Chapter 6 - Chapter 6 The Frequency-response Design Method...

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Chapter 6 The Frequency-response Design Method Problems and Solutions for Section 6.1 1. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G ( s ) U 0 ω s j ω s = j ω = U 0 G ( j ω ) 1 2 j and α 0 = G ( s ) U 0 ω s + j ω s =+ j ω = U 0 G ( j ω ) 1 2 j . (b) By assuming the output can be written as y ( t )= α 0 e j ω t + α 0 e j ω t , derive Eqs. (6.4) - (6.6). Solution: (a) Eq. (6.2): Y ( s α 1 s p 1 + α 2 s p 2 + ••• + α n s p n + α o s + j ω o + α o s j ω o Multiplying this by ( s + j ω ): Y ( s )( s + j ω α 1 s + a 1 ( s + j ω )+ ... + α n s + a n ( s + j ω )+ α o + α o s j ω ( s + j ω ) 283
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284 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD α o = Y ( s )( s + j ω ) α 1 s + a 1 ( s + j ω ) ... α n s + a n ( s + j ω ) α o s j ω ( s + j ω ) α o = α o | s = j ω = Y ( s )( s + j ω ) α 1 s + a 1 ( s + j ω ) ... α o s j ω ( s + j ω ) s = j ω = Y ( s )( s + j ω ) | s = j ω = G ( s ) U o ω s 2 + ω 2 ( s + j ω ) | s = j ω = G ( s ) U o ω s j ω | s = j ω = U o G ( j ω ) 1 2 j Similarly, multiplying Eq. (6.2) by ( s j ω ): Y ( s )( s j ω )= α 1 s + a 1 ( s j ω )+ ... + α n s + a n ( s j ω α o s + j ω ( s j ω α o α o = α o | s = j ω = Y ( s )( s j ω ) | s = j ω = G ( s ) U o ω s 2 + ω 2 ( s j ω ) | s = j ω = G ( s ) U o ω s + j ω | s = j ω = U o G ( j ω ) 1 2 j (b) y ( t α o e j ω t + α o e j ω t y ( t U o G ( j ω ) 1 2 j e j ω t + U o G ( j ω ) 1 2 j e j ω t = U o G ( j ω ) e j ω t G ( j ω ) e j ω t 2 j | G ( j ω ) | = n Re [ G ( j ω )] 2 +Im[ G ( j ω )] 2 o 1 2 = A G ( j ω )=t a n 1 Im[ G ( j ω )] Re [ G ( j ω )] = φ | G ( j ω ) | = n Re [ G ( j ω )] 2 G ( j ω )] 2 o 1 2 = | G ( j ω ) | = n Re [ G ( j ω )] 2 G ( j ω )] 2 o 1 2 = A G ( j ω a n 1 G ( j ω )] Re [ G ( j ω )] =tan 1 G ( j ω )] Re [ G ( j ω )] = φ G ( j ω Ae j φ ,G ( j ω Ae j φ Thus, y ( t U o Ae j φ e j ω t Ae j φ e j ω t 2 j = U o A e j ( ω t + φ ) e j ( ω t + φ ) 2 j y ( t U o A sin( ω t + φ )
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285 where A = | G ( j ω ) | , φ =tan 1 Im[ G ( j ω )] Re[ G ( j ω )] = G ( j ω ) 2. (a) Calculate the magnitude and phase of G ( s )= 1 s +10 by hand for ω = 1, 2, 5, 10, 20, 50, and 100 rad/sec. (b) sketch the asymptotes for G ( s ) according to the Bode plot rules, and compare these with your computed results from part (a). Solution: (a) G ( s 1 s ,G ( j ω 1 10 + j ω = 10 j ω 100 + ω 2 | G ( j ω ) | = 1 100 + ω 2 , G ( j ω tan 1 ω 10 ω | G ( j ω ) | G ( j ω ) 1 2 5 10 20 50 100 0 . 0995 0 . 0981 0 . 0894 0 . 0707 0 . 0447 0 . 0196 0 . 00995 5 . 71 11 . 3 26 . 6 45 . 0 63 . 4 78 . 7 84 . 3
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286 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (b) The Bode plot is : 10 -1 10 0 10 1 10 2 10 3 10 -3 10 -2 10 -1 10 0 Frequency (rad/sec) M agnitude B ode D iagram s 10 -1 10 0 10 1 10 2 10 3 -100 -80 -60 -40 -20 0 20 Frequency (rad/sec) Phase (deg) 3. Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.
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Chapter 6 - Chapter 6 The Frequency-response Design Method...

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