Digital control Engineering lecture notes

digital control engineering analysis and design and digital control engineering analysis and design fadali solution manual pdf free downlaod
DavidCooper Profile Pic
DavidCooper,Singapore,Researcher
Published Date:11-07-2017
Your Website URL(Optional)
Comment
CHAPTER Introduction to Digital Control 1 OBJECTIVES After completing this chapter, the reader will be able to do the following: 1. Explain the reasons for the popularity of digital control systems. 2. Draw a block diagram for digital control of a given analog control system. 3. Explain the structure and components of a typical digital control system. In most modern engineering systems, it is necessary to control the evolution with time of one or more of the system variables. Controllers are required to ensure satisfactory transient and steady-state behavior for these engineering sys- tems. To guarantee satisfactory performance in the presence of disturbances and model uncertainty, most controllers in use today employ some form of negative feedback. A sensor is needed to measure the controlled variable and compare its behavior to a reference signal. Control action is based on an error signal defined as the difference between the reference and the actual values. The controller that manipulates the error signal to determine the desired control action has classically been an analog system, which includes electrical, fluid, pneumatic, or mechanical components. These systems all have analog inputs and outputs (i.e., their input and output signals are defined over a contin- uous time interval and have values that are defined over a continuous range of amplitudes). In the past few decades, analog controllers have often been replaced by digital controllers whose inputs and outputs are defined at discrete time instances. The digital controllers are in the form of digital circuits, digital computers, or microprocessors. Intuitively, one would think that controllers that continuously monitor the output of a system would be superior to those that base their control on sampled values of the output. It would seem that control variables (controller outputs) that change continuously would achieve better control than those that change period- ically. This is in fact true Had all other factors been identical for digital and analog control, analog control would be superior to digital control. What, then, is the reason behind the change from analog to digital that has occurred over the past few decades? Digital Control Engineering, Second Edition. 1 © 2013 Elsevier Inc. All rights reserved.2 CHAPTER 1 Introduction to Digital Control 1.1 Why digital control? Digital control offers distinct advantages over analog control that explain its pop- ularity. Here are some of its many advantages: Accuracy. Digital signals are represented in terms of zeros and ones with typically 12 bits or more to represent a single number. This involves a very small error as compared to analog signals, where noise and power supply drift are always present. Implementation errors. Digital processing of control signals involves addition and multiplication by stored numerical values. The errors that result from digital representation and arithmetic are negligible. By contrast, the processing of analog signals is performed using components such as resistors and capacitors with actual values that vary significantly from the nominal design values. Flexibility. An analog controller is difficult to modify or redesign once implemented in hardware. A digital controller is implemented in firmware or software and its modification is possible without a complete replacement of the original controller. Furthermore, the structure of the digital controller need not follow one of the simple forms that are typically used in analog control. More complex controller structures involve a few extra arithmetic operations and are easily realizable. Speed. The speed of computer hardware has increased exponentially since the 1980s. This increase in processing speed has made it possible to sample and process control signals at very high speeds. Because the interval between samples, the sampling period, can be made very small, digital controllers achieve performance that is essentially the same as that based on continuous monitoring of the controlled variable. Cost. Although the prices of most goods and services have steadily increased, the cost of digital circuitry continues to decrease. Advances in very large-scale integration (VLSI) technology have made it possible to manufacture better, faster, and more reliable integrated circuits and to offer them to the consumer at a lower price. This has made the use of digital controllers more economical even for small, low-cost applications. 