lecture_14 - MIT OpenCourseWare http:/ocw.mit.edu 2.161...

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MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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n n n F ( z ) Y Z Z h n ± ± 1 Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Lecture 14 1 Reading: Proakis & Manolakis, Chapter 3 (The z -transform) Oppenheim, Schafer & Buck, Chapter 3 (The z -transform) The Discrete-Time Transfer Function Consider the discrete-time LTI system, characterized by its pulse response { h n } : c o n v o l u t i o n { f n } L T I s y s t e m { y } = { f Ä h } ( z ) = F ( z ) H ( z ) m u l t i p l i c a t i o n We saw in Lec. 13 that the output to an input sequence { f n } is given by the convolution sum: y n = f n h n = f k h n k = h k f n k , k = −∞ k = −∞ where { h n } is the pulse response. Using the convolution property of the z -transform we have at the output Y ( z )= F ( z ) H ( z ) where F ( z Z{ f n } , and H ( z h n } . Then Y ( z ) H ( z F ( z ) is the discrete-time transfer function , and serves the same role in the design and analysis of discrete-time systems as the Laplace based transfer function H ( s ) does in continuous systems. 1 copyright ± c D.Rowell 2008 14–1
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( ) ( ) 2 In general, for LTI systems the transfer function will be a rational function of z , and may be written in terms of z or z 1 , for example N ( s ) b 0 + b 1 z 1 + b 2 z 2 + ... + b M z M H ( z )= = D ( s ) a 0 + a 1 z 1 + a 2 z 2 + + a N z N where the b i ,i =0 ,...,m , a i ,...,n are constant coefficients. The Transfer Function and the Difference Equation As defined above, let Y ( z ) b 0 + b 1 z 1 + b 2 z 2 + + b M z M H ( z = F ( z ) a 0 + a 1 z 1 + a 2 z 2 + + a N z N and rewrite as a 0 + a 1 z 1 + a 2 z 2 + + a N z N Y ( z b 0 + b 1 z 1 + b 2 z 2 + + b M z M F ( z ) If we apply the z -transform time-shift property Z { f n k } = z k F ( z ) term-by-term on both sides of the equation, (effectively taking the inverse z -transform) a 0 y n + a 1 y n 1 + a 2 y n 2 + + a N y n N = b 0 f n + b 1 f n 1 + b 2 f n 2 + + b M f n M and solve for y n 1 1 y n = ( a 1 y n 1 + a 2 y n 2 + + a N y n N )+ ( b 0 f n + b 1 f n 1 + b 2 f n 2 + + b M f n M ) a 0 a 0 N ± ² M ± ² ³ a i ³ b i = y n i + f n i a 0 a 0 i =1 i =0 which is in the form of a recursive linear difference equation as discussed in Lecture 13. The transfer function H ( z ) directly defines the computational dif- ference equation used to implement a system. Example 1 Find the difference equation to implement a causal system with a transfer function (1 2 z 1 )(1 4 z 1 ) H ( z z (1 1 2 z 1 ) Solution: z 1 6 z 2 +8 z 3 H ( z 1 1 z 1 2 14–2
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from which 1 y n y n 1 = f n 1 6 f n 2 +8 f n 3 , 2 or 1 y n = y n 1 +( f n 1 6 f n 2 f n 3 ) .
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This note was uploaded on 02/27/2012 for the course MECHANICAL 2.161 taught by Professor Derekrowell during the Fall '08 term at MIT.

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lecture_14 - MIT OpenCourseWare http:/ocw.mit.edu 2.161...

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