lecture 22: design of fir / iir filters
TRANSCRIPT
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Selectionβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Selectionβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What condition must be satisfied?
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What condition must be satisfied?
π₯π₯ ππ = Β±π₯π₯ βππ + ππ β 1 = Β±π₯π₯ ππ β 1β ππ Positive: Even symmetry
Negative: Odd symmetry
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response?
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
Odd Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―βππ1π§π§β ππβ2 β ππ0π§π§β ππβ1
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
Odd Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―βππ1π§π§β ππβ2 β ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―βππ1π§π§
1β ππβ12 β ππ0π§π§
β ππβ12
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
Odd Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―βππ1π§π§β ππβ2 β ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―βππ1π§π§
1β ππβ12 β ππ0π§π§
β ππβ12
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πΊπΊ ππ = ππ ππ ππjΞ ππ
πΊπΊ ππ = ππ ππ ππjΞ ππ
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
Odd Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―βππ1π§π§β ππβ2 β ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―βππ1π§π§
1β ππβ12 β ππ0π§π§
β ππβ12
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πΊπΊ ππ = ππ ππ ππjΞ ππ Ξ ππ = οΏ½0 for πΊπΊ ππ > 0ππ for πΊπΊ ππ < 0
πΊπΊ ππ = ππ ππ ππjΞ ππ Ξ ππ = οΏ½0 for πΊπΊ ππ > 0ππ for πΊπΊ ππ < 0
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
Odd Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―βππ1π§π§β ππβ2 β ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―βππ1π§π§
1β ππβ12 β ππ0π§π§
β ππβ12
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β ππ ππ = οΏ½ βππ ππ β 1 /2 for πΊπΊ ππ > 0βππ ππ β 1 /2 + ππ for πΊπΊ ππ < 0
β ππ ππ = οΏ½ βππ ππ β 1 /2 for πΊπΊ ππ > 0βππ ππ β 1 /2 + ππ for πΊπΊ ππ < 0
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
= π§π§βππβ12 οΏ½
ππ=0
ππ/2β1
ππππ π§π§ππβ12 βππ + π§π§β
ππβ12 βππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
= π§π§βππβ12 οΏ½
ππ=0
ππ/2β1
ππππ π§π§ππβ12 βππ + π§π§β
ππβ12 βππ
Notice thatππ π§π§ = π§π§β ππβ1 ππ π§π§β1
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality and Linear Phase Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
= π§π§βππβ12 οΏ½
ππ=0
ππ/2β1
ππππ π§π§ππβ12 βππ + π§π§β
ππβ12 βππ
Pole-zero plot property? ππ π§π§ = π§π§β ππβ1 ππ π§π§β1
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Question: Consider a length-M symmetric, causal filter. What is the phase response? Assume M is even.
Even Symmetryππ π§π§ = ππ0 + ππ1π§π§β1 + ππ2π§π§β2 +β―+ ππ1π§π§β ππβ2 + ππ0π§π§β ππβ1
= π§π§βππβ12 ππ0π§π§
ππβ12 + ππ1π§π§
ππβ12 β1 +β―+ ππ1π§π§
1β ππβ12 + ππ0π§π§
β ππβ12
= π§π§βππβ12 οΏ½
ππ=0
ππ/2β1
ππππ π§π§ππβ12 βππ + π§π§β
ππβ12 βππ
Pole-zero plot property? ππ π§π§ = π§π§β ππβ1 ππ π§π§β1
Causality and Linear Phase
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xx(M-1)
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality Question: How do we describe causal filter magnitude?
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality Question: How do we describe causal filter magnitude?
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Often plotted in dB (decibels)
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Causality Question: How do we describe causal filter magnitude?
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Often plotted in dB (decibels)
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Review Downsampling & Upsamplingβ’ Causality in Filtersβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Selectionβ’ Designing FIR Filters with Equi-ripples
Lecture 21: Design of FIR FiltersFoundations of Digital Signal Processing
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Designing with Windows Question: How can I design an FIR filter from an ideal filter?
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β¦β¦
β±
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Designing with Windows Question: How can I design an FIR filter from an ideal filter?
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β¦β¦
β±Ideal filterNon-causal
Infinite Response
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Designing with Windows Question: How can I design an FIR filter from an ideal filter?
Answer: Window the response!
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β¦β¦
β±Ideal filterNon-causal
Infinite Response
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Question: How can I design an FIR filter from an ideal filter?
Answer: Window the response!
