Circuits 2 Study Guide/Crash Course

29 min readJun 16, 2020

Since I switched to online classes for the 2020 Summer Semester, I’ve been so busy that I’ve been missing content of the course so here is everything I am reviewing for our exams (this was started for the first exam but will be updated until the course is over)

School: Florida Polytechnic University

Textbook: Fundamentals of Electric Circuits 2016 ISBN 978–0–07–802822–9

Overview:

I am currently enrolled as a junior computer engineering student. I have already taken courses like Differential Equations, Circuits 1 and Calc 2 which are needed to take this class.

Circuits 2 is essentially Circuits 1 which involved the learning the behavior/relationships and analysis of current & voltage with different circuit elements such as resistors, capacitors, and inductors (RC, RL, RCL circuits) with the only difference being the current source is primarily AC circuit focused. If you are unfamiliar with those, I will try to explain these topics as if you are very new to the topics.

So to start:

AC vs DC Discussion:

In direct current (DC), the electric charge (current) only flows in one direction. Electric charge in alternating current (AC), on the other hand, changes direction periodically. The voltage in AC circuits also periodically reverses because the current changes direction hence the use of sinusoids.

Now basically the Voltage alternates in the AC circuit and this makes calculations a little more difficult but AC is more widely used in wall outlets and for mass electric hence the usage of Sinusoids in AC circuit calculations to make the math more accurate.

Now that you have an idea of what sinusoids are and the usage in AC circuits, I will try and guide you through the chapters in the textbook covered in circuits 2 exams to the best of my abilities starting with chapter 9. The chapters each have learning objectives and I will try and cover each topic so that it isn’t a mishap on an exam.

Exam 1 Course Content:

Chapter 9: Sinusoids and Phasors

Topic Overview

So far our analysis has been limited for the most part to dc circuits : those circuits excited by constant or time invariant sources. We now begin the analysis of circuits in which in which the source voltage or current is time varying. In this chapter, we are particularly interested in sinusoidally time-varying excitation, or simply, excitation by a sinusoid. We begin with a basic discussion of sinusoids and phasors. We then introduce the concepts of impedance and admittance. The basic circuit laws, Kirchhoff’s and Ohm’s law, introduced for dc circuits, will be applied to ac circuits.

Learning Objectives

1. Better understand sinusoids.

2. Understand phasors.

3. Understand the phasor relationships for circuit elements.

4. Know and understand the concepts of impedance and admittance.

5. Understand Kirchhoff’s laws in the frequency domain.

6. Comprehend the concept of phase-shift.

7. Understand the concept of AC bridges.

To start, a sinusoid is a signal that has the form of the sine or cosine function

A sinusoid current is usually referred to as an AC current

AC Circuits operate using Sinusoidal Currents and are the most commonly used in industries for a number of reasons:

Nature is characteristically Sinusoidal (Vibration through a string, ripples of ocean, etc.)
Sinusoidal Signals are easy to transmit over long distances
Using Fourier analysis shows that any periodic sinusoidal can be represented as a sum of sinusoids

So to start we need to understand Sinusoids and how the voltage is expressed. consider the sinusoidal voltage:

v(t) = (V sub m)sin ωt

where:

V sub m = the amplitude of the sinusoid

ω = the angular frequency in radians/s

ωt = the argument of the sinusoid

The sinusoid repeats itself every T seconds;

thus, T is called t of the sinusoid: T = 2π/ω

The fact that v(t) repeats itself every T seconds is shown by replacing

t by t + T

v(t) = (V sub m)sin ωt

We get

v(t + T) = (V sub m) sin ω(t + T)

= (V sub m) sin ω( t + 2π/ω)

= (V sub m) sin (ωt + 2π)

= (V sub m) sin ωt

= v(t)

Hence,

v(t + T) = v(t)

This is said to be a periodic function which is a function that satisfies

f(t ) = f(t+nT ), for all t and for all integers n.

