Questions with Resolution of Communication - Final Exam | ECET 30400, Exams of Electrical and Electronics Engineering

Download Questions with Resolution of Communication - Final Exam | ECET 30400 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! Name:, ECET 304 Final - 200 pts BF s Se ie Put final ans: in the a: You are allowed 3-3°x5" note cards (both sides) or 1-874°x11" (one side) for anything you can fit on it. Attach the card to your test when you turn it in. Table of Besse! Function Factors Index_| Carrier Side Frequency Pairs a | | 3 | | YY | CO CY, 0.00. 1.00 0.25 0.98 0.12 0.5 0.94 0.24 0,03 - 1.0 0.77 0.44 0.11 0.02 15 0.51 0.56 0.23 0.05, 0.01 7 2.0 0.22 0.58 0.35 0.13 0.03 24 9 0.52 0,43 9,20 0.06 0.02 - yt d(x" )=m""'dx fran n#-l d(e)=e'dx fear=e* (sin x) = cosx-dx foinx-dx=—cosx d(cosx) =—sinx-dx foosx-dr=sin x 2.5 0.05 05 OAS 0.22 0.07 0.02 Fourfer Series Summary Sine-Cosine Form: att) A, +E (4, onlay) 8, anf). = Ia f= z Cosficient | 4 oF ff a0 costae} B= 2 ff xip-sin(nay)ar Table of Fourier Series ‘Signal x0) Fourier series Ll = Full-wave rectified cosine VYVYYY. (ra Zeosay- 2 cosroy + 2c t+. a 1s He 5g OSI 3 Y : Vv 1 i 1 7 {cos ~ zoessaye + poossat — zoos Toe +) 8. (cosa + conden + 2 cossey+ x a) u" 25 oe “Lie | Sawtooth Wave {si ar 1 gin dag + sind - dein Aas... 2 3 4 Page i of 8 O atl emer “=a Spring 2005 Name: dB =10log' Fa am = 10og( 2 K =1.38x10™ J/K Boltzmann's constant P, Im N=KTB Tegan +e +e +e? %THD = (B85 40h oanD= "x10 ra & S/N, - - S.N,= AS, py LN nt F-R+ 1 Ari, Ae-t AN, +N, "SN pu 4 4A, 4A, 87 5.986h NE, = S/N ip, (4B) — SIN, aB Zaz eh| ——— IN sy (4B) S1N rg (28) ° ve, +141 (2) Basic a(t) =Q, = A -10R Q, = 2, = i =0,-R, Xp =—l_ sis ae é "2a 7 taf X, High Pass Configuration: Low Pass Configuration: xs xs Reo R, Rs IH Ry x, x, By oN mma By Ryo Ry xp x, (" l x, Matching Complex Loads — match 2 complex load to a resistive source 1. Absorption — absorb reactance into the matching network. 2. Resonance - compensate for stray reactance with an equal and opposite reactance at the resonant frequency. Low-O and wideband Matching networks: R= RR, o- } OR o- (42)-1 ‘small ¥ High Q and narrow bandwidth matching networks T-Network: ‘Start with the source-side then work the load-side ‘flipped’ R,=R,(Q@' +1) > O= ( x-Network: ‘Start with the load-side then work the source-side ‘flipped’ Page 2 of 8 ST ee Spring 2005 Name: ee) 8. For the AM DSBEC, calculate the following: Box in answers. (9 pts) a Bed (254195) =[13,275y * 175 ¥, & Reg (US- (IS)=Hl LAS {62S -[eoa27 \ . Mt TFE76 1 SOA2R x 9. For the AM DSBFC envelope with M=77.2%, determine the total power into a 75Q load. (10 pts). M k > m ~% Liege 1 Je $ Ate, e = ae Z V1S+2,28) iw bln” © we es > 60 cone, ae \ pe wso( 1) = Sat pe +844 w 10. From the given list of 7 items, write the correct answer