# ch8soln - Chapter 8 Digital Control Problems and Solutions...

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Chapter 8 Digital Control Problems and Solutions 1. The z -transform of a discrete-time ° lter h ( k ) at a 1 Hz sample rate is H ( z ) = 1 + (1 / 2) z 1 [1 (1 / 2) z 1 ][1 + (1 / 3) z 1 ] . (a) Let u ( k ) and y ( k ) be the discrete input and output of this ° lter. Find a di ff erence equation relating u ( k ) and y ( k ). (b) Find the natural frequency and damping coe cient of the ° lter°s poles (c) Is the ° lter stable? Solution: (a) Find a di ff erence equation : H ( z ) = Y ( z ) U ( z ) = 1 + (1 / 2) z 1 [1 (1 / 2) z 1 ] [1 + (1 / 3) z 1 ] = Y ( z ) 1 6 z 1 Y ( z ) 1 6 z 2 y ( z ) = U ( z ) + 1 2 z 1 U ( z ) = y ( k ) 1 6 y ( k 1) 1 6 y ( k 2) = u ( k ) + 1 2 u ( k 1) (b) Two poles at z = 1 / 2 and z = 1 / 3 in z-plane. z = e sT = s = 0 . 693 T and s = 1 . 10 + 3 . 14 j T in s-plane, where T is the sampling period. Since the sample rate is 1 Hz, T = 1 sec. For z = 1 2 , ω n = 0 . 693 T = 0 . 693 rad/sec, ζ = 1 . 0 For z = 1 3 , ω n = 3 . 33 T = 3 . 33 rad/sec, ζ = 0 . 330 547

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548 CHAPTER 8. DIGITAL CONTROL (c) Yes, both poles are inside the unit circle. 2. Use the z -transform to solve the di ff erence equation y ( k ) 3 y ( k 1) + 2 y ( k 2) = 2 u ( k 1) 2 u ( k 2) , where u ( k ) = k, k 0 , 0 , k < 0 , y ( k ) = 0 , k < 0 . Solution: Y ( z ) U ( z ) = 2( z 1 z 2 ) 1 3 z 1 2 z 2 = 2 z 2 u ( k ) = k k 0 0 k < 0 ° = U ( z ) = z ( z 1) 2 Y ( z ) = 2 z 2 ° z ( z 1) 2 = 2 z z 2 2 z z 1 2 z ( z 1) 2 Taking the inverse z-transform from Table 8.1, y ( k ) = 2(2 k 1 k ) ( k 0) 3. The one-sided z -transform is de ° ned as F ( z ) = X 0 f ( k ) z k . (a) Show that the one-sided transform of f ( k + 1) is Z{ f ( k + 1) } = zF ( z ) zf (0). (b) Use the one-sided transform to solve for the transforms of the Fi- bonacci numbers generated by the di ff erence equation u ( k + 2) = u ( k + 1) + u ( k ). Let u (0) = u (1) = 1. [ Hint: You will need to ° nd a general expression for the transform of f ( k + 2) in terms of the transform of f ( k )]. (c) Compute the pole locations of the transform of the Fibonacci num- bers. (d) Compute the inverse transform of the Fibonacci numbers. (e) Show that, if u ( k ) represents the k th Fibonacci number, then the ratio u ( k + 1) /u ( k ) will approach (1 + 5) / 2. This is the golden ratio valued so highly by the Greeks. Solution:
549 (a) Z { f () k + 1) } = P k =0 f ( k + 1) z 1 = P j =1 f ( j ) z 1( j 1) , k + 1 = j = z P 0 f ( j ) z 1 zf (0) = zF ( z ) zf (0) (b) u ( k + 2) u ( k + 1) u ( k ) = 0 We have : Z { f ( k + 2) } = z 2 F ( z ) z 2 f (0) zf (1) Taking the z-transform, z 2 U ( z ) z 2 u (0) zu (1) [ zU ( z ) zu (0)] U ( z ) = 0 = ( z 2 z 1) U ( z ) = ( z 2 z ) u (0) + zu (1) Since u (0) = u (1) = 1, we have : U ( z ) = z 2 z 2 z 1 (c) The poles are at : z = 1 5 2 = 1 . 618 , 0 . 618 , α 1 , α 2 (d) (i) By long division : 1 + z 1 + 2 z 2 + 3 z 3 + • • • 1 z 1 z 2 ) 1 1 z 1 z 2 z 1 + z 2 z 1 z 2 z 3 2 z 2 + z 3 2 z 2 2 z 3 2 z 4 3 z 3 + 2 z 4 • • • u ( k ) = 1 , 1 , 2 , 3 , 5 , • • •

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550 CHAPTER 8. DIGITAL CONTROL (ii) By partial fraction expansion : U ( z ) = 1 1 z 1 z 2 = 1 (1 α 1 z 1 )(1 α 2 z 1 ) = ± α 1 α 1 α 2 1 α 1 z 1 + ± α 2 α 2 α 1 1 α 2 z 1 u ( k ) = α 1 α 1 α 2 α k 1 +
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