optimum demodulation lecture notes 9: intersymbol interference · lecture notes 9: intersymbol...

Lecture Notes 9: Intersymbol Interference

In this lecture we examine optimum demodulation when the transmitted signal is filtered by

the channel and there is additive white Gaussian noise. The optimum demodulator chooses

the possible transmitted vector that would result in the received vector (in the absence of

noise) to be as close as possible (in Euclidean distance) to what was received. This we show

can be implemented by a filter matched to the received signal for a given data symbol

followed by a nonlinear processing via the Viterbi algorithm. The filter is sampled at the data

rate. We also analyze the performance of such a system. The analysis is very similar to that of

convolutional codes. Because the received signal is filtered and sampled, the output of the

filter consists of two components. One due to the transmitted signal and one due to the noise.

The output due to noise is, however, not white. However, in the next section we show that the

output of the matched filter can be whitened. With a whitened matched filter the optimum

receiver (Viterbi algorithm) becomes clear. Finally, in the last section we show how to design

a system to eliminate intersymbol interference.

IX-1

Optimum Demodulation

Consider transmitting data at rate 1 T through a channel with bandwidth W or through adistorting channel. We would like to find the optimum (minimum sequence error probability)receiver. Assume the modulator is a filter acting on a infinite sequence of impulses (at rate1 T with impulse response f t . The channel is characterized by an impulse response of g t

and the receiver is a filter sampled at rate 1 T with impulse response h t .

f t

Modulator

sT t g t

Channel

z t

r t

n t

Data

Rate 1 TThe transmitted signal is of the form

sT u t N

∑m N

um f t mT

IX-2

where um is the data symbol transmitted during the m-th signaling interval assumed to be inthe alphabet A and f t is the waveform used for transmission. We assume a transmission of2N 1 data symbols (think of N as being very large). The output of the channel filter is then

z t ∞ ∞g t τ s τ dτ

∞ ∞g t τ N

∑m N

um f τ mT dτ

N

∑m N

um ∞ ∞g t τ f τ mT dτ

N

∑m N

umh t mT

where

h t ∞ ∞g t τ f τ dτ

The received signal consist of two terms. One due to signal and one due to noise.

r t N

∑m N

umh t mT n t

su t n t

IX-3

where

su t N

∑m N

umh t mT

Since n t is white Gaussian noise the optimum receiver computes for each data sequence v

ΛN v 2 r t sv t sv 2 2 ∞ ∞r t sv t dt ∞ ∞

s2v t dt

2 ∞ ∞r t N

∑k N

vkh t kT dt ∞ ∞∑m

∑k

vmvkh t kT h t mT dt

2N

∑k N

vk ∞ ∞r t h t kT dt ∑

m∑k

vmvk ∞ ∞h t kT h t mT dt

2N

∑k N

vkyk N

∑k N

N

∑m N

vkvmxm k

where

yk ∞ ∞r t h t kT dt

IX-4

xm k ∞ ∞h t kT h t mT dt

Thus the optimum decision rule is

Chose v if ΛN v maxu

ΛN u

Since the decision statistic depends on the received signal only through yk it is clear that yk isa sufficient statistic to implement optimal receiver.

Consider a filter hr t h t . Then if the received sequence is passed through this filter the

output would be

y s ∞ ∞r t hr s t dt

∞ ∞r t h t s dt

The sampled output would be

y kT ∞ ∞r t h t kT dt

IX-5

Since this is yk defined earlier the received signal should be filtered and sampled as shown

below before doing some processing.

f t

Modulator

s t g t

Channel

z t

r t hr t

Demodulator

y t t kT

n t

Data

Rate 1 T yk

IX-6

yk ∞ ∞r t h t kT dt ∞ ∞

N

∑k N

umh t mT h t kT dt ηk

ηk ∞ ∞n t h t kT dt

Now the original continuous time detection problem can be replaced with a discrete time

problem.

yk N

∑m N

umxm k ηk

IX-7

Model

uk 2

uk 1

uk

uk 1

uk 2

+

ηk

+

x 2 x 1 x0 x1 x2

uk 3

yk

IX-8

Model

Note that ηk is Gaussian.

E ηk 0

Var ηk N0

2 ∞ ∞h2 t dt

E ηkηm E

∞ ∞n t h t kT dt ∞ ∞

n s h s mT dt

∞ ∞ ∞ ∞h t kT h s mT E n t n s dtds

N0

2 ∞ ∞h t kT h t mT dt N0

2xk m

However, ηk is not an i.i.d. sequence.

