examII_f96 - BB 428 EXAM II 21 November 1996 L/ This test...

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Unformatted text preview: BB 428 EXAM II 21 November 1996 L/ This test consists of three problems. Answer each problem on the exam itself; if you use additional paper, repeat the identifying information above, and staple it to the rest of your exam when you hand it in. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. Don’t even think about it . i. . Problem 1: (30 points) The system in Fig. 1 contains a unity-feedback loop containing a minor-rate-feedback loop. Y(s) Figure 1: Minor-loop feedback compensation. 1. (3 points) Calculate the transfer function Y(s)/R(s). 2. (2 points) Write the characteristic equation for the system. 3. (15 points) Sketch the root locus of the system for b 2 0 on on the graph provided on page 5. (a) (3 points) Express the characteristic equation as l + K G’(s) : I}, where both the numerator and denominator of G(s) are monic polynomials in s. Specify G(s) and K. (b) (3 points) Calculate the asymptote(s). (c) (3 points) Compute the departure and arrival angles. (d) (6 points) If there are either breakaway or breakin points, calculate their their location. 4. (5 points) Using the root locus sketch, calculate the range of values of b for which the system is BIBO stable. 5. (5 points) Using the root locus sketch, calculate the value of b for which the transient response of the closed-loop system is critically damped. I. ma, transfer 'rbflc'hlon Y‘s-315(5) at the tuner- Qvop 3.: Yrs) _ ___.___L‘_..__—-— _ u or A 56) .s“ :- (4 use): 5'9 6 m characcberis'ac e Vatican 51+ (Husk); + ’6 =0 2.. 19m. 3. 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M Sz=.t63 So :i-u! I A'- (_J—v\ g —‘{So..+l6 70 0.57’ 561-? has a. nuntmum ans-L :6 Jam, 4; fl rave. Sme. e o 5 also Mo 5-5-36 1%» Smce. K“) I". 799. '15 K 9 ‘61:) this means the Saturn :5 012% b>o. 4.: be. 1"ch £64» MS ear”a 190% J.“ rm! .5. For cu G l r J m Pal“ m‘z‘b W aan— Lgfln‘k Lao—Q. . flak (fit-Curr: “Juan ’ fie:ch gab :5,J = -‘i- Um? .J: the. Hawk-Jr: pdtfl'b in, mi: 40005 flo'b K = 155 3 l 615.) [6 .. __L_ 151*127L W1: *131 cat. .1 - _L.. 1" t1 $52 was: Real Axis Figure 2: Root locus for b > 0. Problem 2: (30 points) Eappy Eafley Technology (HVT), a. custom-designer of analog filters, is introducing a novel product line for instrumentation and control applications. HVT anticipates a vast spectrum of marketing opportunities, ranging from biomedical instrumentation to audio recording, and from robotic manipulators to aerospace control applications. The chief engineer of HVT notes that frequency~selective filters can be passiVe or active. Passive filters are traditionally inductor-capacitor (LC) circuits which, particularly at audio frequencies, can be cumbersome and expensive, and display response shapes that are not as elective as desired. Active filters incorporate resistors and capacitors along with operational amplifiers. They are inexpensive, easy to tune, and are insensitive to varying load and source impedances. Furthermore, active filter can be easily cascaded, so that complex filter responses can be factored into simple, non-interacting blocks. To illustrate the product line, the chief engineer focuses on active filter which is characterized by the second-order analog transfer function .92 +w3, T = ———...H_ (a) .9" + 24%;; + w: in which C is the dimensionless damping ratio and w" [radians/ sec] is the underdamped natural frequency. In these filters, C is Leg; than 0.1. The amplitude response measures how effectively the filter transmits the sinusoidal input signal E9 sinwnt. The phase response (Tutu) rmeasures the phase-shift caused by the filter operating on the sinusoidal input Ea sin wot. Design of these filters is specified by the center frequency, peak amplification or attenuation, upper and lower (3 dB amplification or attenuation) frequencies and selectivity or bandwidth. The chief engineer has subcontracted the following tasks to you: 1. (5 points) Find an expression for the amplitude response |T(3w)l and determine the log and high frequency gains of the active filter. 2. (5 points) Find the frequency (or frequencies) at which the amplitude response attains its maximum or minimum values. 