project 1_examples - 1 ME360 - Solved Examples for Project...

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1 ME360 - Solved Examples for Project #1 Problem 1: With your data acquisition system, you need to sample a periodic half-wave rectified sine waveform x p ( t ) (Example 8.2 (f), pp. 203, 204 in Ambardar) 0 0.5 1 1.5 Half-Rectified Sine Wave, xp(t) Time (sec) xp(t) 0 T/2 T 3T/2 -T -T/2 (a) For T=0.1 sec., can you select the sampling period t s of your data acquisition system to avoid aliasing? Explain, using the relationship between time domain and frequency domain for periodic signals. Solution: It is impossible to avoid aliasing, since this is a non-smooth function. Non-smooth functions are not bandlimited. Therefore, it would require an infinitely large sampling rate to avoid aliasing, which is not possible. There is no upper bound, f b ,onthe signal, and the inequality, S f > 2 f b , can never be satisfied. (b) Derive a polar Fourier series representation of x p ( t ) given above for T = 0.1 sec. and draw its single-sided magnitude, phase, and power spectra for the dc component plus the first seven harmonics. Solution: a o = 1 T R T x ( t ) dt = 1 0 . 1 R T 2 0 sin (2 π 10 t ) dt =10 R . 05 0 sin (20 πt ) dt ¡ 1 20 π cos (20 ) ¢ | . 05 0 = 1 2 π ( cos ( π )+cos(0))= 1 2 π (2) = 1 π a k = 2 T R T x ( t )cos(2 πkf o t ) dt = 2 0 . 1 R 0 . 05 0 sin (20 ) cos (20 πkt ) dt
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2 =20 R 0 . 05 0 1 2 [sin (20 πt 20 πkt )+sin(20 +20 )] dt =10 R 0 . 05 0 [sin (20 (1 k )) + sin (20 (1 + k ))] dt h R 0 . 05 0 sin (20 (1 k )) dt + R 0 . 05 o sin (20 (1 + k )) dt i h 1 20 π (1 k ) ( cos (20 (1 k ))) ¯ ¯ ¯ 0 . 05 0 +10 h 1 20 π (1+ k ) ( cos (20 (1 + k ))) ¯ ¯ ¯ 0 . 05 0 = 1 2 π 1 (1 k ) cos ( π (1 k )) + 1 z }| { cos (0) + 1 (1+ k ) cos ( π (1 + k )) + 1 z }| { cos (0) Notice that cos ( π (1 ± k )) = cos ( ± π )= cos ( ) = 1 2 π h 1 (1 k ) (cos ( ) + 1) + 1 (1+ k ) (cos ( )+1) i = = 1 2 π ³ 1 (1 k ) + 1 (1+ k ) ´ (cos ( ) + 1) = 1 2 π ³ (1+ k )+(1 k ) (1+ k )(1 k ) ´ (cos ( ) + 1) = = 1 2 π 2 (1 k 2 ) (cos ( )+1)= = 1 π (1 k 2 ) (cos ( a k = ½ 2 π (1 k 2 ) , even k 0 , odd k b k = 2 T R T x ( t )sin(2 πkf o t ) dt R 0 . 05 0 sin (20 )sin(40 kπt ) dt = R 0 . 05 0 1 2 [cos (20 20 ) cos (20 )] dt = h R 0 . 05 0 cos (20 (1 k )) dt R 0 . 05 0 cos (20 (1 + k )) dt i = h 1 20 π (1 k ) sin (20 (1 k )) ¯ ¯ ¯ 0 . 05 0 10 h 1 20 π (1+ k ) sin (20 (1 + k )) ¯ ¯ ¯ 0 . 05 0 = = 1 2 π ³ 1 1 k 1 1+ k ´ 0 z }| { sin ( π (1 k )) 0 z }| { sin (0) 0 z }| { sin ( π (1 + k )) 0 z }| { sin (0) = =0 b 1 is undefined since 1-k is in denominator ; b 1 = 1 2 π lim k 1 sin( π (1 k )) 1 k = 1 2 π lim k 1 π cos( π (1 k )) 1 = 1 2 b k = ½ 1 2 ,k =1 0 , else Now transform to Polar form: c 1 = p a 2 1 + b 2 1 = 1 2 ; θ 1 =tan 1 ³ b 1 a 1 ´ = tan 1 ¡ 0 . 5 0 ¢ = π 2 ; c 2 = ¯ ¯ ¯ 2 π (1 k 2 ) ¯ ¯ ¯ k =2 = 2 3 π ; θ 2 = π ; c 3 =0 ; θ 3 =0; c 4 = ¯ ¯ ¯ 2 π (1 k 2 ) ¯ ¯ ¯ k =4 = 2 15 π ; θ 4 = π ; c 5 θ 5 c 6 = ¯ ¯ ¯ 2 π (1 k 2 ) ¯ ¯ ¯ k =6 = 2 35 π ; θ 6 = π ; c 7 θ 7 . So, x p ( t c o + 7 P k =1 c k cos (2 o t + θ k ) = 1 π + 1 2 cos ¡ 2 π 10 t + π 2 ¢ + 2 3 π cos (2 π 20 t + π )+ 2 15 π cos (2 π 40 t + π 2 35 π cos (2 π 60 t + π )
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3 0 1 2 3 4 5 6 7 0 0.1 0.2 0.3 0.4 0.5 magnitude 0 1 2 3 4 5 6 7 0 1 2 3 4 phase (rad) harmonic number, k Fig. 1. 0
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This note was uploaded on 04/18/2010 for the course ME 360 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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project 1_examples - 1 ME360 - Solved Examples for Project...

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