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OPTIMUM RECEIVER Simple Pulse Amplitude Modulation Receiver Simple demodulation technique (previous lectures), where the receiver sampled the received signal in the center of bit period If the sampled value is 0 volts; assumed the corresponding data is a 1 If the sampled value is 0 volts; assumed the corresponding data is a 0 (since a “0” from a source would produce a negative pulse from the transmitter Sample in center of each bit period Received LPF signal “1” if sample 0 Comparator “0” if sample 0 f c = bandwidth of transmitted signal Threshold = 0V Fig. 1: A simple PAM receiver The lowpass filter has a bandwidth as the bandwidth of transmitted signal Errors due to channel (attenuation, noise) where the transmitted bit is changed from “0” to “1” or vise versa This inaccuracy/error is quantify by measuring noise margin EEE377 – A: Optimum Receiver 1/36 March 2009 Another powerful tool is to use – probability of bit error Noise Margin The distance between a noiseless received signal and the threshold at the instant of sampling Transmission over perfect channel (no attenuation) and perfect synchronization the noise margin for the third bit – 1 Volts Transmission over channel (50% attenuation) and perfect synchronization the noise margin for the third bit – 0.5 Volts Probability of bit error Let’s quantify accuracy by calculating the probability that a particular bit – say the ith bit is demodulated in error. For the PAM receiver in Fig. 1, the following notations are defined; EEE377 – A: Optimum Receiver 2/36 March 2009 Let bi the ith transmitted bit bi 1, or bi 0 Let st the transmitted signal Let nt noise at the input of the receiver Let r t the received signal, which has been attenuated and corrupted by noise Let z t the voltage at the input of the sampler in the receiver Let no t noise at the output of the receiver’s LPF Source Transmitter Channel Receiver User Received Information bi Transmitted Received information signal st signal r t Noise nt Fig. 2: Block diagram of communication system Sample zt Received signal r t LPF Comparator “1” if sample 0 “0” if sample 0 no t Threshold = 0V Fig. 3: Basic PAM receiver The sampling time for ith bit (center of bit period); Tb t i 1Tb 2 EEE377 – A: Optimum Receiver 3/36 March 2009 The voltage output of sampler for ith bit; T zout t z i 1Tb b 2 We can derive the probability of bit error as follows; P{ith bit =1, but the sample The probability value at the receiver during of the ith bit is the ith bit period < 0} demodulated in = + error P{ ith bit =0, but the sample value at the receiver during the ith bit period 0} Thus T P bi 1, and z i 1Tb b 0 The 2 probability of the ith bit is = T P bi 0, and z i 1Tb b 0 demodulated 2 in error Note that PX a and event Z Pevent ZPX a | event Z EEE377 – A: Optimum Receiver 4/36 March 2009 The T Pbi 1P z i 1Tb b 0 when bi 1 probability of 2 the ith bit is = T Pbi 0P z i 1Tb b 0 when bi 0 (1a) demodulated 2 in error The last two expressions; 1. P z i 1Tb Tb 0 when bi 1 2 T 2. P z i 1Tb b 0 when bi 0 2 are called “conditional probabilities”. We can use the notation of “|” instead of “when”. Thus; T Pbi 1P z i 1Tb b 0 | bi 1 The probability 2 of the ith bit is demodulated in = T error Pbi 0P z i 1Tb b 0 | bi 0 2 (1b) Assume that, 1 Pbi 0 Pbi 1 2 (i.e. “0” and “1” are equiprobable and the value of each bit is independent of the value of all other bits in the data stream) EEE377 – A: Optimum Receiver 5/36 March 2009 Substituiting this value; 1 T The probability P z i 1Tb b 0 | bi 1 2 2 of the ith bit is demodulated in = + (2) 1 T error P z i 1Tb b 0 | bi 0 2 2 Let’s take a look at the input to sampler z(t); z(t) = received signal after filtering But before that, let’s examine the received signal (before filtering or after the channel). Assume the channel has sufficient bandwidth (the channel does not distort the