1. ECE 352 Systems II Manish K. Gupta, PhD Office: Caldwell Lab 278 Email: guptam @ ece. osu. eduguptam @ ece. osu. edu Home Page: http://www.ece.osu.edu/~guptamhttp://www.ece.osu.edu/~guptam TA: Zengshi Chen Email: chen.905 @ osu. eduZengshi Chen Office Hours for TA : in CL 391: Tu & Th 1:00-2:30 pm Home Page: http://www.ece.osu.edu/~chenz/

2. Acknowledgements Various graphics used here has been taken from public resources instead of redrawing it. Thanks to those who have created it. Thanks to Brian L. Evans and Mr. Dogu Arifler Thanks to Randy Moses and Bradley Clymer

3. ECE 352 Slides edited from: Prof. Brian L. Evans and Mr. Dogu Arifler Dept. of Electrical and Computer Engineering The University of Texas at Austin course: EE 313 Linear Systems and Signals Fall 2003

4. Z-transforms

5. For discrete-time systems, z-transforms play the same role of Laplace transforms do in continuous-time systems As with the Laplace transform, we compute forward and inverse z-transforms by use of transforms pairs and properties Bilateral Forward z-transformBilateral Inverse z-transform

6. Region of Convergence Region of the complex z- plane for which forward z-transform converges Im{z} Re{z} Entire plane Im{z} Re{z} Complement of a disk Im{z} Re{z} Disk Im{z} Re{z} Intersection of a disk and complement of a disk Four possibilities (z=0 is a special case and may or may not be included)

7. Z-transform Pairs h[k] =  [k] Region of convergence: entire z- plane h[k] =  [k-1] Region of convergence: entire z- plane h[n-1]  z -1 H(z) h[k] = a k u[k] Region of convergence: |z| > |a| which is the complement of a disk

8. Stability Rule #1: For a causal sequence, poles are inside the unit circle (applies to z-transform functions that are ratios of two polynomials) Rule #2: More generally, unit circle is included in region of convergence. (In continuous-time, the imaginary axis would be in the region of convergence of the Laplace transform.) –This is stable if |a| < 1 by rule #1. –It is stable if |z| > 1 > |a| by rule #2.

9. Inverse z-transform Yuk! Using the definition requires a contour integration in the complex z-plane. Fortunately, we tend to be interested in only a few basic signals (pulse, step, etc.) –Virtually all of the signals we’ll see can be built up from these basic signals. –For these common signals, the z-transform pairs have been tabulated (see Tables)

10. Example Ratio of polynomial z- domain functions Divide through by the highest power of z Factor denominator into first-order factors Use partial fraction decomposition to get first-order terms

11. Example (con’t) Find B 0 by polynomial division Express in terms of B 0 Solve for A 1 and A 2

12. Example (con’t) Express X[z] in terms of B 0, A 1, and A 2 Use table to obtain inverse z-transform With the unilateral z- transform, or the bilateral z-transform with region of convergence, the inverse z-transform is unique.

13. Z-transform Properties Linearity Shifting

14. Z-transform Properties Convolution definition Take z-transform Z-transform definition Interchange summation r = k - m Z-transform definition

15. Example

16. Difference Equations

17. Linear Difference Equations Discrete-time LTI systems can be characterized by difference equations y[k] = (1/2) y[k-1] + (1/8) y[k-2] + f[k] Taking z-transform of the difference equation gives description of the system in the z-domain  f[k]f[k]y[k]y[k] Unit Delay 1/2 1/8 + + + y[k-1] y[k-2]

18. Advances and Delays Sometimes differential equations will be presented as unit advances rather than delays y[k+2] – 5 y[k+1] + 6 y[k] = 3 f[k+1] + 5 f[k] One can make a substitution that reindexes the equation so that it is in terms of delays Substitute k with k-2 to yield y[k] – 5 y[k-1] + 6 y[k-2] = 3 f[k-1] + 5 f[k-2] Before taking the z-transform, recognize that we work with time k  0 so u[k] is often implied y[k-1] = y[k-1] u[k]  y[k-1] u[k-1]