Designing with Windows
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β¦β¦
β±Ideal filterNon-causal
Infinite Response
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Question: How can I design an FIR filter from an ideal filter?
Answer: Window the response!
Designing with Windows
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β¦β¦
β±Ideal filterNon-causal
Infinite Response
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Different Filters
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Different Filters
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Different Filters
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Windowing the sinc impulse response
Designing with Windows
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Samplingβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ = οΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ =1πποΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + οΏ½
ππ= ππ+1 /2
ππβ1
π»π» ππ ππππ2ππππ ππππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ =1πποΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + οΏ½
ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππ ππβππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ =1πποΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + οΏ½
ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππ ππβππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ +π»π» ππ ππππ
2ππππ ππ ππβππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ =1πποΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + οΏ½
ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππ ππβππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ +π»π» ππ ππβππ
2ππππ ππππππππ2ππππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ =1πποΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + οΏ½
ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππ ππβππ
β ππ =1ππ π»π» 0 + οΏ½
ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ +π»π» ππ ππβππ
2ππππ ππππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ =1πποΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + οΏ½
ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππ ππβππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + ππβππ
2ππππ ππππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Option 2: Work backwards with constraints
Consider the DFT:
β ππ =1πποΏ½ππ=0
ππβ1
π»π» ππ ππππ2ππππ ππππ such that π»π» ππ = π»π» ππ β ππ
β ππ =1ππ
π»π» 0 + οΏ½ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππππ + οΏ½
ππ=1
ππβ1 /2
π»π» ππ ππππ2ππππ ππ ππβππ
β ππ =1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππππππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling An inverse DFT that forces time-symmetry
β ππ =1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππππππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling An inverse DFT that forces time-symmetry
β ππ =1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππππππ
Example: Consider the desired 9-sample frequency responsewith the first half defined by [1 1 0 0]
Compute the frequency sampled filter
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling An inverse DFT that forces time-symmetry
β ππ =1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππππππ
Example: Consider the desired 9-sample frequency responsewith the first half defined by [1 1 0 0]
Compute the frequency sampled filter
β ππ =12
1 + 2 cos 2ππ/9 ππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Example: Consider the desired 9-sample frequency response
with the first half defined by [1 1 0 0]
Compute the frequency sampled filter
β ππ =12
1 + 2 cos 2ππ/9 ππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Example: Consider the desired 9-sample frequency response
with the first half defined by [1 1 0 0]
Compute the frequency sampled filter
β ππ =12
1 + 2 cos 2ππ/9 ππ
In practice, this should be circularly shifted so that the maximum is centered.
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling An inverse DFT that forces time-symmetry
β ππ =1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππ
ππ βππ β 1
2ππ
Example: Consider the desired 9-sample frequency responsewith the first half defined by [1 1 0 0]
Compute the frequency sampled filter
β ππ =12
1 + 2 cos 2ππ/9 ππ
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Example: Consider the desired 9-sample frequency response
with the first half defined by [1 1 0 0]
Compute the frequency sampled filter
β ππ =12
1 + 2 cos 2ππ/9 ππ β 8/2
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Example: Consider the desired 17-sample frequency response
with the first half defined by [1 1 1 1 0 0 0 0]
Compute the frequency sampled filter
β ππ =1
171 + 2 cos 2ππ/19 ππππ + 2 cos 4ππ/19 ππππ + 2 cos 6ππ/19 ππππ
ππππ = ππ β162
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Example: Consider the desired 17-sample frequency response
with the first half defined by [1 1 1 1 0 0 0 0]
Compute the frequency sampled filter
β ππ =1
171 + 2 cos 2ππ/19 ππππ + 2 cos 4ππ/19 ππππ + 2 cos 6ππ/19 ππππ
ππππ = ππ β162
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Example: Consider the desired 41-sample frequency response
with the first 10 values defined by 1
Compute the frequency sampled filter
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Example: Consider the desired 401-sample frequency response
with the first 100 values defined by 1
Compute the frequency sampled filter
Note that in practice, this needs to be circularly shifted to the center
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Note: The definition can be slightly modified Our definition:
β ππ =1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππ
ππ βππ β 1
2ππ
=1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππ
ππ βππ2
+12
ππ
=1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
π»π» ππ cos2ππππ
ππ +12
ππ β ππππ
=1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
β1 πππ»π» ππ cos2ππππ
ππ +12
ππ
62
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Frequency Sampling Final Definition
β ππ =1ππ
π»π» 0 + 2 οΏ½ππ=1
ππβ1 /2
β1 πππ»π» ππ cos2ππππ
ππ +12
ππ
Side note: This is very closely related to the discrete cosine transform
63
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Samplingβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
64
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Equi-ripplesPreviously derived:
ππ π§π§ = π§π§βππβ12 οΏ½
ππ=0
ππ/2β1
ππππ π§π§ππβ12 βππ + π§π§β
ππβ12 βππ
ππ ππ = ππβππππππβ12 οΏ½
ππ=0
ππ/2β1
ππππ ππππππππβ12 βππ + ππππβππππ ππβ1
2 βππ
= 2ππβππππππβ12 οΏ½
ππ=0
ππ/2β1
ππππ cos ππππβ 1
2β ππ
65
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Equi-ripples Equi-ripple design
ππ ππ = 2ππβππππππβ12 οΏ½
ππ=0
ππ/2β1
ππππ cos ππππβ 1
2β ππ
Goal: Find the optimal ππππs that satisfies passband / stopband ripple constraints.