1.2 The structure of a digital control system To control a physical system or process using a digital controller, the controller must receive measurements from the system, process them, and then send control signals to the actuator that effects the control action. In almost all applications, both the plant and the actuator are analog systems. This is a situation where the controller and the controlled do not “speak the same language,” and some form of translation is required. The translation from controller language (digital) to1.3 Examples of digital control systems 3 Reference Controlled Input Variable Actuator Computer DAC and Process ADC Sensor FIGURE 1.1 Configuration of a digital control system. physical process language (analog) is performed by a digital-to-analog converter, or DAC. The translation from process language to digital controller language is performed by an analog-to-digital converter, or ADC. A sensor is needed to monitor the controlled variable for feedback control. The combination of the elements discussed here in a control loop is shown in Figure 1.1.Variationson this control configuration are possible. For example, the system could have several reference inputs and controlled variables, each with a loop similar to that of Figure 1.1. The system could also include an inner loop with digital or analog control. 1.3 Examples of digital control systems In this section, we briefly discuss examples of control systems where digital implementation is now the norm. There are many other examples of industrial processes that are digitally controlled, and the reader is encouraged to seek other examples from the literature. 1.3.1 Closed-loop drug delivery system Several chronic diseases require the regulation of the patient’s blood levels of a specific drug or hormone. For example, some diseases involve the failure of the body’s natural closed-loop control of blood levels of nutrients. Most prominent among these is the disease diabetes, where the production of the hormone insulin that controls blood glucose levels is impaired. To design a closed-loop drug delivery system, a sensor is utilized to measure the levels of the regulated drug or nutrient in the blood. This measurement is con- verted to digital form and fed to the control computer, which drives a pump that injects the drug into the patient’s blood. A block diagram of the drug delivery system is shown in Figure 1.2. See Carson and Deutsch (1992) for a more detailed example of a drug delivery system.4 CHAPTER 1 Introduction to Digital Control Drug Tank Computer Regulated Drug Drug or Nutrient Pump Blood Sensor (a) Reference Regulated Blood Drug Level or Nutrient Drug Computer DAC Patient Pump Blood ADC Sensor (b) FIGURE 1.2 Drug delivery digital control system. (a) Schematic of a drug delivery system. (b) Block diagram of a drug delivery system. 1.3.2 Computer control of an aircraft turbojet engine To achieve the high performance required for today’s aircraft, turbojet engines employ sophisticated computer control strategies. A simplified block diagram for turbojet computer control is shown in Figure 1.3. The control requires feedback of the engine state (speed, temperature, and pressure), measurements of the aircraft state (speed and direction), and pilot command. 