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EE 475 : Homework #2 SolutionProf. Uf Tureli, Stevens Institute of Technology

email/tel/fax: utureli@stevens-tech.edu, 201.216.5603/8246

Note: Haykin, Q. 4.1, 4.3,4.6,4.11

1. Question: Matched Filters. Consider the signal shown in Figure P4.1.(a) Determine the impulse response of a filter matched to this signal and sketch it as a

function of time.

Solution: The matched filter impulse response is:

h(t) = S(T t)

(b) Plot matched filter output as a function of time.

Solution: The matched filter output is:

so(t) =

h() S(t )d

(c) What is the peak value of output?

Solution: The matched filter output is maximized at t = Tand equals A2T /4.

2. Question: Matched Filters. Consider a rectangular pulse defined by:

g(t) =

A, 0 t T0, otherwise

It is proposed to approximate the matched filter for g(t) by an ideal low-pass filter ofbandwidth B; maximization of the peak pulse signal-to-noise ratio is the primary objective.

(a) Determine the optimum value ofB for which the ideal low-pass filter provides thebest approximation to the matched filter.

Solution: The transfer function of the matched filter for a rectangular pulse ofduration Tand and amplitude A is given by:

Hopt(f) = sin(f T)exp(jfT)

To approximate the amplitude response, we determine the particular value of band-

width B that will provide the best approximation to the matched filter. The criteriato be used is the output signal to noise ratio at the time of sampling.

The maximum value of the output signal, produced by an ideal low-pass filter inresponse to a rectangular pulse occurs at t = T /2 for BT 1 and is equal to(2A/) sin(BT). The average noise power at the output of the ideal low-pass filter

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is equal to BN0. The maximum output signal-to-noise ratio of the ideal low-passfilter is therefore:

(SN R)

0 =(2A/)sinc(BT)2

BN0

Assuming AT= 1, we get:

(SN R)

0

(SN R)0= 2

2BTsin2(BT)

The peak value occurs at BT= 0.685.

(b) By how many decibels is the ideal low-pass filter worse than the the matched filter?

Solution: The best value or the bandwidth of the ideal low pass filter characteristicof is B = 0.685/T. The maximum signal-to-noise ratio of ideal low pass filter is0.84 dB below that of true matched filter.

3. Question: Probability of Error Calculation. In a binary pulse coded modulation (PCM)system, symbols 0 and 1 have a prioriprobabilities p0 and p1 respectively. The conditional

probability density function of the random variable Y(with sample value y) obtained bysampling the matched filter output in the receiver of Figure 4.4 at the end of a signalingintercal, given that symbol 0 was tranmsitted, is denoted by fY(y|0). Similarly, fY(y|1)denotes the conditional probability density function ofY, given that symbol 1 was trans-mitted. Let denote the threshold used in the receiver, so that if the sample value yexceeds , the receiver decides in favor of symbold 1; otherwise, it decides, in favor ofsymbol 0. Show that the optimum threshold opt , for which the average probability oferror is a minimum, is given by the solution of:

fY(opt|1)fY

(opt|0)

=p0p1

Solution: The average probability of error is:

Pe = p1

fY(y|1)dx + 0

fY(y|0)dx.

An optimum choice of corresponds to minimum Pe. Differentiating Eq. (1) with respectto , we get:

Pe

= p1fY(|1) p0fY(|0)

Setting Pe

= 0, we get the following condition for the optimum value of.

fY(opt|1)fY(opt|0) =

p0p1

,

which is the desired result.

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4. Question: Probability of Error Calculation. In this problem, we revisit the PCM receiverof Figure 4.4., but this time we consider the use of bipolar nonreturn-to-zero signaling, inwhich case the transmitted signal s(t) is defined by:

Binary symbol 1: s(t) = A for 0 < t T

Binary symbol 0: s(t) = 0, 0 < t

T

Determine the average probability of symbol error Pe for this receiver assuming that thebinary symbos 0 and 1 are equiprobable.

Solution: Probability of error for bipolar NRZ signal:

Binary symbol 1: s(t) = ABinary symbol 0: s(t) = 0Energy of symbol 1 = Eb = A

2Tb

The absolute value of the threshold is = 12

Eb = 12

A2Tb.

Pr(error|s = A) = 1N0

exp

(y + Eb)2N0

dy

Let z = (y+Eb)

N0, then:

Pr(error|s = A) = 1N0

+Eb+

Eb

expz2

dz

Similary, Pr(error|s = A) = Pr(error|s = A) and

Pr(error|s = 0) =2

1N0

exp

y2

N0

dy = erfc1

2Eb

N0

The average probability of error is therefore:

Pre = Pr(s = A)Pr(error|sA) + Pr(s = 0)Pr(error|s = 0)= 1

212

erfc

12

EbN0

erfc

34

EbN0

+ 1

2erfc

12

EbN0

= 34erfc

12

EbN0

1

4erfc34

EbN0

3