The period T of the periodic function is the time of one

complete cycle or the number of seconds per cycle. The reciprocal of

this quantity is the number of cycles per second, known as the cyclic

frequency f of the sinusoid. Thus, f =1/T

ω = 2 πf

While ω is in radians per second (rad/s), f is in hertz (Hz).

Let us now consider a more general expression for the sinusoid,

v(t) = Vm sin(ωt + ϕ)

where (ωt + ϕ) is the argument and ϕ is the phase. Both argument and

phase can be in radians or degrees.

Let us examine the two sinusoids

V1(t) = Vm sin ωt

and

V2(t) = Vm sin (ωt + ϕ)

The starting point of v2 occurs first in time.

Therefore, we say that v2 leads v1 by ϕ or that v1 lags v2 by ϕ. If ϕ ≠ 0,

we also say that v1 and v2 are out of phase. If ϕ = 0, then v1 and v2 are

said to be in phase; they reach their minima and maxima at exactly the

same time. We can compare v1 and v2 in this manner because they operate

at the same frequency; they do not need to have the same amplitude.

A sinusoid can be expressed in either sine or cosine form. When

comparing two sinusoids, it is expedient to express both as either sine or

cosine with positive amplitudes. This is achieved by using the following

trigonometric identities:

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

With these identities, it is easy to show that

sin(ωt ± 180°) = −sin ωt

cos(ωt ± 180°) = −cos ωt

sin(ωt ± 90°) = ± cos ωt

cos(ωt ± 90°) = ∓ sin ωt

Using these relationships, we can transform a sinusoid from sine form to

cosine form or vice versa.

Example: Find the amplitude, phase, period, and frequency of the sinusoid

v(t) = 12 cos(50t + 10°) V.

Solution:

The amplitude is Vm = 12 V.

The phase is ϕ = 10°.

The angular frequency is ω = 50 rad/s.

The period T = 2π/ω = 2π/50 = 0.1257 s.

The frequency is f = 1/T = 7.958 Hz

Now recall from circuits 1 that we can analyze a circuit using one of many methods:

Kerchhoffs Current Law
Node Voltage/Mesh Current
Kerchhoffs Voltage Law

We are going to use the methods of analysis again only these equations will now become differential equations with the introduction of derivatives and sinusoidal functions to represent the changing voltage and current.

I will reference this Khan academy class where the start of impedance and phasors are introduced:

In the video the AC circuit drawn with V in, an Inductor, a resistor, and a capacitor.

In the circuit from the video, i is the independent variable since its the only current through the circuit

He looks at the circuit and creates the derivative of the KVL equation for an RLC circuit and proceeds to discuss how to analyze it:

L times the second derivative of the current i

plus

R times the first derivative of the current i

plus

1 over C times the current i

equals Voltage in

The above is a differential equation so we propose the solution for i in the form of

i = Ae^ST

S is the natural frequency*

Natural frequency, also known as eigenfrequency, is the frequency at which a system tends to oscillate in the absence of any driving or damping force. The motion pattern of a system oscillating at its natural frequency is called the normal mode (if all parts of the system move sinusoidally with that same frequency).

You plug this equation into the above equation and if it is indeed a solution you should arrive at

Where you can then solve for A using initial conditions.

Then you take this and solve for V in which you limit using Sinusoids for simplification of the answer hence

V1(t) = Vm sin ωt

and

V2(t) = Vm sin (ωt + ϕ)

To write sinusoidal voltage in phasors, we use

What we can then do is ‘transform’ the circuit in terms of natural frequency*

*I will go further into this but a lot of circuits 2 is memorization

Inductors become SL

Resistors become R

and Capacitors become 1/SC

Now we can go ahead an write our new KVL equation using the natural frequency equivalents and begin to work on finding the impedance or resistance

V in = SLi+Ri+1/SCi

V in = i(SL + R + 1/SC)

