8: IIR Filter Transformations

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Elijah Stevenson
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1 DSP and Digital (5-677) IIR : 8 /

2 Classical continuous-time filters optimize tradeoff: passband ripple v stopband ripple v transition width There are explicit formulae for pole/zero positions. Butterworth: G (Ω) = H(jΩ) = +Ω N Monotonic Ω G(Ω) = ΩN Ω4N + Maximally flat : N derivatives are zero Chebyshev: G (Ω) = +ǫ T N (Ω) where polynomial T N (cosx) = cosnx passband equiripple + very flat at Inverse Chebyshev: G (Ω) = +(ǫ T N (Ω )) stopband equiripple + very flat at Elliptic: [no nice formula] Very steep + equiripple in pass and stop bands N= DSP and Digital (5-677) IIR : 8 /

3 Change variable: z = α+s α s s = αz z+ : a one-to-one invertible mapping R axis (s) R axis (z) I axis (s) Unit circle (z) Proof: z = e jω s = α ejω e jω + = αej ω e j ω e j ω +e j ω = jαtan ω = jω Left half plane(s) inside of unit circle (z) Proof: s = x+jy z = (α+x)+jy (α x) jy x < z < Unit circle (s) I axis (z) = α +αx+x +y α αx+x +y = + 4αx (α x) +y 4 3 α = s-plane α = z-plane ω Ω/α DSP and Digital (5-677) IIR : 8 3 /

4 Take H(s) = s +.s+4 and choose α = Substitute: s = α z z+ [extra zeros at z = ] H(z) = ( z z+) +. z z+ +4 = (z+) (z ) +.(z )(z+)+4(z+) = z +z+ 5.z +6z+4.8 =.9 +z +z +.5z +.9z Frequency response is identical (both magnitude and phase) but with a distorted frequency axis: ω (rad/sample) 3.5 α= Frequency mapping: ω = tan Ω α Ω = [ α α 3α 4α 5α ] ω = [ ] ω Ω/α Choosing α: Set α = Ω to map Ω tan ω ω Set α = f s = T to map low frequencies to themselves DSP and Digital (5-677) IIR : 8 4 /

5 Alternative method: H(s) = s +.s+4 Find the poles and zeros: p s =.±j Map using z = α +s s p z =.58±.77j After the transformation we will always end up with the same number of poles as zeros: Add extra poles or zeros at z = (+z H(z) = g ) (+(.58.77j)z )(+( j)z ) +z = g +z +.5z +.9z Choose overall scale factor, g, to give the same gain at any convenient pair of mapped frequencies: At Ω = s = H(s ) =.5 ω = tan Ω α = z = e jω = H(z ) = g =.5 g =.9 H(z) =.9 +z +z +.5z +.9z s - α= α= - z ω (rad/sample) DSP and Digital (5-677) IIR : 8 5 /

6 We can transform the z-plane to change the cutoff frequency by substituting z = ẑ λ λẑ ẑ = z+λ +λz Frequency Mapping: If z = e jω, then ẑ = z +λz +λz has modulus since the numerator and denominator are complex conjugates. Hence the unit circle is preserved. e jˆω = ejω +λ +λe jω ) (+λ Some algebra gives: tan ω = λ tan ˆω Equivalent to: z s = z z+ λ +ŝ ŝ = +λs ẑ = ŝ Lowpass Filter example: Inverse Chebyshev ω = π λ=.6 =.57 ˆω =.49 H^ z z^ λ = ω (rad/s) λ =.6 ω^ = ω^ (rad/s) DSP and Digital (5-677) IIR : 8 6 /

7 Transform any lowpass filter with cutoff frequency ω to: Target Substitute Parameters Lowpass ˆω < ˆω Highpass ˆω > ˆω Bandpass ˆω < ˆω < ˆω Bandstop ˆω ˆω ˆω z = ẑ λ λẑ z = ẑ +λ +λẑ z = (ρ ) λρẑ +(ρ+)ẑ (ρ+) λρẑ +(ρ )ẑ z = ( ρ) λẑ +(ρ+)ẑ (ρ+) λẑ +( ρ)ẑ λ = sin ( ω ˆω ) sin( ω +ˆω ) λ = cos ( ω+ˆω ) cos( ω ˆω ) λ = cos ( ˆω+ˆω ) cos( ˆω ˆω ) ρ = cot ( ˆω ˆω λ = cos ( ˆω+ˆω ) tan ( ω ) cos( ˆω ˆω ) ρ = tan ( ˆω ˆω ) ) tan ( ω ) Bandpass and bandstop transformations are quadratic and so will double the order: Lowpass Bandpass.5 H^ ω (rad/s) ω^ (rad/s) DSP and Digital (5-677) IIR : 8 7 /

8 Bilinear transform works well for a lowpass filter but the non-linear compression of the frequency distorts any other response. L Alternative method: H(s) h(t) sample h[n] = T h(nt) Z Express H(s) as a sum of partial fractions H(s) = N i= Impulse response is h(t) = u(t) N i= g ie p it Digital filter H(z) T = N i= H(z) g i s p i g i e p i T z has identical impulse response Poles of H(z) are p i = e p it (where T = f s is sampling period) do not map in a simple way Properties: Impulse response correct. No distortion of frequency axis. Frequency response is aliased. Example: Standard telephone filter - 3 to 34 Hz bandpass.5 Analog Filter Frequency (krad/s).5 Bilinear (f s = 8 khz) Matched at 3.4 khz ω (rad/sample) (f s = 8 khz) ω (rad/sample) DSP and Digital (5-677) IIR : 8 8 /

9 Classical filters have optimal tradeoffs in continuous time domain Order transition width pass ripple stop ripple Monotonic passband and/or stopband Bilinear mapping Exact preservation of frequency response (mag + phase) non-linear frequency axis distortion can choose α to map Ω ω for one specific frequency transformations lowpass lowpass, highpass, bandpass or bandstop bandpass and bandstop double the filter order Aliassing distortion of frequency response preserves frequency axis and impulse response For further details see Mitra: 9. DSP and Digital (5-677) IIR : 8 9 /

10 bilinear impinvar butter butterord cheby chebyord cheby chebyord ellip ellipord Bilinear mapping Impulse invariance Analog or digital Butterworth filter Analog or digital Chebyshev filter Analog or digital Inverse Chebyshev filter Analog or digital Elliptic filter DSP and Digital (5-677) IIR : 8 /

LECTURER NOTE SMJE3163 DSP

LECTURER NOTE SMJE3163 DSP LECTURER NOTE SMJE363 DSP (04/05-) ------------------------------------------------------------------------- Week3 IIR Filter Design -------------------------------------------------------------------------