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Transformations. Definition: A mapping of one n-dimensional space onto another k-dimensional space, which could be itself. Example: Mapping a three dimensional object (pre-image) onto a two dimensional plane (image under the transformation)

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## Transformations

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**Transformations**• Definition: A mapping of one n-dimensional space onto another k-dimensional space, which could be itself. • Example: Mapping a three dimensional object (pre-image) onto a two dimensional plane (image under the transformation) • Example: Fourier transform maps an n-dimensional time signal onto an n dimensional (–f/2 to f/2) dimension frequency domain**Why do we need transforms?**• Some problems are easier to solve in a transformed space. • Procedure • Transform the problem • Perform an easier algorithm in the transformed space • Transform back to get the solution • Example: In digital signal processing, it is not so easy to determine the filter coefficients.**Z and Laplace Transforms**• Laplace Transform • Deals with continuous audio signals • Transform time domain continuous signals to s-domain • Laplace (1749-1827) invented this technique to solve differential equations. It takes advantage of the fact that the derivative of ex is itself. • Z-transform • Deals with discrete signals • Transform time domain digital signals to the z-domain • Application to digital signal processing • It is an extension to Fourier Transforms • Enables us to determine the coefficients analog (Laplace) for IIR (Z) filters.**An S-domain Plot**Note The vocal tract can be modeledwith a series of poles placed correctly Note The poles are the peaks The zeroes are the valleys Note: Laplace transforms time domain into a three dimensional space**The Sync Filter Illustration**Transform the rectangular time domain pulse into the frequency domain and Into the s-domain**S-domain for a notch filter**Notation x = pole ○ = zero Fourier frequencies are the points where the real value is zero**Compare S-plane to Z-plane**• Filter designS: analog filters; Z:IIR filters • Equations S:differential equations, Z: difference equations • Filter Points S: rectangular along i axis, Z: polar around unit circle • FrequenciesS: -∞ to ∞ (frequency line). Z: 0 – 2 π (frequency circle) • PlotsS and Z: Upper and lower half are mirror images of each other Note: Nyquist = 2π Note: the Lines to left of s-plane correspond to circles within the unit circle in the z-plane Looking down vertically from the top of the domain**Z- Domain Notch Filter**Z1 = 1.00 ei(π/4), Z2 = 1.00 ei(-π/4) P1=0.9ei(π/4), P2=0.9ei(-π/4) Z1=0.7071+0.7071i Z2=0.7071-0.7071i P1=0.6364+0.6364i P2=0.6364-0.6364i Note: Compare to the s-plane example**Designing a Z-domain filter**• Pick the points for the zeroes and poles • Display a Z-plot using MatLab or Octave • When you are satisfied, convert the polar notation to rectangular form (Me-iθ to x + iy) • Multiply the complex polynomials • Collect terms • Use the coefficients for the filter Note: We’ll describe steps 4 through 6 later Note: An all pole filter is a popular way to model the vocal tract**Stable and unstable IIR filters**• Exponential Decay • Points close to the origin exhibit an exponential decay in the time domain • Points outside the unit circle have time responses that grow exponentially • Points on the unit circle exhibit constant response in the time domain • Filters with resonant points outside this circle are unstable and useless, leading to arithmetic overflows • Filter design involves picking points just inside