Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
ECE 352 Systems II Manish K. Gupta, PhD Office: Caldwell Lab 278 Email: guptam @ ece. osu. edu Home Page: http://www.ece.osu.edu/~guptam TA: Zengshi Chen Email: chen.905 @ osu. edu Office Hours for TA : in CL 391: Tu & Th 1:00-2:30 pm Home Page: http://www.ece.osu.edu/~chenz/ 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 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 Z-transforms Z-transforms • For discrete-time systems, z-transforms play the same role of Laplace transforms do in continuous-time systems Bilateral Forward z-transform H [ z] k hk z k Bilateral Inverse z-transform h[k ] 1 2 j R H [ z ] z k 1dz • As with the Laplace transform, we compute forward and inverse z-transforms by use of transforms pairs and properties Region of Convergence • Region of the complex zplane for which forward z-transform converges • Four possibilities (z=0 is a special case and may or may not be included) Im{z} Entire plane Im{z} Disk Re{z} Re{z} Im{z} Complement of a disk Im{z} Re{z} Intersection of a disk and complement of a disk Re{z} Z-transform Pairs • h[k] = d[k] H [ z] d k z • h[k] = ak u[k] k 0 d k z k 1 H [ z] k k 0 Region of convergence: entire zplane H [ z] 1 z d k 1 z z d k of1 convergence: Region entire zk k k 1 plane h[n-1] z-1 H(z) k a uk z k k k 1 k a a z k 0 k 0 z 1 a if 1 a z 1 z Region of convergence: |z| > |a| which is the complement of a disk k • h[k] = d[k-1] k 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.) a uk k Z 1 1 a z 1 for z a – This is stable if |a| < 1 by rule #1. – It is stable if |z| > 1 > |a| by rule #2. Inverse z-transform c j 1 k 1 f k F z z dz 2j c j • 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) Example z 2 2z 1 X [ z] 3 1 z2 z 2 2 1 2 z 1 z 2 X [ z] 3 1 1 z 1 z 2 2 2 1 2 z 1 z 2 X [ z] 1 1 1 1 z 1 z 2 X [ z ] B0 A1 A2 1 1 1 z 1 1 z 2 • Ratio of polynomial zdomain functions • Divide through by the highest power of z • Factor denominator into first-order factors • Use partial fraction decomposition to get first-order terms Example (con’t) 2 1 2 3 1 z z 1 z 2 2 z 1 1 2 2 z 2 3z 1 2 • Find B0 by polynomial division 5 z 1 1 1 5 z 1 X [ z] 2 1 1 1 1 z 1 z 2 1 2 z 1 z 2 A1 1 z 1 1 2 z 1 z 2 A2 1 1 z 1 2 z 1 2 1 4 4 9 1 2 z 1 1 1 2 1 8 1 2 • Express in terms of B0 • Solve for A1 and A2 Example (con’t) • Express X[z] in terms of B0, A1, and A2 X z 2 9 8 1 1 1 z 1 1 z 2 • Use table to obtain inverse z-transform k 1 xk 2 d k 9 uk 8 uk 2 • With the unilateral ztransform, or the bilateral z-transform with region of convergence, the inverse z-transform is unique. Z-transform Properties • Linearity a1 f1k a2 f 2 k a1F1z a2 F2 z • Shifting f k m uk m z m F z m f k m uk z F z z f k z m k 1 m m Z-transform Properties f1 k f 2 k f m f k m m 1 2 Z f1 k f 2 k Z f1 m f 2 k m m f1 m f 2 k m z k k m f m f k mz m m 1 k k • Convolution definition • Take z-transform • Z-transform definition • Interchange summation 2 f1 m f 2 r z r m • r=k-m k m f1 mz f 2 r z r m k • F1 z F2 z Z-transform definition Example g k k uk uk 6 k uk k uk 6 k uk k 6 uk 6 6 uk 6 Gz z 1 z 1 z 6 6 2 2 6 z z z 1 z 1 z 1 z 1 6 z 12 z 5 z 12 z 5 z 1 z 1 6 z5 z 1 5 z z 12 z 5 z 12 z 1 z 5 z 1 z6 6z 5 5 2 z z 1 Difference Equations Linear Difference Equations • Discrete-time LTI systems can be characterized by difference equations f[k] + y[k] + Unit Delay + 1/2 y[k-1] Unit Delay 1/8 y[k-2] 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 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] 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] 1 1 1 1 1 1 Y z 5 Y z y 1 6 2 Y z y 1 y 2 3 F z f 1 5 2 F z f 1 f 2 z z z z z z 3 5 5 6 11 1 2 Y z 3 z z z z 0.5 z z 0.5 Y z 26 z 7 z 18 z 15 z 0.5 3 z 2 5 z 3 7 k 18 k 26 k yk 0.5 2 3 uk 3 5 15 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 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 nth order difference equation yk an 1 yk 1 a0 yk n bn f k bn 1 f k 1 b0 f k n yk m uk 1 a n 1 1 Y z 0 m z f k m uk 1 F z 0 m z z 1 a1 z1 n a0 z n Y z bn bn 1 z 1 b1 z1 n b0 z n F z Y z Z zero state response bn bn 1 z 1 b1 z1 n b0 z n H z F z Z input 1 an 1 z 1 a1 z1 n a0 z n Y z H z F z Stability • Knowing H[z], we can compute