m 4.6 Signal detection: In co Problem 4.6 Signal detection: In communication cha
ID: 1716165 • Letter: M
Question
m 4.6 Signal detection: In co Problem 4.6 Signal detection: In communication channels, signals are often corrupted by noise which leads to errors in decoding the transmitted message. Message-decoding requires an algorithm to take a noise-corrupted signal measurement, and decide which message, among a finite set of possible messages, was sent. The performance of a decoding algorithm is typically measured in terms of the probability of message decoding error. For instance, if only one out measured in terms ofteprobabilityofessage decoding error. For instance, if only one outExplanation / Answer
(a).When the transmitter encodes the two types of messages with the values X = 0 and 1.
The probability that the decision made by the receiver is wrong when 0 is sent and P(e|1) denotes the probability that the decision made by the receiver is wrong when 1 is sent.
The receiver is wrong when 1 is sent if and only if the input to the threshold device is less than or equal to the threshold . Under this condition that 1 is sent, the input Y (T0) to the threshold device is the random variable Y0(T0). Let X = Y (T0) and fX(x|i) be the pdf of X given that digit i is sent, i = 0, 1. We denote the X being Gaussian with mean µ and variance 2 by N(µ, 2 ). From the previous discussions, we have fX(x|i) is equal to the pdf of a Gaussian random variable with mean sio(T0) and variance 2 , i.e., the pdf of a Gaussian variables with N(s1(T0), 2 ) for i = 0, 1. Therefore,
P(e|1) = P{Y (T0) | 1}
= P{Y1(T0) }
= (( s1o(T0))/)
= Q(s1o(T0) )/).
Similarly, the probability of error when 0 is sent is given bym
P(e|0) = P{Y (T0) > | 0}
= P{Y0(T0) > }
= 1 P{Y0(T0) }
= 1 (( s0o(T0))/)
= Q(( s0o(T0))/).
the fact that (1)i is equal to 1 if i = 0 and is 1 if i = 1, we can summarize the key result as follows. If the transmitted signals are s0(t) and s1(t), and if µi(T0) = sio(T0) is equal to the mean of the signal component of the output of the filter h(t), sampled at time T0 : 0 < T0 T, then probability of error given the signal si(t) is transmitted is ,
P(e|i) = Q[(1)i (sio(T0) )/]
where is the threshold, T0 is the sampling time, and is the standard deviation of the output of the filter.
(b).Assume that the signal selection is antipodal signaling, i.e., s0(t) = s(t) and s1(t) = s(t) where s(t) = A > 0 for 0 t < T, otherwise s(t) = 0. The impulse response h(t) is given by h(t) = s(T t) (it is called a matched filter). If 0 is sent, the input to the threshold device is the random variable
.Y (T0) = Y0(T0) = s0o(T0) + no(T0)
which is Gaussian with mean s0o(T0) and variance 2 = Rno (0) = 1 2N0T. If 1 is sent, the input to the threshold device is the random variable ,
Y (T0) = Y1(T0) = s1o(T0) + no(T0)
which is Gaussian with mean s1(T0) and variance 2 . The difference in the means is
µ1(T0) µ0(T0) = s1o(T0) s0o(T0)
= s1o(T0) [s1o(T0)] = 2s1o(T0)
Since h(t) = s(T t), we have
µ1(T0) µ0(T0) = 2s1o(T0) = 2s(T0) h(T0)
= { 2A^2T0 0 T0 T
{ 2A^2 (T T0) T < T0 2T
{ 0 otherwise
The difference is maximized if the sampling time T0 is equal to T. For certain filters, the transfer function H(f) is easier to work with than the impulse response.
the error probabilities P(e|i), i = 0, 4 when Y<2 AND Y>2, where P(e|0) denotes the probability that the decision made by the receiver is wrong when 0 is sent (i.e., when s1(t) is transmitted), and P(e|1) denotes the probability that the decision made by the receiver is wrong when 4 is sent.
The receiver is wrong when 4 is sent if and only if the input to the threshold device Y is less than or equal to the threshold . Under this condition that 0 is sent, the input Y (T0) to the threshold device is the random variable Y0(T0). Let X = Y (T0) and fX(x|i) be the pdf of X given that digit i is sent, i = 0, 4. We denote the X being Gaussian with mean µ and variance 2 by N(µ, 2 ). From the previous discussions, we have fX(x|i) is equal to the pdf of a Gaussian random variable with mean sio(T0) and variance 2 , i.e., the pdf of a Gaussian variables with N(sio(T0), 2 ) for i = 0, 4. Thus
P(e|4) = P{Y (T0) | 1}
= P{Y1(T0) }
= (( s1o(T0))/)
= Q(s1o(T0) )/).
Similarly, the probability of error when 0 is sent is given by
P(e|0) = P{Y (T0) > | 0}
= P{Y0(T0) > }
= 1 P{Y0(T0) }
= 1 (( s0o(T0))/)
= Q(( s0o(T0))/).
By taking advantage of the fact that (1)i is equal to 1 if i = 0 and is 1 if i = 1, we can summarize the key result as follows. If the transmitted signals are s0(t) and s1(t), and if µi(T0) = sio(T0) is equal to the mean of the signal component of the output of the filter h(t), sampled at time T0 : 0 < T0 T, then probability of error given the signal si(t) is transmitted is
P(e|i) = Q[(1)i (sio(T0) )/]
where is the threshold, T0 is the sampling time, and is the standard deviation of the output of the filter.
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