IX-9

Optimum Receiver

ΛN v 2N

∑k N

vkyk N

∑k N

N

∑m N

vkvmxm k

2N

∑k N

vkyk N

∑k N

x0 vk

2 2N

∑k N

vk

k 1

∑m N

vmxm k

2N

∑k N

vkyk N

∑k N

x0v2k

2N

∑k N

vk

k N

∑j 1

vk jx j

N

∑k N

2vkyk N

∑k N

x0v2k

2N

∑k N

vk

k N

∑j 1

vk jx j

(Note that v j 0 j N). Assume xm 0 m L (finite intersymbol interference)

ΛN v 2N

∑k N

vkyk N

∑k N

x0v2k

2N

∑k N

vk

min L k N

∑j 1

vk jx j

N

∑k N

2vkyk x0v2k

2vk

min L k N

∑j 1

vk jx j

IX-10

The decision rule can be implemented in a Viterbi algorithm like structure. Define the state at

time k to be the last L data symbols. (These are the only symbols that affect the output at time

k).

σk vk L 1 vk

Let

λ σk σk 1 2vk 1yk 1 x0v2k 1

2vk 1

L

∑m 1

vk 1 mxm

Then

ΛN v N 1

∑k N

λ σk σk 1 2v Ny N x0v2 N

Thus we can apply the Viterbi algorithm.

IX-11

Viterbi Algorithm

Let Γ σm be the length (optimization criteria) of the shortest (optimum) path to state σm attime m. Let Ω σm be the shortest path to state σm at time m. Let Γ σm 1 σm be the lengthof the path to state σm 1 at time m that goes through state σm at time m. The algorithm worksas follows.

Storage:

k, time index, Ω σm σm AL Γ σm σm AL

Initialization

k N,

Ω σ N σ N σ N ΦL 1 A (Φ is the empty set).

Γ σ N 2v Ny N x0v2 N

Recursion

Γ σm 1 σm Γ σm λ σm σm 1 Γ σm 1 maxσm Γ σm 1 σm for each σm 1

Let Ωm σm 1 argmaxσmΓ σm 1 σm . Ω σm 1 Ω σm σm 1

IX-12

Example: L 1, x0 0 x1 0 x2 0, vk 1

ΛN v N

∑k N

λ σk σk 1 2y Nv N x0v2 N


2vk 1vkx1

2yk x0 2x1

2yk x0 2x1

2yk x0 2x1

2yk x0 2x1

+1

-1

vk+1

-1

vk 1

IX-13

Example

Consider a channel with x0 0 8 x1 0 2. Consider the following received sequence of

length 5 (N 2).

y y 2 y 1 y0 y1 y2 0 7 0 5 0 9 0 3 0 6

Assume vk 1 . Then the trellis is shown below.

IX-14

Trellis

+1

-1

2y 2 x0 0 6

2y 2 x0 2 2

IX-15

Trellis

+1

-1

0 6

2 2

2y 1 x0 2x1 2

2y 1

x 0 2x 1

0 6

0 4

IX-16

Trellis

+1

-1

0 6

2 2

0 4

2y

1

x0

2x1

1

4

2y 1 x0 2x1 2 2

1 6

IX-17

Trellis

+1

-1

0 6

2 2

0 4

1 6

2y0 x0 2x1 3 0

2y0

x 0

2x

1

2 2

2 6

IX-18

Trellis

+1

-1

0 6

2 2

0 4

1 6

2 6

2y0

x0

2x1

1

4

2y0 x0 2x1 0 61 8

IX-19

Trellis

+1

-1

0 6

2 2

0 4

1 6 2 6

1 8

2y1 x0 2x1 0 6

2y1

x 0

2x

1

02

2 0

IX-20

Trellis

+1

-1

0 6

2 2

0 4

1 6

2 6

1 8

2 0

0 8

2y1

x0

2x1

1

8

2y1 x0 2x1 1 0

IX-21

Trellis

+1

-1

0 6

2 2

0 4

1 6

2 6

1 8

2 0

0 8

2y1 x0 2x1 1 8

2y1

x 0

2x

1

10

1 8

IX-22

Trellis

+1

-1

0 6

2 2

0 4

1 6

2 6

1 8

2 0

0 8

1 8

2y1

x0

2x1

0

2y1 x0 2x1 0 82 0

IX-23

Trellis

-1

+1 0.60

-2.2

0.4

-1.6

-2.6

1.8

2.0

0.8

1.8

2.0

The double lines represent the path chosen by the Viterbi decoder. Thus the Viterbi decoder

would output the sequence v 1 1 1 1 1 .