V in/i = SL + R + 1/SC

V=IR so V/i = R

In the equation V in/i = SL + R + 1/SC

V in/i is a ratio of voltage and current which is known as impedance (Z) or essentially resistance

We also will need to know Euler’s Identity to solve our differential equation above for converting e^jx into sin and cos functions

Using Euler’s equation we will get complex numbers

Heres a nice video representation of the sin and cos functions

If you want I would recommend going through these Khan academy reviews and courses on Trig and complex numbers as these are used heavily in this course

Note that electrical equations use j instead of i due to its use to represent current

This video shows you converting a complex number into exponential form

Chapter 9 Summary

A sinusoid is a signal in the form of the sine or cosine function. It has the general form

v(t) = Vm cos(ωt + ϕ)

where Vm is the amplitude,ω = 2πf is the angular frequency, (ωt + ϕ) is the argument, and ϕ is the phase.

A phasor is a complex quantity that represents both the magnitude and the phase of a sinusoid. Given the sinusoid

v(t) = Vm cos(ωt + ϕ), its phasor V is:

In ac circuits, voltage and current phasors always have a fixed relation to one another at an y moment of time.

If v(t) = Vm cos(ωt + ϕv) represents the voltage through an element and

i(t) = Im cos(ωt + ϕi) represents the current through the element,

then

ϕi = ϕv if the element is a resistor

ϕi leads ϕv by 90° if the element is a capacitor

ϕi lags ϕv by 90° if the element is an inductor.

The impedance Z of a circuit is the ratio of the phasor voltage across it to the phasor current through it:

Z = V/I = R(ω) + jX(ω)

The admittance Y is the reciprocal of impedance:

Y =1/Z=G(ω)+jB(ω)

Impedances are combined in series or in parallel the same w ay as resistances in series or parallel; that is, impedances in series add while admittances in parallel add.

For a resistor Z = R, for an inductor Z = jX = jωL

For a capacitor Z = −jX = 1∕jωC.

Basic circuit laws (Ohm’s and Kirchhoff’s) apply to ac circuits in

the same manner as they do for dc circuits; that is,

V = ZI

Σ I sub k = 0 (KCL)

Σ V sub k = 0 (KVL)

The techniques of voltage/current division, series/parallel combination

of impedance/admittance, circuit reduction, and Y-Δ transformation

all apply to ac circuit analysis.

AC circuits are applied in phase-shifters and bridges.

Chapter 10: Sinusoidal Steady-State Analysis

Learning Objectives

1. Analyze electrical circuits in the frequency domain using

nodal analysis.

2. Analyze electrical circuits in the frequency domain using

mesh analysis.

3. Apply the superposition principle to frequency domain

electrical circuits.

4. Apply source transformation in frequency domain circuits.

5. Understand how Thevenin and Norton equivalent circuits can

be used in the frequency domain.

6. Analyze electrical circuits with op amps.

Steps to Analyze AC Circuits:

1. Transform the circuit to the phasor or frequency domain.

2. Solve the problem using circuit techniques (nodal analysis,

mesh analysis, superposition, etc.).

3. Transform the resulting phasor to the time domain.

Example

Solution:

We first convert the circuit to the frequency domain:

20 cos 4t ⇒ 20 L 0º, Ω = 4 rad/s

1 H ⇒ jΩL = j4

0.5 H ⇒ jΩL = j2

0.1 F ⇒ 1/jΩC

= −j2.5

Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2.

Frequency domain equivalent of the circuit in Fig. 10.1.

Applying KCL at node 1,

(20− V1)/10 = ((V1)/− j2.5) + (V1 − V2)/j4

(1 + j1.5)V1 + j2.5V2 = 20

At node 2,

2Ix + (V1 −V2)/j4= V2/j2

But

Ix = V1∕−j2.5.

