digital modulation 03 3s

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Digital Modulation – Lecture 03

Inter-Symbol InterferencePower Spectral Density

© Richard Harris

Communication Systems 143.332 - Digital Modulation Slide 2

Objectives

• To be able to discuss Inter-Symbol Interference (ISI), its causes and possible remedies.

• To be able to define spectral efficiency• To define what is meant by the Power Spectral Density

(PSD) function• To be able to compute the PSD for different forms of

digital modulation and discuss its properties.


Presentation Outline

• Inter-symbol Interference– What is ISI?– How can it be avoided?

• Power Spectral Density function

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Inter-Symbol Interference – ISI - 1

• Absolute bandwidth of rectangular pulses is infinite• They are filtered as they pass through the

communication system• The pulse for each symbol is spread into adjacent

timeslots• This may cause Intersymbol Interference (ISI)

Baseband Pulse Transmission System


Inter-Symbol Interference – 2

Examples of ISI on received pulses in a binary communication system


Inter-Symbol Interference – 3

• The overall transfer function for the system is– He(f) = H(f).HT(f).HC(f).HR(f)

Where H(f) is the Fourier Transform of the rectangular pulse

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Avoiding Inter-Symbol Interference

• For zero ISI, Nyquist found that He(f) must satisfy the following condition:– For a single pulse, the filtered pulse is non-zero at its own sampling

time, but is always zero at other sampling times.

• Example:

0

0

0

)1( frffB

ffr

+=+=

=

∆

∆

Raised cosine-rolloff – Nyquist filter characteristics


Impulse Response

0

1f


Avoiding ISI – (Contd)

• For a communication system with a raised cosine-roll-off filtering characteristic, the (baud) symbol rate which can be supported by the system is:– 2B/(1+r)

• Where B is the absolute bandwidth of the system and r is the roll-off factor of the filter.– Note that 0 ≤ r ≤ 1 is a requirement

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Spectral Efficiency

• To compare signalling schemes we can define a quantity called Spectral Efficiency:

• For systems operating at the Nyquist rate, BT = r, so that rb/BT = log2M, and for values of M=2,4,8,16 etc, the ideal spectral efficiencies are 1,2,3,4 etc bits/sec/Hz.

• For practical systems, the values are somewhat less than these ideal values.

The ratio of the bit rate (in bits/sec) of the modulating signal to the transmission bandwidth (in Hz) of the modulated signal. ie, rb/BT bits/sec/Hz.


Theoretical Bandwidth Efficiency Limits

8 bits/second/Hz256 QAM




3 bits/second/Hz8PSK

2 bits/second/HzQPSK

1 bit/second/HzBPSK

1 bit/second/HzMSK

Theoretical bandwidthformat efficiency limits

Modulation method

Note: These figures cannot actually be achieved in practical systems as they require perfect modulators, demodulators, filter and transmission paths.


Power Spectral Density Definition

•• Definition: (See Definition: (See WikipediaWikipedia))– The power spectral density (PSD) describes how the power of

a signal or time series is distributed with frequency.• The PSD is the Fourier transform of the

autocorrelation function of the signal if the signal can be treated as a stationary random process.

• The Power Spectral Density of a signal exists if and only if the signal is a wide-sense stationary process.

• Note:– If the signal is not stationary, then the autocorrelation function

must be a function of two variables, so no PSD exists, but similar techniques may be used to estimate a time-varying spectral density.

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Random Process through a Linear Time Invariant Filter - 1

• Time-Invariant Filters– In plain terms, a time-invariant filter (or shift-invariant filter) is one which

performs the same operation at all times. Thus, if the input signal is delayed (shifted) by, say, N samples, then the output waveform is simply delayed by N samples and unchanged otherwise. Thus Y(t), the output waveform from a time-invariant filter, merely shifts forward or backward in time as the input waveform is shifted forward or backward in time.

• Consider a random process X(t) applied as input to a linear time-invariant filter having impulse response h(t) that produces a new random output process Y(t).

• Transmission in this situation is governed by a convolution integral, viz:

Impulse response

h(t)X(t) Y(t)

1 1 1( ) ( ) ( )Y t h X t dτ τ τ∞

−∞= −∫



• We can compute the mean value of Y(t) as

• Now, provided that the expectation E[X(t)] is finite for all t and the system is stable, we can interchange the order of expectation and integration to obtain:

1 1 1

( ) [ ( )]

( ) ( )

Y t E Y t

E h X t d

µ

τ τ τ∞

−∞

=

⎡ ⎤= −⎢ ⎥⎣ ⎦∫

1 1 1

1 1 1

( ) ( ) [ ( )]

( ) ( )

Y

X

t h E X t d

h t d

µ τ τ τ

τ µ τ τ

∞

−∞

∞

−∞

= −

= −

∫∫



• When the input random process X(t) is stationary, the mean µX(t) is a constant µX, so we can further simplify our result as:

– Where H(0) is the zero frequency (DC) response of the system.– This says that the mean of the random process Y(t) produced

at the output of a linear time-invariant system in response to X(t) acting as the input process is equal to the mean of X(t) multiplied by the DC response of the system.

