# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

40 Pages

### F07%20314%20HW%2002%20all-s

Course: EECS 314, Fall 2007
School: Michigan
Rating:

Word Count: 1567

#### Document Preview

314 EECS Fall 2007 HW 02 Solutions Homework 2 Problem 1 Part1: 1) The problem statement gives us the following information: R R X &quot; P ! K! 2 Vout &quot; a ! b! (R In addition, we know because of voltage division that Vout &quot; VS &amp; X &amp;R ' P the given information for R X , you get: ( 1 %( RP % Vout &quot; VS &amp; # &amp; R #&amp; 2 ! K! # \$ ' P \$' ( 1 K! % Vout...

Register Now

#### Unformatted Document Excerpt

Coursehero >> Michigan >> Michigan >> EECS 314

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
314 EECS Fall 2007 HW 02 Solutions Homework 2 Problem 1 Part1: 1) The problem statement gives us the following information: R R X " P ! K! 2 Vout " a ! b! (R In addition, we know because of voltage division that Vout " VS & X &R ' P the given information for R X , you get: ( 1 %( RP % Vout " VS & # & R #& 2 ! K! # \$ ' P \$' ( 1 K! % Vout " VS & ! # &2 R # P \$ ' KVS 1 Vout " VS ! ! 2 RP By matching coefficients, we can see that: 1 a " VS " 6V 2 KV S 3 V V " " 0.12 b" 25 deg deg RP % # . If you substitute # \$ Page 1 of 12 EECS 314 Fall 2007 HW 02 Solutions 2) Plot of Vout vs. Alpha 16 14 12 10 8 12V Vout (V) 6 4 2 0 -2 -4 -80 0V -60 -40 -20 0 alpha (degrees) 20 40 60 80 x=linspace(-75,75,5000); y=6+ 0.12*x; plot(x,y); xlabel('alpha (degrees)'); ylabel('Vout (V)'); grid; 3 ! , the output voltages 25 exceed 0V and 12V at -50 and +50 degrees respectively. Since this isn't possible, it implies that with angles lower than -50 degrees, the output voltage is 0V and with angles greater than 50 degrees, the output voltage is 12V. For the plot, note that if you just plot the equation Vout " 6 ! 3) If we rearrange Vout " 6 ! , 3 25 ! , we get: ! " )Vout + 6* . 25 3 25 50 ! 4 " )4 + 6* " + - +16.67, 3 3 !7 " 25 )7 + 6* " 25 - 8.33, 3 3 , 4) .! " 25 .Vout 3 25 .! " ).010* " .083333, 3 Page 2 of 12 EECS 314 Fall 2007 HW 02 Solutions 5) We can rewrite the error above as: .! " .Vout KVS . We also know that b " . b RP .Vout .Vout / RP When you substitute, you get: .! " " . By inspection, you can KVS K / VS RP see that .! will decrease when R P is decreased, when K is increased, or when VS is increased. Part2: No, you cannot use the device above. The range of the output voltage exceeds the desired range. 1 When ! " 0 , Vout " VS . Since we know that we want Vout " 2.5V when ! " 0 , we 2 have: 1 Vout " VS " 2.5 2 VS " 5V This means that you must decrease VS Next, when ! " 50 , we would like to have Vout " 5V . KVS 1 Vout " VS ! ! 2 RP 1 5K 5 " )5* ! )50* 2 RP K 1 " RP 100 Therefore, as long as you keep the above ratio, there is more than one solution. For example, you could change K to 3 k0 / deg and R P to 300 k0 . However, if K is increased, so must R P and if K is decreased, so must R P . Page 3 of 12 EECS 314 Fall 2007 HW 02 Solutions Problem 2 Potentiometer RP is what effectively limits the range of detection for the circuit. The adjustable tap on the potentiometer gives fractional resistance Rx = RP / 2 + K !. Because the range is limited as Rx ! (0, RP), this puts ultimate limits on the angle variable !, given as: 0 Rx Rp Rp 0 K Rp 2 Rp Rp 2K 2K 40 40 The actuator is turned on when Vout = Vs. The comparator will make the output voltage Vout = Vs whenever the potential V2 at the non-inverting (+) input terminal is greater than the potential V1 at the inverting () terminal. V1 and V2 are given by two voltage dividers: V1 V2 Vs Vs Rx Rp R4 R3 R4 So the actuator turns on when: V1 V2 Rx R4 Rp R3 R4 1 K R4 2 Rp R3 R4 1 2 2.5 k 200k 10 250k 150k 250k And conversely, off when ! > 10. Of course, both on/off angle ranges remain within the above 40 limit. Page 4 of 12 EECS 314 Fall 2007 HW 02 Solutions Problem 3 1. As stated, the resistance RT is decreases by 25 k! at the elevated temperature, so the thermistor has negative temperature coefficient (NTC). 