Substituting this gives

2V1 −j2.5+ (V1−V2)/j4= V2/j2

By simplifying, we get

11V1 + 15V2 = 0

Equations (10.1.1) and (10.1.2) can be put in matrix form as

We obtain the determinants as

Chapter 10 Summary

We apply nodal and mesh analysis to ac circuits by applying KCL

and KVL to the phasor form of the circuits.

In solving for the steady‑state response of a circuit that has independent

sources with different frequencies, each independent source

must be considered separately. The most natural approach to analyzing

such circuits is to apply the superposition theorem. A separate

phasor circuit for each frequency must be solved independently, and

the corresponding response should be obtained in the time domain.

The overall response is the sum of the time domain responses of all

the individual phasor circuits.

The concept of source transformation is also applicable in the frequency

domain.

The Thevenin equivalent of an ac circuit consists of a voltage source

V sub Th in series with the Thevenin impedance Z sub Th.

The Norton equivalent of an ac circuit consists of a current source IN

in parallel with the Norton impedance Z sub N (=Z sub Th).

PSpice is a simple and powerful tool for solving ac circuit problems.

It relieves us of the tedious task of working with the complex numbers

involved in steady‑state analysis.

The capacitance multiplier and the ac oscillator provide two typical

applications for the concepts presented in this chapter. A capacitance

multiplier is an op amp circuit used in producing a multiple of

a physical capacitance. An oscillator is a device that uses a dc input

to generate an ac output.

Chapter 11: AC Power Analysis

Topic Overview

Our effort in ac circuit analysis so far has been focused mainly on calculating voltage and current. Our major concern in this Chapter 11 is power analysis. Power analysis is of paramount importance.

Power is the most important quantity in electric utilities, electronic, and communication systems, because such systems involves transmission of power from one point to another. Also, every industrial and household electrical device — every fan, motor, lamp, pressing iron, TV, personal computer — has a power rating that indicates how much power the equipment requires; exceeding the power rating can do permanent damage to an appliance. The most common form of electric power is 50 — or 60 Hz ac power. The choice of ac over dc allowed high-voltage power transmission from the power generating plant to the consumer.

We will begin by defining and deriving instantaneous power and average power. We will then introduce the power concepts.

Learning Objectives

1. Fully understand instantaneous and average power.

2. Understand the basics of maximum average power.

3. Understand effective or rms values and how to calculate them

and to understand their importance.

4. Understand apparent power (complex power), power, and reactive

power and power factor.

5. Understand power factor correction and the importance of its use.

Chapter 11 Summary

1. The instantaneous power absorbed by an element is the product of

the element’s terminal voltage and the current through the element:

p = vi.

2. Average or real power P (in watts) is the average of instantaneous

power p:

P = 1/T∫ p dt ; 0=<t=<T

If v(t) = (V sub m)cos(ωt + θv)

and

i(t) = (I sub m)cos (ωt + θi),

then

(V sub rms) = (V sub m) ∕ √2 ,

(I sub rms) = (I sub m) ∕ √2 ,

and

P = 1/2 (V sub m) (I sub m) cos(θv − θi)

= (V sub rms) (I sub rms) cos(θv − θi)

Inductors and capacitors absorb no average power, while the average

power absorbed by a resistor is (1∕2)(I² sub m)R= (I² sub rms) R.

Maximum average power is transferred to a load when the load impedance

is the complex conjugate of the Thevenin impedance as

seen from the load terminals, Z sub L = Z* sub Th .

The effective value of a periodic signal x(t) is its root-mean-square

(rms) value.

(X sub eff) = (X sub rms) = √ (1/T(∫ x²dt; 0=<t=<T))

For a sinusoid, the effective or rms value is its amplitude divided

by √2 .

5. The power factor is the cosine of the phase difference between voltage

and current:

pf = cos(θv − θi)

It is also the cosine of the angle of the load impedance or the ratio

of real power to apparent power. The pf is lagging if the current lags

voltage (inductive load) and is leading when the current leads voltage

(capacitive load).