the output given any input F[z] H[z] Y[z] • 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 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] = d[k] Z{h[k]} = H[z] Z{d[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 Complex Exponentials • Complex exponentials have special property when they are input into LTI systems. • Output will be same complex exponential weighted by H[z] yk hk f k hk z k hmz k m m z k h[m]z m m z k H z • When we specialize the z-domain to frequency domain, the magnitude of H[z] will control which frequencies are attenuated or passed. Z and Laplace Transforms Z and Laplace Transforms • Are complex-valued functions of a complex frequency variable Laplace: s = + j 2 f Z: z = e jW • Transform difference/differential equations into algebraic equations that are easier to solve 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. Z f[k] H[z] Laplace y[k] ~ f t H[esT] ~ y t Z and Laplace Transforms ~ f t f k d t kT Z k 0 f[k] ~ y t yk d t kT H[z] y[k] k 0 ~ ~ Y s H s F s yk e s T k k 0 H s f k e k T s k 0 Let z e s T : yk z k o k Laplace H [ z ] f k z Y [ z] H [ z] F[ z] k 0 k ~ f t H[esT] ~ y t Impulse Invariance Mapping • Impulse invariance mapping is z = e s T Im{s} Im{z} 1 -1 1 Re{s} 1 -1 s = -1 j z = 0.198 j 0.31 (T = 1) s = 1 j z = 1.469 j 2.287 (T = 1) Laplace Domain Z Domain Left-hand plane Inside unit circle Imaginary axis Unit circle Right-hand plane Outside unit circle Re{z} Sampling Theorem 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 sk sk Ts s(t) Ts k {…, -2, -1, 0, 1, 2,…} Ts is the sampling period ssampled t t skT d t k T k s s[ k ] s Sampled analog waveform Shannon Sampling Theorem • A continuous-time signal x(t) with frequencies no higher than fmax can be reconstructed from its samples x[k] = x(k Ts) if the samples are taken at a rate fs which is greater than 2 fmax. – Nyquist rate = 2 fmax – Nyquist frequency = fs/2. • What happens if fs = 2fmax? • Consider a sinusoid sin(2 fmax t) – Use a sampling period of Ts = 1/fs = 1/2fmax. – Sketch: sinusoid with zeros at t = 0, 1/2fmax, 1/fmax, … Sampling Theorem Assumptions • The continuous-time signal has no frequency content above the frequency fmax • 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 Why 44.1 kHz for Audio CDs? • Sound is audible in 20 Hz to 20 kHz range: fmax = 20 kHz and the Nyquist rate 2 fmax = 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 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 Aliasing • Analog sinusoid x(t) = A cos(2f0t + f) • Sample at Ts = 1/fs x[k] = x(Ts k) = A cos(2 f0 Ts k + f) • Keeping the sampling period same, sample y(t) = A cos(2(f0 + lfs)t + f) where l is an integer y[k] = y(Ts k) = A cos(2(f0 + lfs)Tsk + f) = A cos(2f0Tsk + 2 lfsTsk + f) = A cos(2f0Tsk + 2 l k + f) = A cos(2f0Tsk + f) = x[k] Here, fsTs = 1 Since l is an integer, cos(x + 2l) = cos(x) • y[k] indistinguishable from x[k] Aliasing • Since l is any integer, an infinite number of sinusoids will give same sequence of samples • The frequencies f0 + l fs for l 0 are called aliases of frequency f0 with respect fs to because all of the aliased frequencies appear to be the same as f0 when sampled by fs 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 fmax, bandwidth is fmax – For a bandpass signal with frequency content on the interval [f1, f2], bandwidth is f2 - f1 Difference Equations and Stability 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 g2 - 0.6 g - 0.16 = (g + 0.2) (g - 0.8) Characteristic equation (g + 0.2) (g - 0.8) = 0 Characteristic roots g1 = -0.2 and g2 = 0.8 Solution y0[k] = C1 (-0.2)k + C2 (0.8)k • Zero-state response k y s k h f k Impulse Response Input Example: Impulse Response • h[k+2] - 0.6 h[k+1] - 0.16 h[k] = 5 d[k+2] with h[-1] = h[-2] = 0 because of causality • In general, from Lathi (3.41), h[k] = (b0/a0) d[k] + y0[k] u[k] • Since a0 = -0.16 and b0 = 0, h[k] = y0[k] u[k] = [C1 (-0.2)k + C2 (0.8)k] u[k] • Lathi (3.41) is similar to Lathi (2.41): y[k ] m[k ] u[k ] y[k 1] m[k 1] u[k 1] m[k 1] u[k ] d [k ] Lathi (3.41) balances m[k 1] u[k ] m[k 1] d [k ] impulsive events at origin Example: Impulse Response • Need two values of h[k] to solve for C1 and C2 h[0] - 0.6 