2. The actuator is turned on when when the MOSFET switch closes; this occurs when the gate voltage is above the stated threshold VT = 3.5 V. The comparator will make the output voltage Vout (connected to the MOSFET gate) equal to Vs whenever the potential V2 greater than V1. Since Vs = 10 V, this is sufficient to turn on the actuator. V1 and V2 are given by two voltage dividers: V1 V2 Vs Vs RT R1 R3 R4 RT R4 Conveniently, R1 = R3, so the on/off transition point occurs whenever RT = R4 = 89k!. 3. Knowing that the thermistor is linear, RT is given by: RT 120k 45k T 100 So solving gives the on/off transition temperature: RT R4 45k T 100 89k 120k T 68.89 4. High temperatures will decrease RT, causing voltage V1 to decrease, causing the comparator to raise Vout, turning on the MOSFET and the actuator. Logically, rising temperatures should activate a cooler. Page 5 of 12 EECS 314 Fall 2007 HW 02 Solutions Problem 3 5. Solve for RT at 55 F to find the necessary R4: RT T 55 F 120k 45k 55 100 95.25k 6. Vs was not used in any of the RT calculations, so no change is necessary. Page 6 of 12 EECS 314 Fall 2007 HW 02 Solutions Solutions to problems 4, 5 Revised Problem 4 Part 1 The output voltages changes from VOUT,HIGH to VOUT,LOW when VIN = V+,HIGH. We can use the equation for V+,HIGH given in the big picture: = +, = , 2 1 + 1 + 2 1 + 2 Plugging in VOUT,HIGH = 10V, VIN = 6.05V, VREF = 6V, R2 = 2k: 6.05 = 10 Solving gives R1 = 158.0 k. 2 1 + 6 1 + 2 1 + 2 To find V+,LOW just use the similar equation given in the big picture. +, = , Substituting values gives: +, = 0 Part 2 use Simply the same method as in part 1 to calculate the resistance for the new conditions: 3.05 = 10 2 1 + 3 1 + 2 1 + 2 2 158 + 6 = . 158 + 2 158 + 2 2 1 + 1 + 2 1 + 2 Solving gives R1 = 278.0 k. The same resistor cannot be used. Part 3 Simply calculate the resistance given the new conditions: 3.15 = 10 2 1 + 3 1 + 2 1 + 2 Page 7 of 12 EECS 314 Fall 2007 HW 02 Solutions Solutions to problems 4, 5 Revised Solving gives R1 = 91.33 k. The resistance cannot be tripled to get the desired effect. In fact, it should be much less than before. Part 4 See part 3. Problem 5 The equation for the RX with the given information included is: = 200 + 1/ 40 = 140 2 Since the voltates over RX and R4 should be equal, both resistors should divide the same amount of the total voltage, VS. Therefore: 4 = 3 + 4 Plugging in values, and cancelling out the source voltage.... 140 4 = 200 100 + 4 Solving gives R4 = 233.3 k. Also, using voltage division: 2 = 4 = 8.4 3 + 4 Part 2 Increasing the angle of the arm will increase the resistance RX, and thus will increase the voltage V1, which is the input voltage to the Schmitt trigger. So we require that the Schmitt trigger switches from high to low when RX reaches the appropriate value determined by the stopping angle. Page 8 of 12 EECS 314 Fall 2007 HW 02 Solutions Solutions to problems 4, 5 Revised First calculate the threshold RX. = 200 + 1/ 42 = 142 2 Now calculate the input voltage to the Schmitt trigger, which is found by voltage division: 1 = 142 = 12 = 8.52 200 This is the input voltage at which the output will switch from high to low, so it must be equal to V+,HIGH. VREF is the V2 that was solved for before. +, = , 8.52 = 12 Therefore R1 = 4.35M 2 1 + 1 + 2 1 + 2 150 1 + 8.4 1 + 150 1 + 150 Now calculate V+,LOW. +, = 0 2 4.35 + 8.4 = 8.12 1 + 2 4.35 + 150 This corresponds to the VIN at which the arm is at the start angle. Using the voltage division formula again, we can find the resistance RX at which this occurs. 