19. Example System described by a difference equation y[k] – 5 y[k-1] + 6 y[k-2] = 3 f[k-1] + 5 f[k-2] y[-1] = 11/6, y[-2] = 37/36 f[k] = 2 -k u[k]

20. Transfer Functions Previous example describes output in time domain for specific input and initial conditions It is not a general solution, which motivates us to look at system transfer functions. In order to derive the transfer function, one must separate –“Zero state” response of the system to a given input with zero initial conditions –“Zero input” response to initial conditions only

21. Transfer Functions Consider the zero-state response –No initial conditions: y[-k] = 0 for all k > 0 –Only causal inputs: f[-k] = 0 for all k > 0 Write general n th order difference equation

22. Stability Knowing H[z], we can compute the output given any input Since H[z] is a ratio of two polynomials, the roots of the denominator polynomial (called poles) control where H[z] may blow up H[z] can be represented as a series –Series converges when poles lie inside (not on) unit circle –Corresponds to magnitudes of all poles being less than 1 –System is said to be stable H[z]H[z] Y[z]Y[z]F[z]F[z]

23. Relation between h[k] and H[z] Either can be used to describe the system –Having one is equivalent to having the other since they are a z-transform pair –By definition, the impulse response, h[k], is y[k] = h[k] when f[k] =  [k] Z{h[k]} = H[z] Z{  [k]}  H[z] = H[z] · 1 h[k]  H[z] Since discrete-time signals can be built up from unit impulses, knowing the impulse response completely characterizes the LTI system

24. Complex Exponentials Complex exponentials have special property when they are input into LTI systems. Output will be same complex exponential weighted by H[z] When we specialize the z-domain to frequency domain, the magnitude of H[z] will control which frequencies are attenuated or passed.

25. Z and Laplace Transforms

26. Are complex-valued functions of a complex frequency variable Laplace: s =  + j 2  f Z: z = e j  Transform difference/differential equations into algebraic equations that are easier to solve

27. Z and Laplace Transforms No unique mapping from Z to Laplace domain or vice-versa –Mapping one complex domain to another is not unique One possible mapping is impulse invariance. –The impulse response of a discrete-time LTI system is a sampled version of a continuous-time LTI system. H[z]H[z] y[k]y[k]f[k]f[k] Z H[e sT ] Laplace

28. Z and Laplace Transforms H[z]H[z] y[k]y[k]f[k]f[k] Z H[e sT ] Laplace

29. Impulse Invariance Mapping Impulse invariance mapping is z = e s T 1 1 Im{s} Re{s} 1 Im{z} Re{z} s = -1  j  z = 0.198  j 0.31 (T = 1) s = 1  j  z = 1.469  j 2.287 (T = 1)

30. Sampling Theorem

31. s(t)s(t) t TsTs Sampling Many signals originate as continuous-time signals, e.g. conventional music or voice. By sampling a continuous-time signal at isolated, equally-spaced points in time, we obtain a sequence of numbers k  {…, -2, -1, 0, 1, 2,…} T s is the sampling period Sampled analog waveform

32. Shannon Sampling Theorem A continuous-time signal x(t) with frequencies no higher than f max can be reconstructed from its samples x[k] = x(k T s ) if the samples are taken at a rate f s which is greater than 2 f max. –Nyquist rate = 2 f max –Nyquist frequency = f s /2. What happens if f s = 2f max ? Consider a sinusoid sin(2  f max t) –Use a sampling period of T s = 1/f s = 1/2f max. –Sketch: sinusoid with zeros at t = 0, 1/2f max, 1/f max, …

33. Sampling Theorem Assumptions The continuous-time signal has no frequency content above the frequency f max The sampling time is exactly the same between any two samples The sequence of numbers obtained by sampling is represented in exact precision The conversion of the sequence of numbers to continuous-time is ideal

