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Unformatted text preview: EECS 215 Fall Term 2005 Midterm Exam II Name: 1 Lecture Section MCAfee Philli s
Rules: 1. Monday, November 14, 2005, 6:00 to 7:30 PM nominal exam time.
DO NOT DISCUSS THIS EXAM WITH ANYONE PRIOR TO 6 PM November 15!! 2. Closed Book, etc.
3. l Page (8.5”xl l”, double sided) of notes allowed
4. Calculators Needed and Allowed. No devices with full alphanumeric keyboards are permitted. 5. Exam is given under the College of Engineering Honor Code principles and
practices. 6. No communications of any kind are allowed. Use of cell phones, cameras, personal
data assistants, computers, or any other electronic devices, besides approved
calculators, will be treated as an Honor Code violation. 7. Work to be done in Exam booklet. Turn in all pages of the exam. Do not unstaple the pages.
8. DO NOT WRITE ON THE BACK OF PAGES. Work on backs of rages will NOT be graded. 9. Show your work and brieﬂy explain major steps to maximize partial credit.
(for example: i3=il+i2, KCL at node A).
NO CREDIT WILL BE GIVEN IF NO WORK IS SHOWN. 10. WRITE YOUR FINAL ANSWERS IN THE AREAS PROVIDED This Exam Contains
4 problems over 21 pages (including ample workspace). Sign the College of Engineering Honor Code Below.
(NO credit will be iven for the exam without a sined lede): I have neither given nor received aid on this examination. Signed: Do not write below this line
>l<>l<>l<**************$$********>l<>i<>l<>l<*>l<>l<>l<>i<*>l<**=i<>i<>l<>l<>l<*>l<>i<************************ Problem I [ a b c ] Problem 2 [ a b c ] Problem 3 [ a b ] Problem4 [ a ] Problem 1:0 Am Circuit Anal sis 25 oints The op—amp circuit shown has five (5) unknown node voltages ( V0, VA, VB, VC, VD ). — Gnd =Reference Node
R1=20KQ R2=4OKQ R3=20KS2 R4=4OKQ RF=50KQ V31 = 4 Volts 152 = 0.1 mA V53 = 4 Volts 154 = 0.1 mA a) Derive the numerical values for the node voltages (V0, VA, VB, Vc, VD). In the box/table
on page 2, enter the values for the node voltages. Round each voltage answer to the
nearest millivolt (mV) and be careful to specify the sign of the voltage. In the box/table
on page 3, CLEARLY indicate the equations you solved to derive the node voltage
values. The only unknowns in the equations should be node voltages (V0, VA, VB, VC,
VD). For each equation Circle whether it is a m (voltage units) or a QC; (current units) equation.
Show your intermediate work on the following pages. In the box/table below, enter the values for the node voltages. Round your answer to the
nearest millivolt (mV) and be careful to specify the sign of the voltage. Workspace for Problem 1(a) In the box/table below, CLEARLY indicate the equations you solved to derive the node
voltage values on page 2. The only unknowns in the equations should be node voltages (V0,
VA, VB, Vc, VD). In the box/table below, enter your equations. For each equation Circle
Whether it is a m (voltage units) or a QC; (current units) equation. Show your
intermediate work on the following pages. Workspace for Problem 1(a) 4/0173 “14+ {4,24% VB‘VD) l/p—l/cz4l/ UnﬁMu/II ﬂan/Q, VI/ﬂgef may be defer/rim ep/ 5 y
WW3”? k6]. egg/4+:an 47‘ V3} 1/0. R1=20KQ R2=40KQ R3=20K§2 R4=40KQ RF=SOKQ
V51=4V01ts 132:0.1mA Vs3=4VOItS 154:0.1mA Workspace for Problem 1(a) \/SI
@ B VB'(‘4) Kev V1514, VB'Va 0 w‘,’ R\ PD. RF R1=ZOKQ R2=40KQ R3220KQ R4=40KQ RF=50KQ
V51 2: 4 Volts 132 = 0.1 mA V33 2 4 VOltS 154 = 0.1 IDA Additional workspace for Problem 1(a) — Gnd=Reference Node
R1=ZOKQ R2=40K£2 R3=20KS2 R4=40KS2 RF=50KQ Vs] = 4 Volts 152 = 0.1 mA V33 1: 4 Volts 134 = 0.1 IDA b) Derive the value of the current 10 ﬂowing into the opamp output terminal? Round your
answer to the nearest microamp (uA) and be careful to specify the sign of the current. — 7 nd = Referee Nod
R1=20KQ R2=40KQ R3=ZOKQ R4=40KS2
V31 2 4 Volts 132 = 0.1 mA V33 = 4 Volts 134 = 0.1 mA RF = Your Redesigned value (not 50 K9) 0) Redesign the oparnp circuit by changing the value of RF (and change only RF) such that
the output voltage is + 3 Volts. Hint: The voltage at node VD is independent of R p. Workspace for Problem 1(0) VD anchmge/ *9 V5 amt/M4754
Ia anehqueq/ Problem 2: FirstOrder Circuits 330 points} A capacitive circuit is shown. Problem 2 has part (a) through part ((1). B*1XwithB=9 Is(t) = 20 + 12 u(t) Amps "'2' (a) Enter values in the table to indicate equilibrium and initial condition values for the circuit.