3. (10 points) Sketch the amplitude response of the active filter as a function of the normalized (dimen- sionless) frequency w 9 5 ~——. “’11 4. (5 points) Classify the active filter (low pass, high pass, notch, comb, bandpass, etc.) and briefly describe the operation and applications of T(s). 5. (5 points) On the amplitude response of part (3), indicate the effect of increasing (and decreasing) the dimensionless damping ratio C and the undamped natural frequency u“. (at-)1 + l l. Ta) :- Ll- TLS) '15 gm. flats-ch any“ can be” Vs'efl é, dim-nuke. “wan-EAL LN! «fin? CampOfl'Gn-bs- J; 6U“ u the, prep,“ 4}: wkfl L [T =— 0) MLJC ‘f Mrm (mg; [5 Lo, w". fl 7’— 5 IML “JbCL't a5 PLOW-n vulva/.2. (J 7 Problem 3: (40 points) A control engineer is designing a sun—seeker control system. Thi feedback system is mounted on a space vehicle so that it will track the sun with high accuracy. A block diagram of the sun-seeker control system is shown in Fig. 3. The system is capable of tracking the sun in one plane. The input 9,. [radiansl is the reference angle of the solar ray, and the output 9° [radians] is the angle of the vehicle axis. The objective of the sun-seeker control system is to maintain near zero the closed-loop error e(t) = as) — 9.0:) between 9,. (t) and 9° Numerical values of the sun-seeker control system parameters are listed in Table 1. The principle elements of the error discriminator are two small rectangular silicon photovoltaic cells which are mounted behind a rectangular slit in an enclosure. The cells are mounted in such a way that when the sensor is pointed at the sun, the beam of light from the slit overlaps both cells. The silicon cells are used as current sources and are connected in opposite polarity to the input of an operational amplifier. Any difference in the short-circuit current of the two cells is sensed and amplified by the operational amplifier. Since the current of each cell is proportional to the illumination on the cell, an error signal appears at the output of the amplifier when the light from the slit is not precisely centered on the cells. The error voltage is fed to the servoamplifier which, in turn, causes the servo to drive the system back into alignment. For small-signal operation, the error discriminator can be approximated by the constant gain K., that is ' M) 3(8) where K. [amperes/ radian] i the sensitivity of the error discriminator. A block representing the cascade controller G43), in case it is needed, is inserted between the operational and servo amplifiers shown in Fig. 3. :.Kg The control engineer specifies the following design requirements for the sun-seeker control system: 1. The steady—state value of e(t) due to a unit ramp function input for 6,.(15) should be less than or equal to 0.01 rad/sec of the final steady-state output velocity. In other words, the steady—state error due to a ramp input should be less than or equal to 1 percent. 2. The peak overshoot should be less than 10 %. 1. (5 points) Find the open-loop transfer function _ 9::(3) GL1) _ E(s of the uncompensated system (Gc(s) = 1) both literally and numerically in terms of the system parameters. 2. (10 points) Find the minimum value of the servo-amplifier gain K for which the closed-loop system with G¢(s) = 1 meets the first design specification. 3. (5 points) For the value of servo-amplifier gain found in part (2), find the undamped natural frequency in... [radians / second], dimensionless damping ratio C and peak overshoot M,D of the closed-loop system. 4. (5 points) Sketch and interpret the corresponding step-response of the closed-loop sun-seeker control system for the value of the servo-amplifier gain found in part (2). 5. (15 points) Doe the closed-loop system with 06(5) 2 1 and the value of the servo-amplifier gain found in part (2) meet the second design specification ? If not, design a compensator of the form 8+;1T 3+5}: GAS) = which is physically realizable with passive analog components. Explain your logic in arriving at numerical values for the parameters a. and T. 10 .8893 3.5qu honOmésm a “o EEmEu M85 Um Emmi “Eu Moan Basin—tume Emu 353$ . 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This note was uploaded on 07/23/2008 for the course EE 428 taught by Professor Schiano during the Fall '07 term at Pennsylvania State University, University Park.

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examII_f96 - BB 428 EXAM II 21 November 1996 L/ This test...

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