1 1( )

(0)Y X

X

h d

H

µ µ τ τ

µ

∞

−∞=

=∫

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• Now consider the auto-correlation function of the output random process Y(t).

– Where t and u are two values of the time at which the output process is observed.

• Using the convolution integral we have:

( , ) [ ( ) ( )]YR t u E Y t Y u=

1 1 1 2 2 2( , ) ( ) ( ) ( ) ( )YR t u E h X t d h X u dτ τ τ τ τ τ∞ ∞

−∞ −∞

⎡ ⎤= − −⎢ ⎥⎣ ⎦∫ ∫



• Now, provided that the expectation E[X2(t)] is finite for all t and the system is stable, we can interchange the order of expectation and integration with respect to both τ1 and τ2 to obtain:

• When the input X(t) is a stationary process, the autocorrelation function of X(t) is only a function of the difference between the observation times t- τ1 and u- τ2. Thus writing τ=t-u we can write:

[ ]1 1 2 2 1 2

1 1 2 2 1 2

( , ) ( ) ( ) ( ) ( )

( ) ( ) ( )( )

Y

X

R t u h d h d E X t X u

h d h d R t u

τ τ τ τ τ τ

τ τ τ τ τ τ

∞ ∞

−∞ −∞

∞ ∞

−∞ −∞

= − −

= − −

∫ ∫∫ ∫

1 2 1 2 1 2( ) ( ) ( ) ( )Y XR h h R d dτ τ τ τ τ τ τ τ∞ ∞

−∞ −∞= − +∫ ∫



• Combining with our earlier result we see that if X(t) is a stationary input process, then Y(t) is also a stationary process.

• Since RY(0) = E[Y2(t)] it follows that we can find this quantity by setting τ = 0 in order to obtain:

• Which is a constant!

21 2 2 1 1 2( ) ( ) ( ) ( )XE Y t h h R d dτ τ τ τ τ τ

∞ ∞

−∞ −∞⎡ ⎤ = −⎣ ⎦ ∫ ∫

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Power Spectral Density - 1

• Consider the characterisation of random processes in linear systems by using frequency domain ideas.

• We are interested in finding the frequency domain equivalent to the result on the previous slide that defines the mean square value of the filter output.

• Now:– The impulse response of a linear time-invariant filter is

equivalent to the inverse Fourier Transform of the frequency response of the system.

• Let H(f) be the frequency response of the system.



• Substituting this expression into our previous result gives:

1 1( ) ( )exp( 2 )h H f j f dfτ π τ∞

−∞= ∫

21 2 2 1 1 2

2 2 2 1 1 1

( ) ( )exp( 2 ) ( ) ( )

( ) ( ) ( )exp( 2 )

X

X

E Y t H f j f df h R d d

dfH f h d R j f d

π τ τ τ τ τ τ

τ τ τ τ π τ τ

∞ ∞ ∞

−∞ −∞ −∞

∞ ∞ ∞

−∞ −∞ −∞

⎡ ⎤⎡ ⎤ = −⎣ ⎦ ⎢ ⎥⎣ ⎦

= −

∫ ∫ ∫

∫ ∫ ∫Define a new variable τ = τ1 – τ2 so that

22 2 2( ) ( ) ( )exp( 2 ) ( )exp( 2 )XE Y t dfH f d h j f R j f dτ τ π τ τ π τ τ

∞ ∞ ∞

−∞ −∞ −∞⎡ ⎤ = −⎣ ⎦ ∫ ∫ ∫



• The middle integral is H*(f) – the complex conjugate of the frequency response of the filter, so simplifying gives:

– Where |H(f)| is the magnitude response of the filter.• Finally we note that the right-most integral is actually

the Fourier transform of the autocorrelation function RX(τ) of the input random process X(t).

22 2 2( ) ( ) ( )exp( 2 ) ( )exp( 2 )XE Y t dfH f d h j f R j f dτ τ π τ τ π τ τ

∞ ∞ ∞

−∞ −∞ ∞⎡ ⎤ = −⎣ ⎦ ∫ ∫ ∫

2 2( ) | ( ) | ( )exp( 2 )XE Y t df H f R j f dτ π τ τ∞ ∞

−∞ ∞⎡ ⎤ = −⎣ ⎦ ∫ ∫

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• Let us now define the quantity

• The function SX(f) is called the Power Spectral Densityor Power Spectrum of the stationary process X(t).

( ) ( )exp( 2 )X XS f R j f dτ π τ τ∞

∞= −∫

2 2( ) | ( ) | ( )XE Y t H f S f df∞

−∞⎡ ⎤ =⎣ ⎦ ∫

The equation marked with says that the mean square value of a stable linear time-invariant filter in response to a stationary process is equal to the integral over all frequencies of the PSDof the input process multiplied by the squared magnitude response of the filter.



