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## Comm and Information Thry

by: Antonina Wuckert

52

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13

# Comm and Information Thry ECE 460

Antonina Wuckert
Mason
GPA 3.94

Bernd-Peter Paris

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COURSE
PROF.
Bernd-Peter Paris
TYPE
Class Notes
PAGES
13
WORDS
KARMA
25 ?

## Popular in ELECTRICAL AND COMPUTER ENGINEERING

This 13 page Class Notes was uploaded by Antonina Wuckert on Monday September 28, 2015. The Class Notes belongs to ECE 460 at George Mason University taught by Bernd-Peter Paris in Fall. Since its upload, it has received 52 views. For similar materials see /class/215015/ece-460-george-mason-university in ELECTRICAL AND COMPUTER ENGINEERING at George Mason University.

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Date Created: 09/28/15
ECE 460 Communication and Information Theory Prof BP Paris Practice Problems Solution Below are solutions to the following problems Chapter 2 Problems 9 34 381 Chapter 3 Problems 19 20 301 Chapter 7 Problems 17 24 25 30 33 11 Problem 29 1 Since a 7 b2 2 0 we have that a2 122 33 with equality if a 121 Let Then substituting aiA for a and for b in the previous inequality we obtain A 2 115139 1 0 12 1 512 if lt if 7 A B i 2 A2 2 B2 with equality if g k or ai k i for all i1 Summing both sides from i l to n we obtain 71 n n 01151 1 042 1 5392 XE S 242 Zi2 i1 i1 i1 1 7 2 1 7 2 1 1 7 7A2 7321 2 2gal2 212 2A2 232 Thus 1 n n i 01151 S 1 01151 S AB Equality holds if ai 165 for i 1111711 i1 2 The second equation is trivial since To see this write 11 and yi in polar coordinates as 11 page 2 an yi pylej 121 Then Vim 1px y 9 9 an7y 11111911 111119211 We turn HOW to prove the rst inequality Let 21 be any complex with real and imaginary components 2133 and 21quot respectively Then7 71 2 71 2 71 2 21 22112 11 11 11 2112211 2112111 2 71 71 2 2112 j Z 21 11 139 1 H Ms 1 H 1 Since 2133sz zm zlJV 2 0 we obtain 21122mR 2112m12 S 21212 21212313 2311 Using this inequality in the previous equation we get 7L 2 71 71 21 Z ZltZiRZmR 2112m1 11 11 7711 71 71 1 1 3 21212 2121 23112 2311i 11 7711 71 71 71 2 1 1 1 ZltZiRZiipgt 2231R2311 gt lt2ltZZRZEI gt 11 7711 11 Thus 71 S i1 n 211 i1 1 1 s 2sz or 11 11 The inequality now follows if we substitute 2i Equality is obtained if fmf kl or 12 sz91 3 From 2 we obtain 71 2 71 11 11 But are real positive numbers so from 1 71 71 1 71 211111 21112 2112 11 11 11 Combining the two inequalities we get 71 1 E 1191 11 1 1 SW 11 2 S E0 From part 1 equality holds if ai k i or lziy ejei Therefore7 the two conditions are lzil klyil Iii Zyi 9 and from part 2 Which imply that for all 239 Il Kyi for some complex constant Ki 3 The same procedure can be used to prove the CauchySchwartz in equality for integrals An easier approach is obtained if one considers the inequality lzt aytl 2 07 for all 1 Then 0 s w W WW w W aylttgtgtltzlttgt wow m warm a zlttgtylttgtdt M m ztytdt W m WMth The inequality is true for ff 1 tytdt 0 Suppose that ff 1 tytdt 0 and se 00 a 7 Leo Wdt ffom1tytdt Then7 2 U5 l1tl2dtl2 ff lWWdt 0 S 1 W 4 and zlttgtylttgtdtl waif Wwf Equality holds if zt 7ayt are for some complex 1 Problem 234 Let yt be the output signal7 Which is the convolution of zt7 and ht7 yt ff h7 zt 7 Td739i Using CauchySchwartz inequality we obtain mm hlt7gtzlttergtdr h7l2dfliMzwi FdTli g 7lxtirl2d7 Squaring the previous inequality and integrating from 700 to 00 we obtain lytl2dt Eh zt7r2drdt But by assumption ff ff lzt 7 Tl2d7dt Eh are finite7 so that the energy of the output signal is nite Consider the LTl system with impulse response ht 227 Ht7 Zn The signal is periodic with period T 2 and the power content of the signal is PH If the input to this system is the energy type signal zt llt7 then 00 m 7 Z w 7 2n