1.3.3 Control of a robotic manipulator Robotic manipulators are capable of performing repetitive tasks at speeds and accuracies that far exceed those of human operators. They are now widely used in manufacturing processes such as spot welding and painting. To perform their tasks accurately and reliably, manipulator hand (or end-effector) positions and velocities are controlled digitally. Each motion or degree of freedom (D.O.F.)1.3 Examples of digital control systems 5 (a) Pilot Aircraft Command Turbojet State Computer DAC Aircraft Engine Engine Engine ADC State Sensors Aircraft ADC Sensors (b) FIGURE 1.3 Turbojet engine control system. (a) F-22 military fighter aircraft. (b) Block diagram of an engine control system. of the manipulator is positioned using a separate position control system. All the motions are coordinated by a supervisory computer to achieve the desired speed and positioning of the end-effector. The computer also provides an interface between the robot and the operator that allows programming the lower-level con- trollers and directing their actions. The control algorithms are downloaded from the supervisory computer to the control computers, which are typically specialized microprocessors known as digital signal processing (DSP) chips. The DSP chips execute the control algorithms and provide closed-loop control for the manipula- tor. A simple robotic manipulator is shown in Figure 1.4a, and a block diagram of its digital control system is shown in Figure 1.4b. For simplicity, only one6 CHAPTER 1 Introduction to Digital Control (a) Reference Trajectory Supervisory Computers DAC Manipulator Computer Position ADC Sensors Velocity ADC Sensors (b) FIGURE 1.4 Robotic manipulator control system. (a) 3-D.O.F. robotic manipulator. (b) Block diagram of a manipulator control system. motion control loop is shown in Figure 1.4, but there are actually n loops for an n-D.O.F. manipulator. Resources Carson, E.R., Deutsch, T., 1992. A spectrum of approaches for controlling diabetes. Control Syst. Mag. 12 (6), 2531. Chen, C.T., 1993. Analog and Digital Control System Design. SaundersHBJ. Koivo, A.J., 1989. Fundamentals for Control of Robotic Manipulators. Wiley. Shaffer, P.L., 1990. A multiprocessor implementation of a real-time control of turbojet engine. Control Syst. Mag. 10 (4), 3842.Problems 7 PROBLEMS 1.1 A fluid level control system includes a tank, a level sensor, a fluid source, 1 and an actuator to control fluid inflow. Consult any classical control text to obtain a block diagram of an analog fluid control system. Modify the block diagram to show how the fluid level could be digitally controlled. 1.2 If the temperature of the fluid in Problem 1.1 is to be regulated together with its level, modify the analog control system to achieve the additional control. (Hint: An additional actuator and sensor are needed.) Obtain a block diagram for the two-input-two-output control system with digital control. 1.3 Position control servos are discussed extensively in classical control texts. Draw a block diagram for a direct current motor position control system after consulting your classical control text. Modify the block diagram to obtain a digital position control servo. 1.4 Repeat Problem 1.3 for a velocity control servo. 1.5 A ballistic missile (see Figure P1.5) is required to follow a predetermined flight path by adjusting its angle of attack α (the angle between its axis and its velocity vector v). The angle of attack is controlled by adjusting the thrust angle δ (angle between the thrust direction and the axis of the missile). Draw a block diagram for a digital control system for the angle of attack, including a gyroscope to measure the angle α and a motor to adjust the thrust angle δ. α Velocity Vector v Thrust Direction δ FIGURE P1.5 Missile angle-of-attack control. 1.6 A system is proposed to remotely control a missile from an earth station. Because of cost and technical constraints, the missile coordinates would be measured every 20 seconds for a missile speed of up to 0.5mm/s. Is such a control scheme feasible? What would the designers need to do to eliminate potential problems? 1 See, for example, Van deVegte, J., 1994. Feedback Control Systems, Prentice Hall.8 CHAPTER 1 Introduction to Digital Control 1.7 The control of the recording head of a dual actuator hard disk drive (HDD) requires two types of actuators to achieve the required high real density. The first is a coarse voice coil motor (VCM) with a large stroke but slow dynamics, and the second is a fine piezoelectric transducer (PZT) with a small stroke and fast dynamics. A sensor measures the head position, and the position error is fed to a separate controller for each actuator. Draw a 2 block diagram for a dual actuator digital control system for the HDD. 1.8 In a planar contour tracking task performed by a robot manipulator, the robot end-effector is required to track the contour of an unknown object with a given reference tangential velocity and by applying a given force to the object in the normal direction. For this purpose a force sensor can be applied on the end-effector, while the end-effector velocity can be deter- mined by means of the joint velocities. Draw a block diagram of the digital 3 control system. 