350 ECE Signals and Transforms Spring 2009 Homework #11
The following problems are to be completed and turned in at the start of class on Friday 17 April 2009. Problems: 1) We have studied Time-Domain multiplexing, and drawn the components multiple times in lecture and lab. If an AM communication system is being used with three input signals, x1(t), x2(t), and x3(t) draw from the start, these three signals, to the end, the received and demodulated signals all the components and the contents of each `wire'. Assume each input signal is band limited with a bandwidth of 1000 rad/sec. The three carrier frequencies should be 1 = 20,000, 2 = 25,000 and 3 = 30,000. Draw and label the modulation stage, the multiplexing stage, the transmission and receiving stage, the demultiplexing stage and the demodulation stage and the final received signals y1(t), y2(t) and y3(t). Design your system to use synchronous demodulation.
2)
Text 8.6 (a) Draw a block diagram showing how g(t) is formed from x(t) (first equation) (b) Draw a block diagram showing how the reconstructed x(t) is formed from g(t) (second equation) (c) Sketch the frequency component of all signals on these block diagrams and determine the value of A such that the reconstructed x(t) is equal to the input x(t). Determine and sketch the values of X(j), G(j), and X2(j) (the x(t) after attempted reconstruction/demodulation).
3)
Text 8.12 (a) complete problem in as book. (b) draw the block diagram that takes the original signals xi(t) and produces the time division multiplexed signal y(t). (c) draw the block diagram that takes the time division multiplexed signal y(t) and reproduces the original signals xi(t).
4)
Text 8.26 (a) complete problem as in book. In addition to the diagrams you need to solve the problem in (a), also list the time domain signal equation for several of the contents of the wires as that will help you solve the problem. (b) draw the block diagram that takes the original signals x(t) and produces the modulated signal y(t).
Totally in the time domain: [x(t)+A].cos(ct+c) Demodulation: Top path: [x(t)+A].cos(ct+c).cos(ct )= [x(t)+A].[ cos(ct+c-ct) + cos(ct+c+ct)] = [x(t)+A].[ cos(c) + cos(2ct+c)] Bottom path: [x(t)+A].cos(ct+c).sin(ct )= [x(t)+A].[ sin(ct+c-ct) + sin(ct+c+ct)] [x...