For the specific time indicated at the top of each column, enter the values for Idt), Vc(t), and d VC(t) /dt. Time (Seconds) —)
1 Variable l I “1 ($38) 0 4 0
($7332) Q 5 Q 5 4' 0
($31??? $35) O 0 ‘5’ O Workspace for Problem 2(a) Is(t) = 20 + 12 u(t) Amps '1' Additional workspace for Problem 2(a) Final Theven/n cfpm: #914 and,“ For Vag/ Rm [Iskl’aqVI’VC QVcsO
Rn 12"?3 3’0 0 ’33
' :
VIC'3’2)+ V6 %*é’o 0 vac v05 5 1" == 25+ ma) a v¢(a')=mo*)=95:
ye (myﬁnrs‘; av B*waithB=9 Is(t) = 20 + 12 u(t) Amps T (b) Derive VC(t) for t > O . —‘t’/ a;
Vc(t) = a 3 Volts for t > 0 Workspace for Problem 2(b) B*IXwithB=9 Is(t) = 20 + 12 u(t) Amps  Additional workspace for Problem 2(b) 11 Is(t) = 20 + 12 u(t) Amps . : (c) Derive the value of time Tp at which VC(Tp) = 32.5 Volts. I Tp = 2 O ' Seconds I Workspace for Problem 2(c) 12 Problem 3: Second Order Circuits 25 Points Consider the circuit shown below. a) For the circuit picture above, find the differential equation that relates vdt) to Vina). In the
space provided, write the equation as an algebraic expression in either of the standard forms:
dzv dv +A—+Bv = (vimt or D C+E—§—+v = 12th dt2 dt C f ) dt2 dz C g< ) vC must be the only unknown variable (assuming Vina) and circuit variables R1, R2, L, C are
known). You may use KVL/KCL/time domain methods or sdomain, but you must clearly show your work to receive full or partial credit.
Warning: Attempts to mix timedomain and sdomain approaches are likely to result in zero credit.
Differential Equation: .L + EL 47?
*(kaC L. Workspace for Problem 3(a)
(/9119 S'ddmm'ﬁ 
.r ' I = ' " l3 14 Vina) Additional workspace foy Problem 3(3) 15 +171 11
Vin“) © R2 C T vC b) Now suppose we have the following component values for this circuit:
R1z3k£2 [€22le L=2mH C=3uF Vina) = 16 + 8 u(t) Determine the values for VC(t), idt), iL(t), and VL(t) at t = 0— and at t = 0+. Enter the
values in spaces provided in the table. Time (Seconds) >
1 Variable l v (t) ( Vilts ) g 8
i (t) ( Afnps ) 0 O art:96?!” 2.57M ( Amps ) VLU) 0 8 ( Volts ) t=0— t=0+ .— Workspace for Problem 3(b) 0”
M R;=3k§2 R2: 1kg LzZmH C=3MF Additional workspace for Problem 303) 17 Problem 4: SecondOrder Circuit Solution 20 Points Consider the circuit shown below. R, L lC "T" +vL + 1L
vino?) © 2,.L R2 0 :: vc .J R1=3l<Q [€22le L=2mH C=3uF Vina) : 32 w 24 u(t) For these component values, the differential equation for this circuit becomes: /) C 7 M
if +(150003X1065_1)£§§~+(1.000x10)5”‘)vc =(5x1085 am)
I. I
01"
7 d2
<1.000><lO"gs") d"; +(i.50003><10'3s“0)——:VC WC = (0.500)vm(t)
l I 4‘ ‘M where 3 denotes seconds, not the Laplace differential operator. Equilibrium and initial condition values are: Time Seconds 9
i \Eariable )i t 2 0+ 1 t a 00
vc(t) (Volts) r 16 4 MG (A & mA) 0 0 mt) (A & mA) 5.333 x 10‘3 A = 16/3 mA 1.333 x 10“3 A = 4/3 mA
vL(t) ( Volts ) — 24 0 (Note that vi” and initial/ﬁnal conditions are different than Problem 3!) a) Derive the complete solution for vdt) for t>O. Write your answer in the space provided. 18 01‘ 2
dvc dz 3 a
d vC dz (1.000x10’052) ,3 dz + (1.50003 ><10"3 5‘0) dvr
dz ' +(1.50003x1063'1)f—ZX€—+(1.000x1095'2)vc =(5x1oss"2)vm(z) ——“« + VC = (0500)Vm (I) Time (Seconds) '9 _ +
L Variable i t" 0 t9 00
vc(t) ( Volts ) 16 4
idt) ( A & mA ) 0 0
iL(t) (A & mA) 5.333 x 10‘3 A = 16/3 mA 1.333 X 10'3 A = 4/3 mA
vL(t) ( Volts ) — 24 0 I
Workspace for Problem 4(3)
5’ ,
M: 7, {a x m 5 ”
4 ' /
(W2: ‘09 r: 3.!ékm 5 51/9“; “@123: 063“%2
5'2": ’6é7'y’6’2 5'!
53.: *3,5’€X}06 5/ 712“): F
7a W <: 74 we 19 dd“? +(1.50003><10"s‘1):VC +(1.000><1095“2)VC : (5x1085‘2)vm(r)
i,” I
—9 2 CZva —3 «0 dVC
(1.000x10 s ) d 7 +(1.50003><10 s )Twc =(O.500)vm(t)
l" 1
Time (Seconds) 9
W t 2 0* t ) 00
i Varlable J,
i
VG“) ( VOItS ) 4
ic(t) (A & mA ) 0 0
iL(t) (A & mA) 5.333x 10‘3A = 16/3mA 1.333x 10’3A =4/3mA
mt) (Volts) —24 0 Additional workspace for Problem 4(a) ’l’EKf)= égeé’fwi’zes’f4'4
112(0):: Am ;+%:/é 52
14”" A“: (l6~4> M )100534
(#2))?
’41., viﬂv,o,005’34 ...
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 Winter '08
 Phillips

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