Homework 3 Solution. Due Friday (5pm), Feb. 14, 2013

Size: px
Start display at page:

Download "Homework 3 Solution. Due Friday (5pm), Feb. 14, 2013"

Transcription

1 University of California, Berkeley Spring 2013 EE 42/100 Prof. K. Pister Homework 3 Solution Due Friday (5pm), Feb. 14, 2013 Please turn the homework in to the drop box located next to 125 Cory Hall (labeled EE 42/100). Make sure to clearly label your Name, Student ID, Class, and Discussion sections on the homework. 1) We start by identifying the extraordinary nodes and choosing the one with the most branches to be ground. The currents in each branch are then labeled as shown below. Since there is no load connected to the end of A&B, V AB =V oc. Substituting the currents gives: (V C 8) 2k A couple notes: The current, I 2 through the resistor above the 4V source is set by the 4V source. The 2mA current source sets the current through the resistor in series with it. We don t know the current through the 4V source so we ll just label it I 3. We could show by KVL: I 1 = V C 8, I 2k 2 = 4, I 1k 4 = V A 8, I 8k 5 = V A 2k KCL at C: I 1 + I 2 + I 3 2mA = k + I 3 2mA = 0 Multiply by 2k and gather terms: 2k I 3 V C = 2mA 2k 8 8 = 12 KCL at A: I 2 + I 3 + I 4 + I 5 = 0 Substituing currents gives: 4 + I 1k 3 + V A 8 + V A = 0 8k 2k Multiply by 8k and gather constants on the right side: 8k I 3 + 5V A = 8 32 = 24 {eqtn1} {eqtn2} There are 2 equations and 3 unknowns: V A, V C, and I x. The 4V source imposes a 4V voltage difference between A and C which gives the 3 rd equation: V A V C = 4V {eqtn3}

2 We can represent the 3 equations in matrix form as: I 3 V A V C Const 2k k We get: V A = V AB = V OC = V th = 4.44V R th is R eq (a,b) looking back at circuit from terminals A & B with the sources zeroed out. The resultant circuit is shown below: Note that both 1k resistors can be removed from the circuit either due to shorting out (by the 4V source) or terminating at an open circuit (due to removing the current source). So we get R th = 890 Ω The Thevenin equivalent circuit thus look like:

3 2) (a) i)the current through the circuit is given by KVL: I s = V Th R th +R L = R L. The maximum current occurs when the load is shorted, R L = 0 so I s (max) = = 50mA. This is the maximum current that the source can provide it is limited by the value of R th. Op amps typically have small output resistances and thus can provide higher currents. ii) This is a voltage divider: V out = 10 R L R L Note that due to the non-zero R th, the voltage across V out can be vastly different than it s open circuit value depending on R L. This is often called loading down similar to the car problem in HW1 in which the voltage across the car dropped from it s open circuit value due to resistance in the wires. (b) i) The ideal op-amp presents an infinite input resistancce to the source circuit meaning, I p = I s = 0. ii) Since I s = 0, there is no voltage drop across R th and thus V p = V th. This could be proved with KVL: V p = V th I s R th = V th 0. We also know V out = V n since they are connected and applying the ideal op amp condition, V p = V n gives, V out = V th = 10V. Now there is 10V across the load regardless of the load resistance. Ideally the source no longer has the potential to be loaded down as the op-amp now provides the necessary load current. Note: Even if the op-amp were non-ideal we would have V out = A(V p V n ) = A(V p V out ) V p = V out (1/A + 1) so feedback imposes, V p V out = V n for reasonable values of A. If A=10 6, we would get V out = 11 = It is silly to carry around so many sig figs. 1/A+1

4 3) (a) Employing the model for the photodiode as a current source gives the figure below: i) From ohms law: V out = I ph R L = R i P opt R L R v = V out P opt = R i R L = = 450[V/W] As long as it s reverse biased, the photodiode acts like a current source converting optical power into current so V out is independent of the 11V source. ii) From KVL: V d = V out 11. The diode will become forward biased when V d > 0. Which occurs when V out > 11V. Setting V out = 11V gives the point at which the diode is no longer reverse biased. So, V out = 11V = R i P opt R L = 0.9 P opt 500 P opt = 11 = 24mW Note1: There can always be a voltage across a current source even zero volts. It is fine to have a short circuited current source just like it s fine to have an open circuited voltage source (like an unused receptacle in your house). Note2: Our current source model of a reverse biased photodiode does break down when the voltage across it becomes positive. At this point, the diode forward current of I d = I s (e V d/v th 1) would flow. However, the forward current reduces V out which in turn reverse biases the diode. This is an interesting problem but you will find that eventually V out is insensitive to P in and V d stays locked at V F. (b) We have the following for the TIA circuit with the current source model for the photodiode employed: iii) 500 = V out I ph G = V out I ph = V out R i P opt = 500 Ω i) We have V p = 0 = V n. Now the anode is at a virtual ground. The diode voltage is still defined as the voltage from the anode to the cathode. KVL gives: V d = 11V Now the reverse bias voltage is indpendent of P in which is great. ii) Note that I 2 = 0 because the voltage on both sides of the 100Ω is the same. KCL at V n with V n = 0 gives: I ph = V out /500 R v = V out P opt = R i 500 = 450 [V/W] So we have achieved a circuit with the same magnitude voltage responsivity as the simple resistive load but that applies a constant reverse bias.