66
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Design with Equi-ripples Equi-ripple design
minππππ
ππ ππ π»π»ππ ππ β 2ππβππππππβ12 οΏ½
ππ=0
ππ/2β1
ππππ cos ππππβ 1
2β ππ
67
Desired frequency response
Equals:πΏπΏ2πΏπΏ1
for ππ in pass band
1 for ππ in stop band
πΏπΏ2 = stopband rippleπΏπΏ1 = passband ripple
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Samplingβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
68
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Designing IIR Filters No easy ways to design digital IIR filters
So let us start from analog filters
69
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Designing IIR Filters No easy ways to design digital IIR filters
So let us start from analog filters
Option 1: Preserve the difference equation!
70
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a derivative in discrete-time? In continuous-timeπππ₯π₯ π‘π‘πππ‘π‘
β π π ππ π π
In discrete-time
71
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a derivative in discrete-time? In continuous-timeπππ₯π₯ π‘π‘πππ‘π‘
β π π ππ π π
In discrete-timeπππ₯π₯ π‘π‘πππ‘π‘
= limΞππβ0
π₯π₯ π‘π‘ β π₯π₯ π‘π‘ β ΞππΞππ
οΏ½πππ₯π₯ π‘π‘πππ‘π‘ π‘π‘=ππππ
=π₯π₯ ππππ β π₯π₯ ππππ β ππ
ππ= π₯π₯ ππ β π₯π₯ ππ β 1
72
T=1
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a derivative in discrete-time? In continuous-timeπππ₯π₯ π‘π‘πππ‘π‘
β π π ππ π π
In discrete-timeπππ₯π₯ π‘π‘πππ‘π‘
= limΞππβ0
π₯π₯ π‘π‘ β π₯π₯ π‘π‘ β ΞππΞππ
οΏ½πππ₯π₯ π‘π‘πππ‘π‘ π‘π‘=ππππ
=π₯π₯ ππππ β π₯π₯ ππππ β ππ
ππ=
1πππ₯π₯ ππ β π₯π₯ ππ β 1
73
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a derivative in discrete-time? In continuous-timeπππ₯π₯ π‘π‘πππ‘π‘
β π π ππ π π
In discrete-timeπππ₯π₯ π‘π‘πππ‘π‘
= limΞππβ0
π₯π₯ π‘π‘ β π₯π₯ π‘π‘ β ΞππΞππ
οΏ½πππ₯π₯ π‘π‘πππ‘π‘ π‘π‘=ππππ
=π₯π₯ ππππ β π₯π₯ ππππ β ππ
ππ=
1πππ₯π₯ ππ β π₯π₯ ππ β 1
οΏ½πππ₯π₯ π‘π‘πππ‘π‘ π‘π‘=ππππ
β1ππ
1β π§π§β1 ππ π§π§
74
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a second-derivative in discrete-time? In continuous-timeππ2π₯π₯ π‘π‘πππ‘π‘2
β π π 2ππ π π
In discrete-timeππ2π₯π₯ π‘π‘πππ‘π‘2
=πππ₯π₯ π‘π‘πππ‘π‘
πππ₯π₯ π‘π‘πππ‘π‘
οΏ½πππ₯π₯ π‘π‘πππ‘π‘ π‘π‘=ππππ
=π₯π₯ ππππ β π₯π₯ ππππ β ππ
ππ
75
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a second-derivative in discrete-time? In continuous-timeππ2π₯π₯ π‘π‘πππ‘π‘2
β π π 2ππ π π
In discrete-timeππ2π₯π₯ π‘π‘πππ‘π‘2
=πππ₯π₯ π‘π‘πππ‘π‘
πππ₯π₯ π‘π‘πππ‘π‘
οΏ½ππ2π₯π₯ π‘π‘πππ‘π‘2
π‘π‘=ππππ