transmitted signal, although it can attenuate the signal); Received signal = attenuated transmitted signal + noise Symbolically; r t st nt where - constant reflecting attenuation of the channel nt - noise at receiver input T T T r i 1Tb b s i 1Tb b ni 1Tb b (3) 2 2 2 EEE377 – A: Optimum Receiver 6/36 March 2009 Effect of filtering to the signal; The filter has the same bandwidth of the transmitted signal, thus st will pass through the filter. Noise has greater bandwidth than the filter, thus we use the variable n0 t to represent the noise the filter output. The voltage at the sampler input (output of filter); zt st no t So T T T z i 1Tb b s i 1Tb b no i 1Tb b (4) 2 2 2 Let’s say the transmitted signal is binary PAM compose of rectangular, sinc-shaped or raised cosine pulses with peak amplitude of A if bi=1 and –A if bi=0. Since there is no ISI in the exact center of bit period; T s i 1Tb b A when bi 1, 2 and T s i 1Tb b A when bi 0 2 EEE377 – A: Optimum Receiver 7/36 March 2009 Substuting into Eq 4, thus the expression is reduced to; T A no i 1Tb b if bi 1 Tb 2 z i 1Tb 2 Tb A no i 1Tb if bi 0 2 Thus, we can express Eq. 2 as; 1 T The probability P A no i 1Tb b 0 2 2 of the ith bit is demodulated in = + (6) 1 T error P A no i 1Tb b 0 2 2 Manipulating Eq (6); Pith bit in error 1 T 1 T (7) P no i 1Tb b A P no i 1Tb b A 2 2 2 2 NOTE For the case of independent bits and equirobable “1”s and “0”s, Eq. 7 tells that the ith bit is demodulated in error:– If at the time of the ith sampling at the receiver, the noise is stronger than the transmitted signal (attenuated) and is of the opposite sign. EEE377 – A: Optimum Receiver 8/36 March 2009 2. Stochastic Mathematics Although we cannot say certainly how much noise will occur in the system at any particular time, but we can determine the probability that the amount of noise at that time will be less than a certain value. T We can’t determine in advance the value of no i 1Tb b , 2 but we can determine the probability that it will be A volts. NOTE 1. This is very different approach (probabilistic/stochastic, mathematics) compared to what we are used to (deterministic mathematic). 2. Deterministic mathematic says – “I can tell you with certainty what the value of a particular variable will be at any time t”. 3. Probabilistic or stochastic, mathematics says- “I don’t know enough about a particular variable (like n(t)) to tell you with certainty its value at particular time in the future (like t i 1Tb Tb ), but I can tell you 2 something about the tendencies of n(t) at time T t i 1Tb b . 2 EEE377 – A: Optimum Receiver 9/36 March 2009 A probabilistic description isn’t as good as a deterministic description, but often it’s the best we can do and is better than nothing. A phenomena that we cannot describe deterministically, but that we can describe probabilistically is called random We don’t know exactly how it will behave at any particular time, but through experimentation and analysis of the physical occurrences that produce the random event, we do know its tendencies. Let’s develop a set of mathematic parameters that we can use to describe the tendencies of a random phenomenon such as noise. Probability distribution function As we saw in calculating probability of error, one parameter we are interested is the probability that n(t) will be less than a certain value A at a certain time ( t i 1Tb Tb ). 