1 = 8.12 = 12 Therefore RX = 135.3k Now solve the arm angle equation 135.3k = And so finally, = . 200 + 1/ 2 200 Page 9 of 12 EECS 314 Fall 2007 HW 02 Solutions Problem 6 I1k 1&2. Write node voltage equation for the non-inverting input terminal and simplify it using the GR#1: I 5k = I 20k " 5 ! V+ V+ ! 0 . Thus, V+ = 4 (V) = 5k 20k 3. From GR#2, V- = V+ = 4V = V1 4,5 & 6. Write node voltage equation for the inverting input terminal and simplify it using the GR#1: I 10 k = I 1 " 6 ! 4 4 ! Vout = . Thus, I1 = I10k = 0.2mA and Vout = -6 (V) 10k 50 k (!6) ! 0 = !6mA 1k 8. KCL: I 1 + I 2 = I 1k " 0.2mA + I 2 = !6mA . Thus, I 2 = ! 6.2 mA 7. I 1k = Page 10 of 12 EECS 314 Fall 2007 HW 02 Solutions Problem 7 Part 1 Vout ,1 = Vs R2 '' R4 , R1 + R2 ' ' R3 + R4 R2 ( R3 + R4 ) / R2 + R3 + R4 where R2 ' ' = R2 || ( R3 + R4 ) = Vout ,1 = Vs R2 R4 R1 ( R2 + R3 + R4 ) + R2 ( R3 + R4 ) Problem 7 Part 2 Using GR#1, we know I1 = I2. R2 Thus, V + = Vs R1 + R2 Using GR#2, we know V+=V-. Vout = V! = Vs R4 R3 + R4 R2 R4 R1 + R2 R3 + R4 Problem 7 Part 3 Plug in R1=10 R2=20 R3=1 R4=1 Vout,1 = (6/7)V = 0.923077 V, and Vout,2 = 4V. Comment by Alexander Ganago: As stated in lectures, node voltage equations are the most powerful tool for solving circuit problems. In Part 1 of this problem, for the node A, where R1, R2 , and R3 are connected, we obtain: ! ! Page 11 of 12 ! EECS 314 Fall 2007 HW 02 Solutions VA " VS VA " 0 VA " 0 + + =0 R1 R2 R3 + R4 \$1 1 1 ' 1 VA # & + + ) = VS # R1 % R1 R2 R3 + R4 ( VA = VS # 1 R1 1 1 1 + + R1 R2 R3 + R4 Also, for the output node, ! VOUT " VA VOUT " 0 + =0 R3 R4 \$1 1 ' 1 VOUT # & + ) = VA # R3 % R3 R4 ( 1 R3 R4 VOUT = VA # = VA # 1 1 R4 + R3 + R3 R4 The last equation ends with the familiar formula for voltage division (note how it is derived from the node voltage equation). ! By comparison, observe that the equation for VA has an extra term 1 R3 + R4 in the denominator; this term represents the loading of the first voltage divider by the second voltage divider (an extra path for current in parallel to R2). In Part 3 of this ! problem, you should notice that the output voltage is grossly decreased due to this ! extra term. Addition of the buffer, or voltage follower, eliminates the extra path for current and algebraically the extra term in the equation. Page 12 of 12 EECS 314 Lecture notes Note that the equation for node A above was written in the assumption that the entire current that flows through R3 also flows through R4, in agreement with the circuit diagram. If we do not make this assumption, the equation for node A should be written as VA " VS VA " 0 VA " VOUT + + =0 R1 R2 R3 \$1 1 1 ' VOUT 1 VA # & + + ) " = VS # R3 R1 % R1 R2 R3 ( The equation for the output node remains as above ! VOUT " VA VOUT " 0 + =0 R3 R4 \$1 1 ' 1 VOUT # & + ) = VA # R3 % R3 R4 ( VOUT = VA # 1 R3 1 1 + R3 R4 = VA # R4 R4 + R3 The expression for VOUT is substituted into the equation for VA Of course, the final result remains the same. ! Alexander Ganago
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Michigan - EECS - 314
EECS 314 Winter 2008 HW 03 Solutions Problem 1Part 1First, before using the golden rule, construct a node voltage equation at the V+ node. This is done by applying KCL to account for all currents exiting or entering the node, followed by rewriting
Michigan - EECS - 314