34. Why 44.1 kHz for Audio CDs? Sound is audible in 20 Hz to 20 kHz range: f max = 20 kHz and the Nyquist rate 2 f max = 40 kHz What is the extra 10% of the bandwidth used? Rolloff from passband to stopband in the magnitude response of the anti-aliasing filter Okay, 44 kHz makes sense. Why 44.1 kHz? At the time the choice was made, only recorders capable of storing such high rates were VCRs. NTSC: 490 lines/frame, 3 samples/line, 30 frames/s = 44100 samples/s PAL: 588 lines/frame, 3 samples/line, 25 frames/s = 44100 samples/s

35. Sampling As sampling rate increases, sampled waveform looks more and more like the original Many applications (e.g. communication systems) care more about frequency content in the waveform and not its shape Zero crossings: frequency content of a sinusoid –Distance between two zero crossings: one half period. –With the sampling theorem satisfied, sampled sinusoid crosses zero at the right times even though its waveform shape may be difficult to recognize

36. Aliasing Analog sinusoid x(t) = A cos(2  f 0 t +  ) Sample at T s = 1/f s x[k] = x(T s k) = A cos(2  f 0 T s k +  ) Keeping the sampling period same, sample y(t) = A cos(2  (f 0 + lf s )t +  ) where l is an integer y[k]= y(T s k) = A cos(2  (f 0 + lf s )T s k +  ) = A cos(2  f 0 T s k + 2  lf s T s k +  ) = A cos(2  f 0 T s k + 2  l k +  ) = A cos(2  f 0 T s k +  ) = x[k] Here, f s T s = 1 Since l is an integer, cos(x + 2  l) = cos(x) y[k] indistinguishable from x[k]

37. Aliasing Since l is any integer, an infinite number of sinusoids will give same sequence of samples The frequencies f 0 + l f s for l  0 are called aliases of frequency f 0 with respect f s to because all of the aliased frequencies appear to be the same as f 0 when sampled by f s

38. Generalized Sampling Theorem Sampling rate must be greater than twice the bandwidth –Bandwidth is defined as non-zero extent of spectrum of continuous-time signal in positive frequencies –For lowpass signal with maximum frequency f max, bandwidth is f max –For a bandpass signal with frequency content on the interval [f 1, f 2 ], bandwidth is f 2 - f 1

39. Difference Equations and Stability

40. Example: Second-Order Equation y[k+2] - 0.6 y[k+1] - 0.16 y[k] = 5 f[k+2] with y[-1] = 0 and y[-2] = 6.25 and f[k] = 4 -k u[k] Zero-input response Characteristic polynomial  2 - 0.6  - 0.16 = (  + 0.2) (  - 0.8) Characteristic equation (  + 0.2) (  - 0.8) = 0 Characteristic roots  1 = -0.2 and  2 = 0.8 Solution y 0 [k] = C 1 (-0.2) k + C 2 (0.8) k Zero-state response

41. Example: Impulse Response h[k+2] - 0.6 h[k+1] - 0.16 h[k] = 5  [k+2] with h[-1] = h[-2] = 0 because of causality In general, from Lathi (3.41), h[k] = (b 0 /a 0 )  [k] + y 0 [k] u[k] Since a 0 = -0.16 and b 0 = 0, h[k] = y 0 [k] u[k] = [C 1 (-0.2) k + C 2 (0.8) k ] u[k] Lathi (3.41) is similar to Lathi (2.41): Lathi (3.41) balances impulsive events at origin

42. Example: Impulse Response Need two values of h[k] to solve for C 1 and C 2 h[0] - 0.6 h[-1] - 0.16 h[-2] = 5  [0]  h[0] = 5 h[1] - 0.6 h[0] - 0.16 h[-1] = 5  [1]  h[1] = 3 Solving for C 1 and C 2 h[0] = C 1 + C 2 = 5 h[1] = -0.2 C 1 + 0.8 C 2 = 3 Unique solution  C 1 = 1, C 2 = 4 h[k] = [(-0.2) k + 4 (0.8) k ] u[k]

