+ - cos ( ω t) V I R 1. V Cos( ω t + φ)

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1 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science í Electronic Circuits Homework è10 Handout F98052 Issued 11è10è98 í Due 11è18è98 Exercise 101: Determine the impedance of the two networks shown below. Also, identify the asymptotic dependence of their impedances on frequency for very low frequencies and for very high frequencies, and explain the dependences physically. L C R 2 R 1 Z 2 = Z C jjz R2 = Z 1 = Z L Z R1 = wl R 1 1 wc 1=R 2 = R 2 1 wr 2 C For the ærst network, the two elements are in series, so their impedances combine like series resistors. At zero frequency èconstant voltageè, the inductor behaves like a short, so the impedance is just R 1. As frequency increases, the rapidly oscillating current induces a larger EMF in the inductor, so less current is required to generate the same voltage. The inductor behaves like an open circuit at high frequency. For the second network, the two elements are in parallel, so their impedances combine like parallel resistors. At zero frequency, the capacitor behaves like an open circuit, so the impedance is just R 2. As frequency increases, the same amount of charge must be transferred between the capacitor plates in a shorter amount of time, so the current increases. The capacitor behaves like a short circuit at high frequency.

2 Exercise 102: Assume that the network shown below is operating in sinusoidal steady state. Determine the amplitude V O and phase ç of the voltage across the capacitor. Hint: see the previous exercise. L V I cos ( ω t) R 1 R 2 C V Cos( ω t φ) o The circuit is operating in sinusoidal steadystate, so we can use impedances. Note that the circuit is a voltage divider with the impedances Z 1 and Z 2 from Exercise è1. Use ~ V to represent the complex amplitude and phase of each voltage. = = Z 2 Z 1 Z 2 = 1 Z 1 Z è1=r 2 èè1 wr 2 Cèè wl R 1 è1 = 1 è1 R 1 =R 2, w 2 LCè wèl=r 2 R 1 Cè Doublecheck that this ratio is unitless, as we expect. Let a = è1 R 1 =R 2, w 2 LCè and b = wèl=r 2 R 1 Cè denote the real and imaginary parts of the denominator. Then we obtain: V O = V I æ ææææ ~ VO ç = 6 æ = V I è ~VO çp a 2 b 2 ç,1 =! =, tan,1 ç b a V I p è1 R1 =R 2, w 2 LCè 2 w 2 èl=r 2 R 1 Cè 2 ç ç ç =, tan,1 wèl=r2 R 1 Cè 1R 1 =R 2, w 2 LC

3 Problem 101: This problems examines the relation between various characteristics of the linear twoport system shown below. One port is driven by the input voltage v IN ètè, while the output voltage v OUT ètè appears at the other port. The system is characterized by its polezero diagram, for which two cases are given below. In addition, in sinusoidal steady state at very low frequencies, it is known that v OUT ètè ç T d dt v INètè, where T is a given time constant. èaè For both cases, sketch and clearly label the Bode diagram of the system. For sketching purposes, assume that ç 3 =10ç 2 = 100ç 1 for Case è1, and that ç 2 =10! 1 = 100ç 1 for Case è2. èbè Assume that the system operates in sinusoidal steady state with v IN = éf ~ VI e j!t g and v OUT = éf ~ VO e j!t g where ~ VI and ~ VO are complex coeæcients. For both cases, determine ~ VO = ~ VI, the system transfer function. ècè Again assume that the system operates in sinusoidal steady state. For both cases, determine V O and ç such that v OUT = V O cosè!t çè when v IN = V I cosè!tè. èdè For both cases, develop a diæerential equation that relates v OUT to v IN. èeè For both cases, derive the ZIR of the system. Each ZIR should have three coeæcients that would be used to match given initial conditions. Vin(t) Linear System Vout(t) Polezero Diagram #1 Im Polezero Diagram #2 Im j ω σ3 σ2 σ1 Re σ 2 σ 1 Re j ω