2 We are interested in knowing the probability that the result of a random phenomena (e.g. noise voltage) is less than or equal to a certain value at certain time. EEE377 – A: Optimum Receiver 10/36 March 2009 Let’s us use the variable X to describe the voltage of noise at a particular time (random variable). The probability distribution function for X is symbolized as Fxa PX a (8) In other words, Fxa is the probability that the random variable X is less than or equal to some specific value a. The probability distribution function has two forms; 1. Discrete variable – tossing coin, rolling die 2. Continuous variable – lifetime of electronic component Probability distribution properties; 1. Fx 0 (9) 2. Fx 1 (10) 3. If b a, then Fx b Fx a (11) 4. The probability that an outcome will have a random variable value x, within the range a x b is given by expression F b F a . X X EEE377 – A: Optimum Receiver 11/36 March 2009 In our case, we need to determine the probability distribution function for the noise. The most prevalent type of noise in the channel (and front end of the receiver too) is noise due to the random motion of electrons within electronic devices – thermal noise, it probability distribution function; x n 2 1 2 2 F x a e n dx 2 n where (mean) n 0 , and (variance) n2 = average normalized noise power. Probability density function It is symbolized as f X x and defined as dFX x f X x dx It can be used to express the probability that the random variable X lies between two values, say, a and b (b>a). Pa X b PX b PX a FX b FX a f X x dx f X x dx b a f X x dx b a EEE377 – A: Optimum Receiver 12/36 March 2009 Properties; 1. f X x 0 for x 2. f X x dx 1 3. FX a f X x dx a Combining property 2 and 3; PX c1 PX c1 F X c 1 f X x dx c f X x dx f X x dx c f X x dx c Thermal noise Earlier the thermal noise is given as x n 2 1 2 2 F x a e n dx 2 n where (mean) n 0 , and (variance) n2 = average normalized noise power. Thus, the probability density function for thermal noise; EEE377 – A: Optimum Receiver 13/36 March 2009 x n 2 1 2 2 f X x e n 2 n This function usually called as –Gaussian probability function, or Gaussian function. Mean (expected value) Defined as; X xf X x dx In other words, X is the average value of random variable X. Often it is also called as the expected value of X and symbolized as EX ; EX X xf X x dx Variance The variance of random variable X is defined as 2 x X f X x dx 2 X Or it also can be expressed as 2 X 2 x X f X x dx E X X 2 Variance is a measure of how much the values of the random variable X fluctuate (or vary) around its average. Thus, variance is a measure of the unpredictability of X. EEE377 – A: Optimum Receiver 14/36 March 2009 NOTE If a random variable has zero mean ( X 0) , then its variance n2 is equal to its average normalized power of X. 2 2 X x f X x dx x 0 f X x dx 2 X x 2 f X x dx Joint Probabilities This is useful when determining how two random variables may be related to each other. The joint probability distribution function of two random variables X and Y is defined as; 2 FX ,Y x, y f X , Y x, y xy Conditional Probabilities The notation P X a | event Z to express “the probability that random variable X is greater than a given that event Z has occurred.” NOTE (We always encounter) PX a and event Z Pevent ZPX a | event Z EEE377 – A: Optimum Receiver 15/36 March 2009 3. Examining Thermal Noise This defines by random motion of electrons within electronic devices. Through experimentation and analysis of the physical events that produce the noise (random motion of electrons), we can determine the following tendencies, or stochastic parameters, for the noise voltage; 1. Its mean (average value) is 0. n 0 2. It has a Gaussian probability density function, x n 2 1 2 2 f X x e n 2 n and thus the probability that the noise voltage nt at time t o will be less than or equal to a certain value, say A , is x n 2 A 1 2 2 Pnto A FX A e n dx 2 n And the probability that nt is greater than a certain value, say A , is; x n 2 1 2 2 Pnto A 1 FX A e n dx (28) A 2 n EEE377 – A: Optimum Receiver 16/36 March 2009 Since the random variable has zero mean ( X 0) , and its variance n2 is equal to the average normalized power of X. A typical Gaussian probability density function is shown below ( X 0 and n2 1) ; 3. The probabilistic properties of nt do not change with time ( nt is said stationary). Thus, Pnto a Pnt1 a Pnt2 a .......... for any to , t1 and t2 .. This is not to say that nt is the same value at times to , t1, and t2 , but merely it has the same probabilistic tendencies. 