Homework 4. Solution. a) Sketch the spectrum of m(t). b) Sketch the spectrum of the DSB-SC signal m(t)cos10,000πt.

Size: px

Start display at page:

Download "Homework 4. Solution. a) Sketch the spectrum of m(t). b) Sketch the spectrum of the DSB-SC signal m(t)cos10,000πt."

Suzan Tyler
5 years ago
Views:

1 CALIFORNIA STATE UNIVERSITY, BAKERSFIELD (CSUB) DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING & COMPUTER SCIENCE ECE 423: DIGITAL COMMUNICATIONS Homework 4 Solution QUESTION 1:(25 POINTS) You are given the baseband signals (i) m(t) = cos1000πt; (ii) m(t) = 2cos1000πt +sin2000πt; (iii) m(t) = cos 1000πt cos 3000πt; (iv) m(t) = exp( 10 t ). For each one, do the following. a) Sketch the spectrum of m(t). b) Sketch the spectrum of the DSB-SC signal m(t)cos10,000πt. c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra. Answer: Chapter 4 (i) m (t) = COS W m t = cos 21rf m t = cos 10007rt -+ f m = 500Hz. M(f) = 0.M(f - 500) + 0.5o(f + 500). (i) M(j) Modulated signal spectrum 1/2 USB \ LSB l/4 LSB \ USB /(Hz) /(khz) (ii) Magnitude IM(j)I Modulated magnitude spectrum 1 1/2 USB LSB 1/2 l/4 LSB USB /(khz) Modulated signal spectrum,-114 USB LSB LSB USB 1/8 (khz) (ii) Phase: L.M(f) 7r /2 ) LS(!! rr/2 ' /(Hz) f (khz) -rr/2 ) -'Tr /2 -- C)

2 , USB LSB LSB USB 1/ (khz) (ii) Phase: L.M(f) 7r /2 ) /(Hz) rr/2-4.5 LS(!! rr/2 ) -'Tr /2 -- ' f (khz) C) m (t) = 2 cos Wm,1t + sin Wm,2t = 2cos 21r J m,1t + sin 21r f m,2t = 2cos 10001rt + sin 20001rt (ii) _. M(f) = o(f ) + o(f ) - o.5jo(f - 500) + o.5jo(f + 500) IM(J)I = o(f ) + o(f ) + o.m(f - 500) + o.m(f + 500) 50 2

3 -7r/2, f = 500 L_M(f) = { 7r /2, f = , else (iii) f m,i = looohz f m,2 = 2000Hz m(t) = cosw m,1t cosw m,2t = cos 10001rt cos30001rt =! (cos21rf m.1t + cos21rf m.2t) =! (cos20001rt + cos 40001rt)--+ (iv) Since m (t) = e-lojtl, we have 20 F(m(t)) = M(f) = r 2 f 2 /(Hz).J JO Spectrum of e co ( TI.) 0 f (Hz) 51 3

4 (b) Sketch the spectrum of the DSB SC signal m( t)cos10000πt. (c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra. 2. (4.2 7)Two signals m QUESTION 2:(25 POINTS) 1 (t) and m 2 (t), both band limited to 5000 Hz, are to be transmitted simultaneously over a channel by the multiplexing scheme shown in Fig. 1. The signal at point b Two signalsis mthe 1 (t) multiplexed and m 2 (t), signal, both which band-limited now modulates toa carrier 5000 Hz, of frequency are to be 20,000 transmitted Hz. The modulated simultaneously over asignal channel at point byc the is transmitted multiplexing over a scheme channel. shown in the following figure. The signal at point b is the (a) multiplexed Sketch signal spectra signal, at which points a, now b, and modulates c. a carrier of frequency 20,000 Hz. The (b) What must be the bandwidth of the distortionless channel? modulated signal at point c is transmitted over a channel. (c) Design a receiver to recover signals m 1 (t) and m 2 (t) from the modulated signal at point c. M 1 (f) m 1 (t) M 2 (f) 5000 f m 2 (t) a Σ b c f 2cos(40000πt) 2cos(20000πt) a) Sketch signal spectra at points a, b, and c. Fig. 1 b) What must be the bandwidth of the distortionless channel? c) Design a receiver to recover signals m 1 (t) and m 2 (t) from the modulated signal at point c. Answer: (b) From the spectrum at point c, it is clear that the channel bandwidth must be at least Hz (from 5000 Hz to Hz). 4

