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Infinite Impulse Response (IIR) Filters Uses both the input signal and previous filtered values y[n] = b 0 x[n] + b 1 x[n-1] + b 2 x[n-2] + … + a 1 y[n-1] + a 2 y[n-2] + a 3 y[n-3] + … The b k coefficients comprise the FIR part; filtering the input signal The a k coefficients comprise the IIR part; additional filtering using previously filtered values This concept has overlap with neural networks, a popular algorithm that statistically “learns” Sometimes called recursive filters

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IIR Filter Code public static double[] convolution(double[] signal, double[] b, double[] a) { double[] y = new double[signal.length + b.length - 1]; for (int i = 0; i < signal.length; i ++) { for (int j = 0; j < b.length; j++) { if (i-j>=0) y[i] += b[j]*signal[i - j]; } if (a!=null) { for (int j = 1; j < a.length; j ++) { if (i-j>=0) y[i] -= a[j] * y[i - j]; } } return y; }

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Characteristics of Recursive Filters Advantages –Powerful filtering with very few parameters –Execute very fast Example 1: b 0 =.15 and a 1 =.85 Example 1: b 0 = 0.93 b 1 = -0.93 a 1 = 0.86 Input Signal Example 1 output Example 2 output 0.0 1.0

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Low and High Pass Recursive Filter Low Pass: b 0 = 1-x, a 1 = x High Pass: b 0 = (1+x)/2, b 1 = -(1+x)/x, a 1 = x 0≤x≤1, which controls the low and high pass boundaries

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011.

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IIR Disadvantages float x = 1 for (int i=0; i

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IIR Moving Average Filter Illustration (Sum of seven previous samples) y[50] = x[47]+x[48]+x[49]+x[50]+x[51]+x[52]+x[53] y[51] = x[48]+x[49]+x[50]+x[51]+x[52]+x[53]+x[54] = y[50] + x[54] – x[47] Example: {1,2,3,4,5,4,3,2,1,2,3,4,5}; M = 4 –Starting filtered values: {¼, ¾, 1 ½, 2 ½, … } –Next value: y 4 = 2 ½ + (5 – 1)/4 = 3 ½ Filter of degree (length) M –Centered Version: y n = y n-1 + x n+(M-1)/2 - x n-(M-1)/2 – 1 –Non Centered Version: y n = y n-1 + x n /M –x n-M /M Two additions per point no matter the filter length Note: Integers work best with this approach to avoid round off drift

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Transforms Procedure –Transform the problem into a different domain –Execute a simpler algorithm in the transformed space –Transform back to get the solution DSP Example: We need a well-defined way to determine IIR filter coefficients that result in a filter that performs properly Image Processing : Mapping a three dimensional image into two dimension Some problems are difficult (or impossible) to solve directly

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Laplace Transform The Fourier domain is a one dimensional space; each point represents a sinusoid of a particular frequency The Laplace domain (S- Domain) is a two dimensional space of complex numbers; each point represents sinusoids of a particular frequency that either amplifies with higher y-axis values or degrades with lower y-axis values Works with continuous functions. Positive frequency sinusoid Negative frequency sinusoid Fourier Domain Laplace Domain Positive frequency sinusoid that amplifies at a positive rate Positive frequency sinusoid that exponentially decay (attenuates) Imaginary axis Real axis

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The S-Domain 1.Each S-Domain point models a basis function 2.The top half is a mirror image of the bottom Real axis Imag axis Note: The waveforms are counterintuitive since filters convolute from the current time backward. Attenuating waves, actually produces unstable filters

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Laplace Transform Example The Fourier transform is the Laplace Transform when σ=0

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Laplace Transform Consider a process whose characteristics change over time Some function governs the changing behavior of the process We observe the process as a signal measured at points of time We want to predict the outputs, when the input system interacts with a filtering system process. We can model this problem with a differential equation, where time is the independent variable: a 1 y ’’’(t) + a 2 y’’(t) + a 3 y’(t) = b 1 x’(t) +b 2 x 2 x(t) A technique for solving differential equations

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Laplace Transform Definition s is a complex number (σ+jω) where σ controls the degree of exponential decay and ω controls the frequency of the basis function e -st = e -(σ+jω)t = e -jωt / e σt The denominator controls the decay rate The numerator controls the frequency The integral starts if we are not interested in negative time (time past) dt

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011. Impact of σ, ω values on the resulting basis function Assuming σ is negative

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Poles and Zeroes The integrations causes S-plane some points to: –Evaluate to 0 (these are zeroes) –Evaluate to ∞ (these are poles) Pole and zero points determine IIR filter coefficients Note: Designing a filter comes down to picking the pole and zero points on the S-plane dt

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An S-domain Plot Note The poles are the peaks The zeroes are the valleys Note: The third dimension is the magnitude of ∫ f(t)e -st dt at s = (σ+jω)

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011.

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Time domain impulse response

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011. Attenuating impulse response Amplifying impulse response

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The Low Pass Filter Illustration Transform the rectangular time domain pulse into the frequency domain and Into the s-domain

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S-domain for a notch filter Fourier frequencies are the points where the real value is zero Notation x = pole ○ = zero

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The z-transform Z-transform is the discrete cousin to the Laplace Transform Laplace Transform (s = σ+jω) – Extends the Fourier Transform, uses integrals and continuous functions – s = e -(σ+jω)t becomes the Fourier transform when ω = 0 – Fourier points fall along the imaginary axis – S-domain stable region is on the negative half of the domain Z-Transform (z = re -2 π k/t s ) – Extends the Discrete Fourier Transform, uses sums and discrete samples – z = e -j2πk/t s becomes the Discrete Fourier Transform (t s =sample size) – Four transform points fall along the unit circle – Z-domain stable domain are those within the unit circle Z{x[n]} =

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Compare S-plane to Z-plane (cont.) Filter design S: analog filters; Z: IIR filters Equations S: differential equations, Z: difference equations Filter Points S: rectangular along i axis, Z: polar around unit circle Frequencies S: -∞ to ∞ (frequency line). Z: 0 to 2 π (frequency circle) Plots S and Z: Upper and lower half are mirror images Looking down vertically from the top of the domain Note: the Lines to left of s-plane correspond to circles within the unit circle in the z-plane Negative frequencies Positive frequencies

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011.