n7oo which is a power type signal with power content PS 1 Problem 238 1 111 Z 71 znT56t7nT5zt Z 71 6t7nT5 11 i 6t72lT57 i mining l7oo l7oo Thus X1f 7 Xf21T Z kiwi 6f7 2936121le 5 1700 1 z 1 z y 1 7 7 7 7 7 7 712W 5T5 2T Z X 2T5 2T5 EGOXU 2T5e 2T 5 1700 7 1 z 1 l l 7 2T 20X f 7 27 7 27 2mm 7 2T5gtlt71gt 1 1 z W175i 2 The spectrum of zt occupies the frequency band 7W7 Suppose that from the periodic spectrum X1 we isolate Xk TisX f 7 2i 7 T357 with a bandpass filter7 and we use it to reconstruct Since occupies the frequency band 216W7 21H71W7 then for all k cannot cover the whole interval 7W Thus at the output of the reconstruction lter there will exist frequency components which are not present in the g 5 input spectrumi Hence the reconstruction lter has to be a timevarying lterl To see this in the time domain note that the original spectrum has been shifted by f 2711 In order to bring the spectrum back to the origin and reconstruct zt the sampled signal 11t has to be multiplied 712 271 s 1 e jMW l However the system described by W emu 16 bye is a timevarying systemi 3 Using a timevarying system we can reconstruct zt as follows Use the bandpass lter T5H to extract the component 7 lnvert Xf 7 2 and multiply the resultant signal by e th l Thus f7W 2W W e jwv f l T5H X1f l Problem 3 19 1 m 7 am Mt amt cos27rf0t bmt cos27rfot2 amt W2 t acos27rf0t 12 cos2 27rf0t 2bmt cos27rf0t 2 The lter should reject the low frequency components the terms of double frequency and pass only the signal with spectrum centered at for Thus the lter should be a BPF with center frequency f0 and bandwidth W such that f0 7 WM gt f0 7 g gt ZWM where WM is the bandwidth of the message signal 3 The AM output signal can be written as 212 ut al cos27rf0t a Since Am we conclude that the modulation index is ZbAm a 7 a Problem 320 1 When USSB is employed the bandwidth of the modulated signal is the same with the bandwidth of the message signall Hence WUSSB W 104 HZ 2 When DSB is used7 then the bandwidth of the transmitted signal is twice the bandwidth of the message signali Thus7 WDSB 2W2 gtlt104 Hz 3 If conventional AM is employed7 then WAM2W2gtlt104HZ 4 Using Carson s rule7 the effective bandwidth of the FM modulated signal is BC 261W 2 1 W 2kf W 140000 Hz 6 Problem 330 1 The instantaneous frequency is given by ii 2 dt A plot of is given in the next gure 1t 1W fa 4250 fa 100mt i i i i ifc 52 i i Ufa 7 l0 t 2 The peak frequency deviation is given by Afmax kfmaxmmw 5 7 Problem 717 1 The output of the integrator is yt Alana Atsi7n7d7 2de new 6 At time t T we have MT ATWW OT WW 7 i T AT new The signal energy at the output of the integrator at t T is 2 5b 55 7 i TT 7 ng whereas the noise power Pn EOT0Tn739nvd7dv T T En7 nvd739dv 0 0 N0 T T N0 7 0 0 67397 vd7 dv 7 T Hence7 the output SNR is 55 7 251 SNRE7VO 2 The transfer function of the RC lter is 1 WWW Thus the impulse response of the lter is 1 t ht e u1t and the output signal is given by 1 H yt E 00 TT 7Wd739 z si7n7e tR5dT 1 70 T 1 t T e Si739 d739 ei n7 e d7 0 700 At time t T we obtain 1 L T L 1 L T L yT e RC 0 SiT RCdT RC nTeRCdT 00 The signal energy at the output of the lter is 55 T T ei siTsZvelce icd7dv 0 0 T 2 T gen The noise power at the output of the lter is Pn Hence7 RlTVE En7nvld7dv 1 E T T N0 74 We RC Dofoo 7677ve RC dev T ied7 1 7E N iltRCgt2 RC 00 2 1 2T N0 2T 7 2R0 amp 2RC2 SNR4I2 lt1767gt2 3 The value of RC that maximizes SNR7 can be found by setting the partial derivative of SNR with respect to RC equal to zero a RC7 then Thus7 if T T T 19SNR 2 77 7 T Vioi ie ige Life lt1Egtl Solving this transcendental equation numerically for a we obtain I71 26gtRCa7 T a 126 i Problem 7 24 The energy of the two signals 81t and 82t is 81 AQT The dimensionality of the signal space is one and by choosing the basis function as g H 1 i 0 S t lt lttgt