5 (c) The following figure shows the receiver diagram to recover both m 1 (t) and m 2 (t) from the modulated signal at point c. 5

6 QUESTION 3:(25 POINTS) An amateur QUESTION audio 3:(25 scrambler/descrambler POINTS) pair is shown in the following figure. An amateur EE456 Digital audiocommunications scrambler/descrambler (Fall 2015) pair is shown in the following figure. ASSIGNMENT M( f ) A mt () LPF 0-20 khz xt () 2cos( ω0t) 10 khz 0 f 10 khz 2cos(40,000 πt) zt () LPF 0-10 khz yt () Figure 4: An audio scrambler considered in Question 6. a) Graphically find and show the spectra of signals x(t), y(t), and z(t) when ω 0 = 20,000π. (b) Graphically find and show the spectra of signals y(t) andz(t) whenω 0 =30, 000π. b) Graphically (b) Show findwhether and show or not theyou spectra can descramble of signalsz(t) y(t), toand recover z(t) m(t). when ω 0 = 30,000π. a) Graphically find and show the spectra of signals x(t), y(t), and z(t) when ω 0 = 20,000π. b) Graphically find and show the spectra of signals y(t), and z(t) when ω 0 = 30,000π. c) Show whether or not you can descramble z(t) to recover m(t). c) Show whether or not you can descramble z(t) to recover m(t). Answer: (a) X(/) ' _,/ /(khz) Y(/1 / o / /(lt z) f',, I ' /(khz) University of Saskatchewan Page 3 3 6

7 (b) (c) We are able to recover m(t) from z(t) in part (a) as shown in the following figure. But we are not able to recover m(t) from z(t) in part (b). 7

8 QUESTION 4(25 POINTS) A modulating signal m(t) is given by: a) m(t) = cos100πt + 2cos300πt b) m(t) = sin100πt sin500πt In each case: i. Sketch the spectrum of m(t). ii. Find and sketch the spectrum of the DSB-SC signal 2m(t)cos1000πt. iii. From the spectrum obtained in part (ii), suppress the LSB spectrum to obtain the USB spectrum. iv. Knowing the USB spectrum in part (ii), write the expression φ USB (t) for the USB signal. v. Repeat part (iii) and (iv) to obtain the LSB signal φ LSB (t). Answer: To generate a DSB-SC signal from m(t), we multiply m(t) by cos(ω c t). However, to generate the signals of the same relative magnitude, it is convenient to multiply m(t) by 2cos(ω c t). This also avoids the nuisance of the fractions 1/2, and yields the DSB-SC spectrum M(ω ω c ) + M(ω + ω c ) We suppress the USB spectrum (above We and below ω c ) to obtain the LSB spectrum. Similarly, to obtain the USB spectrum, we suppress the LSB spectrum (between ω c and ω c ) from the DSB-SC spectrum. The following figures show the three cases. 8

9 I 1 f fi I I -ISO -SO so 150 f J_tJTL.. -4SO -JSO o I I f IP-1/1 J -6SO -5SO JSO -4SO 5SO 6SO j O>,-(Jl l_j J_l -6SO -5SO -4SO -JSO O JSO 4SO f (a) We can express <P LsB (t) = 2 cos (7001rt) + cos (9001rt) and <l> usb (t) = cos (I I001rt) + 2 cos ( 13001rt). (b) We can express: <i>lsb (t) =! [cos (4007Tt) + cos (6007Tt)] and <l> usa (t) =! [cos (14001rt) + cos (16007Tt)]. l '" JOO 700 IIOO I O>,,..(/) 1 I r I I, 11 1 I I 1 '.! JOO I O>,.,.(J) I l l I 114.) l '" I -MOO -700 u I a>,,.(j) l I I <hsa (t) 9 = [cos(100,rt) + 2 cos (3007Tt)] cos (looot) + [sin( I001rt) + 2 sin (3001rt)] sin (looot) = cos ( ,r)t + 2cos ( r)t <l>usa (t), = [cos (1001rt) + 2cos (3007Tt)] cos (IOOOt) - (sin (l001rt) + 2sin (3001rt)) sin (looot) = cos ( ,r)t + 2 cos ( ,r)t 66

HW 6 Due: November 9, 4 PM

Name ID3 ECS 332: Principles of Communications 2018/1 HW 6 Due: November 9, 4 PM Lecturer: Prapun Suksompong, Ph.D. Instructions (a) This assignment has 10 pages. (b) (1 pt) Work and write your answers