ES97H Biomedical Signal Processing Lecturer: Dr Igor Khovanov Office: D207 i.khovanov@warwick.ac.uk Syllabus: Biomedical Signal Processing. Examples of signals.   Linear System Analysis. Laplace Transform. Transfer Function.   Frequency Response. Fourier Transform. Discrete Signal Analysis. Digital (discrete-time) systems. Z-transform. Filtering. Digital Filters design and application. Case Study.

The Z-transform Consider the continuous time signal x(t); sample it with the rate 1/T the discrete-time signal (time-series) Take Laplace transform xn is not a function of t, so it can be considered as a constant. Set ; note that s and z are complex The Z-transform The inverse Z-transform

The Z-transform The Z-transform for discrete-time linear systems plays same role as the Laplace transform for continuous-time linear systems. A general form of a discrete-time linear system corresponds to the linear difference equation It is similar to ODEs, but derivatives are replaced by finite-differences Derivatives are replaced by the time-shift operator z-1 (in Z-space) ai and bi are coefficients xn is the input yn is the output

The Z-transform. Properties Linearity Shift to the right (delay) x[n]=xn x[n-3]=xn-3

The Z-transform. Properties Shift to the left x[n]=xn x[n+3]=xn+3 Remind the Laplace transform of a derivative

The Z-transform. Properties Convolution in the discrete-time domain Summation in the Discrete−Time Domain (similar to integration) Initial value theorem Final value theorem

The Z-transform of common discrete-time function. Delta-function Step-function (it specifies the causality) Here we use Note that the transform exists for |z|<1 only

The Z-transform of common discrete-time function. Geometric sequence (assume the action of u0) we use Discrete-time exponent (assume the action of u0) we use

The Z-transform of common discrete-time function. Discrete-time cosine and sine functions (assume the action of u0) We can use Consider as exercises Note z=esT

Transformation Between s and z Domains Discrete-time Laplace transform T is the sampling period Z-transform So Recall that s is complex and it specifies the system properties. wS is the angular sampling frequency

Transformation Between s and z Domains The polar form relation is The second term defines the unity circle; the unit circle represents frequencies from zero to the sampling frequency and the frequency response is the discrete−time transfer function evaluated on the unit circle. Cases:

Calculation of the Z-transform The connection the Laplace and Z- transforms via the contour integral a contour C enclosing all singularities (poles) of the Laplace transform. The integral can be expressed via Cauchy’s Residue Theorem pk is k-th pole of the Laplace transform Comment. As a rule tables are used.

The inverse Z-transform The inverse transform is Can be found via a. Partial Fraction Expansion b. The Inversion Integral c. Long Division of polynomials The partial Fraction Expansion method is very similar to the partial fraction expansion method that we used in finding the Inverse Laplace transform, that is, we expand into a summation of terms whose inverse is known. The terms are here k is a constant, r and p represent the residues and poles respectively; these can be real or complex.

The inverse Z-transform Example. Partial Fraction Expansion (via residues) Multiply by Next form Calculate residues

The inverse Z-transform Example. Partial Fraction Expansion (via residues) Multiply by Next form Calculate residues Finally,

The inverse Z-transform Example. Partial Fraction Expansion (without residues) Multiply by Next form Need to solve Matlab use syms n z xn=2*(0.5)^n−9*(0.75)^n+8; % This is the answer Xz=ztrans(xn,n,z); simple(Xz) % Verify answer by first taking Z transform of x[n] ans = 8*z^3/(2*z-1)/(4*z-3)/(z-1) iztrans(Xz) % Now, verify that Inverse of F(z) gives back f[n] 2*(1/2)^n-9*(3/4)^n+8

The Z-transform The Z-transform for discrete-time linear systems plays same role as the Laplace transform for continuous-time linear systems. A general form of a discrete-time linear system corresponds to the linear difference equation It is similar to ODEs, but derivatives are replaced by finite-differences Derivatives are replaced by the time-shift operator z-1 (in Z-space) ai and bi are coefficients xn is the input yn is the output

The Transfer Function of Discrete−Time Systems A general form of a discrete-time linear system corresponds to the linear difference equation ai and bi are coefficients xn is the input yn is the output Assuming that all initial conditions are zero, taking the Z transform of both sides Express Y(z) as a function of X(z) The transfer function H(z)

The Transfer Function of Discrete−Time Systems Block diagrams.

Response to unit pulse (impulse) d [n] Consider the input x[n]=d[n]. Since X(z)=1 and the transfer function The discrete−time impulse response h[n] can be found by taking the inverse Z transform of the discrete transfer function H(z) Consider the structure of H(z) Here pi are poles; ri are residues.

Response to unit pulse (impulse) d [n] Consider the structure of H(z) Here pi are poles; ri are residues . Using the Z-transform of the function an Write down the inverse Z-transform, that is the discrete−time impulse response h[n] The unit circle is the boundary between impulse responses that decay with time (when all poles |pi|<1) and those that increase in an unbounded fashion (if at least one |pi|>1). The stable system must have its poles inside (or possibly on) the unit circle.

Response to unit pulse (impulse) d [n] versus poles

Response to unit pulse (impulse) d [n] versus poles

complex-valued exponential signal Frequency response complex-valued exponential signal LTI system LTI system output impulse response Use the convolution theorem Frequency response: By replacing assume the sampling rate is 1, T=1

LTI system output: Frequency response:

LTI system Amplitude (magnitude) response: Phase response: Sinusoid signal LTI system output

constant coefficient linear difference equation Time – Domain: constant coefficient linear difference equation h(n) Z – Domain: transfer function Frequency – Domain: frequency response Z Z-1 FT-1 FT

Example The difference equation describing the input−output relationship of a discrete−time system with zero initial conditions, is Compute: The transfer function H(z) b. The frequency response H(jw) c. Find the poles and zeros and plot them in the complex Z-plane. Comment on the stability of the system.

Compute: The transfer function H(z) Take the Z-transform

Compute: b. The frequency response H(jw) Substitute z=e jw Long calculations (!) So we skip and consider a simpler example

c. Find the poles and zeros and plot them in the complex Z-plane c. Find the poles and zeros and plot them in the complex Z-plane. Comment on the stability of the system. Zeros Poles The system is stable.

Calculating the frequency response. Simpler example.

Example. The impulse response (unit-pulse response) of the system is Determine its transfer function and frequency response