1.9 A typical main irrigation canal consists of several pools separated by gates that are used for regulating the water distribution from one pool to the next. In automatically regulated canals, the controlled variables are the water levels, the manipulated variables are the gate positions, and the fundamental 4 perturbation variables are the unknown offtake discharges. Draw a block diagram of the control scheme. 2 Ding, J., Marcassa, F., Wu, S.-C., Tomizuka, M., 2006. Multirate control for computational saving, IEEE Trans. Control Systems Tech. 14 (1), 165169. 3 Jatta, F., Legnani, G., Visioli, A., Ziliani, G., 2006. On the use of velocity feedback in hybrid force/velocity control of industrial manipulators, Control Engineering Practice 14, 10451055. 4 Feliu-Battle, V., Rivas Perez, R., Sanchez Rodriguez, L., 2007. Fractional robust control of main irrigation canals with variable dynamic parameters, Control Engineering Practice 15, 673686.CHAPTER Discrete-Time Systems 2 OBJECTIVES After completing this chapter, the reader will be able to do the following: 1. Explain why difference equations result from digital control of analog systems. 2. Obtain the z-transform of a given time sequence and the time sequence corresponding to a function of z. 3. Solve linear time-invariant (LTI) difference equations using the z-transform. 4. Obtain the z-transfer function of an LTI system. 5. Obtain the time response of an LTI system using its transfer function or impulse response sequence. 6. Obtain the modified z-transform for a sampled time function. 7. Select a suitable sampling period for a given LTI system based on its dynamics. Digital control involves systems whose control is updated at discrete time instants. Discrete-time models provide mathematical relations between the system variables at these time instants. In this chapter, we develop the mathematical properties of discrete-time models that are used throughout the remainder of the text. For most readers, this material provides a concise review of material covered in basic courses on control and system theory. However, the material is self-contained, and famil- iarity with discrete-time systems is not required. We begin with an example that illustrates how discrete-time models arise from analog systems under digital control. 2.1 Analog systems with piecewise constant inputs In most engineering applications, it is necessary to control a physical system or plant so that it behaves according to given design specifications. Typically, the plant is analog, the control is piecewise constant, and the control action is updated periodically. This arrangement results in an overall system that is conveniently described by a discrete-time model. We demonstrate this concept using a simple example. Digital Control Engineering, Second Edition. 9 © 2013 Elsevier Inc. All rights reserved.10 CHAPTER 2 Discrete-Time Systems EXAMPLE 2.1 Consider the tank control system in Figure 2.1. In the figure, lowercase letters denote perturbations from fixed steady-state values. The variables are defined as � H5steady-state fluid height in the tank � h5height perturbation from the nominal value � Q5steady-state flow rate through the tank � q5inflow perturbation from the nominal value i � q 5outflow perturbation from the nominal value 0 It is necessary to maintain a constant fluid level H by adjusting the fluid flow rate into the tank around Q. Obtain an analog mathematical model of the tank, and use it to obtain a discrete-time model for the system with piecewise constant inflow perturbation q and i output h. Solution Although the fluid system is nonlinear, a linear model can satisfactorily describe the system under the assumption that fluid level is regulated around a constant value. The linearized model for the outflow valve is analogous to an electrical resistor and is given by h5Rq 0 where h is the perturbation in tank level from nominal, q