• Consider the following situation where a random process X(t) is passed through an ideal narrow band filter with magnitude response centred around the frequency fc (see figure below):

| ( )|H f

| | / 21,( )

| | / 20,c

c

f f fH f

f f f± < ∆⎧

= ⎨ ± > ∆⎩

f∆

cfcf−

f∆f0

1.0



• Note that ∆f is the bandwidth of the filter.• Substituting these details into the formula for E[Y2(t)] we have (for

sufficiently small filter bandwidth compared to the mid-band frequency fc with SX(f) a continuous function:

2[ ( )] (2 ) ( )X cE Y t f S f∆

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• When a digital waveform x(t) is modelled by

– where {an} represent data, n is an integer, w(t) is the signalling pulse shape and T is the duration of one data symbol.

• The power spectral density function PSD of x(t) can be shown to be given by

– where k is an integer, σa2 and µa are the variance and mean of the data {an}

and W(f) is the Fourier transform of w(t).

∑∞

−∞=

−=n

n nTtwatx )()(

∑∞

−∞=⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛+=

k

aax T

kfTkW

TfW

TfP δµσ 2

2

22

2

)()(


Properties of the PSD - 1

• The Power Spectral Density Function SX(f) and the autocorrelation function RX(f) of a stationary process X(t) form a Fourier-transform pair with τ and f as the variables satisfying the following:

( ) ( )exp( 2 )

( ) ( )exp(2 )

X X

X X

S f R j f d

R S f j f df

τ π τ τ

τ π τ

∞

−∞

∞

−∞

= −

=

∫∫

(Einstein-Wiener-Khintchine relations)


Properties of the PSD - 2

• Property 1– The zero value of the PSD of a stationary process is equal to the total

area under the graph of the autocorrelation function.• Property 2

– The mean square value of a stationary process is equal to the total area under the graph of the PSD

• Property 3– The PSD of a stationary process is always non-negative

• Property 4– The PSD of a real valued random process is an even function of

frequency; ie SX(-f) = SX(f)• Property 5

– The PSD, appropriately normalised has the properties associated with a Probability Density Function

Attempt these as an exercise in the Tutorial Session.

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Example PSD – Sine wave with random phase - 1

• Consider the following random process– A cos(2πfct+θ) where θ is a uniformly distributed random variable

over the interval [-π, π]• First we determine the autocorrelation function for this random

process:2

2 2

2 2

2

2

( ) [ ( ) ( )]

[ cos(2 2 )cos(2 )]

[cos(4 2 2 )] [cos(2 )]2 2

1 cos(4 2 2 ) cos(2 )2 2 2

0 cos(2 )2

cos(2 )2

X

c c c

c c c

c c c

c

c

R E X t X t

E A f t f f t

A AE f t f E f

A Af t f d f

A f

A f

π

τ τ

π π τ θ π θ

π π τ θ π τ

π π τ θ θ π τπ

π τ

π τ

−

= +

= + + +

= + + +

= + + +

= +

=

∫



• We now take Fourier transforms of both sides of the resulting expression:

[ ]

2

2

2

2

( ) cos(2 )2

( )exp( 2 ) cos(2 )exp( 2 )2

( ) cos(2 )exp( 2 )2

( ) ( )4

X c

X c

X c

c c

AR f

AR j f d f j f d

AS f f j f d

A f f f f

τ π τ

τ π τ τ π τ π τ τ

π τ π τ τ

δ δ

∞ ∞

−∞ −∞

∞

−∞

=

⇒ − = −

= −

= − + +

∫ ∫

∫

(See table of Fourier transform values on next slide)



• We see that the solution to this problem is a pair of delta functions.

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• Note that the total area under a delta function is unity.• Thus, the total area under SX(f) is equal to A2/2 as

expected.

2

( )4 cA f fδ +

2

( )4 cA f fδ −

SX(f)

0-fc fcf


PSD of a Random Binary Wave - 1

• Consider the following random sequence of 1s and 0s:

• It can be shown that the autocorrelation function for this process X(t) is given by:

2 | |1 ,| |( )

0, | |X

A TR T

T

τ ττ

τ

⎧ ⎛ ⎞− <⎪ ⎜ ⎟= ⎝ ⎠⎨⎪ ≥⎩

+A

-Atd

T

A2

( )XR τ

+T-T 0



• Taking Fourier transforms to obtain the PSD once again we find:

• Conversion back from tables of Fourier transforms we find the sinc2(fT) function:

• Where

2 | |( ) 1 exp( 2 )T

X TS f A j f d

Tτ π τ τ

−

⎛ ⎞= − −⎜ ⎟⎝ ⎠∫

2 2( ) sinc ( )XS f A T fT=

1 if 0sinc sin otherwise

xx x

x

=⎧⎪= ⎨⎪⎩

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• Plotting the Power Spectral Density function in this case gives:

sinc2(x)

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-8.00 -6.00 -4.00 -2.00 0.00 2.00 4.00 6.00 8.00

sinc(x)

digital modulation 03 3s

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