IX-24

Example: L 2, x0 0, x1 0, x2 0, vk 1

ΛN v N 1

∑k N

λ σk σk 1 2v Ny N x0v2 N


2vk 1vkx1 2vk 1vk 1x2

IX-25

vkvk 1 vk 1vk

-1 -1 -1 -1

-1 +1 -1 +1

+1 -1 +1 -1

+1 +1 +1 +1

IX-26

vk vk 1 vk 1 vk Metric

1 1 1 1 2yk 1 x0 2x1 2x2 1 1 1 1 2yk 1 x0 2x1 2x2 1 1 1 1 2yk 1 x0 2x1 2x2 1 1 1 1 2yk 1 x0 2x1 2x2 1 1 1 1 2yk 1 x0 2x1 2x2 1 1 1 1 2yk 1 x0 2x1 2x2 1 1 1 1 2yk 1 x0 2x1 2x2 1 1 1 1 2yk 1 x0 2x1 2x2

IX-27

Error Probability

Now consider the performance of the above maximum likelihood sequence detector (MLSD).

We will evaluate the union upper bound to the error probability. To do this we need to

determine the pairwise error probability between two sequences. This is the probability that

sequence v is demodulated given sequence u is transmitted for a system with only two

possible transmitted sequences (u v). Let P u v denote the conditional error probability

given u transmitted. Then

P u v P ΛN v ΛN u u Q

sv t su t

2N0

e

sv t su t 2 4N0

As expected the pairwise error probability depends only on the square Euclidean distance

between the signals su and sv.

sv t su t 2 ∞ ∞

sv t su t 2 dt

IX-28

∞ ∞

N

∑k N

vk uk h t kT

2

dt

∞ ∞

N

∑k N

N

∑m N

vk uk vm um h t kT h t mT dt

N

∑k N

N

∑m N

vk uk vm um ∞ ∞h t kT h t mT dt

N

∑k N

N

∑m N

vk uk vm um xk m

Let εk 12 vk uk

1 vk 1 uk 1

0 vk uk 1 vk 1 uk 1

sv t su t 2 4N

∑k N

N

∑m N

εkεmxk m

4N

∑k N

ε2kx0 4

N

∑k N 1

2k 1

∑m N

εkεmxk m

IX-29

4

N

∑k N

ε2kx0 2

k N

∑j 1

εkεk jx j

4

N

∑k N

ε2kx0 2

L

∑j 1

εkεk jx j

P u v exp

1

N0

N

∑k N

ε2kx0 2

L

∑m 1

εkεk mxm

P e

Thus the incremental Euclidean distance between two paths for a given time index k isdepends on the past L errors. The error state is defined to be the last L errors εk εk L 1 .The all zero error state corresponds to the past L symbols being correctly demodulated.

An error event is defined to be an error sequence that diverges once from the all zero state andthen remerges later. Since a necessary condition for an error of a particular type (first eventerror or symbol error) is that an error event occurs that causes the demodulator/decoder tofollow a path that diverges and then at some later time remerges we can calculate the errorprobability for a particular node by counting the number of paths (and their distance) thatdiverge and remerge. We can use the state diagram to determine the number of errorsequences with a particular distance.

IX-30

Let e ε N εN wH e Hamming weight of e (number of nonzero terms)

PE m First event error probability P at time m decoder is not at correct state for the first time

Pb Bit error probability P bit error occurs for symbol m

The union bound on the probability of error at time 0 is

PE m ∑u vε 0

P u v P u

∑e

ε 0

∑u v

e 12 v u

P u v P u

where the sum is over all sequences that diverge from the all zero state and then remerge later.

Each of the 22N 1 u sequences are equally likely. In each position where εk 0 the

components of the sequences u and v are determined. If εk 1 then vk 1 and uk 1.

Similarly if εk 1 then vk 1 and uk 1. The components e where εk 0 there are two

choices for uk and vk (uk vk 1 or uk vk 1). Since there are 2N 1 wH e places

IX-31

where εk 0 there are 22N 1 wH e such sequences u and v. Hence

PE f ∑e

ε0 0

P e 2 2N 1 wH e 2

2N 1

∑e

ε0 0

2

wH e P e

The bit error probability is bounded by

Pb ∑e

w e

2w e

P e

P e 2w e

N

∏k N

1

2wH εk

exp

1

N0

ε2kx0 2

L

∑m 1

εkεk mxm

For L 1P e 2wh e

N

∏k N

1

2wH εk

exp

1

N0

ε2kx0 2εkεk 1x1

We calculate these union bounds by enumerating the sequences that diverge from the all zeroerror state and remerge (error events) that correspond to a given Euclidean distance betweentwo data sequences and has a given number of nonzero terms (or is a given length). To do thiswe draw a state diagram (similar to that for convolutional codes) and label each path with

IX-32

DxNlMl where x is the incremental Euclidean distance squared (divided by 4) in going from

one state to another and l is 1 if the error path is nonzero and is zero if the error is zero. (This

redundant use of l will be explained when we determine the bit error probability).

