intersymbol interference · 2005-01-04 · m l (ﬁnite intersymbol interference) Λn v 2 n ∑ k n...

Chapter 10

Intersymbol Interference

In this chapter we examine optimum demodulation when the transmitted signal is filtered by the channel and thereis additive white Gaussian noise. The optimum demodulator chooses the possible transmitted vector that wouldresult in the received vector (in the absence of noise) to be as close as possible (In Euclidean distance) to whatwas received. This we show can be implemented by a filter matched to the received signal for a given data symbolfollowed by a nonlinear processing via the Viterbi algorithm. The filter is sampled at the data rate. We also analyzethe performance of such a system. The analysis is very similar to that of convolutional codes. Because the receivedsignal is filtered and sampled, the output of the filter consists of two components. One due to the transmitted signaland one due to the noise. The output due to noise is, however, not white. However, in the next section we showthat the output of the matched filter can be whitened. With a whitened matched filter the optimum receiver (Viterbialgorithm) becomes clear. Finally, in the last section we show how to design a system to eliminate intersymbolinterference.

1. Optimum Demodulation

Consider transmitting data at rate 1T through a channel with bandwidth W or through a distorting channel. We

would like to find the optimum (minimum sequence error probability) receiver. Assume the modulator is a filteracting 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 1T with impulse response h t .

f t Modulator

sT t g t Channel

z t r t n t

Data

Rate 1T

The transmitted signal is of the form

sT t u N

∑m N

um f t mT where um is the data symbol transmitted during the m-th signaling interval assumed to be in the alphabet A andf t is the waveform used for transmission. We assume a transmission of 2N 1 data symbols (think of N as beingvery large). The output of the channel filter is then

z t ∞ ∞g t τ s τ dτ

10-1

10-2 CHAPTER 10. INTERSYMBOL INTERFERENCE

∞ ∞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

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

xm k ∞ ∞h t kT h t mT dt

Thus the optimum decision rule isChose v if ΛN v max

uΛN u

Since the decision statistic depends on the received signal only through yk it is clear that yk is a sufficient statisticto 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

1. 10-3

The sampled output would be

y kT ∞ ∞r t h t kT dt

Since this is yk defined earlier the received signal should be filtered and sampled as shown below before doingsome processing.

f t Modulator

s t g t Channel

z t r t hr t

Demodulator

y t t kT

n t Data

Rate 1T yk

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

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.

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 2

N

∑k N

vk

k N

∑j 1

vk jx j

N

∑k N

2vkyk N

∑k N

x0v2k 2

N

∑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 2

N

∑k N

vk

min L k N ∑j 1

vk jx j

N

∑k N

2vkyk x0v2

k 2vk

min L k N ∑j 1

vk jx j The decision rule can be implemented in a Viterbi algorithm like structure. Define the state at time k to be the lastL 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.Let Γ σm be the length (optimization criteria) of the shortest (optimum) path to state σm at time m. Let σm

be the shortest path to state σm at time m. Let Γ σm 1 σm be the length of the path to state σm 1 at time m thatgoes through state σm at time m. The algorithm works as 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

1. 10-5

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

λ σk σk 1 2vk 1yk 1 x0v2k 1 2vk 1vkx1

2yk x0 2x1

2yk x0 2x1

2yk x0 2x1

2yk x0 2x1

+1

-1

vk+1

-1

vk 1

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.

-1

+1 0.60

-2.2

0.4

-0.8

-2.6

1.8

2.0

0.0

-4.4

2.8The double lines represent the path chosen by the Viterbi decoder. Thus the Viterbi decoder would output thesequence v 1 1 1 1 1 .


Example L 2 x0 0 x1 0 x2 0vk 1

ΛN v N 1

∑k N

λ σk σk 1 2v Ny N x0v2 N

λ σk σk 1 2vk 1yk 1 x0v2k 1 2vk 1vkx1 2vk 1vk 1x2

vkvk 1 vk 1vk

-1 -1 -1 -1

-1 +1 -1 +1

+1 -1 +1 -1

+1 +1 +1 +1

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

2. Error Probability

Now consider the performance of the above maximum likelihood sequence detector (MLSD). We will evaluate theunion upper bound to the error probability. To do this we need to determine the pairwise error probability betweentwo sequences. This is the probability that sequence v is demodulated given sequence u is transmitted for a systemwith only two possible transmitted sequences (u v). Let P u v denote the conditional error probability given

2. 10-7

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 suand sv.

sv t su t 2 ∞ ∞

sv t su t 2 dt

∞ ∞ 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 4

N

∑k N

N

∑m N

εkεmxk m

4N

∑k N

ε2kx0 4

N

∑k N 1

2k 1

∑m N

εkεmxk m

4

N

∑k N ε2

kx0 2k N

∑j 1

εkεk jx j 4

N

∑k N ε2

kx0 2L

∑j 1

εkεk jx j P u v

exp 1N0

N

∑k N ε2

kx0 2L

∑m 1

εkεk mxm P e

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

An error event is defined to be an error sequence that diverges once from the all zero state and then remergeslater. Since a necessary condition for an error of a particular type (first event error or symbol error) is that anerror event occurs that causes the demodulator/decoder to follow a path that diverges and then at some later timeremerges we can calculate the error probability for a particular node by counting the number of paths (and theirdistance) that diverge and remerge. We can use the state diagram to determine the number of error sequences witha particular distance.


