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Z Transform Signal and System Analysis z Transform Signal and System Analysis Block Diagrams and Transfer Functions Just as with CT systems, DT systems are conveniently described by block diagrams and transfer functions can be determined from them. For example, from this DT system block diagram the difference equation can be determined. 1 yxx[]nnn= 21[]−−[]−− y[] n1 2 5/10/04 M. J. Roberts - All Rights Reserved 2 Block Diagrams and Transfer Functions From a z-domain block diagram the transfer function can be determined. −−1 YX()zzzzzz= 2 ()− 11 X()− Y() 2 Y()z 2 − z−1 21z − H()z = = = () 1 − 1 X z 1+ z 1 z + 2 2 5/10/04 M. J. Roberts - All Rights Reserved 3 Block Diagram Reduction All the techniques for block diagram reduction introduced with the Laplace transform apply exactly to z transform block diagrams. 5/10/04 M. J. Roberts - All Rights Reserved 4 System Stability A DT system is stable if its impulse response is absolutely summable. That requirement translates into the z-domain requirement that all the poles of the transfer function must lie in the open interior of the unit circle. 5/10/04 M. J. Roberts - All Rights Reserved 5 System Interconnections Cascade Parallel 5/10/04 M. J. Roberts - All Rights Reserved 6 System Interconnections Feedback Y()z H ()z H ()z H()z = = 1 = 1 () + () () + () X z 11HH12zz T z ()= () () THHzzz12 5/10/04 M. J. Roberts - All Rights Reserved 7 Responses to Standard Signals N()z If the system transfer function is H () z = the z D()z z N()z transform of the unit-sequence response is Y()z = z −1 D()z which can be written in partial-fraction form as N ()z z Y()zz= 1 + H()1 D()z z −1 N ()z If the system is stable the transient term, z 1 , dies out D()z z and the steady-state response is H1 () . z −1 5/10/04 M. J. Roberts - All Rights Reserved 8 Responses to Standard Signals Kz Let the system transfer function be H()z = zp− z Kz K z pz Then Y()z = = − z −−11zp −−p z 1zp− K + and yu[]n = ()1− pnn 1 [] 1− p Let the constant, K be 1 - p. Then yu[]npn=−()1 n+1 [] 5/10/04 M. J. Roberts - All Rights Reserved 9 Responses to Standard Signals Unit Sequence Response One-Pole System 5/10/04 M. J. Roberts - All Rights Reserved 10 Responses to Standard Signals Unit Sequence Response Two-Pole System 5/10/04 M. J. Roberts - All Rights Reserved 11 Responses to Standard Signals N()z If the system transfer function is H () z = the z transform D()z of the response to a suddenly-applied sinusoid is N()z zz[]− cos()Ω Y()z = 0 () 2 − ()Ω + D z zz21cos 0 Ω = j 0 Let pe 1 . Then the system response can be written as N ()z []= −1 1 + () (()Ω +∠ ()[] y nzZ Hpnp10 cos H 1u n D()z and, if the system is stable, the steady-state response is () ()Ω +∠ ()[] Hpnpn10 cos H 1 u a DT sinusoid with, generally, different magnitude and phase. 5/10/04 M. J. Roberts - All Rights Reserved 12 Pole-Zero Diagrams and Frequency Response For a stable system, the response to a suddenly-applied sinusoid approaches the response to a true sinusoid (applied for all time). 5/10/04 M. J. Roberts - All Rights Reserved 13 Pole-Zero Diagrams and Frequency Response Let the transfer function of a DT system be z z H()z = = 2 z 5 ()zpzp− ()− z −+ 12 2 16 12+ j 12− j p = p = 1 4 2 4 e jΩ H()e jΩ = jjΩΩ−− epep12 5/10/04 M. J. Roberts - All Rights Reserved 14 Pole-Zero Diagrams and Frequency Response 5/10/04 M. J. Roberts - All Rights Reserved 15 The Jury Stability Test N()z Let a transfer function be in the form, H()z = D()z ()=+D D−1 +++L where D zazazD D−1 aza10 Form the “Jury” array L 1 aaa012 aaaDDD−− 21 L 2 aaDD−−12 a D a 210 aa L 3 bbb012 bbDD−− 21 L 4 bbbDDD−−−123 bb 10 L 5 ccc012 cD− 2 L 6 cccDDD−−−234 c 0 MMMMN − 23Dsss012 5/10/04 M. J. Roberts - All Rights Reserved 16 The Jury Stability Test The third row is computed from the first two by aa aa aa aa ==0 D 01D−− =02D L =01 b0 ,,b1 b2 ,,bD−1 aaD 0 aaD 1 aaD 2 aaDD−1 The fourth row is the same set as the third row except in reverse order. Then the c’s are computed from the b’s in the same way the b’s are computed from the a’s. This continues until only three entries appear. Then the system is stable if D1()> 0 ()−110D D()− > >> >L > aabbDDD00,,,,−− 10 cc 2 ss 0 2 5/10/04 M. J. Roberts - All Rights Reserved 17 Root Locus Root locus methods for DT systems are like root locus methods for CT systems except that the interpretation of the result is different. CT systems: If the root locus crosses into the right half-plane the system goes unstable at that gain. DT systems: If the root locus goes outside the unit circle the system goes unstable at that gain. 5/10/04 M. J. Roberts - All Rights Reserved 18 Simulating CT Systems with DT Systems The ideal simulation of a CT system by a DT system would have the DT system’s excitation and response be samples from the CT system’s excitation and response. But that design goal is never achieved exactly in real systems at finite sampling rates. 