is the perturbation in the outflow 0 from the tank from a nominal level Q, and R is the fluid resistance of the valve. Assuming an incompressible fluid, the principle of conservation of mass reduces to the volumetric balance: rate of fluid volume increase5rate of fluid volume in 2 rate of fluid volume out: dCðh1HÞ 5ðq 1QÞ2ðq 1QÞ i o dt where C is the area of the tank or its fluid capacitance. The term H is a constant and its derivative is zero, and the term Q cancels so that the remaining terms only involve perturba- tions. Substituting for the outflow q from the linearized valve equation into the volumetric 0 fluid balance gives the analog mathematical model dh h q i 1 5 dt τ C q i h H q o FIGURE 2.1 Fluid level control system.2.2 Difference equations 11 where τ5RC is the fluid time constant for the tank. The solution of this differential equation is ð t 1 2ðt2tÞ=τ 2ðt2λÞ=τ 0 hðtÞ5e hðtÞ1 e qðλÞdλ 0 i C t 0 Let q be constant over each sampling period T—that is, q(t)5q(k)5constant for t in i i i the interval kT,(k11) T. Then we can solve the analog equation over any sampling period to obtain 2T=τ 2T=τ hðk11Þ5e hðkÞ1R½12e qðkÞ; k50; 1; 2; ... i where the variables at time kT are denoted by the argument k. This is the desired discrete- time model describing the system with piecewise constant control. Details of the solution are left as an exercise (Problem 2.1). The discrete-time model obtained in Example 2.1 is known as a difference equation. Because the model involves a linear time-invariant analog plant, the equation is linear time invariant. Next, we briefly discuss difference equations, and then we introduce a transform used to solve them. 2.2 Difference equations Difference equations arise in problems where the independent variable, usually time, is assumed to have a discrete set of possible values. The nonlinear differ- ence equation yðk1nÞ5f½yðk1n21Þ; yðk1n22Þ; ... ; yðk11Þ; yðkÞ; uðk1nÞ; (2.1) uðk1n21Þ; ... ; uðk11Þ; uðkÞ with forcing function u(k) is said to be of order n because the difference between the highest and lowest time arguments of y(.) and u(.) is n. The equations we deal with in this text are almost exclusively linear and are of the form yðk1nÞ1a yðk1n21Þ1?1a yðk11Þ1a yðkÞ n21 1 0 (2.2) 5b uðk1nÞ1b uðk1n21Þ1?1b uðk11Þ1b uðkÞ n n21 1 0 We further assume that the coefficients a, b, i50, 1, 2,..., are constant. The i i difference equation is then referred to as linear time invariant, or LTI. If the forcing function u(k) is equal to zero, the equation is said to be homogeneous. EXAMPLE 2.2 For each of the following difference equations, determine the order of the equation. Is the equation (a) linear, (b) time invariant, or (c) homogeneous? 1. y(k12)10.8y(k11)10.07y(k)u(k) 2. y(k14)1sin(0.4k)y(k11)10.3y(k)50 2 (k) 3. y(k11)520.1y12 CHAPTER 2 Discrete-Time Systems Solution 1. The equation is second order. All terms enter the equation linearly and have constant coefficients. The equation is therefore LTI. A forcing function appears in the equation, so it is nonhomogeneous. 2. The equation is fourth order. The second coefficient is time dependent, but all the terms are linear and there is no forcing function. The equation is therefore linear time varying and homogeneous. 3. The equation is first order. The right-hand side (RHS) is a nonlinear function of y(k), but does not include a forcing function or terms that depend on time explicitly. The equation is therefore nonlinear, time invariant, and homogeneous. Difference equations can be solved using classical methods analogous to those available for differential equations. Alternatively, z-transforms provide a convenient approach for solving LTI equations, as discussed in the next section. 2.3 The z-transform The z-transform is an important tool in the analysis and design of discrete-time systems. It simplifies the solution of discrete-time problems by converting LTI difference equations to algebraic equations and convolution to multiplication. Thus, it plays a role similar to that served by Laplace transforms in continuous-time pro- blems. Because we are primarily interested in application to digital control systems, this brief introduction to the z-transform is restricted to causal signals (i.e., signals with zero values for negative time) and the one-sided z-transform. The following are two alternative definitions of the z-transform. DEFINITION 2.1 Given the causal sequence u , u , u , ... , u , ..., its z-transform is defined as 0 1 2 k 21 22 2k UðzÞ5u 1u z 1u z 1?1u z 0 1 2 k N X (2.3) 2k 5 u z k k50 21 The variable z in the preceding equation can be regarded as a time delay operator. The z-transform of a given sequence can be easily obtained as in the following example. DEFINITION 2.2 Given the impulse train representation of a discrete-time signal,  uðtÞ5u δðtÞ1u δðt2TÞ1u δðt22TÞ1?1u δðt2kTÞ1... 