=π₯π₯ ππππ β π₯π₯ ππππ β ππ /ππ β π₯π₯ ππππ β ππ β π₯π₯ ππππ β 2ππ /ππ
ππ
76
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a second-derivative in discrete-time? In continuous-timeππ2π₯π₯ π‘π‘πππ‘π‘2
β π π 2ππ π π
In discrete-timeππ2π₯π₯ π‘π‘πππ‘π‘2
=πππ₯π₯ π‘π‘πππ‘π‘
πππ₯π₯ π‘π‘πππ‘π‘
οΏ½ππ2π₯π₯ π‘π‘πππ‘π‘2
π‘π‘=ππππ
=π₯π₯ ππππ β 2π₯π₯ ππππ β ππ + π₯π₯ ππππ β 2ππ
ππ2 βπ₯π₯ ππ β 2π₯π₯ ππ β 1 + π₯π₯ ππ β 2
ππ2
77
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a second-derivative in discrete-time? In continuous-timeππ2π₯π₯ π‘π‘πππ‘π‘2
β π π 2ππ π π
In discrete-timeππ2π₯π₯ π‘π‘πππ‘π‘2
=πππ₯π₯ π‘π‘πππ‘π‘
πππ₯π₯ π‘π‘πππ‘π‘
οΏ½ππ2π₯π₯ π‘π‘πππ‘π‘2
π‘π‘=ππππ
=π₯π₯ ππππ β 2π₯π₯ ππππ β ππ + π₯π₯ ππππ β 2ππ
ππ2 βπ₯π₯ ππ β 2π₯π₯ ππ β 1 + π₯π₯ ππ β 2
ππ2
οΏ½πππ₯π₯ π‘π‘πππ‘π‘ π‘π‘=ππππ
β1ππ2
1β 2π§π§β1 + π§π§β2 ππ π§π§ =1ππ2
1β π§π§β1 2ππ π§π§
78
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a derivative in discrete-time? Translate continuous-time to discrete-time
πππππ₯π₯ π‘π‘πππ‘π‘ππ
β π π ππππ π π
οΏ½πππππ₯π₯ π‘π‘πππ‘π‘ππ
π‘π‘=ππππ
β1ππ
1β π§π§β1 ππππ π§π§
79
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a derivative in discrete-time? Translate continuous-time to discrete-time
πππππ₯π₯ π‘π‘πππ‘π‘ππ
β π π ππππ π π
οΏ½πππππ₯π₯ π‘π‘πππ‘π‘ππ
π‘π‘=ππππ
β1ππ
1β π§π§β1 ππππ π§π§
π π β1ππ
1β π§π§β1
80
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π π β 1
ππ1β π§π§β1
Use the derivative conversion to transform the following biquadfilter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9
81
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π π β 1
ππ1β π§π§β1
Use the derivative conversion to transform the following bi-quad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9
π»π» π§π§ =1
1ππ 1β π§π§β1 + 0.1
2+ 9
=ππ2
ππ2 1ππ 1β π§π§β1 + 0.1
2+ 9
=ππ2
1β π§π§β1 + 0.1ππ 2 + 9ππ2=
ππ2
1 + 0.1ππ β π§π§β1 2 + 9ππ2
82
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π π β 1
ππ1β π§π§β1
Use the derivative conversion to transform the following bi-quad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9
π»π» π§π§ =ππ2
1 + 0.1ππ β π§π§β1 2 + 9ππ2
1 + 0.1ππ β π§π§β1 2 + 9ππ2 = 0
1 + 0.1ππ β π§π§β1 2 = β9ππ2
1 + 0.1ππ β π§π§β1 = Β±3ππππ
π§π§β1 = 1 + 0.1ππ β 3ππππ π§π§ =1
1 + 0.1β 3ππ ππ83
Finding poles
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π π β 1
ππ1β π§π§β1
Use the derivative conversion to transform the following bi-quad filter into the discrete-time domain.