4. Thermal noise has the average normalized power spectral density Gn(f) [Volts2/Hz] as constant from dc to approximately 1012 Hz . EEE377 – A: Optimum Receiver 17/36 March 2009 This power spectral density, shown in Fig.4-4, means that thermal noise contains equal amount of all frequency components in the band from dc to 1012 Hz . As an analogy, consider the color white that is composed of all equal component colors. Similarly, thermal noise is composed of the sum of equal strength components of all frequencies. Thermal noise is often called as white noise or more specifically additive white Gaussian noise. Returning to our discussion on the receiver – we know the stochastic properties of nt - the thermal noise at the input of the receiver, therefore we can find the stochastic properties of output noise after LPF is no t . If the LPF has an ideal frequency response and cutoff frequency of f c , then the average normalized power spectrum of the noise at the filter output is as shown in Fig. 4-5. EEE377 – A: Optimum Receiver 18/36 March 2009 If a random signal with a Gaussian probability density function is passed through a linear, time-invariant filter, then the output is a random signal that also has a Gaussian probability density function. Thus no t is still has a Gaussian probability density function. The mean of no t is zero, but different value for variance n2 Using Eq. 28, the probability that no t is greater than some constant a at time t to can be expressed as; x o 2 1 2 2 Pno to a e o dx (29) a 2 o Where o 0 and 2 o average normalized power [Volts2] at the filter output. As defined in lecture notes (Signal Analysis – Review) where the average normalized power [Volts2] as Ps, we can express o2 as; EEE377 – A: Optimum Receiver 19/36 March 2009 fc No o 2 Ps G no f df df N o f c fc 2 [Volts2] (30) NOTE The area under Fig 4-5 which is o2 N o f c (average normalized power after filtering) is significantly smaller than n2 (area under Fig. 4-4 due to extra frequencies after f c ) which signifies average normalized power before filtering. The smaller the value of o2 , the smaller the value of integral in Eq. 29 - thus, the smaller the probability that the noise will exceed a given threshold a. Let’s come back to Eq. 7. Such a probability is related to the probability that the transmitted signal is received in error. Pith bit in error 1 T 1 T P no i 1Tb b A P no i 1Tb b A 2 2 2 2 1 1 Pno to A Pno to A (31) 2 2 For any t o since the noise is stationary. EEE377 – A: Optimum Receiver 20/36 March 2009 From Eq. 29, we know that x o 2 1 2 2 Pno to A e o dx A 2 o Since the probability density function is continuous and does not contain any discrete terms, we can write; x o 2 1 2 2 Pno to A e o dx A 2 o By inspecting Fig. 4-3, we see that the Gaussian probability density function is symmetric about its mean, so x o 2 x o 2 A 1 2 2 1 2 2 Pno to A e o dx e o dx 2 o A 2 o Pno to A Substituting into Eq. 31; EEE377 – A: Optimum Receiver 21/36 March 2009 Pith bit in error 1 T 1 T P no i 1Tb b A P no i 1Tb b A 2 2 2 2 1 1 Pno to A Pno to A 2 2 x o 2 1 2 2 A 2 o e o dx This is often called probability of bit error and denoted Pb EXAMPLE (Probability of bit error for a simple PAM receiver) A transmitter uses raised cosine pulse shaping with pulse amplitudes of +3 and -3 volts. By the time the signal arrives at the receiver, the received signal voltage has been attenuated to half of the transmitted signal, and the signal has been corrupted with additive white Gaussian noise. The average normalized power at the output of the receiver’s filter is 0.36 volts2 and the mean µo= 0. Find Pb assuming perfect synchronization. SOLUTION EEE377 – A: Optimum Receiver 22/36 March 2009 5. Gaussian Probability Density Function As has been established the