5 4.) V OUT1 Assuming op-amp is ideal. You could start by labeling all the nodes (i.e. V P, V N ) of the two op-amps. Then, then perform KCL at nodes (i.e. V N ) of each op-amp. Invoke ideal op-amp assumption that I N = 0, I P =0, and V P =V N to simplify the equations. Then finally relate V out to V in1 and V in2. Alternatively, if you recognize that this circuit is simply two cascading op-amps, where the first stage is a non-inverting amplifier and the second stage is a difference (of subtracting) amplifier. For a non-inverting amplifier, V out = +R 2 V in R 2, where is a resistor in the negative feedback path and R 2 is connected from V n to ground. Thus, for the 1 st stage of this circuit, we have: V out1 V in1 = 40k + 10k 10k Or 40k + 10k V out1 = V 10k in1 Now, looking at the 2 nd stage of the circuit, we also recognize that this is just a summing amplifier. Recall that the output voltage of a summing amplifier has the form: V out = R f V 1 R f R 2 V 2 R f R 3 V 3,where R f is a resistor in the negative feedback path and,r 2,R 3 are resistors connected to the input voltages V 1, V 2, and V 3, respectively. For our circuit, there are only two input voltage sources: V in2 and V out1, which is the output voltage from the 1 st stage of the amplifier. Thus, we have: V out = 50k 50k V out1 50k 50k V in2

6 Substitute the expression we derived previously for V out1, we have: V out = 50k + 10k 40k V 50k 10k in1 50k 50k V in2 V out = 5 V in1 V in2

7 5.)

8 6.) R 2 v 1 R 3 v 2 v 0 R 4 For a linear circuit, superposition principle can be applied when analyzing the circuit. To do this, we will have to consider the output response of the circuit under the influence of each input sources at a time. The final output will be equal to the summation of the output responses due to each input sources. What follows below is a detailed analysis, but first let s just try by inspection. If we ground V2, then we have an op-amp in a standard inverting configuration, and the gain should be -R2/R1 from V1 to Vout. If we ground V1, then we have the op-amp in a standard non-inverting configuration, and the gain from V+ to Vout should be (R1+R2)/R2. The gain from V2 to V+ is just a voltage divider. Often superposition lets you simplify a circuit enough that you can analyze it by inspection. Now for the more complicated version Let s first consider voltage source V 1. Since we are only focusing on V 1, all the other independent sources in the circuit have to be deactivated. To deactivate or turn-off the V 2 voltage source, we will need to short it out. Thus, we have a new circuit shown below: v 1 R v 2 n I n = 0 R 3 v p I p = 0 v 0 R 4

9 Assuming ideal op-amps, we have: I n = 0, I p = 0 and V p = V n. Since I p = 0, we see that no current can flow through the parallel combination of R 3 and R 4. Therefore, V p = 0 Next, we can apply KVL at node V n. Summing the current flowing out from this node (assume current flowing out is positive), we arrive at: V n V 1 + I n + V n V o = 0 R 2 Since V p = V n amd V p was found to be equal to 0 for this particular circuit, we see that V n = 0. Also, I n = 0 by assuming ideal op-amps. We can simplify the expression to: Solving for V o gives: 0 V V o = 0 R 2 V o = R 2 V 1 (Note: An astute student might also recognize that this just an expression for an inverting amp.) Let s now consider the V 2 voltage source. We will turn off V 1 by shorting it out. We have the following circuit schematic: R v 2 n I n = 0 R 3 v p I p = 0 v 0 v 2 R 4

10 At the positive input terminal, we realize that V p is a voltage divider of V 2. We have: We can perform KCL at node V n, giving us: Since I n = 0, we have: Solving for V o, we get: Since V n = V p = R 4 R 4 +R 3 V 2, we have: R 4 V p = V R 4 + R 2 3 V n + I R n + V n V o = 0 1 R 2 V n V n V o = 0 R 2 V o = + R 2 V R n 1 V o = + R 2 Invoking the superposition principle, we finally have: V o = V o + V o = R 2 V 1 + +R 2 R 4 R 4 +R 3 V 2 R 4 V R 4 + R 2 3 As expected, this circuit solving technique gives you the same result as what you would have gotten had you solved the original circuit using node analysis/kcl.

11 7) In this approximation, when a diode is on, the voltage across it is V d = V F where V d is the voltage from the anode to the cathode. When it is off, the diode acts like an open circuit. To determine if the diode is off we check if I d 0 or V d V F. If the diode is indeed off we set I d =0 and re-analyze. The circuit has been labeled below. We could use KVL to show that when each diodes is on : I 1 = V s 0.7 V A, I 1k 2 = V A 0.7, I = V A 150 There are three possible configurations for the two diodes: 1) D1 and D2 are off: I 1 = I 2 = 0. gives I 3 = 0 so V A = 0. {eqtn1} 2) D1 and D2 are on: V d1 = V d2 = 0.7. I 1 I 2 I 3 = 0 V s 0.7 V A V A 0.7 V A 1k = 0 Multiplying by 6k and simplifying gives: 76 V A = 6 V s {eqtn2} 3) D1 is on and D2 is off: V d1 = 0.7, I 2 = 0: I 1 I 3 = 0 V s 0.7 V A V A 1k 150 = 0 Multiplying by 3k and simplifying gives: 23 V A = 3 V s 2.1 {eqtn3} Note that when D1 is off, D2 is automatically off since no current would flow through the circuit. Thus, there is no case when D1=Off and D2=on.. (a) V s =0.5V: Since V s < 0.7 there is no way that D1 could be on so configuration 1 is met and D1 = off, D2 = off, V A = 0. (b) V s =3V: Since V s 0.7V, D1 must be on. Assuming configuration 2 is true, we get from {eqtn2}, V A = 0.46V which gives I 2 = 1.2mA, our assumption was wrong since D2 is off. Knowing D1 = On, D2 = Off use {eqtn3} to get V A = 0.3V. (c) V s =5V: Since V s 0.7V, D1 must be on. Assuming configuration 2 is true, we get from {eqtn2}, V A = 0.62V which gives I 2 < 0, our assumption was wrong since D2 is off. Knowing D1 = On, D2 = Off use {eqtn3} to get V A = 0.56V. (d) V s =7V: Since V s 0.7V, D1 must be on. Assuming configuration 2 is true, we get from {eqtn2}, V A = 0.77V which gives I 2 > 0, our assumptions are correct. We know D1 = On, D2 = On

Homework 2. Due Friday (5pm), Feb. 8, 2013

Homework 2. Due Friday (5pm), Feb. 8, 2013 University of California, Berkeley Spring 2013 EE 42/100 Prof. K. Pister Homework 2 Due Friday (5pm), Feb. 8, 2013 Please turn the homework in to the drop box located next to 125 Cory Hall (labeled EE