4 To simplify matters, the two cases will be analyzed separately, and somewhat out of the order in which the problem is posed. Case 1: The transfer function èparts B and Cè If we look at the polezero plot, we may immediately write down the following: s èsè =K èsç 1 èès ç 2 èès ç 3 è where K is an undetermined constant èwe can't read K directly oæ the polezero plotè. At very low frequencies, we know that v OUT ètè ç T d dt v INètè, i.e. ~VO VI ç T w. Thus K = T èç ~ 1ç 2 ç 3 è. When the system is operating at sinusoidal steadystate, we obtain the following transfer function: èpart Bè w è wè =K è w ç 1 èè w ç 2 èè w ç 3 è This gives us the answer for Part ècè directly: ç = 6 = 6 V O = V I æ ææææ ~ VO w è w ç 1 èè w ç 2 èè w ç 3 è = ç 2, tan,1 èw=ç 1 è, tan,1 èw=ç 2 è, tan,1 èw=ç 3 è æ = V ITç 1 ç 2 ç 3 w æè w ç 1 èè w ç 2 èè w ç 3 è æ = V ITç 1 ç 2 ç 3 w q q w 2 ç 2 1 w 2 ç 2 2 q w 2 ç 2 3 The Bode plot èpart Aè Using the fact that ç 3 =10ç 2 = 100ç 1,we can rewrite the system equation using the normalized frequency çw = w=ç 1 : çw è çwè =ètç 1 è è çw 1èè çw çw 1èè 1è The great thing about a properly normalized transfer function is that you can always use the same straightline approximations í just relabel the axes when you're ænished! Here are some convenient points at the frequencies where the straightline approximations intercept: w çw log 10 çw j VO ~ = VI ~ æ ètç 1 èj 6 è~ VO =~ VI è èradiansè Near DC Near DC,1 0 ç=2 ç ç 1= p 2 ç=4 ç ç 1= p 2,ç=4 ç ç 1=è10 p 2è,5ç=4 Note that ç and j VO ~ j for Part ècè can be read oæ the Bode plot directly. ~VI The diæerential equation and ZIR èparts D and Eè We know from the polezero plot that the following equation holds: èsèëèsç 1 èès ç 2 èès ç 3 èë = ~ VI èsèëtèç 1 ç 2 ç 3 èsë

5 However, this is directly related to the diæerential equation for the system, since it is obtained by assuming a certain form of the input and output èe st è: h i èsè s 3 s 2 èç 1 ç 2 ç 3 èsèç 1 ç 2 ç 2 ç 3 ç 3 ç 1 èèç 1 ç 2 ç 3 è = VI ~ èsèëtèç 1 ç 2 ç 3 èsë d 3 v OUT dt 3 èç 1 ç 2 ç 3 è d2 v OUT dt 2 The ZIR is also simple to obtain èpart Eè: èç 1 ç 2 ç 2 ç 3 ç 3 ç 1 è dv OUT dt èç 1 ç 2 ç 3 èv OUT =èç 1 ç 2 ç 3 èt dv IN dt v OUT ètè =A 1 e,ç 1t A 2 e,ç 2t A 3 e,ç 3t Case 2: The transfer function èparts B and Cè Again, if we examine the polezero plot, we can write down the following directly: s èsè =K èsç 1, æèès ç 1 æèès ç 2 è where K is an undetermined constant ènote the use of æ í this is simply to avoid confusion with frequency w, which looks a lot like pole coordinate!è. At very low frequencies, we know that the system's transfer function looks like T w, which implies that K = T èç 2 1 æ 2 èç 2. At sinusoidal steadystate èpart Bè, the transfer function becomes: w è wè =K è w ç 1, æèè w ç 1 æèè w ç 2 è This gives us the answer for Part ècè directly: ç = 6 = 6 w è w ç 1, æèè w ç 1 æèè w ç 2 è ç = ç 2, tan,1 è w, æ è, tan,1 è w æ è,tan,1 è w è ç 1 ç 1 ç 2 æ ææææ VO ~ V O = V I æ = V IT èç 2 1 æ 2 èç 2 w æè w ç 1, æèè w ç 1 æèè w ç 2 èæ V O = V I T èç 2 1 æ 2 èç 2 w q q q èw,æè 2 ç 2 1 èw Omegaè 2 ç 2 1 w 2 ç 2 2 The Bode plot èpart Aè The normalization trick also works here. Let çw = w=ç 1. We can rewrite the transfer function as follows: è! ç 2 è çwè =T 1æ 2 çw ç 1 è çw1, 10èè çw 1 10èè çw 1è 100