EECS 314 Fall 2007 HW 04 solutionsProblem 1This problem asks you to find the Thevenin and Norton equivalent circuits of the circuit above using source transformations and parallel/series reductions. Begin by transforming the 16V voltage source and
Michigan - ME - 235
HOMEWORK SET 4b 6.9 In a boiler you vaporize some liquid water at 100 kPa flowing at 1 m/s. What is the velocity of the saturated vapor at 100 kPa if the pipe size is the same? Can the flow then be constant P? The continuity equation with average val
Michigan - ME - 235
HOMEWORK SET 1 2.22 A car of mass 1775 kg travels with a velocity of 100 km/h. Find the kinetic energy. How high should it be lifted in the standard gravitational field to have a potential energy that equals the kinetic energy? Solution: Standard kin
Michigan - ME - 235
HOMEWORK SET 7b 9.44 A small pump is driven by a 2 kW motor with liquid water at 150 kPa, 10C entering. Find the maximum water flow rate you can get with an exit pressure of 1 MPa and negligible kinetic energies. The exit flow goes through a small ho
Michigan - EECS - 314
EECS 314Winter 2008Homework set 7Student's name _ Discussion section # _ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required field
Michigan - EECS - 314
EECS 314 Winter 2008 HW 10 Solutions Problem 1Part 1Determine the intermediate logic signals to solve the problem incrementally, and verify the net operation: Input A 0 1 0 1 Output A 1 0 1 0 Input B 0 0 1 1 Output B 1 1 0 0 Output C 0 1 1 1The b
Michigan - EECS - 314
EECS 314Winter 2008Homework set 9Student's name _ Discussion section # _ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required field
Michigan - EECS - 314
EECS 314 Winter 2008 HW 04 Solutions Problem 1Part 1 &amp; 21A 2 8 2 48V 5.2 50V 5 b 2.5a2A2 6A = 2A 2 82V5.2 50V 52.57.2 a = 4A 1.62V 2.5 50V 5 abb7.2 = 1.6 6.4V2V 2.5 10A a = 5 b 2.5 2.5 10.5A = 5 b 5.69 = 33.5V a = 5.89A 5.69
Michigan - EECS - 314
EECS 314Winter 2008Homework set 5Student's name _ Discussion section # _ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required field
Michigan - EECS - 314
W08 EECS 314 Homework 6 Solutions Problem 1Part 1The problem asks to find VOUT in terms of VIN, so we start out the problem by doing Node Voltage Analysis. IR and IC are labeled on the diagram and following the Passive Sign Convention, this in tur
Michigan - EECS - 314
EECS 314 Winter 2007 HW 10 Problem 1Student's name _ (Last name, first name, IN INK) Discussion section # _The big pictureAs you know from lecture notes, music on CD is recorded with 16- or 20-bit resolution, at the sampling rate of ~ 44 kHz. As
Michigan - EECS - 314
EECS 314 Winter 2007 HW 07 Overview OverviewIn this HW 07, we keep studying steady-state responses of circuits that include capacitors and inductors to sinusoidal input signals, both at varied frequencies and at fixed frequencies. The main analytic
Michigan - EECS - 314
Sample solutionsEECS 314 Fall 2007 Exam 1Instructor: Alexander Ganagoganago@umich.eduWednesday October 10, 2007, starts at 5:00 PMExam rooms, according to the first letter of students' last names:A through K 220 Chrysler Auditorium (our lec
Michigan - EECS - 314
Homework 2 Solutions Problem 1EECS 314 Winter 2008Part1: 1) The problem statement gives us the following information: R RX = P + K 2 Vout = a + b In addition, we know because of voltage division that Vout = VS the given information for R X , you
Michigan - EECS - 314
EECS 314 Fall 2007 HW 11 For extra credit Problem 1Student's name _ (Last name, first name, IN INK) Discussion section # _For the Big Picture, see the file 2007 Analog and Digital posted on the web as part of lecture notes for November 27, 2007.
