数字信号处理—基于计算机的方法第7章答案.pdf

7-1In a binary communication system the receiver test statistic,ro(t0) =ro, consists of a polar signal plus noise. The polar signal has valuess s s so1o1o1 o1= = = = + + + +A A A Aands s s so2o2o2o2= = = = A A A A. . . . Assume that the noise has a Laplacian distribution, which is 2|/ 1 () 2 oo n o o f ne = whereois the rms value of the noise. (a) Find the probability of errorPeas a function ofA/ofor the case of equally likely signaling andVThaving the optimum value. (b) PlotPeas a function ofA/odecibles. Compare this result with that obtained for Gaussian noise as given by Eq.(7-26a). Solution:Solution:Solution:Solution: (a) 1 2 , sent , sent o o o Ans r Ans + = + 2| 1 1 (|) 2 o o rA o o f rse = 2| 2 1 (|) 2 o o rA o o f rse + = Using (7-8) MMMMro1ro1ro1 ro1=A, =A,=A,=A, MMMMro2 ro2ro2ro2= = = = A A A A, , , , the source probabilities are equally likely, and the conditional probabilities have symmetrical shapes about A. 2|2| 1111 2222 oo T oo T rArA v eoo v oo Pedredr + =+ ThusVT= 0 2|2| 0 0 2()2() 0 0 1 2 2 1 2 2 oo oo oo oo rArA eoo o rArA oo o Pedredr edredr + + =+ =+ 12 12 2()2() and 22 oo oo oo oo rArA Let xx dxdrdxdr + = = 12 12 2 / 12 2 / 2 /2 / 12 1 2 222 1 4 o o oo A xx oo e A o AA xx Pedxedx e dxe dx =+ =+ 2/ 2/ 11 22 o o A A xx e dxe = 2 /2 / 11 22 oo AA e eeeP = = Pemuch larger for Laplacian Noise. 7-4Awhole binary communication system can be modeled as an information channel, as shown in Fig. P7-4. Find equations for the four transition probabilities (|)P m m , where both m andmcan be binary ls or binary 0s. Assume that the test statistic is a linear function of the receiver input and that additive white Gaussian noise appears at the receiver input. Hint: Look at Eq. (7-15). Figure P7-4 Solution:Solution:Solution:Solution: 1 2 oo o oo sn r sn + = + ()2 1 2 2 1 1 (|) 2 oo o rs o o f rse = ()2 2 2 2 2 1 (|) 2 oo o rs o o f rse = 1 1 (0/1)(/) T V To oo o Vs Pf rs drQ + = 1 (1/1)1(0/1)1 To o Vs PPQ + = = 2 2 (1/0)(/) T To oo V o Vs Pf rs drQ = 22 (0/0)1(1/0)1 TooT oo VssV PPQQ = = = 7-7Examine how the performance of a baseband digital communication system is affected by the receiver filter. Equation (7-26a) describes the BER when a low-pass filter is used and the bandwidth of the filter is large enough that the signal level at the filter output iss s s so1o1o1 o1= = = = + + + +A A A Aors s s so2o2o2o2= = = = A A A A. Instead, suppose that a RC low-pass filter with a restricted bandwidth is used whereT=1/f0=2RC.Tis the duration (pulse width) of one bit, andf0is the 3-dB bandwidthoftheRClow-passfilteras described by Eq. (2-147). Assume that the initial conditions of the filter are reset to zero at the beginning of each bit interval. (a) Derive an expression forPeas a function ofEb/N0. Solution:Solution:Solution:Solution: (a) ( ) 1 1 1 1 j C Hf j RC R j C = + + ( ) () 2 2 1 1 Hf RC = + ( ) 2 01H= () () 2 02 0 0 0 11 2 12 1 3 2 1 22 Hf f RC fdB RC T RC f = + = = 带宽 ( ) ()() 2 2222 0 111 1 121/ Hf f T fRCff = + + () 2 0 22 0 0 00 2 0 |( )| 1 |(0)| 1/ 12 122 eq H fdf Bdf H ff f dxf xTT = + = + 2 00 2 22 oeq NN B T = RC 电路的冲击响应： ( ) 0 0 0 1 ,0 0,0 t et h tRC t = ( ) /2tT s tA T = ( )( )( )( ) () 00 0 0 0 1 1 t o tt t sts th tsh td AedAe = = ( )() 0 2 1 111 T T o RC o ssAeAeAeT = = Using (7-17) where12oo ss= (ie. Polar signaling) we get 22 11 22 222 0 22 0 (2)() 4 (1) /2 2 (1) oo e oo b ss PQQ Ae Q NT E Qe N = = = 2 12bbb EEEA T= For (b), plot: 22 0 (1) 2 b e Ee PQ N = For(c), plot (7-216) ,MF： () 0 2/ b PeQEN= 7-15Equally likely polar signaling is used in a baseband communication system. Gaussian noise having a PSD ofN0/2 W/Hz plus a polar signal with a peak level ofAvolts is present at thereceiverinput.Thereceiverusesa matched-filter circuit having a voltage gain of 1,000. (a) Find the expression forPeas a function ofA,N0,T, andVT, whereR=1/Tis the bit rate andVTis the threshold level. (b) PlotPeas a function ofVTfor the case ofA=810 3 V,N0/2=410 9 W/Hz,and R=1200 bits/sec. Solution:Solution:Solution:Solution: (a) Using (7-15) and12oo ss= 11 11 22 ToTo e oo VSVs PQQ + =+ 0 1 10001000 T o AdtsAT= ( ) () ( )() 2 2 622 1000 2 1000 10 T j a jfT a a T HTse TsfT e HT sfT = = = ( ) ( ) 2 62 2 0 0 262 sin 10 1 2 0 0 11 22 eq fT Tdf Hdf fT R T B T H T = = ( ) 2 62 0 2 0 0 6 1 0210 22 10 2 oeq N HBN N T T T = 32 6 0 32 6 0 (10)1 210/2 (10)1 210/2 T e T ATV PQ N T ATV Q N T = + + 7-35 An AM transmitter is