6 The straightline approximation works okay for frequencies below ç 1 or above ç 2, but the secondorder eæect around the intermediate frequency æ can muddle things up. The best way to deal with this when making hand plots is to note that the system will do one of two things: display a resonant peak around the damped frequency p æ 2, ç 2, or behave as though there were two poles at the damped frequency p ç 2, æ 2. So calculate a few test points in the frequency band of interest. Let's take a quick peek inside MatLab to see if we can get a better picture. First oæ, we need to deæne our ësystem" in MatLab. Expand the numerator and denominator of our sdomain transfer function into their polynomial forms èdon't worry, wehave to do this for parts èdè and èeè anywayè: N = s 0 D=èsç 1, æèès ç 1 æèès ç 2 è=ès 2 2ç 1 sç 2 1æ 2 èès ç 2 è D = s 3 s 2 è2ç 1 ç 2 èsè2ç 1 ç 2 ç 2 1 æ 2 èç 2 èç 2 1æ 2 è For convenience, we will normalize the frequency by setting ç 1 = 1. Then N = s 0 and D = s 3 102s 2 301s Now we can tell MatLab the following: éé system2 = tfè ë1 0ë, ë ë è This tells MatLab to make a transfer function with the numerator N and the denominator D by specifying the polynomial coeæcients. Don't worry, you'll get the hang of it. Note that we have explicitly set K to unity... we can set it to some other value by simply scaling all of the numerator coeæcients. Now just type in the command: éé bodeèsystem2è Nifty, eh? We can look at the system response to a unit step at the input with the command ëstepèsystem2è". The diæerential equation and ZIR èparts D and Eè By inspection: i hs 3 s 2 è2ç 1 ç 2 èsè2ç 1 ç 2 ç 2 1 æ 2 èç 2 èç1æ 2 2 è i= VI ~ ëskë = VI ~ hst èç 21 æ 2 èç 2 d 3 v OUT è2ç dt 3 1 ç 2 è d2 v OUT è2ç dt 2 1 ç 2 ç1æ 2 2 è dv OUT dt The ZIR of the system is given by: ç 2 èç 2 1 æ 2 èv OUT = Tèç 2 1 æ 2 èç 2 dv IN dt v OUT ètè =A e,ç 1t e æt A, e,ç 1t e, æt Be,ç 2t

7 Bode Diagrams Phase (deg); Magnitude (db) Frequency (rad/sec)

8 Bode Diagrams Phase (deg); Magnitude (db) Frequency (rad/sec)

9 Problem 102: A transformer is a twoport electrical circuit element that comprises two magneticallycoupled windings. The symbol for an ideal transformer and a model for that transformer are given below. In this transformer, Winding è2 has N times as many turns as does Winding è1, as indicated. Ideal Transformer Symbol i 1 i 2 v 1 Winding Winding v 2 #1 #2 1 : N Ideal Transformer Model i 1 i 2 v 1 Ni 2 NV 1 v 2 R s Part D R s Part E Vs cos( ω s t) R L L R L 1 : N 1 : N Vs cos( ω s t) Magnetizing Inductance èaè Using the ideal transformer model, determine the relation between v 1 and v 2, and the relation between i 1 and i 2. Examine the dependent sources: i 1 =,Ni 2 and v 2 = Nv 1. èbè Show that power is conserved across the ideal transformer. That is, show that v 1 i 1 v 2 i 2 =0. Using the relationship in part èaè: v 1 i 1 v 2 i 2 = v 1 è,ni 2 èènv 1 èi 2 =0 Power is conserved. ècè Suppose that a network having impedance Z were connected across Winding è2. Determine the impedance that would be observed at the terminals of Winding è1. If we connect an impedance Z across Winding è2, then v 2 =,i 2 Z by Ohm's Law ènote the negative sign; v 2 and i 2 are antiassociated variables for the impedance!è. Substitute in the relationships from part èaè: ènv 1 è=è,1=ni 1 èè,zè. The eæective impedance seen at Winding è1 is Z=N 2. èdè Transformers are commonly used to maximize the power transfered from a source to a load.