More information

Midterm Exam (closed book/notes) Tuesday, February 23, 2010

Midterm Exam (closed book/notes) Tuesday, February 23, 2010 University of California, Berkeley Spring 2010 EE 42/100 Prof. A. Niknejad Midterm Exam (closed book/notes) Tuesday, February 23, 2010 Guidelines: Closed book. You may use a calculator. Do not unstaple

More information

Designing Information Devices and Systems I Fall 2018 Lecture Notes Note Introduction: Op-amps in Negative Feedback

Designing Information Devices and Systems I Fall 2018 Lecture Notes Note Introduction: Op-amps in Negative Feedback EECS 16A Designing Information Devices and Systems I Fall 2018 Lecture Notes Note 18 18.1 Introduction: Op-amps in Negative Feedback In the last note, we saw that can use an op-amp as a comparator. However,

More information

Problem Set 4 Solutions

Problem Set 4 Solutions University of California, Berkeley Spring 212 EE 42/1 Prof. A. Niknejad Problem Set 4 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different

More information

EE-201 Review Exam I. 1. The voltage Vx in the circuit below is: (1) 3V (2) 2V (3) -2V (4) 1V (5) -1V (6) None of above

EE-201 Review Exam I. 1. The voltage Vx in the circuit below is: (1) 3V (2) 2V (3) -2V (4) 1V (5) -1V (6) None of above EE-201, Review Probs Test 1 page-1 Spring 98 EE-201 Review Exam I Multiple Choice (5 points each, no partial credit.) 1. The voltage Vx in the circuit below is: (1) 3V (2) 2V (3) -2V (4) 1V (5) -1V (6)

More information

P1: Basics - Things you now know that you didn t know you knew (25 pts)

P1: Basics - Things you now know that you didn t know you knew (25 pts) P1: Basics - Things you now know that you didn t know you knew (25 pts) a) Birds routinely land and relax on power lines which carry tens of thousands of volts of electricity. Explain why these birds do

More information

Notes for course EE1.1 Circuit Analysis TOPIC 3 CIRCUIT ANALYSIS USING SUB-CIRCUITS

Notes for course EE1.1 Circuit Analysis TOPIC 3 CIRCUIT ANALYSIS USING SUB-CIRCUITS Notes for course EE1.1 Circuit Analysis 2004-05 TOPIC 3 CIRCUIT ANALYSIS USING SUB-CIRCUITS OBJECTIVES 1) To introduce the Source Transformation 2) To consider the concepts of Linearity and Superposition

More information

The equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A =

The equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A = The equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A = 10 10 4. Section Break Difficulty: Easy Learning Objective: Understand how real operational

More information

EE 321 Analog Electronics, Fall 2013 Homework #3 solution

EE 321 Analog Electronics, Fall 2013 Homework #3 solution EE 32 Analog Electronics, Fall 203 Homework #3 solution 2.47. (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by + [ Rf v N + R f v N2 +... + R ] f v Nn R N R N2 R [

More information

Chapter 5. Department of Mechanical Engineering

Chapter 5. Department of Mechanical Engineering Source Transformation By KVL: V s =ir s + v By KCL: i s =i + v/r p is=v s /R s R s =R p V s /R s =i + v/r s i s =i + v/r p Two circuits have the same terminal voltage and current Source Transformation

More information

D is the voltage difference = (V + - V - ).

D is the voltage difference = (V + - V - ). 1 Operational amplifier is one of the most common electronic building blocks used by engineers. It has two input terminals: V + and V -, and one output terminal Y. It provides a gain A, which is usually

More information

POLYTECHNIC UNIVERSITY Electrical Engineering Department. EE SOPHOMORE LABORATORY Experiment 2 DC circuits and network theorems

POLYTECHNIC UNIVERSITY Electrical Engineering Department. EE SOPHOMORE LABORATORY Experiment 2 DC circuits and network theorems POLYTECHNIC UNIVERSITY Electrical Engineering Department EE SOPHOMORE LABORATORY Experiment 2 DC circuits and network theorems Modified for Physics 18, Brooklyn College I. Overview of Experiment In this

More information

ECE2262 Electric Circuits

ECE2262 Electric Circuits ECE2262 Electric Circuits Equivalence Chapter 5: Circuit Theorems Linearity Superposition Thevenin s and Norton s Theorems Maximum Power Transfer Analysis of Circuits Using Circuit Theorems 1 5. 1 Equivalence

More information

Series & Parallel Resistors 3/17/2015 1

Series & Parallel Resistors 3/17/2015 1 Series & Parallel Resistors 3/17/2015 1 Series Resistors & Voltage Division Consider the single-loop circuit as shown in figure. The two resistors are in series, since the same current i flows in both

More information

EE100Su08 Lecture #9 (July 16 th 2008)

EE100Su08 Lecture #9 (July 16 th 2008) EE100Su08 Lecture #9 (July 16 th 2008) Outline HW #1s and Midterm #1 returned today Midterm #1 notes HW #1 and Midterm #1 regrade deadline: Wednesday, July 23 rd 2008, 5:00 pm PST. Procedure: HW #1: Bart

More information

E1.1 Analysis of Circuits ( ) Revision Lecture 1 1 / 13

E1.1 Analysis of Circuits ( ) Revision Lecture 1 1 / 13 RevisionLecture 1: E1.1 Analysis of Circuits (2014-4530) Revision Lecture 1 1 / 13 Format Question 1 (40%): eight short parts covering the whole syllabus. Questions 2 and 3: single topic questions (answer

More information

Experiment #6. Thevenin Equivalent Circuits and Power Transfer

Experiment #6. Thevenin Equivalent Circuits and Power Transfer Experiment #6 Thevenin Equivalent Circuits and Power Transfer Objective: In this lab you will confirm the equivalence between a complicated resistor circuit and its Thevenin equivalent. You will also learn

