7 - 04/23/08 ECE 483 Homework #7 Solution Problem 8-2...

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04/23/08 ECE 483 Homework #7 Solution Problem 8-2 Solution: First, we obtain G ( z ) as G ( z ) = z 1 z Z b 1 s 2 ( s + 1) B = ( T 1 + e T ) z + 1 e T Te T ( z 1)( z e T ) = 0 . 3679 z + 0 . 2642 ( z 1)( z 0 . 3679) . Transforming it to the w -plane, we get G ( w ) = G ( z ) v v z = 2+ w 2 - w = 0 . 1037 w 2 1 . 0568 w + 2 . 5284 2 . 7358 w 2 + 2 . 5284 w , whose frequency response is plotted in Fig. 1. Alternatively, the data in Table P8-2 can be used to plot approximately the frequency response. -40 -30 -20 -10 0 10 20 30 40 50 60 Magnitude (dB) 10 -2 10 -1 10 0 10 1 10 2 10 3 135 180 225 270 Phase (deg) Bode Diagram Frequency (rad/sec) Figure 1: Frequency response of the uncompensated system in Problem 8-2. (a) With D ( z ) = 1, from the frequency response of the uncompensated system in Table P8-2, we ±nd that the 0 dB crossover frequency of | G ( w ) | occurs at approximately ω w = 0 . 8, and at this frequency, G ( w ) = 149 (or 211 in Fig. 1). Thus, the phase margin is approximately φ m = 149 ( 180 ) 31 . (b) If D ( z ) = K , then as we adjust the gain K > 0, the phase plot of the compensated frequency response D ( w ) G ( w ) will remain the same, while the magnitude plot 1
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will shift by 20 log K dB. To have a 45 phase margin, we need to place the new 0 dB crossover frequency at ω w = 0 . 56 rad/s where D ( w ) G ( w ) = G ( w ) = 180 + 45 = 225 , i.e. 20 log | D ( w ) G ( w ) | ω w =0 . 56 = 20 log K + 20 log | G ( j 0 . 56) | = 0 . Since 20 log | G ( j 0 . 56) | = 3 . 98 dB, we must have 20 log K = 3 . 98 dB, i.e., K = 0 . 8195. 0 5 10 15 20 25 0 0.2 0.4 0.6 0.8 1 1.2 1.4 n (samples) Amplitude Step Response Figure 2: Step response of part (b) of Problem 8-2 of Problem 8-2. (c) See Fig. 2 for the unit response of part (b). The rise time is approximately 1.5 seconds,
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This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue.

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7 - 04/23/08 ECE 483 Homework #7 Solution Problem 8-2...

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