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Analog Filter Example σ=(A-3)/2RC; ω=±(A 2 +6A-5) ½ /2RC Sallen-Key Filters on a circle of 1/RC radius A = amplification R= resistance C = capacitance

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ButterWorth Filters ButterWorth poles are equally spaced around the left side of a circle

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Example Low Pass Filters Butterworth poles are equally spaced on a circle Chebshev poles are equally spaced on an ellipse Elliptic poles are on an ellipse; zeroes added to the stopband north and south of the poles

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Examples rφPolesa1a2 0.80.750.58±0.54j1.17-1.64 0.81.00.43±0.67j0.86-0.64 0.81.250.25±0.76j0.50-0.64 0.81.50.056±0.78j0.11-0.64 x x x x x x x x FreqBandwthrφPoles F13002500.950.120.963±0.116j F222002500.950.860.619±0.719j F330002500.951.170.370±0.874j x x x x x x There are tables of pole/zero points and their effects

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Pole Placement Poles characterized by: – Amplitude: height of the resonance – Frequency: placement in the spectrum – Bandwidth: Sharpness of the pole Place pole at re iφ – Amplitude and bandwidth shrink as r approaches the origin 3-db down (half power) estimate = -2 ln(r) or 2(1-r)/r ½ – ω controls the resonant frequency When ω ≠0, IIR coefficients will be complex numbers IIR coefficients real if there are conjugate pairs (re iφ,re -iφ ) If r < 0.5, the relationship between φ and frequency breaks down because the pole “skirts” cause results to be less precise

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Transfer Function 1.IIR Definition: y n = b 0 x n + b 1 x n-1 +…+ b M x n-M + a 1 y n-1 +…+ a N y n-N 2.Z transform both sides: Y z =Z{b 0 x n +b 1 x n-1 +…+b M x n-M +a 1 y n-1 +…+a N y n-N } 3.Linearity Property: Y z = Z{b 0 x n }+Z{b 1 x n-1 }+…+Z{b M x n-M }+Z{a 1 y n-1 }+…+Z{a N y n-N } 4.Time delay property (Z{x n-k } = x z z -k ) Y z = b 0 X z + b 1 X z z -1 +…+b n-M X z z -M + a 1 Y z z -1 +…+a N Y z z -N 5.Gather Terms Y z - a 1 Y z z -1 -…- a N Y z z -N = b 0 X z + b 1 X z z -1 +…+ b n-M X z -M Y z (1- a 1 z -1 -…- a N z -N ) = X z (b 0 + b 1 z -1 +…+ b n-M z -M ) 6.Divide to get transfer function Y z /X z = H z = (b 0 + b 1 z -1 +…+ b n-M z -M )/(1- a 1 z -1 -…- a N z -N ) Y z = X z (Y z /X z ) = X z (H z ) // H z is the transform function for the filter 7.Perform Inverse Z transform: y[n] = x[n] * filter[n] // where * is convolution Because Z domain multiplication is time domain convolution A polynomial equation that defines filter coefficients for particular Z,S domain setting Starting with the IIR definition, derive Z-Transform Transfer Function Z{x[n]} =

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011.

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Transfer Function Example Suppose – b 0 =0.389, b 1 =-1.558, b 2 =2.338, b 3 =-1.558, b 4 =0.389 – a 1 =2.161, a 2 =-2.033, a 3 =0.878, a 4 =-0.161 Transfer Function H[z] = 0.389 - 1.558z -1 +2.338z -2 -1.558z -3 +0.389z -4 / (1-2.161z -1 + 2.033z -2 -0.878z -3 +0.161z -4 ) = 0.389z 4 - 1.558z 3 +2.338z 2 -1.558z 1 +0.389 / (z 4 -2.161z 3 + 2.033z 2 -0.878z 1 +0.161) Find the roots = (z-z 1 )(z-z 2 )(z-z 3 )(z-z 4 ) / (z-p 1 )(z-p 2 )(z-p 3 )(z-p 4 ) Note: The bottom form tells us where the zeroes and poles are

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011.

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Notch Filter Z 1 = 1.00 e i(π/4), Z 2 = 1.00 e i(-π/4) P 1 =0.9e i(π/4), P 2 =0.9e i(-π/4) Z 1 =0.7071+0.7071i Z 2 =0.7071-0.7071i P 1 =0.6364+0.6364i P 2 =0.6364-0.6364i 1.Convert to rectangular form 2.Multiply the complex polynomials 3.Collect terms 4.Use the coefficients for the filter Note: Compare to the s-plane example

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011. Second-order low pass IIR filter examples

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011.

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Linear filters can be combined into parallel or serial systems

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011. Note: It’s easier to design multiple two tap systems than a larger multiple tap system

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Understanding Digital Signal Processing, Third Edition, Richard Lyons (0-13-261480-4) © Pearson Education, 2011. Original Gain: (b 0 +b 1 )/(1 - a 1 ) = 0.22 + 0.25 / (1-0.87) = 3.614 Do normalize gain, divide coefficients by 3.64

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