0 1 2 k N X (2.4) 5 u δðt2kTÞ k k502.3 The z-transform 13 the Laplace transform of (2.4) is  2sT 22sT 2ksT UðsÞ5u 1u e 1u e 1?1u e 1... 0 1 2 k N X 2sT k (2.5) 5 uðe Þ k k50 Let z be defined by sT z5e (2.6) Then, substituting from (2.6) in (2.5) yields the z-transform expression (2.3). EXAMPLE 2.3 N Obtain the z-transform of the sequencefug 5f1;3;2;0;4;0;0;0;...g. k k50 Solution 21 22 24 Applying Definition 2.1 gives U(z)5113z 12z 14z . Although the preceding two definitions yield the same transform, each has its advantages and disadvantages. The first definition allows us to avoid the use of impulses and the Laplace transform. The second allows us to treat z as a complex variable and to use some of the familiar properties of the Laplace transform, such as linearity. Clearly, it is possible to use Laplace transformation to study discrete time, continuous time, and mixed systems. However, the z-transform offers significant simplification in notation for discrete-time systems and greatly simplifies their analysis and design. 2.3.1 z-Transforms of standard discrete-time signals Having defined the z-transform, we now obtain the z-transforms of commonly used discrete-time signals such as the sampled step, exponential, and the discrete- time impulse. The following identities are used repeatedly to derive several important results: n n11 X 12a k a 5 ; a6¼ 1 12a k50 (2.7) N X 1 k a 5 ; jaj,1 12a k50 EXAMPLE 2.4: UNIT IMPULSE Consider the discrete-time impulse (Figure 2.2)  1; k50 uðkÞ5δðkÞ5 0; k6¼ 014 CHAPTER 2 Discrete-Time Systems 1 k 1 –1 0 FIGURE 2.2 Discrete-time impulse. Applying Definition 2.1 gives the z-transform UðzÞ51 Alternatively, one may consider the impulse-sampled version of the delta function  u (t)5δ(t). This has the Laplace transform  UðsÞ51 Substitutionfrom(2.6)hasnoeffect.Thus,thez-transformobtainedusingDefinition2.2 isidenticaltothatobtainedusingDefinition2.1. EXAMPLE 2.5: SAMPLED STEP N Consider the sequencefug 5f1; 1; 1; 1; 1; 1;...g of Figure 2.3. Definition 2.1 gives k k50 the z-transform 21 22 23 2k UðzÞ511z 1z 1z 1?1z 1... N X 2k 5 z k50 Using the identity (2.7) gives the following closed-form expression for the z-transform: 1 UðzÞ5 21 12z z 5 z21 Note that (2.7) is only valid forjzj,1. This implies that the z-transform expression we obtain has a region of convergence outside which is not valid. The region of convergence 1 … k –1 0 12 3 FIGURE 2.3 Sampled unit step.2.3 The z-transform 15 must be clearly given when using the more general two-sided transform with functions that are nonzero for negative time. However, for the one-sided z-transform and time functions that are zero for negative time, we can essentially extend regions of convergence and use 1 the z-transform in the entire z-plane. 1 The idea of extending the definition of a complex function to the entire complex plane is known as analytic continuation. For a discussion of this topic, consult any text on complex analysis. EXAMPLE 2.6: EXPONENTIAL Let  k a ; k0 uðkÞ5 0; k,0 Figure 2.4 shows the case where 0,a,1. Definition 2.1 gives the z-transform 21 2 22 k 2k UðzÞ511az 1a z 1?1a z 1... Using (2.7), we obtain 1 UðzÞ5 12ða=zÞ z 5 z2a As in Example 2.5, we can use the transform in the entire z-plane in spite of the validity condition for (2.7) because our time function is zero for negative time. 1 … a 2 a 3 a k –1 0 12 3 FIGURE 2.4 Sampled exponential. 