Apparent power S (in VA) is the product of the rms values of voltage

and current:

S = (V sub rms) (I sub rms)

It is also given by S = ∣S∣ = √(P² + Q²)

where P is the real power and Q is reactive power.

Reactive power (in VAR) is:

Q=1/2(V sub m)(I sub m)sin(θv − θi)=(V sub rms)(I sub rms) sin(θv−θi)

Complex power S (in VA) is the product of the rms voltage phasor

and the complex conjugate of the rms current phasor. It is also the

complex sum of real power P and reactive power Q.

S = (V sub rms) (I* sub rms) = (V sub rms) (I sub rms) L θv − θi = P + jQ

Also,

S = I² sub rms= Z = (V² sub rms) /Z*

The total complex power in a network is the sum of the complex

powers of the individual components. Total real power and reactive

power are also, respectively, the sums of the individual real powers

and the reactive powers, but the total apparent power is not calculated

by the process.

Power factor correction is necessary for economic reasons; it is the

process of improving the power factor of a load by reducing the

overall reactive power.

The wattmeter is the instrument for measuring the average power.

Energy consumed is measured with a kilowatt-hour meter.

Exam 1: Chapters 9–11 See above

Actual Exam 1 Questions :

Problem 1–25 points — CLO 1

a. (15 Points) Evaluate the following complex numbers and express your results in rectangular form:

b. (10 Points) Using the phasor approach, determine the current,

in a circuit described by the following integrodifferential equation

Problem 2–25 points — CLO 2

Determine the current

in the circuit of following Figure 1 using mesh analysis

Problem 3–25 points — CLO 2

Determine the Norton equivalent of the circuit of Figure 2 as seen from terminals Use the equivalent to find I sub 0

Problem 4–25 points — CLO 2

a. (15 Points) Determine the load impedance,

that maximize the average power drawn from the circuit of Figure 3. What is the maximum average power?

b. (10 Points) Determine the rms value of the current waveform of Figure 4. If the current is passed through a

resistor, find the average power absorbed by the resistor.

Exam 2 Course Content

Chapter 12: Three Phase Circuits

So far the topics discussed in the previous weeks have dealt with single-phase circuit.

A single-phase ac power system consists of a generator connected through a pair of wires (a transmission line) to a load.

What is more common in practice is a single-phase three-wire system. For example, the normal household system is a single — phase three-wire system because the terminal voltages have the same magnitude and the same phase. Such a system allows the connection of both 120- and 240-V appliance.

Circuits or systems in which the AC sources operate at the same frequency but different phases are known as polyphase.

A three phase system is produced by a generator consisting of three sources having the same magnitude and frequency but out of phase with each other by 120 degree.

Because the three-phase system is by far the most prevalent and most economical polyphase system, discussion in this chapter is mainly on three-phase systems.

Three -phase systems are important for at least three reasons:

First, nearly all electric power is generated and distributed in three phase at operated frequency of 60 Hz in the United States or 50Hz in some other parts of the world. When one-phase or two phase inputs are required, they are taken from the three-phase system rather than generated independently. Even when more than three phases are needed — such as in the aluminum industry, where 48 phases are required for melting purposes-they can be provided by manipulating the three phase supplied.
Second, the instantaneous power in a three-phase system can be constant. This results in uniform power transmission and less vibration of three-phase machines.
Third, for the same amount of power, the three-phase system is more economical than the single phase. The amount of wire required for a three-phase system is less than that required for an equivalent single phase system.

In this chapter 12, we will discuss balanced three-phase voltages. Then we analyze each of the four possible configurations of balanced three phase systems.

