EECE 340 Homework, Spring 2002

American University of Beirut · Signals and Systems · 11 pages

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EECE 340- Signals and Systems Homework # 2 Problem # 1 Consider the instantaneous sampling of the following sinusoidal signal: FQ) = 2c0s(@yt) It is desired to produce a DSB-SC signal from /(t) by sampling this signal and then using an ideal BPF of bandwidth equal to 2@,,. The desired carrier frequency of the DSB-SC signal is 500KHz, which is considered much bigger than f,,, Determine the smallest sampling period that can be used to obtain the DSB-SC signal. Problem #2 Determine the spectrum F(«) of the signal given below with > being a constant phase. S(t) = Acos(@@t + #). Problem # 3 Consider the following DSB-SC signal generated by the transmitter of a communication system: s(t) = f(P)cos(a@gft) . The signal f(t) is the information signal with bandwidth equal to J/OKHz. Due to time delay caused by signal propagation between transmitter and receiver, the signal at the receiver input is given by 5(0) = f@)cos(wgt + 4). Assume that ¢ is independent of time (constant). Determine the minimum bandwidth, BW, of the band pass filter that needs to be used in the receiver to pick s,(t). Use the result of Problem # 11 and the correspondence between time domain multiplication and frequency domain convolution to obtain the spectrum of s,(#). Page | of 3 Problem #4 Consider the following signal: t f®O=4 rea — 704 } Let f(t) be present at the input of a LPF of bandwidth B Hz. The LPF output is sampled at the rate of (//T) samples/sec, Assume that the LPF is ideal and the sampling is instantaneous. Let B Hz be also the bandwidth of f(t) defined by the first zero-crossing of the spectrum of f(t) with the frequency axis. Determine the sampling rate that permits the LPF output signal to be retrieved from its sampled version by ideal low-pass filtering. Problem #5 Consider the following signal: t-10° t) = 2 rect . fO : x | Let f(t) be sampled at the rate of 50,000 samples/sec to obtain the discrete time signal f(a). Determine the magnitude of the Fourier transform of f(n); i.e., \F(@)h Problem # 6 Consider the periodicity of the Fourier transform, F(@), of a discrete time signal, ffn), and the frequency shift property of F(a); i.e. f(n)e/*"” <> F(a). Determine all the values of @ that make F(@-@)=F(a). Problem #7 Consider a DSB-SC signal p(t) given as follows: pP(t)= A, f(t)cos(@,t). The signal f(f) is of low-pass type and having no frequency components above BHz. The DSB-SC signal is present at the input of an ideal BPF having a center frequency a, bandwidth equal to 2B Hz and frequency response H(@). The height of the magnitude frequency response of the BPF is 1 and its phase frequency response is equal to zero, (a) Determine the output of the BPF. (b) Draw the block diagram of a system that can be used to demodulate the DSB-SC signal. Page 2 of 3 Problem # 8 Consider the following low-pass signal: fse* ult). The signal /(t) is sampled at a rate equal to //T , where T is the sampling period. Plot the magnitude spectrum of the sampled signal and determine the bandwidth of an ideal LPF that picks the frequency components in f(t) within its half-power bandwidth. Problem # 9 Let a discrete-time signal fn) be given by: f(n)=e°"" u(n)=a" u(n) The signal f(n) is obtained by sampling f(t) =@7 7 u(t )at t=nT. (a) Apply