the unit circle for zeroes and poles**Fourier verses Laplace**• The basis functions • Fourier: Sinusoids • Laplace: Sinusoids and exponentials • Transformation • Fourier: frequencies • Laplace: frequencies and exponential constants • Fourier Transform: X(ω) = ∫-∞,∞x(t) e-i ωt dt • Equations • Fourier:X(ω) = ∫-∞,∞x(t) e-i ωt dt • Laplace:X(σ,ω)=∫-∞,∞[x(t) e-σt]e-iωt dt=∫-∞,∞x(t)e-(σ+iω)tdt=∫-∞,∞x(t)e-stdt**Example of Laplace Transform**Laplace Transform tables can be used to eliminate the calculations The Fourier transform is the Laplace Transform when σ=0**The S-Domain**• Each S-Domain point has a basis function • Like Fourier Transform, the top half is a mirror image of the bottom Real part basis functions Re X(σ=1.5,ω=±40) = ∫-∞,∞ x(t)cos(40t)e-1.5t dt ImX(σ=1.5,ω=±40) = ∫-∞,∞ x(t)sin(40t)e-1.5t dt**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**The z-transform**• Definition: X(z) = ∑n=- ∞,∞x[n]z-n where z is a complex variable • Notes • If variable z = ej2πk/Tsit becomes the Fourier Transform (ts=sample size) • Using all of the other z points maps to the full Z-domain • Linear Time Invariant (LTI) Characteristics • Linear: Z{axn + byn} = Z{axn} + Z{byn} • Time Delay: if ym = sn-k then Yz =z-kXzNote: m = n-k and n = m+k Yz=∑n=-∞,∞ s[n-k]z-n =∑n=-∞,∞ s[m]z-(m+k) =z-k ∑n=-∞,∞ s[m]z-m = z-k Xz • Note:The ym signal is the xn signal delayed by k samples. The result is that the Z transform of y is the same of the Z transform of x, but multiplied by z-k. • Importance: Time delay and linear properties helps us find the linear coefficients**Deriving the Transfer Function**• IIR Definition: yn = b0xn + b1xn-1 +…+ bMxn-M + a1yn-1 +…+ aNyn-N • Z transform both sides: Yz=Z{b0xn+b1xn-1+…+bMxn-M+a1yn-1+…+aNyn-N} • Linearity: Yz = Z{b0xn}+Z{b1xn-1}+…+Z{bMxn-M}+Z{a1yn-1}+…+Z{aNyn-N} • By time delay property (Z{xn-k} = xzz-k)Yz = b0Xz + b1Xzz-1 +…+bn-MXzz-M + a1Yzz-1+…+aNYzz-N • Gather TermsYz - a1Yzz-1-…- aNYzz-N = b0Xz + b1Xzz-1 +…+ bn-MXz-M Yz(1- a1z-1-…- aNz-N ) = Xz(b0 + b1z-1 +…+ bn-Mz-M) • Divide to get transfer functionYz/Xz= Hz= (b0 + b1z-1 +…+ bn-Mz-M)/(1- a1z-1-…- aNz-N ) Yz = Xz (Yz/Xz) • Frequency domain multiply = time domain convolution: yn = xn*hn**Transfer Function Example**• Suppose • b0=0.389, b1=-1.558, b2=2.338, b3=-1.558, b4=0.389 • a1=2.161, a2=-2.033, a3=0.878, a4=-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)= 0.389z4 - 1.558z3+2.338z2-1.558z1+0.389 /(z4-2.161z3 + 2.033z2 -0.878z1+0.161) • Find the roots • H[z] = (z-z1)(z-z2)(z-z3)(z-z4) / (z-p1)(z-p2)(z-p3)(z-p4) Note: The roots tell us where the zeroes and poles are**Modeling the Vocal Tract**• Numerator and Denominator of a transfer function • Each have a polynomial of some degree d having d roots • The roots of the numerator are called zeroes • The roots of the denominator are called poles • Formant frequencies of the vocal tract • They do not affect F0 • The alter the harmonics of F0 • Correctly placed poles within the ZPlane unit circle mimic the formants**Pole Placement**• Poles characterized by: • Amplitude: height of the resonance • Frequency: placement in the spectrum • Bandwidth: Sharpness of the pole • Place pole at reiφ • Amplitude and bandwidth shrink as r approaches the origin3-db down (half power) estimate = -2 ln(r) or 2(1-r)/r½ • φ controls the resonant frequency • When φ≠0, filter coefficients will be complex numbers • Filter pairs will be real if they are conjugate pairs (reiφ,re-iφ) • If r < 0.5, the relationship between φ and frequency breaks down because of the “skirts” of the poles**Examples**x x x x x x x x x x x x x x Note: Normalized Angular Frequencies range from 0 to 2π

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