Learning Outcomes

After successful completion of these activities, you will be able to:

Understand balanced three — phase voltages.
Analyze balanced wye-wye circuits
Understand and analyze balanced wye-delta circuits
Analyze balanced delta-delta circuits
Under and analyze balanced delta-wye circuits

Chapter 12 Summary

1. The phase sequence is the order in which the phase voltages of a

three-phase generator occur with respect to time. In an abc sequence

of balanced source voltages, V sub an leads V sub bn by 120°, which in turn

leads V sub cn by 120°. In an acb sequence of balanced voltages, V sub an

leads V sub cn by 120°, which in turn leads V sub bn by 120°.

2. A balanced wye- or delta-connected load is one in which the three-phase

impedances are equal.

3. The easiest way to analyze a balanced three-phase circuit is to transform

both the source and the load to a Y-Y system and then analyze

the single-phase equivalent circuit. Table 12.1 presents a summary

of the formulas for phase currents and voltages and line currents and

voltages for the four possible configurations.

4. The line current I sub L is the current flowing from the generator to the

load in each transmission line in a three-phase system. The line

voltage V sub L is the voltage between each pair of lines, excluding the

neutral line if it exists. The phase current I sub p is the current flowing

through each phase in a three-phase load. The phase voltage V sub p is the

voltage of each phase. For a wye-connected load,

V sub L = √3 * V sub p

and

I sub L = I sub p

For a delta-connected load,

V sub L = V sub p

and

I sub L = √3 I sub p

5. The total instantaneous power in a balanced three-phase system is

constant and equal to the average power.

6. The total complex power absorbed by a balanced three-phase

Y-connected or Δ-connected load is

S = P + jQ = √3 (V sub L)*(I sub L)< θ

where θ is the angle of the load impedances.

7. An unbalanced three-phase system can be analyzed using nodal or

mesh analysis.

8. PSpice is used to analyze three-phase circuits in the same way as it

is used for analyzing single-phase circuits.

9. The total real power is measured in three-phase systems using either

the three-wattmeter method or the two-wattmeter method.

10. Residential wiring uses a 120/240-V, single-phase, three-wire system.

Chapter 14: Frequency Response

Topic Overview

In our sinusoidal circuit analysis, we have learned how to find voltages and currents in a circuit with a constant frequency source. If we let the amplitude of the sinusoidal source remain constant and vary the frequency, we obtain the circuit’s frequency response.

The frequency response may be regarded as a complete description of the sinusoidal steady-state behavior of a circuit as a function of frequency.

The sinusoidal steady-state frequency responses of circuits are of significance in many applications, especially in communications and control systems. A specific application is in electric filters that block out or eliminate signals with unwanted frequencies and pass signals of the desired frequencies. Filters are used in radio, TV, and telephone systems to separate one broadcast frequency from another.

We begin this chapter (Chapter 14) by considering the frequency response of simple circuits using their transfer functions. We then consider Bode plots, which are the industry standard way of presenting frequency response. We also consider series and parallel resonant circuits and encounter important concepts such as resonance, quality factor, cut off frequency, and bandwidth. We also discuss different types of active and passive filters.

Learning Outcomes

After successful completion of chapter 14, you will be able to:

Understand what transfer functions are and how to determine them.
Understand the decibel scale, why we use it and how to use it.
Understand Bode plots, and know why we use them and how to determine them.
Understand series and parallel resonance , why they are important, and how to find them
Understand passive filters
Understand active filters

Chapter 14 Summary

1. The transfer function H(ω) is the ratio of the output response Y(ω)

to the input excitation X(ω); that is, H(ω) = Y(ω)∕X(ω).

2. The frequency response is the variation of the transfer function with

frequency.

3. Zeros of a transfer function H(s) are the values of s = jω that make

H(s) = 0, while poles are the values of s that make H(s) → ∞.

4. The decibel is the unit of logarithmic gain. For a voltage or current

gain G, its decibel equivalent is GdB = 20 log10 G.

5. Bode plots are semi-log plots of the magnitude and phase of the

transfer function as it varies with frequency. The straight-line

approximations of H (in dB) and ϕ (in degrees) are constructed using

the corner frequencies defined by the poles and zeros of H(ω).