to the following criteria in the blank provided (answers may be used more than once) (10 pts) | ASK ESK BPSK QPSK QAM None ll a. Digital modulation using amplitude only;__ ASI b. Digital modulation using sroplitude and phase:_ PAM, li] c. Digital modulation using four phases only: OPS ie L 4. Uses a tibit or quadbit to set the digital content:_ CpAM 0 @ e. Also known as IQ modulation: ALL 1. Determine the baud sate of a FSK signal with a f,=50kHz, {7-41 and a bit rate of 8kbps? (4pts) a. 12kbaud c. Adkbaud Uck-SdL) + 2(ge) = UtIb Yee b. 20kbaud 4 112kbaud board. © A 20% bad 12, Calculate the min, bandwidth for a BPSK modulator with f(=150MHz and input bit rate of 250kbps. (4pts) a, 125kHz ¢. 75MHz Zeotbeps = BO. 25016 Ch. 250KHE> . 1S0MHz ! 13. How many symbols are shown in the QAM constellation diagram shown to the right? (4pts) ° a 2 c 6 Meesnreseneee Page 5 of 8 ee _ oN 14, AFM short wave receiver uses fip=5.5MHz, has a preselector with a Q=55 and is tuned to 43.5MHz. i) Using high-side injection, determine: (10pts) @ a. The image frequency of the receiver. Fimege © fue *2he =4ESmt late Somage™ Su. Sake b, The value of the IFRR. Tree = liege =Yiigsrluwy® « 25,00 Spring 2005 Name: WS gs =| € > aE 7 es ZS7A 9487 YSU IFRR= ZS ' 15, The following system has S/N j.,, = 504B, Nia =1* 107" W , and a 50Q load, determine: (10 pts) Amp tp 312, S/N, —> Ct | [> 5/N, Ny =98¥ © Sty (4) whe 250d@ ® Ib ”°=le0,c00 OQ Biv lee coo inj W) = OOM W Sya=_ 100 Aw 4. SIN gay, (4B) e _ beSi - guzal teoeW) = 2b 0ce. ox = bedithe Grez.sy( ici) 440w ‘ 4 - 44 dBelo leg Caeee0.9) = 4U.15 SIN sap 41S & NF © SK, @3)- Si (ds) = S0-Hhis =S8S NF~_ $B w 16. Compute the trace width required for a 502 characteristic impedance (Za) using a 30 mil thick, TMMA- rca ee Ne (12 pts) h=3o. 1 (ERY bh in (oe Seve (30) bey vm 5 n - b..: tenn CRW tt | Gon Eerie , i = (S.6mi) {79, So >25.98S In (aes) er" = as OO TEWNS Gswrise) = SS— 2747 )* Cae we (B26 a Ge il -BWwlb.e 2 W=)3.26 — pagesors - . Spring 2005 Name, e@ 17, Match a 302 source to a 502 load at 70MHz. Use a single L-network and low pass configuration. Draw y} the complete circuit schematic. (Opts) “2-1 = SIS % = (pb, 2 S165 BO) # BAAS Le BRET = 85"oh os B.S 2 *P oe” ses © ot Mow B20. — ..65.Mnlt Cc entyoryei24) = BFF 18. Match a 1502 source to a 75Q load with a loaded Q=15 at 125MHz. Use a high pass configuration. Draw the complete circuit schematic. (25pts) . Source lary , load small DN network, hgh pass & ns. ReASon Xs, Xe we. TW. a \*Q Leis® 339 5 head~side Bost: % Bt aS BeBe C) Kg, > @ly= 15 (389 = 498.0 ° : Ry Ke Cia ikgs” maCiesmy(4FE) 7 285: &pF Kee ee ee 1s moe Tee 5 a = = ganna L> (am) nero Souece - SIDE (Fge®)D 7 . Gale) = 2124 “3B Xs,= BQ = SBAL DI) = 1.05. C= satemtte® = [8O.beF Xps Fee E06 *Q Bia = 206 4nk f vo 2n(nsm) BMAn Qe Page 7 of 8