Telecommunications Networking IDigital Communication SystemPoint-to-Point Digital Communications: The Concept of “Errors”What Causes Errors?Tradeoffs Between Misses and False AlarmsTrading Off “Miss” and “False Alarm” ErrorsClassical Additive Noise Detection ProblemMatched FilterMatched Filter Output: rSlide 10Calculating the Error ProbabilitiesCalculating ERFC*(x) (“earf-see-star”); also known as the Q function: Q(x)ExampleExample (cont’d)Slide 15Slide 16Copyright 1998, S.D. Personick. All Rights Reserved.Telecommunications Networking ILecture 9Quantifying the Performance of Communication Systems Carrying Digital InformationCopyright 1998, S.D. Personick. All Rights Reserved.Digital Communication SystemData input: 1011101 Data output: 1011101Communication linkCopyright 1998, S.D. Personick. All Rights Reserved.Point-to-Point Digital Communications: The Concept of “Errors”Input information: …1 0 1 0 1 1 ...Output information: …1 0 1 0 0 1time 1 0 1 0 1 1 1 0 1 0 0 1Copyright 1998, S.D. Personick. All Rights Reserved.What Causes Errors?•Errors are caused by noise, distortion, and interference in the communications link, and can also be caused by “synchronization” errors and “congestion” •We can have two types of errors:- “miss” : we mistake a “1” for a “0”- “false alarm” : we mistake a “0” for a “1”Copyright 1998, S.D. Personick. All Rights Reserved.Tradeoffs Between Misses and False Alarms•We can reduce the “miss” probability to zero by declaring that each output bit value will be a “1”; but then we will get a “false alarm” every time the true value of the corresponding input bit is a “0”•We can reduce the “false alarm” probability to zero by declaring that each output bit value will be a “0”; but then we will get a “miss” every time the true value of the corresponding input bit is a “1”Copyright 1998, S.D. Personick. All Rights Reserved.Trading Off “Miss” and “False Alarm” ErrorsProbability of a “Miss”Probability of a “False Alarm”110Receiver operating characteristic (ROC)Copyright 1998, S.D. Personick. All Rights Reserved.Classical Additive Noise Detection Problems(t) = 1 volt peakn(t)= white, Gaussian noise +r(t)Copyright 1998, S.D. Personick. All Rights Reserved.Matched FilterMatched FilterSample herer = s + na(t)=pulse shapeh(t)=a(-t)Copyright 1998, S.D. Personick. All Rights Reserved.Matched Filter Output: rr = s + n (volts)“s” is equal to either: S or 0 (volts); corresponding to a digital “1” or a digital “0”n is a Gaussian random variable with variance N (volts**2)Copyright 1998, S.D. Personick. All Rights Reserved.Matched Filter Output: r0 voltsS voltsDecision Threshold = D voltsr= s + nIf r > D, then we will guess that s=S (I.e., that we have a “1”.If r < D, then we will guess that s=0 (I.e., that we have a “0”Copyright 1998, S.D. Personick. All Rights Reserved.Calculating the Error Probabilities 0 D SThe probability of a miss is given by:ERFC* [ (S-D)/(N**0.5)]; where ERFC* [ ] is called the “error function complement star”For example ERFC*[6] = 10**-9Copyright 1998, S.D. Personick. All Rights Reserved.Calculating ERFC*(x) (“earf-see-star”); also known as the Q function: Q(x)ERFC* (x) = the integral from x to infinity of 1/[(2 pi)**(0.5)] [exp -(y**2)/2] dy = Q(x)Copyright 1998, S.D. Personick. All Rights Reserved.Example •We have a point-to-point metallic cable system•r(t)= s(t) + n(t) (volts) ; where:s(t) is a sequence of pulses, modulated on or off, where the pulse shape is a(t)n(t) is white Gaussian noise with spectral density, N, equal to 2kTR (volts**2/Hz)Copyright 1998, S.D. Personick. All Rights Reserved.Example (cont’d)•R is the impedance of the metallic cable over which s(t) is being communicated; kT = 4 x 10**-21 Joules•We will assume that the energy in a single received pulse a(t) = integral of {[a(t)]**2 dt}/R = E Joules•If we pass r(t) through a matched filter with impulse response h(t) = C a(-t), where C is a constant, then …..Copyright 1998, S.D. Personick. All Rights Reserved.Example (cont’d)•…. The output of the matched filter will ber = s + n; where: •s= S or 0 volts; n is a Gaussian random variable with variance N= 2kTR(C**2)ER (volts**2); and S = CER (volts)•The ratio of (S/2)/(N**0.5)= [E/(8kT]**0.5•If we wish to have an error rate of 10**-9, for both misses and false alarms, then...Copyright 1998, S.D. Personick. All Rights Reserved.Example (cont’d)•….we require:[E/(8kT)] = 36 or more•Since 8kT ~ 3.2 x 10**-20, we require the received pulse energy E to equal ~ 10**-18 (Joules) or more. •If the transmitted pulse is 1 volt peak, and 100ns in duration; and the cable has an impedance of 50 ohms, then the transmitted pulse energy is: .02 x 10**-7 = 2 x 10**-9
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