IX-33

Error State Diagram L 1

MNDx0

MNDx0

1

1

MNDx0 2x1 MNDx0 2x1

MNDx0 2x1

MNDx0 2x1

-1

+1

a

c

b

d0 0

IX-34

Transfer Function

The transfer function is calculated by solving the following equations for Td Ta.

Td Tc Tb

Tb MNDx0 2x1Tb MNDx0 2x1Tc MNDx0Ta

Tc MNDx0 2x1Tc MNDx0 2x1Tb MNDx0Ta

Adding the last two equations and solving for Tb Tc and substituting the result into the first

equation yields

Td

Ta

D N M 2NMDx0

1 NMDx0 2x1 NMDx0 2x1 2NMDx0 2N2M2D2x0 2x1 2N2M2D2x0 2x1

Thus there are two paths with 1 error and Euclidean distance squared of 4x0. There are two

IX-35

paths with two errors and Euclidean distance squared of 4x0 4x1 and so on.

PE T D N M D e

1

N0 N 1 M 1 2

e x0 N0

1 12 e x0 N0 e2x1 N0 e 2x1 N0

Pb ∂T D N M

∂N D e

1

N0 N 1 M 1 2

2MDx0

1 NM Dx0 2x1 Dx0 2x1

2 D e

1

N0 N 1 M 1 2 e x0 N0

1 12 e x0 N0 e2x1 N0 e 2x1 N0

2

For large SNR this is the same as no ISI! Just as with convolutional codes this

Union-Bhattacharyya bound can be improved by using the exact error probability for the first

few terms and then upper bounding the error probability for higher order terms with the

Bhattacharyya bound.

IX-36

Union Bound

The union bound can be tightly bounded by

Pb

J

∑j 5

w j Q 2Eb j5N0

D j 5 w D D exp x0 N0 IX-37

Energy

Notice that the energy of the signal at the output of the channel is

E Ex ∞ ∞s2 t dt

∞ ∞Ex N

∑k N

ukh t kT 2dt

∞ ∞Ex

N

∑k N

ukh t kT N

∑l N

ulh t lT

∞ ∞

N

∑k N

N

∑l N

Ex ukul h t lT h t kT dt

where Ex denotes expectation with respect to the data bit. Because the data are assumed

independent, identically distributed with Ex ukul 1i f k l and is 0 otherwise the energy of

the signal is

E ∞ ∞

N

∑k N

ukh2 t kT dt

IX-38

N

∑k N

∞ ∞h2 t kT dt

N

∑k N

x0

2N 1 x0

The energy per bit is Eb E 2N 1 x0.

IX-39

0 2 4 6 8 10 12 14 16 1810

−10

10−9

10−8

10−7

10−6

10−5

10−4

10−3

10−2

10−1

100

Eb/N

0 (dB)

Pb

Union−Bhattacharrya Bound

Hard Decision Performance

Simulation

Union Bound

AWGN

Figure 106: Performance of Optimal Receiver (x0 1 0 x1 0 2)

IX-40

Maple Code

x0:=1.00;

x1:=0.20;

with(linalg);

a:=M*N*Dˆ(x0-2*x1);

b:=M*N*Dˆ(x0+2*x1);

c:=M*N*Dˆx0;

m:=matrix(3,3,[[-1, -1, 1],[1-a, -b, 0],[-b, 1-a, 0]]);

d:=[0, M*N*Dˆx0, M*N*Dˆx0];

fxx1:=linsolve(m,d);

fxx2:=fxx1[3];

fxx3:=diff(fxx2, N);

fxx4:=eval(fxx3, N=1);

fxx5:=eval(fxx4, M=0.5);

fxx6:=simplify(fxx5);

fxx7:=series(fxx6, D=0, 5);

IX-41

Bounds (x0 1 x1 0 2)