Michigan - EECS - 314
EECS 314 Winter 2008 Problem 1HW 01SolutionsKCL, KVL, and the Passive Sign ConventionConsider the circuit shown on this diagram. Note that nodes A, C and B are all connected with wires thus they are in fact one node! Some currents and some vo
Michigan - EECS - 314
EECS 314Winter 2008Midterm exam 2The Key A B C D E 1 3 6 2 4 5 10 12 11 8 7 9Problem # Correct answer 1 2 3 4 5 6 7 8 9 10 11 12 A D B E A C A E E B D CInstructor: Alexander GanagoEECS 314Problem 1Winter 2008Midterm exam 2In the ci
Michigan - EECS - 314
EECS 314Winter 2008Midterm exam 1The Key A B C D E 6 2 1 4 3 7 8 5 10 9 12 11Problem # Correct answer 1 2 3 4 5 6 7 8 9 10 11 12 C B E D C A A B E D D CInstructor: Alexander GanagoEECS 314 Problem 1Winter 2008Midterm exam 1In the ci
Michigan - EECS - 314
EECS 314Winter 2008Homework set 8Student's name _ Discussion section # _ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required field
Michigan - EECS - 314
EECS 314 Winter 2007 HW 09 Problem 1Student's name _ (Last name, first name, IN INK) Discussion section # _The big pictureThis problem introduces several important aspects of rectifier circuits: the actual voltages across the load resistance (the
Michigan - EECS - 314
EECS 314 Fall 2007 HW 10 Problem 1Student's name _ (Last name, first name, IN INK) Discussion section # _The big pictureAs you know from lecture notes, music on CD is recorded with 16- or 20-bit resolution, at the sampling rate of ~ 44 kHz. As yo
Michigan - ME - 235
HOMEWORK SET 4a 5.86 A piston/cylinder contains 1.5 kg of air at 300 K and 150 kPa. It is now heated up in a two-step process. First constant volume to 1000 K (state 2) and then followed by a constant pressure process to 1500 K, state 3. Find the hea
Michigan - ME - 235
HOMEWORK SET 2 3.32 Water at 120oC with a quality of 25% has its temperature raised 20oC in a constant volume process. What is the new quality and pressure? Solution: State 1 from Table B.1.1 at 120oC v = vf + x vfg = 0.001060 + 0.25 0.8908 = 0.2237
Michigan - ME - 235
HOMEWORK SET 6 8.17 Find the missing properties and give the phase of the ammonia, NH3. a. T = 65C, P = 600 kPa s=?v=? b. T = 20C, P = 100 kPa v=?s=? x=? 3/kg s = ? x = ? P = ? c. T = 50C, v = 0.1185 m a) B.2.2 average between 60C and 70C v = (0.2598
Michigan - ME - 235
HOMEWORK SET 3 4.35 A piston cylinder contains 1 kg of liquid water at 20oC and 300 kPa. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 3 MPa with a volume of 0.1 m3. a) Find the final temperatu
Michigan - ME - 235
HOMEWORK SET 7a 8.105 A piston/cylinder has ammonia at 2000 kPa, 80oC with a volume of 0.1 m3. The piston is loaded with a linear spring and outside ambient is at 20oC, shown in Fig. P8.71. The ammonia now cools down to 20oC at which point it has a q
Michigan - ME - 235
HOMEWORK SET 5 6.57 A small stream with 20oC water runs out over a cliff creating a 100 m tall waterfall. Estimate the downstream temperature when you neglect the horizontal flow velocities upstream and downstream from the waterfall. How fast was the
UMiami - MTH - 112
SECTION TAYLOR MACLAURIN 8.7 AND SERIEStr253By 39, Example 7,tan-1, : i (-1)' #forwehave : ta&quot;-'(r7al@ro, lrl &lt; 1.Inparticular,&quot; -ft,. ir, , /+) : i (_r),!44)&quot;*' t (-t)&quot; (.i/)'+=1. znT!-: i!o' ,/i2n+L'r-1)'(-1)&quot; 6 -&amp; sor:ft Affi-2v3E