More information

Electric Circuits I. Nodal Analysis. Dr. Firas Obeidat

Electric Circuits I. Nodal Analysis. Dr. Firas Obeidat Electric Circuits I Nodal Analysis Dr. Firas Obeidat 1 Nodal Analysis Without Voltage Source Nodal analysis, which is based on a systematic application of Kirchhoff s current law (KCL). A node is defined

More information

ECE2262 Electric Circuits. Chapter 5: Circuit Theorems

ECE2262 Electric Circuits. Chapter 5: Circuit Theorems ECE2262 Electric Circuits Chapter 5: Circuit Theorems 1 Equivalence Linearity Superposition Thevenin s and Norton s Theorems Maximum Power Transfer Analysis of Circuits Using Circuit Theorems 2 5. 1 Equivalence

More information

EIT Review. Electrical Circuits DC Circuits. Lecturer: Russ Tatro. Presented by Tau Beta Pi The Engineering Honor Society 10/3/2006 1

EIT Review. Electrical Circuits DC Circuits. Lecturer: Russ Tatro. Presented by Tau Beta Pi The Engineering Honor Society 10/3/2006 1 EIT Review Electrical Circuits DC Circuits Lecturer: Russ Tatro Presented by Tau Beta Pi The Engineering Honor Society 10/3/2006 1 Session Outline Basic Concepts Basic Laws Methods of Analysis Circuit

More information

MAE140 - Linear Circuits - Fall 14 Midterm, November 6

MAE140 - Linear Circuits - Fall 14 Midterm, November 6 MAE140 - Linear Circuits - Fall 14 Midterm, November 6 Instructions (i) This exam is open book. You may use whatever written materials you choose, including your class notes and textbook. You may use a

More information

EE292: Fundamentals of ECE

EE292: Fundamentals of ECE EE292: Fundamentals of ECE Fall 2012 TTh 10:00-11:15 SEB 1242 Lecture 4 120906 http://www.ee.unlv.edu/~b1morris/ee292/ 2 Outline Review Voltage Divider Current Divider Node-Voltage Analysis 3 Network Analysis

More information

E40M Review - Part 1

E40M Review - Part 1 E40M Review Part 1 Topics in Part 1 (Today): KCL, KVL, Power Devices: V and I sources, R Nodal Analysis. Superposition Devices: Diodes, C, L Time Domain Diode, C, L Circuits Topics in Part 2 (Wed): MOSFETs,

More information

INTRODUCTION TO ELECTRONICS

INTRODUCTION TO ELECTRONICS INTRODUCTION TO ELECTRONICS Basic Quantities Voltage (symbol V) is the measure of electrical potential difference. It is measured in units of Volts, abbreviated V. The example below shows several ways

More information

0 t < 0 1 t 1. u(t) =

0 t < 0 1 t 1. u(t) = A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 13 p. 22/33 Step Response A unit step function is described by u(t) = ( 0 t < 0 1 t 1 While the waveform has an artificial jump (difficult

More information

ECE 220 Laboratory 4 Volt Meter, Comparators, and Timer

ECE 220 Laboratory 4 Volt Meter, Comparators, and Timer ECE 220 Laboratory 4 Volt Meter, Comparators, and Timer Michael W. Marcellin Please follow all rules, procedures and report requirements as described at the beginning of the document entitled ECE 220 Laboratory

More information

Operational amplifiers (Op amps)

Operational amplifiers (Op amps) Operational amplifiers (Op amps) v R o R i v i Av i v View it as an ideal amp. Take the properties to the extreme: R i, R o 0, A.?!?!?!?! v v i Av i v A Consequences: No voltage dividers at input or output.

More information

Designing Information Devices and Systems I Spring 2018 Homework 7

Designing Information Devices and Systems I Spring 2018 Homework 7 EECS 6A Designing Information Devices and Systems I Spring 08 Homework 7 This homework is due March, 08, at 3:59. Self-grades are due March 5, 08, at 3:59. Submission Format Your homework submission should

More information

Problem Set 5 Solutions

Problem Set 5 Solutions University of California, Berkeley Spring 01 EE /0 Prof. A. Niknejad Problem Set 5 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different

More information

R 2, R 3, and R 4 are in parallel, R T = R 1 + (R 2 //R 3 //R 4 ) + R 5. C-C Tsai

R 2, R 3, and R 4 are in parallel, R T = R 1 + (R 2 //R 3 //R 4 ) + R 5. C-C Tsai Chapter 07 Series-Parallel Circuits The Series-Parallel Network Complex circuits May be separated both series and/or parallel elements Combinations which are neither series nor parallel To analyze a circuit

More information

University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences. EECS 40 Midterm II

University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences. EECS 40 Midterm II University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences EECS 40 Midterm II Spring 2001 Prof. Roger T. Howe April 11, 2001 Name: Last, First Student

More information

Designing Information Devices and Systems II Fall 2016 Murat Arcak and Michel Maharbiz Homework 0. This homework is due August 29th, 2016, at Noon.

Designing Information Devices and Systems II Fall 2016 Murat Arcak and Michel Maharbiz Homework 0. This homework is due August 29th, 2016, at Noon. EECS 16B Designing Information Devices and Systems II Fall 2016 Murat Arcak and Michel Maharbiz Homework 0 This homework is due August 29th, 2016, at Noon. 1. Homework process and study group (a) Who else

More information

EE 40: Introduction to Microelectronic Circuits Spring 2008: Midterm 2

EE 40: Introduction to Microelectronic Circuits Spring 2008: Midterm 2 EE 4: Introduction to Microelectronic Circuits Spring 8: Midterm Venkat Anantharam 3/9/8 Total Time Allotted : min Total Points:. This is a closed book exam. However, you are allowed to bring two pages

More information

Operational Amplifiers

Operational Amplifiers Operational Amplifiers A Linear IC circuit Operational Amplifier (op-amp) An op-amp is a high-gain amplifier that has high input impedance and low output impedance. An ideal op-amp has infinite gain and

More information

Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer

Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer J. McNames Portland State University ECE 221 Circuit Theorems Ver. 1.36 1

