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1 Name: Student ID: ECE 457 FINAL EXAM April 30, 003. No textbooks, notes or HW solutions.. One page of hand-written notes.. Calculators are allowed.. Exam is hours.. To maximize your score on this exam, read the questions carefully and write legibly. For those problems that allow partial credit, show your work clearly.. Good luck. I. [5] Determine whether the following statements are true or false. If true, explain why. If false, either give a counterexample or show it quantitatively. a) [5] The input to an envelope detector is IOcos(0m)cos(8000m) + IOsin(0m)sin(8000m). The output of the envelope detector will be approximately equal to IOcos(0m). FCAI l'ne 0 IJ "'\.1:\ WO,\\ 'oe 1'1 «QJ.. to Jl \ 0 U)S ( C>lH-) ) L l I( =O b) [5] The output SNR for DSB modulation is 40 db. If it is desired to achieve the same SNR for SSB modulated signal, then the transmitted power for SSB modulation should be greater than the transmitted power for DSB modulation. FCA'Se.- e SN R) \)SB = tsl\.h<.)ss IS "\Y"sm\-\t. {)(J\Jjec,s'i-ne me \r\ s.
2 c) [5] A DSB signal can be demodulated using carrier reinsertion followed by an envelope detector and a DC block. (fu.. Ac m loongos\fl,--'-f ) + \.( CDS \ nlc-\) } h\ct ijl -{eqfe (' Acm) +k \DCI A.) -7 c ry1t-r d) [5] A signal with power spectral density Sx(f) = Af,-W :::;f :::;W is transmitted through a baseband communication channel with white noise, Sn(f) = o. The signal plus noise is passed through an ideal lowpass filter with cutoff frequency AW W. The SNR at the output of the filter is. 3N -r(e 0 s,ig('qt power: ) A-Rd.f -= A W -w 3- I\J"0'ISe..pGUJe(: No, w =--:AI \f\1 SNR::-.3 tv-=> e) [5] The bandwidth for the narrowband set) = Ac cos(0,ooojzt + O.Olsin(0JZt))is 0Hz. FM modulated signal ) :J.. AW'- lnae.- :: 0.0\ 'N-= L(?>t- I.--\-PM '=... C.i) (a) '0H.
3 . " / [5] Consider the FM signal s(t) A, cos( nfj + nfj.!m( T)d T). where m( t) is the tone signal met) = Amsin(7ifmt).The following parameters are given: Ac = 0V,ld = 10KHz /Volt,Am = 1V,lm = 8kHz a) [5] Find the peak frequency deviation. b) [5] Find the modulation index fj. c) [8] This signal is transmitted through a channel with additive white noise with power spectral density, S n(i) = No = 10-5 W / Hz. Assume that we use a standard FM receiver without deemphasis filter, what's the SNR in db at the output? d) [7] Suggest a way of increasing the SNR by 5dB. Hint: This will require you to add in a deemphasis filter. Specify the transfer function of the filter. 0-) rj) = -R mt-+) c;t1 cl+ c?c\k -freq -d.q.\.;\cd,dn z 10k t{t. " / b) (3= Am d ::: (t) \ 0 'L\-t ) :::.t. -PM - l %ki{:t,) c) CSrJR. ) M :: 3 \ d\ M) 1-\'\A(\?l+) No PT W :: 3. ( ). (' /;,J lac '-/- ) t ::?- '><\0-5 X )( \Cr ::: q K\o6<1 ' \('\des d) If1creG\f)e SN w'd"0 aleempms\s l'\\e..re+o(e / -me... - A0 Lo'3t. 0 l 'L C\ 0..Gq) -= e.. SNR. bl( d B :A..E;1- d B -7". z.. - PT -d. =-,S \t) l\a,;'l) t-j(;)w q Z-b '-I. l(.?).(!3 =. s-g'k.h'"t. cl..eem()y=?i5.:f;\iex is \ \-\\ \. -=- 16\..('R'. Jlf (1.) w, th 3::: S'i k: H 'L L: 4'3 ),-,
