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Unformatted text preview: EE264: Lecture 6 IIR Filters cont. Differential equation to Difference equation Transformation When a system has a transfer function H ( s ) of rational form, the system relates its output to its input through a Linear Constant Coefficient Differential Equation. Approximating derivatives with (discrete-time) differ- ences allows us to translate the differential equation to a difference equation that defines a Discrete-time system which will approximate the differential equation if the sampling period is small enough. Example: Let x ( t ) and y ( t ) be the input and output of a continuous-time system with the following transfer function: H c ( s ) = s- 1 ( s + 1)( s + 2) = s- 1 s 2 + 3 s + 2 then Y ( s ) X ( s ) = s- 1 s 2 + 3 s + 2 Y ( s )( s 2 + 3 s + 2) = X ( s )( s- 1) d 2 y ( t ) dt 2 + 3 dy ( t ) dt + 2 y ( t ) = dx ( t ) dt- x ( t ) Let x [ n ] = x ( nT ) , we can approximate derivatives at times nT as follows: dx ( t ) dt vextendsingle vextendsingle vextendsingle vextendsingle t = nT ≈ x [ n + 1]- x [ n ] T forward difference dx ( t ) dt vextendsingle vextendsingle vextendsingle vextendsingle t = nT ≈ x [ n ]- x [ n- 1] T backward difference dx ( t ) dt vextendsingle vextendsingle vextendsingle vextendsingle t = nT ≈ x [ n + 1]- x [ n- 1] 2 T symmetrical difference d 2 x ( t ) dt 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle t = nT ≈ x [ n +1]- x [ n ] T- x [ n ]- x [ n- 1] T T = x [ n + 1]- 2 x [ n ] + x [ n- 1] T 2 Using the forward difference approximation we obtain: y [ n + 1]- 2 y [ n ] + y [ n- 1] T 2 + 3 y [ n + 1]- y [ n ] T + 2 y [ n ] = x [ n + 1]- x [ n ] T- x [ n ] Taking the z-transform on both sides of the difference equation we get: H ( z ) = Y ( z ) X ( z ) = 1 T- ( 1 T + 1) z- 1 ( 1 T 2 + 3 T )- ( 2 T 2 + 3 T- 2) z- 1 + 1 T 2 z- 2 Copyright c circlecopyrt 1995–1996 by G. Plett. Copyright c circlecopyrt 1998–2004 by M. Kamenetsky. Copyright c circlecopyrt 2005 by A. Flores 65 66 IIR Filters cont. Lect. 6 Bilinear (but not linear) Transformation To avoid aliasing, we need a one-to-one mapping from the s-plane to the z-plane. Consider s = 2 T parenleftBigg 1- z- 1 1 + z- 1 parenrightBigg ⇒ H ( z ) = H c parenleftBigg 2(1- z- 1 ) T (1 + z- 1 ) parenrightBigg which defines a bilinear transformation z = 1 + T 2 s 1- T 2 s ⇔ s = 2(1- z- 1 ) T (1 + z- 1 ) Q: Is this a legal transformation? A: Let s = j Ω . Then z = 1 + j Ω T 2 1- j Ω T 2 = radicalBig 1 + Ω 2 T 2 4 negationslash tan- 1 parenleftBig Ω T 2 parenrightBig radicalBig 1 + Ω 2 T 2 4 negationslash tan- 1 parenleftBig- Ω T 2 parenrightBig = 1 negationslash 2 tan- 1 parenleftbigg Ω T 2 parenrightbigg So, | z | = 1 . Also, if σ < , | z | < 1 , so OK....
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This note was uploaded on 10/24/2009 for the course SCE sysc5602 taught by Professor El-tanany during the Winter '09 term at Carleton CA.
- Winter '09