Gaussian probability Density function is expressed as; x 2 1 2 2 f X x e (19R) 2 Thus, if a random variable X can be described using the Gaussian probability density function; x o 2 value of Gaussian 1 2 2 P random var iable a a 2 o e o dx (29R) Many random occurrences can be expressed using the Gaussian function; thermal noise (zero mean and variance 2 as the average normalized power) IQ results (mean = 100, and variance 2 of 256) The Gaussian function is described by mean and standard deviation . Changing mean corresponds to shifting the graph along the x-axis. Changing affects the “spread” of the function. EEE377 – A: Optimum Receiver 23/36 March 2009 As seen in the previous example, solving the integral of Gaussian function (Eq. 29R) requires numerical analysis using computer program. Is there any easier solution? 1. We create a single table that calculate the value of integral for various value of a with 0 and 1. This has the following integral, or defines as Q(a); x 0 2 x2 1 2 12 1 Qa e dx e 2 dx (35a) a 2 1 a 2 Fig. 4-7 shows the Q(a) integral graphically. Table 4- 1 shows the values for Q(a) for various values of a. EEE377 – A: Optimum Receiver 24/36 March 2009 EEE377 – A: Optimum Receiver 25/36 March 2009 NOTE a. Q(0.53) = 0.2981 b. For a 3 correspond to Qa 0.0014 a2 1 Qa e 2 for a 3 (36) a 2 2. What if is not zero? Consider the integral; x 2 1 a 2 e 2 dx Changing variable, Let w x ; thus, dw dx Lower limit: w x ; when x a; w a Thus, x 2 w2 1 1 e 2 dx e 2 dw Qa (38) a 2 a 2 Fig. 4-8a and 4-8b show the equivalency. EEE377 – A: Optimum Receiver 26/36 March 2009 Geometrically, Eq. 38 corresponds to sliding the Gaussian probability density function to the left by . Table 4-1 is suitable to solve integral of a Gaussian function with any value of a , , but only for 1. 3. What if is not 1? Consider the integral; x 2 1 2 2 a 2 e dx EEE377 – A: Optimum Receiver 27/36 March 2009 Changing variable, Let w x ; thus, dw dx Lower limit: w x ; when x a; w a Thus, x 2 w2 1 1 a 2 e 2 2 dx a 2 e 2 2 dw Now let’s perform a second change in variable; Let z w ; thus, dz dw Lower limit: z w x ; when x a; z a Thus, x 2 w2 z2 1 1 1 a 2 e 2 2 dx a 2 e 2 2 dw a 2 e 2 dz Or x 2 z2 1 1 a a 2 e 2 2 dx a 2 e 2 dz Q (42) Now our table 4-1 can be used to solve Eq. 28 for any value of a , , and . Fig. 4-9a to Fig. 4-9c shows the equivalency after changing the variables. EEE377 – A: Optimum Receiver 28/36 March 2009 NOTE The combination of expending the y axis by and compressing the x axis by (Fig. 4-9c) produces no net effect on the total shaded area under the curve. EEE377 – A: Optimum Receiver 29/36 March 2009 EXAMPLE (Revisit) (Probability of bit error for a simple PAM receiver) A transmitter uses raised cosine pulse shaping with pulse amplitudes of +3 and -3 volts. By the time the signal arrives at the receiver, the received signal voltage has been attenuated to half of the transmitted signal, and the signal has been corrupted with additive white Gaussian noise. The average normalized power at the output of the receiver’s filter is 0.36 volts2 and the mean µo= 0. Find Pb assuming perfect synchronization. SOLUTION Simplifying the Expression of Bit Error The probability of bit error for PAM receiver is given by; Pith bit in error 1 T 1 T P no i 1Tb b A P no i 1Tb b A 2 2 2 2 1 1 Pno to A Pno to A 2 2 x o 2 1 2 2 A 2 o e o dx EEE377 – A: Optimum Receiver 30/36 March 2009 Since µo=0, using the Q function we can simplify this equation to; x o 2 2 2 A o Pith bit in error 1 e o dx Q A 2 o o A 0 Q o (43) A Q o The physical interpretation; A Pith bit in error Q o received signal strength of sampled value Q (44) o received signal strength of sampled value Q ave normalized noise power at filter output Consider the following observations; 1. The larger the value of a, the smaller the value Q(a) (see Figure 4-7, 4-8 etc). Thus, increasing the argument of function Q, decreases the probability of bit error. 