1. 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

2. 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.

3. 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 e x 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.

4. An S-domain Plot Note: Laplace transforms time domain into a three dimensional space Note The poles are the peaks The zeroes are the valleys Note The vocal tract can be modeled with a series of poles placed correctly

5. The Sync Filter Illustration Transform the rectangular time domain pulse into the frequency domain and Into the s-domain

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

7. Compare S-plane to Z-plane 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 – 2 π (frequency circle) Plots S and Z: Upper and lower half are mirror images of each other 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 Note: Nyquist = 2π

8. Z- Domain Notch Filter Z 1 = 1.00 ei(π/4), Z 2 = 1.00 ei(-π/4) P 1 =0.9ei(π/4), P 2 =0.9ei(-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 Note: Compare to the s-plane example

9. Designing a Z-domain filter 1.Pick the points for the zeroes and poles 2.Display a Z-plot using MatLab or Octave 3.When you are satisfied, convert the polar notation to rectangular form (Me -iθ to x + iy) 4.Multiply the complex polynomials 5.Collect terms 6.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

10. 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

11. 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)t dt=∫ -∞,∞ x(t)e -st dt

12. Example of Laplace Transform The Fourier transform is the Laplace Transform when σ=0 Laplace Transform tables can be used to eliminate the calculations

13. The S-Domain 1.Each S-Domain point has a basis function 2.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

14. 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

15. The z-transform Definition: X(z) = ∑ n=- ∞,∞ x[n]z -n where z is a complex variable Notes – If variable z = e j2πk/Ts it becomes the Fourier Transform (t s =sample size) – Using all of the other z points maps to the full Z-domain Linear Time Invariant (LTI) Characteristics – Linear: Z{ax n + by n } = Z{ax n } + Z{by n } – Time Delay: if y m = s n-k then Y z =z -k X z Note: m = n-k and n = m+k Y z = ∑ n=-∞,∞ s[n-k]z -n = ∑ n=-∞,∞ s[m]z -(m+k) = z -k ∑ n=-∞,∞ s[m]z -m = z -k X z – Note: The y m signal is the x n 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

16. Deriving the Transfer Function 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 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 } Linearity: 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 } By 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 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 ) 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 ) Frequency domain multiply = time domain convolution: y n = x n *h n

17. 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 H[z] = (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 roots tell us where the zeroes and poles are

18. 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

19. 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 – φ controls the resonant frequency  When φ≠0, filter coefficients will be complex numbers  Filter pairs will be real if they are conjugate pairs (re iφ,re -iφ )  If r < 0.5, the relationship between φ and frequency breaks down because of the “skirts” of the poles

20. 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 Note: Normalized Angular Frequencies range from 0 to 2π