Pb D D85 0 75D

115 D

125 0 5D

145 1 5D3

0 3125D175 1 5D

185 0 75D

195 0 1875D4 1 25D

215 1 5D

225 0 109375D

235

0 9375D245 1 875D5 0 5625D

265 0 65625D

275 1 875D

285 1 28515625D

295

0 4375D6 1 640625D315 1 89453125D

325 0 59375D

335 1 3125D

345

2 198242188D7 D exp 1 N0 ∞

∑j 15

w jDj 5

D exp 1 N0

w D D exp 1 N0

IX-42

Signal Design for Filtered Channels

Because the complexity of the Viterbi algorithm grows as the A

L where A is the alphabet

size and L is the memory of the channel it is desirable to design a system with zero

intersymbol interference. So consider transmitting data at rate 1 T through a channel with

bandwidth W . At what rate is this possible without creating intersymbol interference?

Assume the modulator is a filter acting on a infinite sequence of impulses (at rate 1 T with

impulse response f t . The channel is characterized by an impulse response of g t and the

receiver is a filter sampled at rate 1 T with impulse response h t .

IX-43

f t Modulator

s(t) g t

Channel

z t

r t h t

Demodulator

y t t kT

n t

Data

Rate 1 T yk

The transmitted signal is of the form

s t ∞

∑m ∞

um f t mT IX-44

The output of the received filter is then

yk N

∑m N

umxk m ηk

In order that there be no intersymbol interference we require that

xm

1 m 0

0 m 0

Let

x t ∞ ∞h τ h τ t dt

∞ ∞h τ h t τ dt

where h t h t . Then x t is the convolution of h t with h t and x nT xn. Thus

X f H f H f H f

2. If the (absolute) bandwidth of the channel is W then X f also

has bandwidth W . That is

X f 0 f W

IX-45

Thus by the sampling theorem

x t ∞

∑n ∞

x n2W sin 2πW t n

2W

2πW t n2W

∞

∑n ∞

x n2W φn W t

where

φn W t sin 2πW t n2W

2πW t n2W

If W 1 2T then

x t ∞

∑n ∞

x nT sin π t nT T

π t nT TFor no intersymbol interference we require that x nT 0 for n 0. Let x 0 1 then

x t sin π t nT T

π t nT T

IX-46

which implies that

X f

T f 12T

0 f

12T

Thus

H f

2

T f 12T

0 f

12T

Thus 2W pulses per second can yield zero intersymbol interference. It is easy to see that by

signaling faster than rate 2W we can not guarantee that there is no intersymbol interference.

IX-47

4 6 8 10 12 14 16−1

−0.5

0

0.5

1

time

x1(t

)

Pulse Shape

0 0.5 1 1.5 2 2.5

x 104

0

1

2

3

4

5

6x 10

−5

frequency

X1(

f)

Spectrum

Figure 107: Nyquist Pulse Shape and Spectrum

IX-48

0 5 10 15 20 25 30 35 40−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

time

x1(t

)

Data Waveform

Figure 108: Nyquist Waveform

IX-49

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−3

−2

−1

0

1

2

3

Figure 109: Nyquist Eye Diagram

IX-50

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−4

−3

−2

−1

0

1

2

3

4

Figure 110: Nyquist Eye Diagram

IX-51

Nyquist Pulses

Problems with this pulse shape are: It is hard to generate

A slight timing error results in infinite series decaying as 1 t for intersymbol interference.

Solutions

Signal slower

Allow intersymbol interference in a controlled fashion.

IX-52

Intersymbol-Interference Free Pulse Shapes

Consider slower signaling first. Consider 1T 2W , W 1 2T . (This implies aliasing at the

receiver. Since W 1 2T we can divide the interval W W into segments of length 1 T .

Let N 2WT 1 2 be the number of such segments. When the signal is sampled these

segments get moved to the origin and cause aliasing.

x kT W WX f e j2π f kT d f

N

∑n N 2n 1 2T

f 2n 1 TX f e j2π f kT d f

N

∑n N

1 2T

f 1 2TX f n

T e j2π f nT kT d f

1 2T

f 1 2T

N

∑n N

X f nT e

j2π f kT d f

IX-53

Let

Xeq f N

∑n N

X f nT

Then

x kT 1 2T

f 1 2T

Xeq f e j2πukT d f

Clearly we can have zero intersymbol interference if

Xeq f

T f 12T

0 f

12T

Examples (1)

IX-54

f

Xeq f

12T

12T

(2)

X f

T 0 f 1 α2T

T2 1 sin πT f 1

2T α 1 α2T f 1 α

2T

IX-55

x t sin πt

πtcos απt T

1 4α2t2 T 2

Notice that the pulse decays as 1 t3 instead of 1 t for the Nyquist pulse. The parameter α is

called the rolloff factor.

