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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 unityfeedback loop containing a minorratefeedback loop. Y(s) Figure 1: Minorloop 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 closedloop system is critically damped. I. ma, transfer 'rbﬂc'hlon Y‘s315(5) at the tuner Qvop 3.: Yrs) _ ___.___L‘_..__—— _ u or A
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1" t1 $52 was: Real Axis Figure 2: Root locus for b > 0. Problem 2: (30 points) Eappy Eaﬂey Technology (HVT), a. customdesigner 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 inductorcapacitor (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, noninteracting blocks. To illustrate the product line, the chief engineer focuses on active ﬁlter which is characterized by the
secondorder 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 phaseshift 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. ﬂatsch any“ can be” Vs'eﬂ é,
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(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 sunseeker 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 sunseeker control system is to maintain near zero the closedloop error e(t) = as) — 9.0:) between 9,. (t) and 9° Numerical values of the sunseeker 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 shortcircuit 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 smallsignal 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 sunseeker 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 steadystate 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 openloop 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 servoampliﬁer gain K for which the closedloop system
with G¢(s) = 1 meets the ﬁrst design speciﬁcation. 3. (5 points) For the value of servoampliﬁer gain found in part (2), ﬁnd the undamped natural frequency
in... [radians / second], dimensionless damping ratio C and peak overshoot M,D of the closedloop system. 4. (5 points) Sketch and interpret the corresponding stepresponse of the closedloop sunseeker control
system for the value of the servoampliﬁer gain found in part (2). 5. (15 points) Doe the closedloop system with 06(5) 2 1 and the value of the servoampliﬁer 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
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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.
 Fall '07
 SCHIANO

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