43. Example: Solution Zero-state response solution (Lathi, Ex. 3.13) y s [k] = h[k] * f[k] = {[(-0.2) k + 4(0.8) k ] u[k]} * (4 -k u[k]) y s [k] = [-1.26 (4) -k + 0.444 (-0.2) k + 5.81 (0.8) k ] u[k] Total response: y[k] = y 0 [k] + y s [k] y[k] = [C 1 (-0.2) k + C 2 (0.8) k ] + [-1.26 (4) -k + 0.444 (-0.2) k + 5.81 (0.8) k ] u[k] With y[-1] = 0 and y[-2] = 6.25 y[-1] = C 1 (-5) + C 2 (1.25) = 0 y[-2] = C 1 (25) + C 2 (25/16) = 6.25 Solution: C 1 = 0.2, C 2 = 0.8

44. Repeated Roots For r repeated roots of Q(  ) = 0 y 0 [k] = (C 1 + C 2 k + … + C r k r-1 )  k Similar to the continuous-time case Continuous Time Discrete TimeCase non-repeated roots repeated roots

45. Stability for an LTID System Asymptotically stable if and only if all characteristic roots are inside unit circle. Unstable if and only if one or both of these conditions exist: –At least one root outside unit circle –Repeated roots on unit circle Marginally stable if and only if no roots are outside unit circle and no repeated roots are on unit circle (see Figs. 3.17 and 3.18 in Lathi) 1    Unstable Stable Marginally Stable Lathi, Fig. 3.16 Re  Im 

46. Stability in Both Domains Discrete-Time Systems Marginally stable: non-repeated characteristic roots on the unit circle (discrete-time systems) or imaginary axis (continuous- time systems) 1 Unstable Stable Marginally Stable Re  Im  UnstableStable Marginally Stable Re Im Continuous-Time Systems

47. Frequency Response of Discrete-Time Systems

48. Frequency Response For continuous-time systems the response to sinusoids are For discrete-time systems in z-domain For discrete-time systems in discrete-time frequency

49. Response to Sampled Sinusoids Start with a continuous-time sinusoid Sample it every T seconds (substitute t = k T) We show discrete-time sinusoid with Resulting in Discrete-time frequency is equal to continuous- time frequency multiplied by sampling period

50. Example Calculate the frequency response of the system given as a difference equation as Assuming zero initial conditions we can take the z-transform of this difference equation Since

51. Example Group real and imaginary parts The absolute value (magnitude response) is

52. Example The angle (phase response) is where 0 comes from the angle of the nominator and the term after – comes from the denominator of Reminder: Given a complex number a + j b the absolute value and angle is given as

53. Example We can calculate the output of this system for a sinusoid at any frequency by substituting  with the frequency of the input sinusoid.

54. Discrete-time Frequency Response As in previous example, frequency response of a discrete-time system is periodic with 2  –Why? Frequency response is function of the complex exponential which is periodic with 2  : Absolute value of discrete-time frequency response is even and angle is odd symmetric. –Discrete-time sinusoid is symmetric around 

55. Aliasing and Sampling Rate Continuous-time sinusoid can have a frequency from 0 to infinity By sampling a continuous-time sinusoid, Discrete-time frequency  unique from 0 to  –We only can represent frequencies up to half of the sampling frequency. –Higher frequencies exist would be “wrapped” to some other frequency in the range.

56. Re Im Effect of Poles and Zeros of H[z] The z-transform of a difference equation can be written in a general form as We can think of complex number as a vector in the complex plane. –Since z and z i are both complex numbers the difference is again a complex number thus a vector in the complex plane.

57. Re Im x x oo Effect of Poles and Zeros of H[z] Each difference term in H[z] may be represented as a complex number in polar form –Magnitude is the distance of the pole/zero to the chosen point (frequency) on unit circle. –Angle is the angle of vector with the horizontal axis.

58. Effect of Poles/Zeros (Lathi) x --  -- -  /2  -- x o x o x x x o    