H f

T 0 f 1 α2T

T2 1 sin πt f 1

2T α 1 α2T f 1 α

2T

h t sin π 1 α t 4αt cos π 1 α t

π 1 4αt 2 t

IX-56

4 6 8 10 12 14 16−1

−0.5

0

0.5

1

time

x(t)

Pulse Shape

0 0.5 1 1.5 2 2.5

x 104

0

2

4

6

8x 10

−5

frequency

X1(

f)

Spectrum

Figure 111: Raised Cosine Pulse Shape and Spectrum (α 0 5)

IX-57

0 5 10 15 20 25 30 35 40−1.5

−1

−0.5

0

0.5

1

1.5

time

x(t)

Data Waveform

Figure 112: Raised Cosine Waveform (α 0 5)

IX-58

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−3

−2

−1

0

1

2

3

time

x(t)

Raised Cosine Eye Diagram

Figure 113: Raised Cosine Eye Diagram α 0 5

IX-59

0 5 10 15 20 25 30 35 40−1.5

−1

−0.5

0

0.5

1

1.5

time

Rea

l(x(t

))

Data Waveform

0 5 10 15 20 25 30 35 40−1.5

−1

−0.5

0

0.5

1

1.5

time

Imag

(x(t

))

Data Waveform


IX-60

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−3

−2

−1

0

1

2

3

Figure 115: Raised Cosine Waveform Eye Diagram (α 0 5)

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−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2


IX-62

0 5 10 15 20 25 30 35 40−2

−1

0

1

2

time

Rea

l(x(t

))

Data Waveform

0 5 10 15 20 25 30 35 40−2

−1

0

1

2

time

Imag

(x(t

))

Data Waveform

Figure 117: Raised Cosine Waveform π

4 QPSK (α 1 0)

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0 0.5 1 1.5 2 2.5 3 3.5 4−3

−2

−1

0

1

2

0 0.5 1 1.5 2 2.5 3 3.5 4−3

−2

−1

0

1

2

Figure 118: Raised Cosine Waveform Eye Diagram π

4 QPSK (α 1 0)

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−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2


4 QPSK (α 1 0)

IX-65

0 5 10 15 20 25 30 35 40−2

−1

0

1

2

time

Rea

l(x(t

))

Data Waveform

0 5 10 15 20 25 30 35 40−2

−1

0

1

2

time

Imag

(x(t

))

Data Waveform

Figure 120: Raised Cosine Constellation π

4 QPSK (α 0 5)

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0 0.5 1 1.5 2 2.5 3 3.5 4−3

−2

−1

0

1

2

0 0.5 1 1.5 2 2.5 3 3.5 4−3

−2

−1

0

1

2


4 QPSK (α 0 5)

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−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2


4 QPSK (α 0 5)

IX-68

0 5 10 15 20 25 30 35 40−2

−1

0

1

2

time

Rea

l(x(t

))

Data Waveform

0 5 10 15 20 25 30 35 40−1.5

−1

−0.5

0

0.5

1

1.5

2

time

Imag

(x(t

))

Data Waveform


4 QPSK (α 0 35)

IX-69

0 0.5 1 1.5 2 2.5 3 3.5 4−3

−2

−1

0

1

2

3

0 0.5 1 1.5 2 2.5 3 3.5 4−3

−2

−1

0

1

2

3


4 QPSK (α 0 35)

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−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2


4 QPSK (α 0 35)

Because we do not have any intersymbol interference the performance of this method for

IX-71

avoiding intersymbol interference has the same performance as BPSK. The difference is that

this modulation scheme requires absolute bandwidth of W 1 α2T . However, since this does

not result in a constant envelope signal for applications requiring constant envelope

transmission this modulation scheme is not acceptable.

IX-72

Controlled Intersymbol Interference: Partial-Response

The second method is to allow some intersymbol interference. This intersymbol interference

is allowed in a controlled fashion. We still signal at rate 1 T 2W but do not require zero

intersymbol interference.

x t ∞

∑n ∞

x nT sin π t nT T

π t nT T

X f

∑∞n ∞ x nT e jnπ f

W

2W f W

0 f W

Example (1):

x nT

1 n 0 1

0 n 0 1

IX-73

This is called Duobinary Transmission (also called partial response class I).