More information

Notes for course EE1.1 Circuit Analysis TOPIC 10 2-PORT CIRCUITS

Notes for course EE1.1 Circuit Analysis TOPIC 10 2-PORT CIRCUITS Objectives: Introduction Notes for course EE1.1 Circuit Analysis 4-5 Re-examination of 1-port sub-circuits Admittance parameters for -port circuits TOPIC 1 -PORT CIRCUITS Gain and port impedance from -port

More information

Chapter 5 Solution P5.2-2, 3, 6 P5.3-3, 5, 8, 15 P5.4-3, 6, 8, 16 P5.5-2, 4, 6, 11 P5.6-2, 4, 9

Chapter 5 Solution P5.2-2, 3, 6 P5.3-3, 5, 8, 15 P5.4-3, 6, 8, 16 P5.5-2, 4, 6, 11 P5.6-2, 4, 9 Chapter 5 Solution P5.2-2, 3, 6 P5.3-3, 5, 8, 15 P5.4-3, 6, 8, 16 P5.5-2, 4, 6, 11 P5.6-2, 4, 9 P 5.2-2 Consider the circuit of Figure P 5.2-2. Find i a by simplifying the circuit (using source transformations)

More information

4/27 Friday. I have all the old homework if you need to collect them.

4/27 Friday. I have all the old homework if you need to collect them. 4/27 Friday Last HW: do not need to turn it. Solution will be posted on the web. I have all the old homework if you need to collect them. Final exam: 7-9pm, Monday, 4/30 at Lambert Fieldhouse F101 Calculator

More information

UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences

UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences EECS 40 Spring 2000 Introduction to Microelectronic Devices Prof. King MIDTERM EXAMINATION

More information

Notes for course EE1.1 Circuit Analysis TOPIC 4 NODAL ANALYSIS

Notes for course EE1.1 Circuit Analysis TOPIC 4 NODAL ANALYSIS Notes for course EE1.1 Circuit Analysis 2004-05 TOPIC 4 NODAL ANALYSIS OBJECTIVES 1) To develop Nodal Analysis of Circuits without Voltage Sources 2) To develop Nodal Analysis of Circuits with Voltage

More information

Chapter 2. Engr228 Circuit Analysis. Dr Curtis Nelson

Chapter 2. Engr228 Circuit Analysis. Dr Curtis Nelson Chapter 2 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 2 Objectives Understand symbols and behavior of the following circuit elements: Independent voltage and current sources; Dependent voltage and

More information

Chapter 10 AC Analysis Using Phasors

Chapter 10 AC Analysis Using Phasors Chapter 10 AC Analysis Using Phasors 10.1 Introduction We would like to use our linear circuit theorems (Nodal analysis, Mesh analysis, Thevenin and Norton equivalent circuits, Superposition, etc.) to

More information

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science : Circuits & Electronics Problem Set #1 Solution

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science : Circuits & Electronics Problem Set #1 Solution Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.2: Circuits & Electronics Problem Set # Solution Exercise. The three resistors form a series connection.

More information

20.2 Design Example: Countdown Timer

20.2 Design Example: Countdown Timer EECS 16A Designing Information Devices and Systems I Fall 018 Lecture Notes Note 0 0.1 Design Procedure Now that we ve analyzed many circuits, we are ready to focus on designing interesting circuits to

More information

Basics of Network Theory (Part-I)

Basics of Network Theory (Part-I) Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]

More information

Operational amplifiers (Op amps)

Operational amplifiers (Op amps) Operational amplifiers (Op amps) Recall the basic two-port model for an amplifier. It has three components: input resistance, Ri, output resistance, Ro, and the voltage gain, A. v R o R i v d Av d v Also

More information

UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS

UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 1.0 Kirchoff s Law Kirchoff s Current Law (KCL) states at any junction in an electric circuit the total current flowing towards that junction is equal

More information

Two-Port Networks Admittance Parameters CHAPTER16 THE LEARNING GOALS FOR THIS CHAPTER ARE THAT STUDENTS SHOULD BE ABLE TO:

Two-Port Networks Admittance Parameters CHAPTER16 THE LEARNING GOALS FOR THIS CHAPTER ARE THAT STUDENTS SHOULD BE ABLE TO: CHAPTER16 Two-Port Networks THE LEARNING GOALS FOR THIS CHAPTER ARE THAT STUDENTS SHOULD BE ABLE TO: Calculate the admittance, impedance, hybrid, and transmission parameter for two-port networks. Convert

More information

Simultaneous equations for circuit analysis

Simultaneous equations for circuit analysis Simultaneous equations for circuit analysis This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/,

More information

In this lecture, we will consider how to analyse an electrical circuit by applying KVL and KCL. As a result, we can predict the voltages and currents

In this lecture, we will consider how to analyse an electrical circuit by applying KVL and KCL. As a result, we can predict the voltages and currents In this lecture, we will consider how to analyse an electrical circuit by applying KVL and KCL. As a result, we can predict the voltages and currents around an electrical circuit. This is a short lecture,

More information

ECE 1311: Electric Circuits. Chapter 2: Basic laws

ECE 1311: Electric Circuits. Chapter 2: Basic laws ECE 1311: Electric Circuits Chapter 2: Basic laws Basic Law Overview Ideal sources series and parallel Ohm s law Definitions open circuits, short circuits, conductance, nodes, branches, loops Kirchhoff's

More information

DC motors. 1. Parallel (shunt) excited DC motor

DC motors. 1. Parallel (shunt) excited DC motor DC motors 1. Parallel (shunt) excited DC motor A shunt excited DC motor s terminal voltage is 500 V. The armature resistance is 0,5 Ω, field resistance is 250 Ω. On a certain load it takes 20 A current

More information

Thevenin equivalent circuits

Thevenin equivalent circuits Thevenin equivalent circuits We have seen the idea of equivalency used in several instances already. 1 2 1 2 same as 1 2 same as 1 2 R 3 same as = 0 V same as 0 A same as same as = EE 201 Thevenin 1 The