2. The argument of Q is the ratio of received signal strength versus noise o . The larger this signal-to- noise- ratio (SNR), the lower the probability bit error. 3. The LPF use in the receiver is to reduce the average noise power at the receiver, while not affecting the EEE377 – A: Optimum Receiver 31/36 March 2009 strength of received signal (compatible to the bandwidth of signal). Thus, increases the value of SNR, and reducing the probability of bit error. OBSERVE 1. Pb can be interpreted as the probability that a particular transmitted bit is demodulated in error. 2. Pb can also be interpreted as the percentage of all transmitted bits, which on average will be demodulated in error. EXAMPLE A transmitter uses raised cosine pulse shaping with pulse amplitudes of +1 and -1 volts. By the time the signal arrives at the receiver, the received signal voltage has been attenuated to half of the transmitted signal, and the signal has been corrupted with additive white Gaussian noise. The average normalized power at the output of the receiver’s filter is 0.035 volts2 and the mean µo= 0. Find Pb assuming perfect synchronization. SOLUTION A = 1 volts = fraction of the transmitted signal voltage = 0.5 EEE377 – A: Optimum Receiver 32/36 March 2009 2 o = average normalized power = 0.035 volts 2 received signal strength sampled value Pb Q o A 0.51 Q Q Q2.67 0.0038 0.035 0.035 Thus, on average, 0.38% of the transmitted bits will be received in error or using different interpretation, on average 3.8 out of every thousand bits will be received in error. Optimum Receiver Sample zt Received signal r t LPF Comparator “1” if sample 0 “0” if sample 0 no t Fig. 3: Basic PAM receiver Let develop a receiver with better performance Can we think about other than just sampling to detect the signal (previous notes). Consider a general block diagram for demodulation process; EEE377 – A: Optimum Receiver 33/36 March 2009 Restrict the “processing” part to be linear, thus it can be represented as transfer function H(f) as shown below; Received signal and Processed signal noise z t r t * ht r t st nt z iTb Threshold comparison “1” if z iTb 0; (Threshold Hf “0” if z iTb 0; z t Sample at = 0 volts) t iT Fig. 4-11: Receiver for implementing general demodulation process Let; r t - the received signal which can be decomposed into a component due to the attenuated signal st , and a component due to noise nt EEE377 – A: Optimum Receiver 34/36 March 2009 ht - the impulse response corresponding to the transfer function H(f); ht 1H f zt - the processed received signal z t r t * ht st nt * ht st * ht nt * ht OBSERVE 1. nt has a Gaussian probability distribution, so nt * ht has also Gaussian probability distribution. 2. Processing of the received signal for the ith bit occurs during the time interval i 1Tb t iTb . At the end of this period, the processing produces a single value z(iTb ) , which is compared to a threshold to determine whether the ith bit is demodulated as a “1” or “0”. For a simple PAM receiver, the probability bit error A Pith bit in error Q o received signal strength of sampled value Q ave normalized noise power at filter output EEE377 – A: Optimum Receiver 35/36 March 2009 Since nt * ht (processed noise) exhibits a zero-mean Gaussian probability distribution; received signal strength of sampled processed signal Pb , simple receiver Q ave normalized noise power at filter output siTb * hiTb Q p (47) Where 2 represents the average normalized power of the p processed noise (and the variance of its probability distribution) QUESTION What type of processing will minimize the probability of bit error for receiver Fig. 4-11? This is equivalent of asking what expression for H(f) will optimized receiver accuracy. Minimizing probability of bit error corresponds to maximizing the argument within Q of Eq. 47. EEE377 – A: Optimum Receiver 36/36 March 2009