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 ndimensional 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 modeled
with 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 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
Note: the Lines to left
of s-plane correspond
to circles within the unit
circle in the z-plane
Note: Nyquist = 2π
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
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
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
1. Each S-Domain point has a basis function
2. Like Fourier Transform, the top half is a
mirror image of the bottom
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
Real part basis functions
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/Ts it 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-kXz Note: 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 Terms
Yz - 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 function
Yz/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 origin
3-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
x x
x
x x
x
Poles
x
Examples
x
x
x
x
x
x
r
φ
a1
a2
0.8
0.75 0.58±0.54j
1.17 -1.64
0.8
1.0
0.43±0.67j
0.86 -0.64
0.8
1.25 0.25±0.76j
0.8
1.5
x
Freq
Bandwth r
φ
Poles
F1
300
250
0.95 0.12 0.963±0.116j
0.50 -0.64
F2
2200 250
0.95 0.86 0.619±0.719j
0.056±0.78j 0.11 -0.64
F3
3000 250
0.95 1.17 0.370±0.874j
Note: Normalized Angular Frequencies range from 0 to 2π