More information

ECE2262 Electric Circuits. Chapter 4: Operational Amplifier (OP-AMP) Circuits

ECE2262 Electric Circuits. Chapter 4: Operational Amplifier (OP-AMP) Circuits ECE2262 Electric Circuits Chapter 4: Operational Amplifier (OP-AMP) Circuits 1 4.1 Operational Amplifiers 2 4. Voltages and currents in electrical circuits may represent signals and circuits can perform

More information

Lecture 1. EE70 Fall 2007

Lecture 1. EE70 Fall 2007 Lecture 1 EE70 Fall 2007 Instructor Joel Kubby (that would be me) Office: BE-249 Office Hours: M,W,F 2-3 PM or by appointment Phone: (831) 459-1073 E-mail: jkubby@soe.ucsc.edu Teaching Assistant Drew Lohn

More information

Delta & Y Configurations, Principles of Superposition, Resistor Voltage Divider Designs

Delta & Y Configurations, Principles of Superposition, Resistor Voltage Divider Designs BME/ISE 3511 Bioelectronics - Test Three Course Notes Fall 2016 Delta & Y Configurations, Principles of Superposition, esistor Voltage Divider Designs Use following techniques to solve for current through

More information

Voltage Dividers, Nodal, and Mesh Analysis

Voltage Dividers, Nodal, and Mesh Analysis Engr228 Lab #2 Voltage Dividers, Nodal, and Mesh Analysis Name Partner(s) Grade /10 Introduction This lab exercise is designed to further your understanding of the use of the lab equipment and to verify

More information

MAE140 HW3 Solutions

MAE140 HW3 Solutions MAE40 HW Solutions.7) Method: Remove load resistor and find Thevenin equivalent circuit VOC 600 450 00 450 To find Isc, use a voltage divider at a a 00 450 00 450 00 600 0 0 900 6 Isc Isc 00 6*00 00 Amps

More information

Designing Information Devices and Systems I Fall 2015 Anant Sahai, Ali Niknejad Homework 10. This homework is due November 9, 2015, at Noon.

Designing Information Devices and Systems I Fall 2015 Anant Sahai, Ali Niknejad Homework 10. This homework is due November 9, 2015, at Noon. EECS 16A Designing Information Devices and Systems I Fall 2015 Anant Sahai, Ali Niknejad Homework 10 This homework is due November 9, 2015, at Noon. 1. Homework process and study group Who else did you

More information

Chapter 5: Circuit Theorems

Chapter 5: Circuit Theorems Chapter 5: Circuit Theorems This chapter provides a new powerful technique of solving complicated circuits that are more conceptual in nature than node/mesh analysis. Conceptually, the method is fairly

More information

Homework 6 Solutions and Rubric

Homework 6 Solutions and Rubric Homework 6 Solutions and Rubric EE 140/40A 1. K-W Tube Amplifier b) Load Resistor e) Common-cathode a) Input Diff Pair f) Cathode-Follower h) Positive Feedback c) Tail Resistor g) Cc d) Av,cm = 1/ Figure

More information

Review of Circuit Analysis

Review of Circuit Analysis Review of Circuit Analysis Fundamental elements Wire Resistor Voltage Source Current Source Kirchhoff s Voltage and Current Laws Resistors in Series Voltage Division EE 42 Lecture 2 1 Voltage and Current

More information

Errors in Electrical Measurements

Errors in Electrical Measurements 1 Errors in Electrical Measurements Systematic error every times you measure e.g. loading or insertion of the measurement instrument Meter error scaling (inaccurate marking), pointer bending, friction,

More information

ELEC 250: LINEAR CIRCUITS I COURSE OVERHEADS. These overheads are adapted from the Elec 250 Course Pack developed by Dr. Fayez Guibaly.

ELEC 250: LINEAR CIRCUITS I COURSE OVERHEADS. These overheads are adapted from the Elec 250 Course Pack developed by Dr. Fayez Guibaly. Elec 250: Linear Circuits I 5/4/08 ELEC 250: LINEAR CIRCUITS I COURSE OVERHEADS These overheads are adapted from the Elec 250 Course Pack developed by Dr. Fayez Guibaly. S.W. Neville Elec 250: Linear Circuits

More information

ECE 212H1F Circuit Analysis October 20, :15-19: Reza Iravani 02 Reza Iravani 03 Ali Nabavi-Niaki. (Non-programmable Calculators Allowed)

ECE 212H1F Circuit Analysis October 20, :15-19: Reza Iravani 02 Reza Iravani 03 Ali Nabavi-Niaki. (Non-programmable Calculators Allowed) Please Print Clearly Last Name: First Name: Student Number: Your Tutorial Section (CIRCLE ONE): 01 Thu 10:00 12:00 HA403 02 Thu 10:00 12:00 GB412 03 Thu 15:00 17:00 GB412 04 Thu 15:00 17:00 SF2202 05 Fri

More information

EE 330 Lecture 22. Small Signal Modelling Operating Points for Amplifier Applications Amplification with Transistor Circuits

EE 330 Lecture 22. Small Signal Modelling Operating Points for Amplifier Applications Amplification with Transistor Circuits EE 330 Lecture 22 Small Signal Modelling Operating Points for Amplifier Applications Amplification with Transistor Circuits Exam 2 Friday March 9 Exam 3 Friday April 13 Review Session for Exam 2: 6:00

More information

Sirindhorn International Institute of Technology Thammasat University at Rangsit

Sirindhorn International Institute of Technology Thammasat University at Rangsit Sirindhorn International Institute of Technology Thammasat University at Rangsit School of Information, Computer and Communication Technology COURSE : ECS 304 Basic Electrical Engineering Lab INSTRUCTOR

More information

UC DAVIS. Circuits I Course Outline

UC DAVIS. Circuits I Course Outline UC DAVIS Circuits I Course Outline ENG 17 Professor Spencer Fall 2010 2041 Kemper Hall Lecture: MWF 4:10-5:00, 1003 Giedt Hall 752-6885 Discussion Section 1: W 1:10-2:00, 55 Roessler CRN: 61417 Discussion

More information

Designing Information Devices and Systems I Spring 2019 Homework 7

Designing Information Devices and Systems I Spring 2019 Homework 7 Last Updated: 2019-03-16 22:56 1 EECS 16A Designing Information Devices and Systems I Spring 2019 Homework 7 This homework is due March 15, 2019 at 23:59. Self-grades are due March 19, 2019, at 23:59.

