sec2-student-solns

# sec2-student-solns - ECE 2200 Signals and Information –...

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Unformatted text preview: ECE 2200 Signals and Information – Section 2 Solutions (Week 3 Feb 8-11)Page 1 1. Problem P-2.21 in text (page 34). Solution: (i) The time delays are the distances traveled by each signal divided by the speed of light ( c = 3 × 10 8 m/s): vextendsingle vextendsingle vextendsingle t 1 ( x ) = d 1 ( x ) c = √ x 2 + 10 6 3 × 10 8 sec vextendsingle vextendsingle vextendsingle t 2 ( x ) = d 2 ( x ) c = radicalbig ( x − 55) 2 + 10 6 + 55 3 × 10 8 sec (ii) The transmitted signal is s ( t ) = cos(300 × 10 6 πt ). If we consider the reflected signal as a function of both time and position, then vextendsingle vextendsingle vextendsingle r ( t, x ) = s ( t − t 1 ( x )) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright direct + s ( t − t 2 ( x )) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright reflection . = 1 . 3888 cos(300 × 10 6 πt − . 8032) meaning the amplitude of the received signal at x = 0 is A = 1 . 3888, the phase is φ = − . 8032 and the (radian) frequency is ω = 300 × 10 6 π . (iii) The general expression for the received signal in terms of the mobile position x is vextendsingle vextendsingle vextendsingle r ( t, x ) = s ( t − t 1 ( x )) + s ( t − t 2 ( x )) = 2 cos parenleftBig ω 2 c ( d 2 − d 1 ) parenrightBig cos parenleftBig ωt + ω 2 c ( d 2 + d 1 ) parenrightBig = 0 , when parenleftBig ω 2 c ( d 2 − d 1 ) parenrightBig = π 2 (iv) MATLAB script and plot: t = 0:200; x = -100:100; c = 3e8; t_1 = sqrt(x.^2+10^6)/c; t_2 = (sqrt((x-55).^2+10^6)+55)/c; r = cos(300e6*pi*(t-t_1)) + cos(300e6*pi*(t-t_2)); plot(x,r,’k’,’linewidth’,2) xlabel(’x’), ylabel(’r’), grid on ECE 2200 Signals and Information – Section 2 Solutions (Week 3 Feb 8-11)Page 2 . 100 . 80 . 60 . 40 . 20 20 40 60 80 100 . 2 . 1.5 . 1 . 0.5 0.5 1 1.5 2 r ( t,x ) x (mobile position) Figure 1: Signal strength example. 2. Consider the system that with input u u ( t ) = cos( at ) produces the output y y ( t ) = u ( t )(cos( bt )) . Given that the magnitude spectrum of y has lines at only four frequencies: ± 16 . 8 MHz and ± 17 . 6 MHz, determine a and b . Solution (read “Section 3-1, 3-2”): a + b = 2 π 17 . 6 × 10 3 rad sec a − b = 2 π 16 . 8 × 10 3 rad sec 3. The signal x ( t ) has the two-sided spectrum representation shown in Figure 1. (i) Write an equation for x ( t ) as a sum of cosines. (ii) Is x ( t ) a periodic signal? If so, determine its fundamental period and its fundamental frequency....
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sec2-student-solns - ECE 2200 Signals and Information –...

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