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 ∑

eε 0

∑u v

e 12v 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 the22N 1 u sequences are equally likely. In each position where εk 0 the components of the sequences u and v aredetermined. If εk 1 then vk 1 and uk 1. Similarly if εk 1 then vk 1 and uk 1. The components ewhere 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 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 ∑

eε0 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 1N0 ε2

kx0 2L

∑m 1

εkεk mxm For L 1

P e 2wh e N

∏k N

1

2wH εk exp

1N0 ε2

kx0 2εkεk 1x1 We calculate these union bounds by enumerating the sequences that diverge from the all zero error state and

remerge (error events) that correspond to a given Euclidean distance between two data sequences and has a givennumber of nonzero terms (or is a given length). To do this we draw a state diagram (similar to that for convolutionalcodes) and label each path with DxNlMl where x is the incremental Euclidean distance squared (divided by 4) ingoing from one state to another and l is 1 if the error path is nonzero and is zero if the error is zero. (This redundantuse of l will be explained when we determine the bit error probability).

Error State Diagram L 1

2. 10-9

MNDx0

MNDx0

1

1

MNDx0 2x1MNDx0 2x1

MNDx0 2x1

MNDx0 2x1

0 0

1

-1

a

d

b

c

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

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 paths with two errorsand Euclidean distance squared of 8x0 8x1 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 Dx0 1 N 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 beimproved by using the exact error probability for the first few terms and then upper bounding the error probabilityfor higher order terms with the Bhattacharyya bound.

For L 2 the error state diagram is shown below.

3. 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 memoryof the channel it is desirable to design a system with zero intersymbol interference. So consider transmitting dataat 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 1T with impulse

response f t . The channel is characterized by an impulse response of g t and the receiver is a filter sampled atrate 1

T with impulse response h t .

f t Modulator

s(t) g t Channel

z t r t h t

Demodulator

y t t kT

n t Data

Rate 1T yk

The transmitted signal is of the form

s t ∞

∑m ∞

um f t mT 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 00 m 0

let

x t ∞ ∞h τ h τ t dt

∞ ∞h τ h t τ dt

3. 10-11

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

Thus by the sampling theorem

x t ∞

∑n ∞

x n2W sin 2πW t n

2W 2πW t n

2W ∞

∑n ∞

x nT φn W t where

φn W t sin 2πW t n2W T

2πW t n2W

If W 12T then

x t ∞

∑n ∞

x nT sin π t nT T π t nT T

For 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

which implies that

X f T f 1

2T0 f 1

2T

Thus H f 2 T f 1

2T0 f 1

2T

Thus 2W pulses per second can yield zero intersymbol interference. It is easy to see that by signaling faster thanrate 2W we can not guarantee that there is no intersymbol interference.

Problems with this pulse shape are: (1) It is hard to generate and (2) a slight timing error results in infiniteseries decaying as 1

t for intersymbol interference.

Solutions (a) signal slower or (b) allowed intersymbol interference in a controlled fashion.

1. Intersymbol-Interference Free Pulse Shapes

Consider slower signaling first. Consider 1T 2W , W 1

2T . This implies aliasing at the receiver. Since

W 12T 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 2TX f e j2π f kT d f

N

∑n N

1 2T

f 1 2TX f n

T e j2π f n

T kT d f

1 2T

f 1 2T N

∑n N

X f nT e j2π f kT d f


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 10.1: Nyquist Pulse Shape and Spectrum

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 10.2: Nyquist Waveform

3. 10-13

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 10.3: Nyquist Eye Diagram

Let

Xeq f N

∑n N

X f nT

Then

x kT 1 2T

f 1 T Xeq f e j2πukT d f

Clearly we can have zero intersymbol interference if

Xeq f T f 1

2T0 f 1

2T Examples (1)

f

Xeq f

12T

12T


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

(2)

X f T 0

f 1 α2T

T2 1 sin πT f 1

2T α 1 α2T

f 1 α2T

x t sin πt πt

cos απtT

1 4α2t2T 2

Notice that the pulse decays as 1t3 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 α2T

h t sin π 1 α t 4αt cos π 1 α t π 1 4αt 2 t

Because we do not have any intersymbol interference the performance of this method for avoiding intersymbolinterference has the same performance as BPSK. The difference is that this modulation scheme requires absolutebandwidth of W 1 α

2T . However, since this does not result in a constant envelope signal for applications requiringconstant envelope transmission this modulation scheme is not acceptable.

2. Controlled Intersymbol Interference: Partial-Response

The second method is to allow some intersymbol interference. This intersymbol interference is allowed in acontrolled 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

3. 10-15

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 10.5: Raised Cosine Waveform (α 0 5)

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 10.6: Raised Cosine Eye Diagram α 0 5


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


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 10.8: Raised Cosine Waveform Eye Diagram (α 0 5)

3. 10-17

−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


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 10.10: Raised Cosine Waveform π4 QPSK (α 1 0)


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 10.11: Raised Cosine Waveform Eye Diagram π4 QPSK (α 1 0)

−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


3. 10-19

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 10.13: Raised Cosine Constellation π4 QPSK (α 0 5)

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



−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


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


3. 10-21

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


−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



X f ∑∞n ∞ x nT e jnπ f W

2W f W0 f W

Example (1):

x nT 1 n 0 10 n 0 1

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 10 n 0 1

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

X f j

W sin π fW 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.

intersymbol interference · 2005-01-04 · m l (ﬁnite intersymbol interference) Λn v 2 n ∑ k n...

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