More information

Designing Information Devices and Systems I Spring 2018 Lecture Notes Note 20

Designing Information Devices and Systems I Spring 2018 Lecture Notes Note 20 EECS 16A Designing Information Devices and Systems I Spring 2018 Lecture Notes Note 20 Design Example Continued Continuing our analysis for countdown timer circuit. We know for a capacitor C: I = C dv

More information

Discussion Question 6A

Discussion Question 6A Discussion Question 6 P212, Week 6 Two Methods for Circuit nalysis Method 1: Progressive collapsing of circuit elements In last week s discussion, we learned how to analyse circuits involving batteries

More information

Chapter 4: Techniques of Circuit Analysis

Chapter 4: Techniques of Circuit Analysis Chapter 4: Techniques of Circuit Analysis This chapter gies us many useful tools for soling and simplifying circuits. We saw a few simple tools in the last chapter (reduction of circuits ia series and

More information

ENGG 225. David Ng. Winter January 9, Circuits, Currents, and Voltages... 5

ENGG 225. David Ng. Winter January 9, Circuits, Currents, and Voltages... 5 ENGG 225 David Ng Winter 2017 Contents 1 January 9, 2017 5 1.1 Circuits, Currents, and Voltages.................... 5 2 January 11, 2017 6 2.1 Ideal Basic Circuit Elements....................... 6 3 January

More information

ENGR 2405 Class No Electric Circuits I

ENGR 2405 Class No Electric Circuits I ENGR 2405 Class No. 48056 Electric Circuits I Dr. R. Williams Ph.D. rube.williams@hccs.edu Electric Circuit An electric circuit is an interconnec9on of electrical elements Charge Charge is an electrical

More information

Designing Information Devices and Systems I Spring 2019 Midterm 2

Designing Information Devices and Systems I Spring 2019 Midterm 2 1 EECS 16A Designing Information Devices and Systems I Spring 2019 Midterm 2 1. How is your semester so far? (1 point) 2. Do you have any summer plans? (1 point) Do not turn this page until the proctor

More information

Sinusoidal Steady State Analysis (AC Analysis) Part I

Sinusoidal Steady State Analysis (AC Analysis) Part I Sinusoidal Steady State Analysis (AC Analysis) Part I Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/

More information

Designing Information Devices and Systems I Spring 2019 Midterm 2. Exam Location: Cory 521 (DSP)

Designing Information Devices and Systems I Spring 2019 Midterm 2. Exam Location: Cory 521 (DSP) 1 EECS 16A Designing Information Devices and Systems I Spring 2019 Midterm 2 Exam Location: Cory 521 (DSP) PRINT AND SIGN your name:, (last name) (first name) (signature) PRINT time of your Monday section

More information

Chapter 10 Sinusoidal Steady State Analysis Chapter Objectives:

Chapter 10 Sinusoidal Steady State Analysis Chapter Objectives: Chapter 10 Sinusoidal Steady State Analysis Chapter Objectives: Apply previously learn circuit techniques to sinusoidal steady-state analysis. Learn how to apply nodal and mesh analysis in the frequency

More information

Lecture 5: Using electronics to make measurements

Lecture 5: Using electronics to make measurements Lecture 5: Using electronics to make measurements As physicists, we re not really interested in electronics for its own sake We want to use it to measure something often, something too small to be directly

More information

ESE319 Introduction to Microelectronics. BJT Biasing Cont.

ESE319 Introduction to Microelectronics. BJT Biasing Cont. BJT Biasing Cont. Biasing for DC Operating Point Stability BJT Bias Using Emitter Negative Feedback Single Supply BJT Bias Scheme Constant Current BJT Bias Scheme Rule of Thumb BJT Bias Design 1 Simple

More information

Tutorial #4: Bias Point Analysis in Multisim

Tutorial #4: Bias Point Analysis in Multisim SCHOOL OF ENGINEERING AND APPLIED SCIENCE DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING ECE 2115: ENGINEERING ELECTRONICS LABORATORY Tutorial #4: Bias Point Analysis in Multisim INTRODUCTION When BJTs

More information

EE40 Midterm Review Prof. Nathan Cheung

EE40 Midterm Review Prof. Nathan Cheung EE40 Midterm Review Prof. Nathan Cheung 10/29/2009 Slide 1 I feel I know the topics but I cannot solve the problems Now what? Slide 2 R L C Properties Slide 3 Ideal Voltage Source *Current depends d on

More information

Active loads in amplifier circuits

Active loads in amplifier circuits Active loads in amplifier circuits This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/,

More information

Outline. Week 5: Circuits. Course Notes: 3.5. Goals: Use linear algebra to determine voltage drops and branch currents.

Outline. Week 5: Circuits. Course Notes: 3.5. Goals: Use linear algebra to determine voltage drops and branch currents. Outline Week 5: Circuits Course Notes: 3.5 Goals: Use linear algebra to determine voltage drops and branch currents. Components in Resistor Networks voltage source current source resistor Components in

More information

A two-port network is an electrical network with two separate ports

A two-port network is an electrical network with two separate ports 5.1 Introduction A two-port network is an electrical network with two separate ports for input and output. Fig(a) Single Port Network Fig(b) Two Port Network There are several reasons why we should study

More information

Designing Information Devices and Systems I Spring 2016 Elad Alon, Babak Ayazifar Midterm 2. Exam location: 145 Dwinelle, last SID# 2

Designing Information Devices and Systems I Spring 2016 Elad Alon, Babak Ayazifar Midterm 2. Exam location: 145 Dwinelle, last SID# 2 EECS 16A Designing Information Devices and Systems I Spring 2016 Elad Alon, Babak Ayazifar Midterm 2 Exam location: 145 Dwinelle, last SID# 2 PRINT your student ID: PRINT AND SIGN your name:, (last) (first)

More information

ENGG 1203 Tutorial. Solution. Op Amps 7 Mar Learning Objectives. Determine V o in the following circuit. Assume that the op-amp is ideal.