X f

e jπ f

2W

W cos π f2W f W

0 f W

Example (2):

x nT

1 n 1 1

0 n 0 1

This is called Modified Duobinary Transmission (also called Partial Response Class IV).

X f

jW sin π f

W f W

0 f W

This is used in magnetic recording (with maximum likelihood decoding).

For these systems with controlled intersymbol interference we still need a way to detect the

data. One method is decision feedback. The other method is precoding.

IX-74

Ortogonal Frequency Division Multiplexing (OFDM)

Orthogonal frequency division multiplexing (OFDM) is an alternative method of eliminatingintersymbol interferece.

Let b b0 bNbp denote the packet of encoded data. This packet is serial-to-parallelconverted into M streams of data each of length L Nbp M. The sequences are b0 bM 1

where

b0 b0 bM b2M b L 1 M

b1 b1 bM 1 b2M 1 b L 1 M 1

bM 1 bM 1 b2M 1 bLM 1

Equivalently the data streams can be written as a sequence of L column vectors of length Mcomponents.

b0 b0 b1 bM 1 TThe data vector

b0 is augmented with zeros to generate the data column vector C0 where

C0 0 0 0 b0 bM 1 0 0 0 TIX-75

where the leading and trailing zeros are used to form a guardband around the signal as will be

described subsequently. The vector C0 has length Md . The vector C0 is the input to an IFFT

which produces a sequence of time samples of length Md . These time domain samples

correspond to samples in time with a sampling rate of T Md . Let c0 denote the output of the

IFFT when C0 is the input. Now for purposes of mitigating intersymbol interference prepend

a number of these samples.

c0 cMd ν cMd ν 1 cMd 1 c0 cM 1

The length of

c is Md ν where ν represents the length of the channel memory (including any

filtering in the transmitter and receiver). Since the sample rate is Ts Md the duration

corresponding to

c is Ts Md ν Md. The data rate is then M Md Ts Md ν coded

symbols/sec. For Md ν the data rate is approximately M Ts symbols per second. The

sequence of samples after the prefix insertion for a packet then is given by

d

c0

c1

c2

c3

cL

These samples then represent the signal at a sample rate of M T . Thus

d lT M dl l 0 1 M ν L 1

IX-76

In the frequency domain this signal is

S f

M 1

∑l 0

clsinc f l T T

The desired transmitted signal is

s t M 1

∑l 0

cos 2π fH t t φl t

where fH t is a frequency hopping pattern given by

fH t Nhp 1

∑m 0

fm pTh t mTh

Here Nhp denotes the number of hops per packet, Th denotes the duration of a dwell (hop)interval and pT t 1 for 0 t T and zero elsewhere. The information bearing phase,φl t , is given by

φl t Nsp 1

∑m 0

2π flt bl m pTs t mTs where fl is the l-th carrier and bl m is the data symbol for the l th carrier during the m-thsymbol interval which takes values kπ 2 π 4 k 0 1 2 3. The number of OFDM symbols

IX-77

per packet is denoted by Nsp. The desired RF signal is obtained by mixing a baseband signalsb t up to a carrier frequency via

s t Re sb t exp j2π fct Re sb t cos j2π fct Im sb t sin j2π fct

The necessary baseband transmitted signal is

sb t M 1

∑l 0

exp φl t M 1

∑l 0

Nsp 1

∑m 0

pT t mTs exp j2π flt jbl m

In order for the signal to have M orthogonal components it is necessary that any pair offrequencies used fm and fl satisfy

fm fl n Ts

for some integer n. We will assume fl l Ts. consider the signal transmitted during a singlesymbol interval. If the signal sb t is sampled at spacings ∆t Ts M then the samples are

sb m∆t M 1

∑l 0

bl pT m∆t exp j2π flm∆t

M 1

∑l 0

bl pTs m∆t exp j2π flmTs M 0 m M 1

IX-78

M 1

∑l 0

bl exp j2πlm M 0 m M 1

MIFFT b m 0 1 M 1

where the IFFT is the inverse Fourier transform.

IX-79

Receiver

At the receiver the received signal is

rb t sb t n t

The optimal receiver does a complex correlation to determine the data.

Zl T

0sb t exp j2π flt dt

T

0sb t exp j2π flt dt

Suppose that the sampling rate at the receiver is δt T ML . Then

Zl

N 1

∑n 0

s nT ML exp j2π flnT ML T ML

ML 1

∑n 0

s nT ML exp j2πln ML T ML

ML 1

∑n 0

M 1

∑k 0

bk exp j2π fknT ML exp j2πln ML T ML

IX-80

ML 1

∑n 0

M 1

∑k 0

bk exp j2π k T nT ML exp j2πln ML T ML

ML 1

∑n 0

M 1

∑k 0

bk exp j2π k l n ML T ML

ML 1

∑n 0

blT ML

blT

The calculation of Zl can be done via the FFT.