2.3.2 Properties of the z-transform The z-transform can be derived from the Laplace transform as shown in Definition 2.2. Hence, it shares several useful properties with the Laplace trans- form, which can be stated without proof. These properties can also be easily proved directly, and the proofs are left as an exercise for the reader. Proofs are provided for properties that do not obviously follow from the Laplace transform.16 CHAPTER 2 Discrete-Time Systems Linearity This equation follows directly from the linearity of the Laplace transform. ZfαfðkÞ1βfðkÞg5αFðzÞ1βFðzÞ (2.8) 1 2 1 2 EXAMPLE 2.7 Find the z-transform of the causal sequence fðkÞ5231ðkÞ14δðkÞ; k50; 1; 2;... Solution Using linearity, the transform of the sequence is 2z 6z24 FðzÞ5Zf231ðkÞ14δðkÞg52Zf1ðkÞg14ZfδðkÞg5 145 z21 z21 Time delay This equation follows from the time delay property of the Laplace transform and equation (2.6). 2n Zffðk2nÞg5z FðzÞ (2.9) EXAMPLE 2.8 Find the z-transform of the causal sequence  4; k52; 3;... fðkÞ5 0; otherwise Solution The given sequence is a sampled step starting at k52 rather than k50 (i.e., it is delayed by two sampling periods). Using the delay property, we have 4z 4 22 22 FðzÞ5Zf431ðk22Þg54z Zf1ðkÞg5z 5 z21 zðz21Þ Time advance Zffðk11Þg5zFðzÞ2zfð0Þ (2.10) n n n21 Zffðk1nÞg5z FðzÞ2z fð0Þ2z fð1Þ2?2zfðn21Þ2.3 The z-transform 17 PROOF Only the first part of the theorem is proved here. The second part can be easily proved by induction. We begin by applying the z-transform Definition 2.1 to a discrete-time function advanced by one sampling interval. This gives N X 2k Zffðk11Þg5 fðk11Þz k50 N X 2ðk11Þ 5z fðk11Þz k50 Now add and subtract the initial condition f(0) to obtain () " N X 2ðÞ k11 Zffðk11Þg5zfð0Þ1 fkðÞ 11 z 2fðÞ 0 k50 Next, change the index of summation to m5k11 and rewrite the z-transform as () " N X 2m Zffðk11Þg5z fðmÞz 2fð0Þ m50 5zFðzÞ2zfð0Þ EXAMPLE 2.9 Using the time advance property, find the z-transform of the causal sequence ffðkÞg5f4; 8; 16;...g Solution The sequence can be written as k12 fðkÞ52 5gðk12Þ; k50; 1; 2;... where g(k) is the exponential time function k gðkÞ52 ; k50; 1; 2;... Using the time advance property, we write the transform z 4z 2 2 2 2 FðzÞ5z GðzÞ2z gð0Þ2zgð1Þ5z 2z 22z5 z22 z22 Clearly, the solution can be obtained directly by rewriting the sequence as ffðkÞg54f1; 2; 4;...g and using the linearity of the z-transform. Multiplication by exponential 2k Zfa fðkÞg5FðazÞ (2.11)18 CHAPTER 2 Discrete-Time Systems PROOF N N X X 2k 2k 2k LHS5 a fðkÞz 5 fðkÞðazÞ 5FðazÞ k50 k50 EXAMPLE 2.10 Find the z-transform of the exponential sequence 2αkT fðkÞ5e ; k50; 1; 2;... Solution Recall that the z-transform of a sampled step is 21 21 Þ FðzÞ5ð12z and observe that f(k) can be rewritten as αT 2k fðkÞ5ðe Þ 31; k50; 1; 2;... Then apply the multiplication by exponential property to obtain  z αT 2k αT 21 21 Z ðe Þ fðkÞ 5½12ðe zÞ  5 2αT z2e This is the same as the answer obtained in Example 2.6. Complex differentiation  m  d m Z k fðkÞ52z FðzÞ (2.12) dz PROOF To prove the property by induction, we first establish its validity for m51. Then we assume its validity for any m and prove it for m11. This establishes its validity for 11152, then 21153, and so on. For m51, we have   N N X X d 2k 2k kfðkÞz 5 fðkÞ 2z ZfkfðkÞg5 z dz k50 k50     N X d d 2k 5 2z fðkÞz 5 2z FðzÞ dz dz k50 Next, let the statement be true for any m and define the sequence m f ðkÞ5k fðkÞ; k50; 1; 2;... m2.3 The z-transform 19 and obtain the transform N X 2k Zfkf ðkÞg5 kf ðkÞz m m k50   N X d 2k 5 f ðkÞ 2z z m dz k50     N X d d 2k 5 2z f ðkÞz 5 2z F ðzÞ m m dz dz k50 Substituting for F (z), we obtain the result m  m11   d m11 Z k fðkÞ 5Z kf ðkÞ52z FðzÞ m dz EXAMPLE 2.11 Find the z-transform of the sampled ramp sequence fðkÞ5k; k50; 1; 2;... Solution Recall that the z-transform of a sampled step is z FðzÞ5 z21 and observe that f(k) can be rewritten as fðkÞ5k31; k50; 1; 2;... Then apply the complex differentiation property to obtain  d z ðz21Þ2z z Zfg k3152z 5ð2zÞ 5 2 2 dz z21 ðz21Þ ðz21Þ 2.3.3 Inversion of the z-transform Because the purpose of z-transformation is often to simplify the solution of time domain problems, it is essential to inverse-transform z-domain functions. As in the case of Laplace transforms, a complex integral can be used for inverse transforma- tion. This integral is difficult to use and is rarely needed in engineering applications. Twosimplerapproachesforinversez-transformationarediscussedinthissection. Long division This approach is based on Definition 2.1, which relates a time sequence to its z-transform directly. We first use long division to obtain as many terms as desired of the z-transform expansion; then we use the coefficients of the expansion to