UMiami - MTH - 112
9 trPARAMETRIC EQUATIONS POLAR AND COORDINATES9,1 Parametric Gurves1.r:l+ft,a:t2-4t,tT0&lt;r&lt;5234501r2r+ \O2.4rr+Jl2.733r+J63.24a0-3-4-3053 .r : 5 s i n t , A : t 2 ,-r -r/21 t 1 n 0 r/2 7rt r-7ra0 n2 s.87n
Michigan - ASIAN - 126
ASIAN LAN 1261 10Habitual action with ACTIVITY verbsV-te form +topicalso expresses a habitualaction / repeated action. V te-formMr. Tanaka eats vegetables everyday.Habitual action with ACTIVITY verbs Practicete-form Adjective/Nounte
University of Texas - MIS - 311F
Outline Internet Security Definition How We Connect to the Net Why Use the Net Needs, Risks, Issues Analysis Solutions / OptionsInternet Security DefinitionInternet Security Definition Protection of: InformationT ATategy trS Against th
University of St Andrews - IR - 2003
School of International Relations IR2003: Power and Violence in World Politics Semester 1: January 2006 Time Allowed: 2 hours Answer 3 questions: Answer question 1 Answer one question from Section A Answer one question from Section B1. `It is unrea
University of Texas - BIO - 311C
Enzymes1Last TimeMeasuring the energy that can perform work in the cell - Free energy - G Free energy and metabolism - exergonic and endergonic reactions Coupling exergonic reactions with endergonic reactions to do work2EnzymesEnzymes are p
University of Texas - BIO - 311C
Respiration: electron transferhttp:/www.cbs.dtu.dk/staff/dave/roanoke/mitochondriadrawing.gif1Last TimeEnzymes are proteins speed up exergonic chemical reactions Enzyme activity is affected by the cellular environment pH and temperature Regula
University of Texas - CH - 301
McCord Vanden BoutWhat sections?THERMODYNAMICSExam 4 Review TopicsFall 2007Chapter 9 sections 1-6 and Chapter 10 sections 1-9. Plus heating curve problems Chapter 16 section 10 and Bond Energies (Chapter 13, Tables 13.6 and 13.7 - don't memor
UCSB - ECON - 100a
Econ 100A - Problem Set 1 I. Math Reviewdy ? dx y 2. y = zx3/4 + z ln d. What is the expression for ? x y 3. y = 7k ln( x ). What is the expression for ? x dy x 4. E y ,x = . Let y=2x2. What is the expression for Ey,x? dx y1. y = 5x 2 / 3 + 7x +
UCSB - ECON - 100a
Econ 100A - Problem Set 2 Utility Maximization and Demand 1. George's utility is given by U(x1, x2) = 6x1*x22. Use calculus to work through the utility maximization problem and derive his demand functions x1(p1, p2, I) and x2(p1, p2, I). 2. Ted's dem
UCSB - ECON - 100a
Problem Set 3 Comparative Statics of Demand 1. Good 1 is a normal good, Good 2 is an inferior good. Using 3 budget lines and 2 indifference curves, illustrate the effect of an increase in p2 on the consumption of both x1 and x2. Label income and subs
UCSB - ECON - 100a
Econ 100A - Problem Set 4 Supply of Labor 1. Let U(Le, I) = Le2 I. What is the individual's labor supply function (La(w, I0)? How many hours does the individual work if the wage (w) is 10\$/hr and non-wage income (I0) is 320\$/wk? 2. Illustrate the inc
UCSB - ECON - 100a
Problem Set 5 THEORY OF PRODUCTION 1. a. If a firm's production function is given by F(L,K) = 3 18 L + 9 K , which is larger APL(8, 8) or APK(8, 8)? 2. a. What is the MRTS of the function F(L,K)= L2 + K? b. Does this function exhibit diminishing MRTS
UCSB - ECON - 100a