5/10/04 M. J. Roberts - All Rights Reserved 19 Simulating CT Systems with DT Systems One approach to simulation is to make the impulse response of the DT system be a sampled version of the impulse response of the CT system. []= () hhnnTs With this choice, the response of the DT system to a DT unit impulse consists of samples of the response of the CT system to a CT unit impulse. This technique is called impulse-invariant design. 5/10/04 M. J. Roberts - All Rights Reserved 20 Simulating CT Systems with DT Systems []= () When hh nnT s the impulse response of the DT system is a sampled version of the impulse response of the CT system but the unit DT impulse is not a sampled version of the unit CT impulse. A CT impulse cannot be sampled. First, as a practical matter the probability of taking a sample at exactly the time of occurrence of the impulse is zero. Second, even if the impulse were sampled at its time of occurrence what would the sample value be? The functional value of the impulse is not defined at its time of occurrence because the impulse is not an ordinary function. 5/10/04 M. J. Roberts - All Rights Reserved 21 Simulating CT Systems with DT Systems In impulse-invariant design, even though the impulse response is a sampled version of the CT system’s impulse response that does not mean that the response to samples from any arbitrary excitation will be a sampled version of the CT system’s response to that excitation. All design methods for simulating CT systems with DT systems are approximations and whether or not the approximation is a good one depends on the design goals. 5/10/04 M. J. Roberts - All Rights Reserved 22 Sampled-Data Systems Real simulations of CT systems by DT systems usually sample the excitation with an ADC, process the samples and then produce a CT signal with a DAC. 5/10/04 M. J. Roberts - All Rights Reserved 23 Sampled-Data Systems An ADC simply samples a signal and produces numbers. A common way of modeling the action of a DAC is to imagine the DT impulses in the DT signal which drive the DAC are instead CT impulses of the same strength and that the DAC has the impulse response of a zero-order hold. 5/10/04 M. J. Roberts - All Rights Reserved 24 Sampled-Data Systems The desired equivalence between a CT and a DT system is illustrated below. () () The design goal is to make y d t look as much like y c t as possible by choosing h[n] appropriately. 5/10/04 M. J. Roberts - All Rights Reserved 25 Sampled-Data Systems Consider the response of the CT system not to the actual signal, x(t), but rather to an impulse-sampled version of it, ∞ ()= ()δ()− = () () xδ tnTtnTtfft∑ xss xss comb n=−∞ The response is ∞ ∞ ()= ()∗ ()= ()∗ []δ()− = []()− yhxttδ tt h∑∑ x mtmTmtmTs xh s m=−∞ m=−∞ []= () where xx nnT s and the response at the nth multiple of Ts ∞ is ()= []()()− yxhnTss∑ m n m T m=−∞ []= () The response of a DT system with hh nnT s to the excitation, xx [] nnT = () is s ∞ yxh[]nnn= []∗ []= ∑ xh[] mnm[]− m=−∞ 5/10/04 M. J. Roberts - All Rights Reserved 26 Sampled-Data Systems The two responses are equivalent in the sense that the values at corresponding DT and CT times are the same. 5/10/04 M. J. Roberts - All Rights Reserved 27 Sampled-Data Systems Modify the CT system to reflect the last analysis. Then multiply the impulse responses of both systems by Ts 5/10/04 M. J. Roberts - All Rights Reserved 28 Sampled-Data Systems In the modified CT system, ∞ ∞ ()= ()∗ ()= ()δ()− ∗ () = ()()− yxtδ t Tsss h t ∑∑ x nT t nT h t Tssss x nT h t nT T n=−∞ n=−∞ In the modified DT system, ∞ ∞ []= [][]− = [] ()()− yxhxhn∑∑ m nm mTnmTss m=−∞ m=−∞ []= () where hh nTnT ss and h(t) still represents the impulse response of the original CT system.
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