability vs Transmit PowerThe probability of error for binary PAM decays rapidly as snr ↑:• Q(x) ≈ e−x2/2 for large x > 0 ⇒ Pe ≈ e−snr/2

Can we set the snr Ebσ2 to be as high as we want, by increasing Eb?

(i.e., by increasing A since Eb = Es = A2 for binary PAM)

• The problem is that transmitted power also increases!

• Intuition: 1 symbol with transmitted every T seconds withaverage energy Es ⇒ transmit power is Es/T

• Thus as you increase the snr, you battery drains faster!

xb(t) =∑

k

Xk p(t − kT )

For any PAM constellation the power of the baseband PAM signalxb(t) is Es

T = Eb log2 MT , where

• Es is the average symbol energy of the constellation.

• Eb is the average energy per bit31 / 36

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Power of PAM signal

Consider the PAM signal from time −nT to nT , carrying symbolsX−n, X−n+1, . . . , Xn drawn randomly from the constellation

PPAM = limn→∞

1

2nTE∫ nT

−nT

(n∑

k=−n

Xk p(t − kT )

)2

dt (2)

where the expectation is needed because the symbols {Xk} arerandom symbols drawn uniformly from the PAM constellation.

We write( n∑

k=−n

Xk p(t − kT ))2

=n∑

k=−n

X 2k p2(t − kT ) +

n∑

k=−n

∑

j ̸=k

XkXj p(t − kT )p(t − jT )

Plug back into (2), integrate each of these sums separately . . .

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First term = limn→∞

1

2nT

n∑

k=−n

E[X 2k ]

∫ nT

−nTp2(t − kT ) dt

(a)= lim

n→∞Es

2nT

n∑

k=−n

∫ nT

−nTp2(t − kT ) dt

(b)=

Es

Tlim

n→∞2n + 1

2n=

Es

T

(a) holds because E[X 2k ] = Es , the average symbol energy. (b)

holds because the pulse shape p(t) has unit energy.

The second term is

limn→∞

1

nT

n∑

k=−n

∑

j ̸=k

E[XkXj ]

∫ nT

−nTp(t − kT )p(t − jT ) dt = 0

because: a) E[XkXj ] = E[Xk ]E[Xj ] = 0 as the symbols Xj , Xk areindependent and the symbol constellation is symmetric around 0;b) further the orthogonal shifts property of p(t) implies that theintegral also → 0.Hence PPAM = Es/T .

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Pulse Amplitude Modulation - The Key Points

10110100

Modulator +xb(t)

Demodulatory(t)

10100100

n(t)

PAM is a way to map a sequence of information bits to acontinuous-time baseband waveform

1. Pick a constellation, map the information bits to symbolsX1, X2, . . . in the constellation

2. These symbols then modulate the amplitude of a pulse shapep(t) to generate the baseband waveform xb(t)

xb(t) =∑

k

Xkp(t − kT )

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Desirable properties of the pulse shape p(t):

• p(t) should decay quickly in time; its bandwidth W shouldn’tbe too large

• Orthonormal shifts property for simple and reliable decoding

At the receiver, first demodulate then detect:

• The demodulator is a matched filter with IR h(t) = p(−t)

• Matched filter output is sampled at times . . . , 0, T , 2T , . . ..At time mT , the output is

Ym = Xm + Nm

Nm is Gaussian noise with zero mean and variance σ2 thatcan be empirically estimated

• Detection rule: X̂m = the constellation symbol closest to Ym

Probability of detection error can be calculated:

• Decays exponentially with snr Es/σ2

• Es is average energy/symbol of the constellation; power ofPAM waveform xb(t) is Es/T

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