π§π§ =1
1 + 0.1 + 3ππ ππ
84
Poles
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π π β 1
ππ1β π§π§β1
π§π§ =1
1 + 0.1 + 3ππ ππ
85
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Question: What is a derivative in discrete-time? Translate continuous-time to discrete-time
π π β1ππ
1β π§π§β1
Pros: Relatively simple
Cons: Very limiting
Stable continuous-time poles can only be mapped to low frequencies
86
1
Stable IIR analog filters map here
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Samplingβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
87
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Impulse Invariance Designing IIR Filters No easy ways to design digital IIR filters
So let us start from analog filters
Option 2: Preserve the impulse response!
88
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Impulse Invariance Question: How else can I represent my transfer function?
π»π» π π = οΏ½ππ=1
πΎπΎ1
π π β ππππ
89
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Impulse Invariance Question: How else can I represent my transfer function?
π»π» π π = οΏ½ππ=1
πΎπΎ1
π π β ππππ= οΏ½
ππ=1
πΎπΎ
πππππππππππ‘π‘
90
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Impulse Invariance Question: How else can I represent my transfer function?
π»π» π π = οΏ½ππ=1
πΎπΎ1
π π β ππππ= οΏ½
ππ=1
πΎπΎ
πππππππππππ‘π‘
β π‘π‘ = οΏ½ππ=1
πΎπΎ
πππππππππππ‘π‘
91
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Impulse Invariance Question: How else can I represent my transfer function?
π»π» π π = οΏ½ππ=1
πΎπΎ1
π π β ππππ= οΏ½
ππ=1
πΎπΎ
πππππππππππ‘π‘
β π‘π‘ = οΏ½ππ=1
πΎπΎ
πππππππππππ‘π‘
β ππππ = β ππ = οΏ½ππ=1
πΎπΎ
ππππππππππππππ = οΏ½ππ=1
πΎπΎ
ππππ ππππππππ ππ
92
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Impulse Invariance Question: How else can I represent my transfer function?
π»π» π π = οΏ½ππ=1
πΎπΎ1
π π β ππππ= οΏ½
ππ=1
πΎπΎ
πππππππππππ‘π‘
β π‘π‘ = οΏ½ππ=1
πΎπΎ
πππππππππππ‘π‘
β ππππ = β ππ = οΏ½ππ=1
πΎπΎ
ππππππππππππππ = οΏ½ππ=1
πΎπΎ
ππππ ππππππππ ππ
π»π» π§π§ = οΏ½ππ=1
πΎπΎππππ
1β πππππππππ§π§β1
93
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π»π» π§π§ = βππ=1πΎπΎ ππππ
1βπππππππππ§π§β1
Use impulse invariance to transform the following biquad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9
94
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π»π» π§π§ = βππ=1πΎπΎ ππππ
1βπππππππππ§π§β1
Use impulse invariance to transform the following biquad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9
Poles:π π + 0.1 2 + 9 = 0π π = Β±3ππ β 0.1
95
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π»π» π§π§ = βππ=1πΎπΎ ππππ
1βπππππππππ§π§β1
Use impulse invariance to transform the following biquad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9=
1/2π π + 3ππ + 0.1
+1/2
π π β 3ππ + 0.1
Poles:π π + 0.1 2 + 9 = 0π π = Β±3ππ β 0.1
96
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π»π» π§π§ = βππ=1πΎπΎ ππππ
1βπππππππππ§π§β1
Use impulse invariance to transform the following biquad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9=
1/2π π + 3ππ + 0.1
+1/2
π π β 3ππ + 0.1
π»π» π§π§ =1/2
1β ππ β3ππβ0.1 πππ§π§β1+
1/21β ππ 3ππβ0.1 πππ§π§β1
97
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design from Derivatives Example: π»π» π§π§ = βππ=1πΎπΎ ππππ
1βπππππππππ§π§β1
π»π» π§π§ =1/2
1β ππ β3ππβ0.1 πππ§π§β1
+1/2
1β ππ 3ππβ0.1 πππ§π§β1
98
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Samplingβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
99
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Bilinear Transform Designing IIR Filters No easy ways to design digital IIR filters
So let us start from analog filters
Option 3: Preserve the definition of z!