Zl ML 1

∑n 0

sn exp j2πln ML T ML

FFT sl T ML

where FFT is the fast Fourier transform. The first M values from the FFT are the desired

decision variables.

IX-81

Consider a channel with intersymbol interference. The time domain data signal is

ck M 1

∑l 0

blej2πlk M k 0 1 M 1

The cyclic prefix guard interval forces c l cM l l 1 ν. The received signal (withintersymbol interference) is

rk ν

∑m 0

ck mhm

where hm m 0 ν is the channel impulse response. The receiver processes the receiveddata via

zn M 1

∑k 0

rke

j2πnk M

M 1

∑k 0

ν

∑m 0

ck mhme

j2πnk M

ν

∑m 0

hm

M 1

∑k 0

ck me

j2πnk M

ν

∑m 0

hm

M 1 m

∑u m

cue

j2πn u m M

IX-82

ν

∑m 0

hme

j2πnm MM 1 m

∑u m

M 1

∑l 0

blej2πlu M

e

j2πnu M

ν

∑m 0

hme

j2πnm MM 1

∑l 0

bl

M 1 m

∑u m

e j2π l n u M

ν

∑m 0

hme

j2πnm MbnM

HnbnM

where

Hn ν

∑m 0

hme

j2πnm M

IX-83

Design Example

As a possible design example consider a system where the carriers are spaced 1 T 25kHz

apart and there are Md 40 carriers with a guard band on 12 channels on each side. In this

case M 64. Ignoring the prefix insertion we obtain a signal as shown in Figure 126. The

signal is upsampled by inserting zeros between the samples of the original data signal and

filtering. In Figure 127 we show the case where the oversampling is by a factor of 8. The

resulting data signal is then filtered (using an ideal brick wall filter) and results in the desire

spectrum as seen in Figure 128. The data rate possible can be calculated as follows. First, the

data is encoded. Typically the code rate, r,is on the order of 1/4-1/2. Assume QPSK is used

on each of the Md carriers and each symbol has duration T seconds. The data rate is then

R 2 Md r T

MM ν

The bandwidth that is used depends on the filtering done. In theory the bandwidth could be as

small as M T Hz.

IX-84

Spectrum

0 1 2 3 4

x 10−5

−1

−0.5

0

0.5

1Time domain signal (real and imaginary)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 106

−100

−50

0

10*lo

g10(

|S(f)

|)

Spectrum of Signal

0 2 4 6 8 10 12 14 16

x 105

−4

−2

0

2

4Phase of Signal

Figure 126: Output of IFFT

IX-85

0 0.5 1 1.5 2 2.5

x 106

−160

−140

−120

−100

−80

−60

−40

−20

0

20Spectrum of Oversampled Signal


IX-86

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 10−4

−1

−0.5

0

0.5

1Bandlimited version of Oversampled Signal

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 106

−150

−100

−50

0

10*lo

g10(

|X(f)

|)

Bandlimited version of Oversampled Signal


IX-87

Transmitter Block Diagram

TI DSPTMS 320C6711

Analog Devices 9857 (DDS/IQ Mod)

bi

FECb

S/P Zero PadC

IFFTc

Cyclic Prefixc

Oversample Filter IQ Mod D/A

IX-88

System Parameters

The DSP generates a packet of coded bits as follows. A packet of information bits of length

20480 is used to as the input to a rate 1/3 code (convolutional, turbo or LDPC) producing

61440 coded bits. Depending on the multipath of the channel the codeword is punctured to

fewer bits increasing the effective rate of the code. Consider the case of no multipath initially.

Then the coded bits are partitioned into blocks of length 64 which is used to determine 32

complex (frequency domain) samples. The frequency domain samples are padded with zeros

on either side (16 on either side in this example). The resulting 64 complex samples are used

as input to an IFFT which produces 64 complex samples. For multipath channels a cyclic

prefix is added which allows us to deal with the intersymbol interference.

IX-89

Eye Diagrams

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 10−4

−20

−15

−10

−5

0

5

10

15

20

Figure 129: Eye Diagram

IX-90

Eye Diagram

1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3

x 10−4

−20

−15

−10

−5

0

5

10

15

20

Figure 130: Eye Diagram

IX-91

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