10 In this problem, the source is a Thevenin network having voltage V S cosè! S tè and resistance R S, as shown below. The load is a single resistor having resistance R L. Find the timeaverage power delivered to, or dissipated in, the load, where the time average is taken over the period 2ç=! S. Also ænd the turns ratio N for which the timeaverage power delivered to the load is maximized. For this special case of N, ænd the equivalent resistance of the transformer and load resistor as viewed into transformer Winding è1. First, ænd the current in Winding è1. Looking into the terminals of Winding è1, we see an eæective impedance of R L =N 2, and this impedance is in series with R S : ~ i1 = ~V S R S R L =N 2 Next, ænd the voltage across Winding è1. By using a voltage divider, we see that ~v 1 = ~ VS R L =N 2 R S R L =N 2 The average power supplied by Winding è1 must be dissipated in the load èaccording to part Bè. To ænd the average power supplied by Winding è1, we need to integrate the instantaneous power v 1 i 1 over a full period T =2ç=w: çp = 1 T Z T 0 j~vj cosèwt 6 ~vèj~ ij cosèwt 6 ~ ièdt Mathematica enjoyed doing that integral a lot more than I would: çp = 1 T j~vjj ~ ij ç w cosè 6 ~v, 6 ~ iè= j~vjj~ ij 2 cosè6 ~v, 6 ~ iè This is actually the general formula for the average power dissipated by an circuit element operating at sinusoidal steadystate. In our case, the current and voltage are in phase, since the network is purely resistive. çp LOAD = 1 V S 2 R S R L =N æ V R L =N 2 2 S R S R L =N = V 2 S R L N 2 1 èr S R L =N 2 è 2 To maximize the average power delivered to the load, minimize the denominator of PLOAD ç çn 2 R 2 S 2R S R L R 2 =N2 =2NR 2 S,2R 2 L =N3 =0 Solving for N, we ænd N max = p R L =R S. Then the eæective impedance of Winding è1 is R S. èeè One transformer nonideality is ænite magnetizing inductance. A transformer with ænite magnetizing inductance may be modeled as an ideal transformer in parallel with an inductor, as shown below. If N is chosen to maximize power transfer to a load resistor following the results of Part D, over what range of! S will the power dissipated in the load be approximately that found in Part D given the presence of the nonideality. Deæne Z L = wl èthe impedance of the magnetization inductanceè, and Z W 1 = R L =N 2 = R S èthe impedance seen looking into Winding è1è. When jz L j ç jz W 1 j, i.e. w ç R S =L, very little current will æow through Z L, and the power dissipated in the load will approach the ideal value.