100
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Bilinear Transform Question: How are z and s related? From continuous-time to discrete-timeπ π = ππΞ©πππ π π‘π‘ = πππ π ππππ = π§π§ππ
π§π§ = πππ π ππ
Building an approximation (πππ₯π₯ β 1 + π₯π₯)
π§π§ =πππ π ππ2
ππβπ π ππ2β
1 + π π ππ/21β π π ππ/2
101
Taylor Series Expansion / Approximation
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Bilinear Transform Question: How are z and s related? From continuous-time to discrete-timeπ π = ππΞ©πππ π π‘π‘ = πππ π ππππ = π§π§ππ
π§π§ = πππ π ππ π π =1ππ
ln π§π§
Building an approximation
π π β2πππ§π§ β 1π§π§ + 1
102
BilinearExpansion / Approximation
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Bilinear Transform The Bilinear Transform
Continuous-time to discrete-time
π π β2ππ
1β π§π§β1
1 + π§π§β1
Discrete-time to continuous-time
π§π§ β1 + π π ππ/21β π π ππ/2
103
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Bilinear Transform
Example: π π β 2ππ1βπ§π§β1
1+π§π§β1, π§π§ β 1+π π ππ/2
1βπ π ππ/2
Use the bilinear transform to transform the following biquad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9
104
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Bilinear Transform
Example: π π β 2ππ1βπ§π§β1
1+π§π§β1, π§π§ β 1+π π ππ/2
1βπ π ππ/2
Use the bilinear transform to transform the following biquad filter into the discrete-time domain.
π»π» π π =1
π π + 0.1 2 + 9
π»π» π§π§ =1
2ππ
1β π§π§β11 + π§π§β1 + 0.1
2+ 9
=1 + π§π§β1 2
2ππ 1β π§π§β1 + 0.1
2+ 9 1 + π§π§β1 2
105
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
IIR Filter Design by Bilinear Transform
Example: π π β 2ππ1βπ§π§β1
1+π§π§β1, π§π§ β 1+π π ππ/2
1βπ π ππ/2
π»π» π§π§ =1 + π§π§β1 2
2ππ 1β π§π§β1 + 0.1
2+ 9 1 + π§π§β1 2
106
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Outlineβ’ Designing FIR Filters with Windowsβ’ Designing FIR Filters with Frequency Samplingβ’ Designing FIR Filters with Equi-ripplesβ’ Designing IIR Filters with Discrete Differentiationβ’ Designing IIR Filters with Impulse Invarianceβ’ Designing IIR Filters with the Bilinear Transformβ’ Related Analog Filters
Lecture 22: Design of FIR / IIR FiltersFoundations of Digital Signal Processing
107
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Multi-pole Filters Butterworth: Maximally flat passband
Chebyshev: Faster cutoff with passband ripple
Elliptic: Fastest cutoff with passband and stopband ripple
108
Frequency [Hz]
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Mag
nitu
de R
espo
nse
0
0.5
1
1.5
Frequency [Hz]
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Phas
e R
espo
nse
-30
-20
-10
0
10
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Butterworth Filter Butterworth Filter of order N
109
π»π» ππππ =1
βππ=1ππ π π β π π ππ π π ππ = ππππ 2ππ+ππβ1 ππ
2ππ
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Butterworth Filter Butterworth Filter of order N N equally spaced poles on a circle on the left-hand-side of the s-
plane
110
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Butterworth Filter Properties of the Butterworth Filter It is maximally flat at ππ = 0
It has a cutoff frequency π»π» ππ = 12
at ππ = ππππ
For large ππ, it becomes an ideal filter
111
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Chebyshev Filter Chebyshev Filter of order N
112
π»π» ππππ =1
1 + ππ2πΆπΆππ2 ππ
πΆπΆππ2 ππ is an nth-order Chebyshev Polynomial ππ2 controls ripple
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Chebyshev Filter Chebyshev Filter of order N
113
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Chebyshev Filter Properties of the Chebyshev Filter It has ripples in the passband and is smooth in the stopband.
The ratio between the maximum and minimum ripples in the passband is
(1 + ππ2)β 1/2
If Ο΅ is reduced (i.e., the ripple size is reduced), then the stopband attenuation is reduced.
It has a sharper cut-off than a Butterworth filter, but at the expense of passband rippling
114
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Elliptic Filter Elliptic Filter of order N
115
π»π» ππππ =1
1 + ππ2π π ππ2 ππ
π π ππ2 ππ is an nth-order elliptic function ππ2 controls ripple
Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Elliptic Filter Elliptic Filter of order N
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Lecture 22: Designing FIR / IIR FiltersFoundations of Digital Signal Processing
Elliptic Filter Properties of the Elliptic Filter It has ripples in the passband and the stopband
The ratio between the maximum and minimum ripples is larger than the Chebyshev filter, but it has an even quicker transition from passband to stopband
It has poles and zeros, but they are much more difficult to compute compared with the Butterworth and Chebyshev filters
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