modulated 40% by a sine-wave audio test tone. This AM signal is transmitted over an additive white Gaussian noise channel. Evaluate the noise performance of this system and determine by how many decibels this system is inferior to a DSB-SC system. Solution: ( ) 2 2 0.4 0.4sin0.08 2 m m ttm= ForAM, with product detector, use (7-90) () () 2 2 / 0.08 0.074111.3 /1.08 1 out base S N m dB S N m = + For DSB-SC, with product detector, use (7-98) () () /The AM system is 10 inferior by 11.3dB/ out base S N dB S N = 7-39 Compare the performance of AM, DSB-SCandSSBsystemwhenthe modulatingsignal ( )m t isGaussian random process. Assume that the Gaussian modulation has a zero mean value and a peak value of 1 p V= , where 4 pm V , compare the noise performance of these three system by plotting()() / outbase S NS N . (a) TheAM syetem. (b) The DSB-SC syetem. (C) The SSB syetem. Solution: (a) 22 141/41/16 pmmm Vm= = () () 2 2 / 1/16 0.058812.3 /1 1/16 1 out base S N m dB S N m = + + 12.3dB worse than baseband and 2 T BB= (b) () () /Equivalent to baseband noise performance, 1 but uses the BW()/2 out b se T a BB S N S N= = (c) () () /Equivalent to baseband noise performance, 1 but uses the BW()/ out bas T e B N NB S S= = 7-2 Using Eq.(7-8),show that the optimum threshold level for the case of antipodal signaling with additive white gaussian noise is () () 2 2 11 ln 2 o T o P s sent V sP s sent = Here the receiver filter has an output with a variance of 2 o .1o s is the value of the sampled binary 1 signal at the filter output. ()() 12 and P s sentP s sent are the probabilities of transmitting a binary 1 and a binary 0, respectively. Solution:The conditional pdf are: ()2 2 1 2 1 2 1 (|) 2 ooo rs o o f rse = ()2 2 1 2 212 2 1 (|) 2 ooo rs ooo o ssf rse + = = ()() ()() ()() () () () () 22 22 11 1122 1122 1122 22 12 22 (|)(|) (|)1(|) (|)(|)0 11 22 T T TT TooToo V eoooo V VV oooo e TT T VsVs oo PP s sentf rs drP s sentf rs dr P s sentf rs drP s sentf rs dr dP P s sent f VsP s sent f Vs dV P s senteP s sente + =+ =+ = = () () () () ()()() () ()()() () () () () () 22 22 11 22 12 22 112 2 1 22 112 2 1 2 1 2 1 2 2 11 exp 2 ln 2 4 ln 2 ln 2 TooToo VsVs ToTo o ToTo o To o o T o P s sent eP s sent e VsVsP s sent P s sent VsVsP s sent P s sent P s sentV s P s sent P s sent V sP s sent + = + = + = = = 7-5Abaseband digital communication system usesunipolarsignaling(rectangularpulse shape) with matched filter detection. The data rate is 9,600/secRbits= . (a) Find an expression for bit error rate (BER), Pe,as an function of(S/N)in. (S/N)inis the signal-to-noise power ratio at the receiver input where the noise is measured in a bandwidth corresponding to the equivalent bandwith of the matched filter. Hint: First find an expression of Eb/N0in terms of (S/N)in. (b) Plot Pevs. (S/N)inin dB units on a log scale over a rang of (S/N)infrom 0 to 15 dB. Solution: () min 0 Q7.24 b e E Pb N = 0 0 0 / 1 2 2 /12 2 bbb ineq eq eq in b iinn ETE RS NN B NB B SRS RNRN ES NN = = = min 0 1 QQ 2 b e in ES P NN = ( ) ( ) ( ) ( ) ( ) 0 2 * 0 2 0 2 02 0 2 2 / 2 sin2 2 2 sin sin 2222 , /2 s 20 in b j t b j T b b b b b b q b b e b b b Hfdf B KSf e HftT N fT S fTe fTK Tdf NfT KT N fT x df dx fT f R T T H x T = = = = = = = = 7-13 for bipolar signaling, the discussion leading up to Eq.(7-28) indicated that the optimumthresholdatthereceiveris 2 ln2 2 o Topt A V A =+ (a) Prove that this is the optimum threshold values. (b) Show that A/2 approximates the optimum threshold if 3 10 e P Solution:见课件 7-14Forunipolarbasebandsignalingas described by Eq.(7-23), (a) Find the matched-filter frequency response and show how the filtering operation can be implemented by using an integrate-and-dump filter. (b) Show that the equivalent bandwidth of the matched filter is ()1/ 2/2 eq BTR= . Solution: (a) ( ) () ( )() 2 2 22 2 T j a jfT a a T HTse TsfT e HT sfT = = = (b) ( ) ( ) 2 2 2 0 0 22 sin 0 11 222 eq fT Tdf Hdf fT T R T H T B = = 7-23ABPSK signal is given by ()( )si1n,0 ccp s tAttT=+ The binary data are represent by ()1 , where (+1)is used to transmit a binary 1 and (-1) is used to transmit a binary 0. p is the phase modulation indexas defined by Eq.(5-47). (a)For /2 p = , show that is BPSK signal becomes the BPSK signal as described by Eq.(7-34). (b)For 0/2 p , show that a discrete carrier term is present in addition to the BPSK signal as described by Eq.(7-34). Solution: (a) sin/20sent1 ( ) sin/20sent 0