h[-1] - 0.16 h[-2] = 5 d[0] h[0] = 5 h[1] - 0.6 h[0] - 0.16 h[-1] = 5 d[1] h[1] = 3 • Solving for C1 and C2 h[0] = C1 + C2 = 5 h[1] = -0.2 C1 + 0.8 C2 = 3 Unique solution C1 = 1, C2 = 4 • h[k] = [(-0.2)k + 4 (0.8)k] u[k] Example: Solution • Zero-state response solution (Lathi, Ex. 3.13) ys[k] = h[k] * f[k] = {[(-0.2)k + 4(0.8)k] u[k]} * (4-k u[k]) ys[k] = [-1.26 (4)-k + 0.444 (-0.2)k + 5.81 (0.8)k] u[k] • Total response: y[k] = y0[k] + ys[k] y[k] = [C1(-0.2)k + C2(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] = C1 (-5) + C2(1.25) = 0 y[-2] = C1(25) + C2(25/16) = 6.25 Solution: C1 = 0.2, C2 = 0.8 Repeated Roots • For r repeated roots of Q(g) = 0 y0[k] = (C1 + C2 k + … + Cr kr-1) gk • Similar to the continuous-time case Continuous Time Discrete Time e t u (t ) g k u[k ] t m e t u (t ) k mg k u[k ] Case non-repeated roots repeated roots 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 Im g Marginally Stable Unstable g |g| b -1 Re g 1 Stable Lathi, Fig. 3.16 • 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) Stability in Both Domains Im Im g Marginally Stable Marginally Stable Unstable -1 1 Re Re g Stable Stable Discrete-Time Systems Unstable Continuous-Time Systems Marginally stable: non-repeated characteristic roots on the unit circle (discrete-time systems) or imaginary axis (continuoustime systems) Frequency Response of Discrete-Time Systems Frequency Response • For continuous-time systems the response to sinusoids are e j t H j e j t cos t H j cos t H j • For discrete-time systems in z-domain z k H z z k • For discrete-time systems in discrete-time frequency e j W k H e j W e j W k cosW k H e j W cos W k H e j W Response to Sampled Sinusoids • Start with a continuous-time sinusoid cos t • Sample it every T seconds (substitute t = k T) cos k T • We show discrete-time sinusoid with cosW k cos k T • Resulting in W T • Discrete-time frequency is equal to continuoustime frequency multiplied by sampling period Example • Calculate the frequency response of the system given as a difference equation as y[k 1] 0.8 y[k ] x[k 1] • Assuming zero initial conditions we can take the z-transform of this difference equation Y [ z ]z 0.8 X [ z ]z • Since z e Y z z 1 H z 1 X z z 0 . 8 1 0 . 8 z jW He jW 1 1 jW 1 0.8e 1 0.8cos W j sin W Example • Group real and imaginary parts He jW 1 1 0.8 cos W j 0.8 sin W • The absolute value (magnitude response) is 1 0.8 cos W j0.8 sin W He 1 jW 1 1 0.8 cos W2 0.8 sin W2 1 1.64 1.6 cos W Example • The angle (phase response) is H e jW 0.8 sin W 0 tan 1 0.8 cos W 1 where 0 comes from the angle of the nominator and the term after – comes from the denominator of H e jW • Reminder: Given a complex number a + j b the absolute value and angle is given as a j b a 2 b 2 b a j b tan 1 a Example • We can calculate the output of this system for a sinusoid at any frequency by substituting W with the frequency of the input sinusoid. H e jW H e jW 5 2 3 2 3 W 53.13 W 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 : e j W e j W 2 m • Absolute value of discrete-time frequency response is even and angle is odd symmetric. – Discrete-time sinusoid is symmetric around cosW 2m k cosWk 2mk cos Wk cos W x k cosW x k cosk sin W x k sin k cosW x k cosk cos W x cos W x Aliasing and Sampling Rate • Continuous-time sinusoid can have a frequency from 0 to infinity • By sampling a continuous-time sinusoid, sample cos t cos k T cosW k t kT • Discrete-time frequency W unique from 0 to 0 T 0 f s 0 2f f s 0 f f s / 2 T – 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. Effect of Poles and Zeros of H[z] • The z-transform of a difference equation can be written in a general form as H z bn z z1 z z2 z zm z g 1 z g 2 z g m • We can think of complex number as a vector in the complex plane. Im – Since z and zi are both complex numbers the difference is again a complex number thus a vector in the complex plane. zi z zi z Re 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. jfm jf1 jf2 r e r e r e H e jW bn 1 j1 2 j 2 m j m d1e d 2e d m e Im 1 g1 x r z2 1 f2 o 2 g2 x r1r2 rm j f1 f2 fm 1 2 m bn e d1d 2 d m d1 d2 r2 f z1 1 o Re Effect of Poles/Zeros (Lathi) H H T x T o - -/2 H H H H T o x x o T x x x - - H H