the Discrete-Time Fourier transform; i.e. F(@) = vf(n) el? to fn) n=—00 and determine F(a). (b) Determine and plot the magnitude spectrum of /(n); i.e., |F(a,]| , and say if this is consistent with the plot you obtained in Problem 8. Page 3 of 3 / EECE 34.0 — Homiwork #- 2 ReluJon ( £ onary 4 stems | Frebhm#é1 j : The ccpeetrim a the tnslartaa eres ly Sampled Visseey ff 4) a tren by 4 po Figg= =D F(w-4). Sebo Whe F(a) = LTS (W- Wy)-4- 27 § (id + Udy), “= aT / 73 and Ts e's the period f Sampling f PE Bo) i BPF — ded, =, —Uy 0 Un L LJ a? -— > -— > <—> Ay aw Awd, ™ Wy Th Smallest Sampleig petiad Cn be o blacn ed Fram the Ca cil Sanplig ee, ae Hence, Seance Pe Ager t= o0kKHz, aS deter mined oom fh bon fisure, thu Ts yy Smallest = 8 fx /5 eer x Leo, boax /o* Of cote the. Samp leig ra Zz a Cdn ha Small Hay 7 Saakhe. Fer Kemple, AGO, 15, 6,5 KHe,... tay aber be 14.7, But. LL mush be grecly Han Le}, - Troblem #2 WA The olanding signal es conpeteled : f= Aws(ut +) p is a Cons lait phace, vod Srctten av i = A re, jue ver vat rw= Adt OTe cy) + A AuarS lor) = _ Are 15 (o-03) + AT EP (e040) Fro bhm #3 5.4) = $lt) cos (ut +d) S (os) = oF lux f Bfer(aee b)} le F (w- ~wW,) + LE Posey) Fe) A QD ane Th. uate 2 neti fan BPE node “Zeal? = y, t £ 1% fart aa fp fod N - te. Liohlen # A Ss f Gé) =4.2ut [- ap) ) [Wal | PY «¢ £.Wltded| LY so cara te f Ue) ts 2. (edaug lat ? hae haveng a. Atlalisn. equal GT=4*1o Face. Hence he spetrim ff PEL Ley Fide a Se funn The fuse Zea fas st of Fly wtp pi di ate dsy Et, Me bend of Pte fais ge by Ba -! = £5700 Hz. 4xig7 Hyeece. yp hee first LPF produces a hand hinttad. syyrall Z y hotcng a bordut dl epual é B=.2500 Hz, Now, ade be Ahab potrime $b) from Ls 7 Sampled Vel sre; tt, Z, © fy heal pees . #: / “4? lt apply Hee ln CHIL. ZIAL! Svat plex yea 4B =2x#500 2. 7 oO T 7 a = 4000 samples figec. Fiobhm #5 4/9 Fb) = dpel [ £1077 ] y 5 ply X10 0 JEXlor? > [ _ . - = 50, 000 samples [fee => T=2X/0 Ppt, a, n=0,| Hen a = aa ro. 1° BLardshee FmM= > ple?’ = FO +pye? N=- 00 ' — gor a APéinatively, the folladeag can be wWre'Hen: Fie) = ai4e) = J (I+ esl [F-(w)| = 2 [ lit 05 (iT) y Sen ‘an! =2[a(i+ stat] ” = 4 [4o8 27 wT le = 25 = 4 | oot 5/3 Fo hbem te 6 . F @& = > fw 0 is pertiada wth petisd Zt, N=-00 r Hence, any frpeny Shift of F (wy by an algal miltiple. xf ts petdod; . ey by LRT ww hue R=0+| +00 prenoup E(w). qT — Thelefse,, F (0-04) =F (ed) of t= a ushete faotl t+2,t3--- Mee beig tay bled be din av pls: F (w-4) = > fae Er" As — 02 2k py —yleonT” n. = foe EX Az=-wo =| per k=42)44 =f (id). [roblem #7 pl)= A. Pb) 6 (ub). DS8-Se, Q) le cleat BPE hat Phe. Same bardevi'dify (2B) and Qutlir regione (4)ac the DsB— Se. Lyra perio, Wee | Hl] = lar LHle). = 0 wifes Me BPE bead — will, Hey the BPE output tis the. Same ob He. DSB-Se. Wy Aas within the LB bandurclh, we car wile, Bld) = Plo) Hl) {LHe = Pla) [Hey € = Pl) x |x? = Pl) phe BPE ontpet BO =pO=A.f Beer (ust) (b) ee cos (ett) pli}cos alt) = A, fH tof (ut) =tA LO + tA. fb Ut) A LPF of bandenth BHz peeks LALE), We teed &f.-B > B ov fee >B, Tt pracis, He. cartier frequen ds seunlly much ut than tee bavdinr th af fhe. mescage. egal f) far speed sigsahy fer eanple, Bits chhe-erte f 4kHz, whi he. 23 vk the order gf 1000 Zz KHz qe uc the CHz, a Lew be he. tarelusech fh of fhe. L pp that picks The fi Lene Conprnncs def (t/ wifhin As tab af handure ltt, Wbsaueck Hits Pt pW Hd Wat fper op il