j I xT 2 i i we nd the vector representation of the signals as 313 iAT n With n a zeromean Gaussian random variable of variance The prob ability of error for antipodal signals is given by where 55 A2T1 Hence 2E1 2A2T P 7 e Q 1 No Q N0 9 Problem 725 The three symbols A 0 and 7A are used With equal probability Hence A the optimal detector uses two thresholds Which are 3 an 7 an bases its decisions on the criterion A A z T gt i 2 A A 0 if lt T lt i 2 2 A 7A z T lt 77 2 If the variance of the AWG noise is 072 then the average probability of error is A 2 P 1 1 ifvwzd 1 1 1 737d 6 7 76 2 T7 7 76 2 T 3Do x27T0392 3 x27T0392 1 A 1 A 1 A EQ lal 1 l2a l Q l2a l 4 A EQ 10 Problem 730 The vector r T1 T2 at the output of the integrators is r mm 0134de mm If 31 t is transmitted7 then Erma A1395Sltntdt1A1395ntdt 1n1 amt 12sltntdtA2ntdt Where n1 is a zeromean Gaussian random variable With variance 15 15 NO 15 021 E n7 nvd7 dv d7 15 0 0 0 and n2 is is a zeromean Gaussian random variable With variance 02 E1212n7nvd7dv 12d71 Thus the vector representation of the received signal at the output of the integrators is r 1 n1 n2 Similarly we nd that if 82t is transmitted7 then r 05 n11 n2 Suppose now that the detector bases its decisions on the rule 31 7 1 7 7 2 2 T 82 The probability of error Pel31 is obtained as Pel81 P7 l 7 7 2 lt Tlsl Pln17n2ltTPn17n2ltT7l PnltT Where the random variable n n1 7 n2 is zeromean Gaussian With vari ance an 021 022 7 2En1n2 15 N ailaiz72 ng 1 15172X0515 10 H Hence 1 Pelsl 7 W e ndz Similarly we nd that P l2 P015 n1 717 n2 gt T Pn1 ing gt T015 12 72 dz 1 oo 7 e M27702 T05 The average probability of error is 1 1 ma Pelsl Pels2 12 Wdz 1 T71 72 1 oo 7 e on dz 7 e 2x27T0392 foo 2x27T0392 T05 To nd the value of T that minimizes the probability of error we set the derivative of Pe with respect to T equal to zero Using the Leibnitz rule for the differentiation of de nite integrals we obtain 19Pe 1 71 71 7 e 0n 7 e 0n 19T 2x27T0392 T 71 T 0152 j T 0125 Thus the optimal decision rule is new 2 025 l Problem 733 1 The dimensionality of the signal space is two An orthonormal basis set for the signal space is formed by the signals w1lttgt 03 w2lttgt 9 0 0 otherwise otherwise 2 The optimal receiver is shown in the next gure 11 Select the largest 3 Assuming that the signal 81t is transmitted7 the received vector at the output of the samplers is A2T r T n17n2l N where n1 n2 are zero mean Gaussian random variables with variance 7 The probability of error Pel31 is Pel31 Pn727n1gt 12 ZNOdI Q 7 AQT 2N0 where we have used the fact the n n2 7 n1 is a zeromean Gaussian random variable with variance Nor Similarly we nd that Pel31 Q 4 7 so that 136 Pel31Pels2 Q 1 0 x27TN0 L2T e AQT 2N0 4 The signal waveform 7 t matched to 1t is exactly the same with the signal waveform 2 T 7 t matched to 2 That is7 17t 2T7t 1t 0 ogtlt otherwise Thus7 the optimal receiver can be implemented by using just one lter followed by a sampler which samples the output of the matched lter at t 3 and t T to produce the random variables 7 1 and 7 2 respectively 5 If the signal 31 t is transmitted7 then the received signal Tt is 1 T W sllttgt isle 7 5 W 12 The output of the sampler at t g and t T is given by 1 gage mm HEW T4 2 T4 2 8 A 2T 1 AQT T2 5 Tlm V77m If the optimal receiver uses a threshold V to base its decisions that is 81 7 1 7 7 2 2 V 52 then the probability of error P l81 is P l81 Pn2 inl gt 24 g 7 V Q If 32 t is transmitted then 2 AZTi V 8N0 MNO Mt 320 5 7 g W The output of the sampler at t g and t T is given by 7 1 n1 T2 A zd E Ier T4 2 T4 7 5 E 7 The probability of error P l82 is 5 A2T P l82 Pn17n2 gt E TV Q Thus the average probability of error is given by E L 2 8N0 We 5 A2T V 2 8N0 We 1 1 W Plte181gt Plte152gt 2 L 8N0 N0 13 1 1 EQ Q

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