11 Problem 103: This problem develops an electric analog for the electromechanical system shown below. The system involves an electric actuator which drives a mechanical load, and it has many of the features of a permanentelectret speakerèmicrophone. At its electric port, the actuator may be modeled by the network shown below comprising a capacitor having capacitance C in parallel with a resistor having resistance R and a dependent current source. The strength of the current source is Gu, where u is the velocity of the actuator at its mechanical port, and G is a coeæcient. Electric energy into the current source represents energy that is converted to mechanical energy and output through the mechanical port of the actuator. At its mechanical port, the actuator drives a load having mass M. The mass is connected to mechanical ground through a spring having stiæness K and a damper having viscous friction coeæcient B. The force f produced by the actuator is Gv where v is the voltage across the electrical port of the actuator. v i R C Gu v i Electric Actuator f = Gv K B M x u = dx/dt Electromechanical System èaè With the force from the actuator acting as an input, develop an electric analog for the mechanical elements of the system. In doing so, represent the force of the actuator by a voltage source, the mass by an inductor, the spring by a capacitor, and the damper by a resistor. In addition, state how the values of the elements in the electric analog are related to the values of the mechanical elements in the electromechanical system. Finally, identify the variable in the network that represents the velocity u of the actuator and mass. Assume that force and voltage will be analagous quantities. This immediately lets us create a onetoone map between electrical elements and quantities, and mechanical elements and quantities èsee the ægure belowè. The velocity u is given by the current u 0 in the network. force, æævoltage velocity, ææcurrent spring K, capacitor èæ=æèc damper B, resistor èæ=æèr mass M, inductor èæ=æèl èbè Show that energy is conserved through the electricaltomechanical energy conversion process of the actuator. That is, show that the power into the electric port of the actuator is equal to the power out of the mechanical port. Because of this, an ideal transformer may be used to represent energy conversion through the actuator, and hence to couple the network which models the electric side of the actuator to the electric analog for the mechanical elements of the system. Using the model for an ideal transformer given in Problem 102, couple the two electric networks to develop one complete electrical analog for the electromechanical system.

12 In the mechanical circuit, all of the elements share a common velocity. Since velocity maps to current, all of these elements must be in series. A positive applied force will create a positive velocity, sou 0 must have the polarity shown in the ægure below. Note that all of the elements will oppose the applied force. The ideal transformer connecting the mechanical and electrical circuits is just the pair of dependent sources, one in each circuit. What should we do with the conversion factor G? When we originally speciæed G in the problem statement, we implicitly gave it a funny set of units, such that the units of G depended on whether it was being used on the mechanical or electrical side. By creating our electrical analog of the mechanical circuit and hooking it up with an ideal transformer, we are essentially getting rid of this whole dilemma by making G into a unitless quantity. The power into the electrical port is v æ èi PORT è = vè,gu 0 è, and the power out of the mechanical port is f 0 u 0 = ègvèu 0. Obviously these are matched, so energy is conserved in the electricalmechanical interface. Note that the electromechanical system as originally drawn does not necessarily conserve energy; we must select the constants æ and æ correctly for this to be the case èyou can check that we must have ææ = 1è. Note: The prime è'è on each variable in the mechanical analog denotes the presence of the scaling factors æ and æ, which have been omitted for convenience èeg. f 0 = f=æ, B 0 =èæ=æèb, etc.è. p f df/dt l v dv/dt K f=kx f=ma v=cq v=ldi/dt B f=bu v=ri f u x u a q i di/dt M Mechanical Elements Electrical Elements Analog of Mechanical Circuit ècè Derive the diæerential equations which describe the dynamics of the electromechanical system in total. Also, derive the diæerential equations which describe the dynamics of the electrical analog. Show that the two sets of diæerential equations describe identical dynamics as viewed from the input current i. On the electrical side èusing KCLè: i = v R C dv èg=æèu = vè1=r scèèg=æèu dt è1è

13 On the mechanical side èusing KVLè: f = M du Z t dt Bu K udt =èsm B K=sèu,1 è2è In addition, we know that f = èg æ æèv and i PORT = èg=æèu. These are four diæerential equations in four unknowns, and hence provide a complete description ètogether with a set of initial conditionsè for the electromechanical dynamics. If we couple the electrical system and èthe electrical analog ofè the mechanical system using an ideal transformer, we obtain the equations: i = v R C dv dt Gu0 = V è1=r scègu 0 è3è f 0 = M 0 du0 Z t dt B0 u 0 K 0 u 0 dt =èsm 0 B 0 K 0 =sèu 0,1 The systems described by equations è2è and è4è must have identical dynamics èthey are identical by constructionè, and we know that Gu 0 = Gèu=æè =èg=æèu, so equations è1è and è3è must also described identical dynamics. è4è