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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 8 Digital Control Problems and Solutions 1. The z -transform of a discrete-time & lter h ( k )ata1 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 ad i f erence equation relating u ( k 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 f erence equation : H ( z Y ( z ) U ( z ) = / 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 / 2and z = 1 / 3inz-p lane . 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 sec. For z = 1 2 , ω n = 0 . 693 T =0 . 693 rad/sec, ζ . 0 z = 1 3 , ω n = 3 . 33 T =3 . 33 rad/sec, ζ . 330 547
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
548 CHAPTER 8. DIGITAL CONTROL (c) Yes, both poles are inside the unit circle. 2. Use the z -transform to solve the di f 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 kk 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 f 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)intermsofthe 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 )rep re sen t sthe k th Fibonacci number, then the ratio u ( k /u ( k ) will approach (1 + 5) / 2. This is the golden rat iova luedsoh igh lybytheGreeks . Solution:
Background image of page 2
549 (a) Z{ f () k +1) } = P k =0 f ( k 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 ) (0) (b) u ( k +2) u ( k u ( k )=0 We have : f ( k } = z 2 F ( z ) z 2 f (0) (1) Taking the z-transform, z 2 U ( z ) z 2 u (0) zu (1) [ zU ( z ) (0)] U ( z = ( z 2 z 1) U ( z )=( z 2 z ) u (0) + (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 z 4 u ( k )=1 , 1 , 2 , 3 , 5 ,
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 + α 2 α 2 α 1 α k 2 = ˆ 5+ 5 10 1+ 5 2 !
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/17/2009 for the course MEEM 4700 taught by Professor Staff during the Spring '08 term at Michigan Technological University.

Page1 / 70

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online