ENGG 1203 Tutorial. Solution. Op Amps 7 Mar Learning Objectives. Determine V o in the following circuit. Assume that the op-amp is ideal. ENGG 03 Tutorial Q Op Amps 7 Mar Learning Objectives Analyze circuits with ideal operational amplifiers News HW Mid term Revision tutorial ( Mar :30-6:0, CBA) Ack.: MIT OCW 6.0 Determine V o in the following

More information

Ohm's Law and Resistance

Ohm's Law and Resistance Ohm's Law and Resistance Resistance Resistance is the property of a component which restricts the flow of electric current. Energy is used up as the voltage across the component drives the current through

More information

Systematic methods for labeling circuits and finding a solvable set of equations, Operational Amplifiers. Kevin D. Donohue, University of Kentucky 1

Systematic methods for labeling circuits and finding a solvable set of equations, Operational Amplifiers. Kevin D. Donohue, University of Kentucky 1 Systematic methods for labeling circuits and finding a solvable set of equations, Operational Amplifiers Kevin D. Donohue, University of Kentucky Simple circuits with single loops or node-pairs can result

More information

Capacitors. Chapter How capacitors work Inside a capacitor

Capacitors. Chapter How capacitors work Inside a capacitor Chapter 6 Capacitors In every device we have studied so far sources, resistors, diodes and transistors the relationship between voltage and current depends only on the present, independent of the past.

More information

Capacitance. A different kind of capacitor: Work must be done to charge a capacitor. Capacitors in circuits. Capacitor connected to a battery

Capacitance. A different kind of capacitor: Work must be done to charge a capacitor. Capacitors in circuits. Capacitor connected to a battery Capacitance The ratio C = Q/V is a conductor s self capacitance Units of capacitance: Coulomb/Volt = Farad A capacitor is made of two conductors with equal but opposite charge Capacitance depends on shape

More information

Homework 1 solutions

Homework 1 solutions Electric Circuits 1 Homework 1 solutions (Due date: 2014/3/3) This assignment covers Ch1 and Ch2 of the textbook. The full credit is 100 points. For each question, detailed derivation processes and accurate

More information

ENGG 1203 Tutorial. Op Amps 10 Oct Learning Objectives. News. Ack.: MIT OCW Analyze circuits with ideal operational amplifiers

ENGG 1203 Tutorial. Op Amps 10 Oct Learning Objectives. News. Ack.: MIT OCW Analyze circuits with ideal operational amplifiers ENGG 1203 Tutorial Op Amps 10 Oct Learning Objectives Analyze circuits with ideal operational amplifiers News Mid term Revision tutorial Ack.: MIT OCW 6.01 1 Q1 This circuit is controlled by the charge

More information

1. Review of Circuit Theory Concepts

1. Review of Circuit Theory Concepts 1. Review of Circuit Theory Concepts Lecture notes: Section 1 ECE 65, Winter 2013, F. Najmabadi Circuit Theory is an pproximation to Maxwell s Electromagnetic Equations circuit is made of a bunch of elements

More information

THERE MUST BE 50 WAYS TO FIND YOUR VALUES: AN EXPLORATION OF CIRCUIT ANALYSIS TECHNIQUES FROM OHM S LAW TO EQUIVALENT CIRCUITS

THERE MUST BE 50 WAYS TO FIND YOUR VALUES: AN EXPLORATION OF CIRCUIT ANALYSIS TECHNIQUES FROM OHM S LAW TO EQUIVALENT CIRCUITS THERE MUST BE 50 WAYS TO FIND YOUR VALUES: AN EXPLORATION OF CIRCUIT ANALYSIS TECHNIQUES FROM OHM S LAW TO EQUIVALENT CIRCUITS Kristine McCarthy Josh Pratti Alexis Rodriguez-Carlson November 20, 2006 Table

More information

ECS 40, Fall 2008 Prof. Chang-Hasnain Test #3 Version A

ECS 40, Fall 2008 Prof. Chang-Hasnain Test #3 Version A ECS 40, Fall 2008 Prof. ChangHasnain Test #3 Version A 10:10 am 11:00 am, Wednesday December 3, 2008 Total Time Allotted: 50 minutes Total Points: 100 1. This is a closed book exam. However, you are allowed

More information

Chapter 4 Circuit Theorems

Chapter 4 Circuit Theorems Chapter 4 Circuit Theorems 1. Linearity and Proportionality. Source Transformation 3. Superposition Theorem 4. Thevenin s Theorem and Norton s Theorem 5. Maximum Power Transfer Theorem Mazita Sem 1 111

More information

Department of Electrical Engineering and Computer Sciences University of California, Berkeley. Final Exam Solutions

Department of Electrical Engineering and Computer Sciences University of California, Berkeley. Final Exam Solutions Electrical Engineering 42/00 Summer 202 Instructor: Tony Dear Department of Electrical Engineering and omputer Sciences University of alifornia, Berkeley Final Exam Solutions. Diodes Have apacitance?!?!

More information

OPERATIONAL AMPLIFIER APPLICATIONS

OPERATIONAL AMPLIFIER APPLICATIONS OPERATIONAL AMPLIFIER APPLICATIONS 2.1 The Ideal Op Amp (Chapter 2.1) Amplifier Applications 2.2 The Inverting Configuration (Chapter 2.2) 2.3 The Non-inverting Configuration (Chapter 2.3) 2.4 Difference

More information

ENGG 1203 Tutorial_9 - Review. Boolean Algebra. Simplifying Logic Circuits. Combinational Logic. 1. Combinational & Sequential Logic

ENGG 1203 Tutorial_9 - Review. Boolean Algebra. Simplifying Logic Circuits. Combinational Logic. 1. Combinational & Sequential Logic ENGG 1203 Tutorial_9 - Review Boolean Algebra 1. Combinational & Sequential Logic 2. Computer Systems 3. Electronic Circuits 4. Signals, Systems, and Control Remark : Multiple Choice Questions : ** Check

More information