4 ' [5] Consider the modulation scheme given below, where met) is the message signal, net) is white Gaussian noise with power spectral density Sn(I) = No. The input message signal, met) = 4cos(t) + 4cos(7ift) with J; = 1. The filter response for HI (I) is shownin Figure, with the followingspecifications 0, 1 f 1= fe - f 4 ' 1 HI (f) = 1 ' 3 4' If I = fe - fl If I = fe If I = fe + fl 1, 1 f 1= fe + f The modulated signal is transmitted through a channel with additive white noise n(t), with power spectral density No. The received signal r(t) is demodulated as shown in Figure 1. '--" Figure 1: <-- met) {/x ) H, (f) </-" C +) r(t) T Cos(nfcJ) 1 net) r(t) Figure : lm f <7) cos(nfct) H,(f) H3(f) yet) -fc-f -fc+f fc-f fc+f f
5 ' / a) [6] Sketch the spectrum of the signal at the output of HI (I). What type of modulation scheme does this correspond to? b) [4] Statethe cutofffrequenciesof the idealbandpassfilter,h (I), such that it bandlimits the noise and passes the signal undistorted. c) [7]Findthe SNRat the outputof the filterh (I). d) [8]Assumethat H3(I) is an ideallowpass filterwith cutoff frequency 1' Find the SNR at the output of H 3 (I). Q) me-\-\ -;; \.( CDs l:t'i< \ -{-C03(?1\ z t) \r\pw k> 0C(') -LL-L \ --Je.. c.. r \ -.k:+sz.. LL-r-l--l-- rc -4 L l'c.j' I Cc-.('c...f:, -(.'c.:t-+z Ou- -\-pu-r- \{l ) : ' ' ", 3/l{ Sc-.t-t- --kc-.t\ -j::c. -tl( -.tc'si \ 3/t{ \..<.{.Q. -k-s I.(' c -\\:- +.,C: ( 'c...-t+ L. V s: IS b) t--\ 1.C{:). 1 i'c- \ \ S"c++z.. 0, l j c -\\. Lt\ \ 0 o.w. d) c.) S; oral \)()oj«o.."te o""'i>u.\ at \\...l+): I" ::',15,,)..,.5 : l-kf+ (%)1.+ i ) w N>1- I'1D\Oe. 1>D(J.JeI'.. ) U:L-t s\) = 10", l-e, +) :3 'rj z. I SN I 3 N \'),f.. Chj.u.\ cn,. 3 \ ): l \ \ \ s 13I"'Ol c l.j,jef l\ -l- -J: I $, +-l..c
6 Extra Sheet for Question 3:./ t-j '\. pwe.r 80X'f'e us bot\ c\::. h 0 \'i.e wer 3NL - SN-.: y- 3Nz. '- -= L/,/ or. - - tj.{?l
7 '--" 4. [5] A sinusoidal message signal whose frequency is less than 1000 Hz, modulates the carrier 10-3 cos(7ifet). The modulation scheme is conventional AM and the modulation index is 0.5. The channel noise is additive white with power spectral density of No = 10-1W / Hz. At the receiver, the signal is processed as shown in the figure. H(f) is a noise-limitingfilter centered at fe with bandwidth 3000 Hz as shown in the figure below. Received Signa1 Hit) I 1>1 X LPF I BW=1000H1 - cos(n(j) H(f) '--" 0.5 -fc fc -fc1500 fc f " " G\) a) [8] Find the signal power and the noise power at the output of the noise-limiting filter, H(f). b) [5] Compute the SNR in db at the output of the noise-limiting filter, H(!). c) [6] Compute the SNR in db at the output. d) [6] Compute the detection gain. How does this compare to the AM receiver that uses an ideal bandpass noise-limiting filter, like the system we discussed in class? Si9ra1 pd"'-jec Ac... [, N \:),S/L p u.:er: +Cl?!\' l-t\j Ol 1 -TW po-6) r\ + (O,C:;Y-l}z.) J :: 5.6- X\0 "hil IJ'I" \ // CL--.:. ( r'h (.-. u,/\-- N \) \ " 00) + Nu \. 000) ). -=:: '<)0 W't) 9" z, -q w -;;;: usx' \0./?
8 "-../ Extra Sheet for Question 4: b)snq {D \()Ha S',6Z '1<10- t ctq..5x 10 -q ::.? 0, '1{, d B Z\ d g. c) 8 i9na..l CJ\JJ a..."\ -ItuL 0 u..\"))<.lt. Ac? a ' "l( -\-) -::: {, zs 'A\ 0 -=1'vV f\j01 \\Uex cd 0 Ll.A. l'j') No;) I tj/j \'5uO --:r J '---' Gtoao) N.. -::. It )(10-9 W -T Sf\\'R-: \D\og.-\Q \,'')lu ::: n.\,'15db\<)db U Xll)-q ) Deh ()ui() (S;NR )pu's.t C SN) p 3 \. l 5"' -:: O. ::> I.;.Ll.T I(\ it\e... s to..'(\dor <i ArJv \-e.c ver : uek }y'",jr c"){"\ = f).. = 6t (0.')) - lo S) \ + (O" \ 1\.0.':) a.'". "-../
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