Formulas and Definitions Utility Maximization max U(x1, x2) s.t. p1x1 + p2x2= I F.O.C.s MU1/MU2 = p1/p2 p1x1 + p2x2= I Labor Supply max U(Le, I) s.t. I + wLe = I0 + 168w F.O.C.'s MULe/MUI = w I + wLe = I0 + 168w Capital Supply Max U(c1,c2) s.t. c1+c2
UCSB - ECON - 100a
Problem Set 6 Long Run Cost 1. Let F(L,K)=L1/3K2/3. What are the firm's output-constrained factor demand functions, L*(Q, w, r) and K*(Q,w, r)? What is the firm's long run cost minimizing input bundle when w=4, r=1, and Q=4? 2. A firm is producing an
UCSB - ECON - 100a
Problem Set 7 Profit Maximization and Supply 1. Suppose LTC(Q, w, r)= wrQ2 a. Find LMC(Q, w,r) b. Find the firm's supply function Q(P, w, r). c. Let w=2 and r=2 and P=4. What is the firms' profit maximizing level of production? 2. LMC(Q, w, r) = .5 (
UCSB - ECON - 100a
REVIEW QUESTIONS 1. George's and Fred's utility functions over goods 1 and 2 are given by: UG(x1, x2)= 3x1 + x2 UF(x1, x2)= 9x1 + 4x2 , respectively. Is it reasonable to say that Fred likes good 1 more than George does? Explain and discuss. 2. Give a
UCSB - ECON - 100a
REVIEW QUESTIONS1. Good 1 is normal, good 2 is normal and the two goods are substitutes (but not perfect substitutes).Using budget lines and indifference curves, illustrate the effect of an increase in p2 on the consumption of both x1 and x2. Labe
BU - PY - 211
PY211 MIDTERM EXAM 2 PRACTICE 1. For the multiple choice questions below, place your letter answer on the line next to each question. _ A children's slide is to be built 3 m high. The slide can be built so that it is either a straight inclined plane
BU - PY - 211
Practice PY211 MIDTERM EXAM 2 G = 6.67 x 10-11 Nm2/kg2 g = 9.8 m/s2 REarth = 6.38 x 103 km MEarth = 5.97 x 1024 kg RMoon = 1.74 x 103 km MMoon = 7.35 x 1022 kg Earth-Moon Distance = 384 x 103 kmUseful Numbers:1. Write your answer in the space pro
BU - PY - 211
Starting from the second problem:
BU - PY - 211
Assignment #3 SolutionsQuestions 14 and 15 are the quick quiz 2 and 4 respectively from the book and the answers are in there at the end of chapter 4.
BU - PY - 211
BU - PY - 211
Assignment 10 Solutions
BU - PY - 211
Assignment 9 Solutions
BU - PY - 211
Assignment 6 Solutions
BU - PY - 211
Assignment 5 Solutions:
BU - PY - 211
Assignment 7 Solutions:
BU - PY - 211
Assignment 8 Solutions
Ohio State - CSE - 755
CIS 755 Programming Languages Neelam Soundarajan Computer Sc. and Engineering Dreese Lab 579 Tel: 292 1444 Please send corrections by e-mail to neelam@cse1Main Topic: Ways to formally define syntax and semantics of programming languages. Plus: A
Ohio State - CSE - 755
Operational Semantics: Specify semantics of a language by specifying how each command in the language is to be executed. One possible approach for O.S. of L: Define an interpreter for L. Metacircular interpreter for L: The interpreter is defined in L
Ohio State - CSE - 755
CIS 755, Assignment #2 Due: 24 April '06 1. (4 points). Consider the IMP language whose translational semantics we defined in class. Suppose we revise IMP to be a block structured language (but no procedures or functions). So a prog is a block (rathe
Ohio State - CSE - 755
Individual steps: (p3 x = 100) p4 : easy. {p1} read x {p2} : easy from the read axiom. p2 p3 : easy from math logic. {p3 x = 100} write x; read x {p3} : {p3} write x; read x {p3} where p3 [OU T^x^head(IN )^tail(IN ) = 1, 2, . . . , 100 ] and it