6. The resonant frequency is that frequency at which the imaginary part

of a transfer function vanishes. For series and parallel RLC circuits.

ω sub 0 = 1√LC

7. The half-power frequencies (ω1, ω2) are those frequencies at which

the power dissipated is one-half of that dissipated at the resonant

frequency. The geometric mean between the half-power frequencies

is the resonant frequency, or

ω sub 0 = √(ω sub 1)(ω sub 2)

8. The bandwidth is the frequency band between half-power

frequencies:

B =(ω sub 2 − ω sub 1)

9. The quality factor is a measure of the sharpness of the resonance

peak. It is the ratio of the resonant (angular) frequency to the bandwidth,

Q = (ω sub 0/ B)

10. A filter is a circuit designed to pass a band of frequencies and reject

others. Passive filters are constructed with resistors, capacitors, and

inductors. Active filters are constructed with resistors, capacitors,

and an active device, usually an op amp.

11. Four common types of filters are low-pass, high-pass, band-pass, and

band-stop. A low-pass filter passes only signals whose frequencies are

below the cutoff frequency ωc. A high-pass filter passes only signals

whose frequencies are above the cutoff frequency ωc. A band-pass

filter passes only signals whose frequencies are within a prescribed range:

(ω1 < ω < ω2)

A band-stop filter passes only signals whose

frequencies are outside a prescribed range (ω1 > ω > ω2).

12. Scaling is the process whereby unrealistic element values are

magnitude-scaled by a factor K sub m and/or frequency-scaled by a factor

K sub f to produce realistic values.

R′ = K sub mR,

L′ = (K sub m/K sub f)L

C′ = 1/(K sub m)(K sub f)C

13. PSpice can be used to obtain the frequency response of a circuit if a

frequency range for the response and the desired number of points

within the range are specified in the AC Sweep.

14. The radio receiver — one practical application of resonant circuits —

employs a band-pass resonant circuit to tune in one frequency

among all the broadcast signals picked up by the antenna.

15. The touch-tone telephone and the crossover network are two typical

applications of filters. The touch-tone telephone system employs

filters to separate tones of different frequencies to activate electronic

switches. The crossover network separates signals in different frequency

ranges so that they can be delivered to different devices such

as tweeters and woofers in a loudspeaker system.

Review for Exam 2

So far:

A sine wave or sinusoid is a mathematical curve that describes a smooth periodic oscillation.

All AC Circuits operate using sinusoids as the input meaning they’re a sin or cos function that represents the voltage and current.

With Euler’s Identities we can convert our sinusoids into complex exponential notion of functions of just sin or cos

Omega is frequency

T is time

j is an imaginary number that essentially just represents the rotation of the sinusoidal function.

If we use e^jωt as the voltage or current we can start our circuit analysis using the original circuit element equations and finding the impedance of each

Resistor: V = IR becomes Z sub R = R

Inductor: V = L di/dt becomes Z sub L = jωL

Capacitors: i = C dv/dt becomes Z sub C = 1/jωC

In series you can just add these elements together as you normally would

Exam 2: Chapters 12 + 14 See above

Actual Exam 2 Questions :

Problem 1–25 points — CLO 3

A Y-Connected balanced three-phase generator with an impedance of 0.4 + 𝑗0.3 Ω per phase is

connected to a Y-connected load with an impedance of 24 + 𝑗19 Ω per phase. The line joining

the generator and the load has an impedance of 0.6 + 𝑗0.7 Ω per phase. Assuming a positive

sequence for the source voltages and that 𝑉!” = 120∠30# 𝑉. Find:

(a) the line voltages

(b) the line currents

Problem 2–25 points — CLO 3

A balanced delta-connected source has phase voltage Vab = 416 ‹30° V and a positive phase

sequence. If this is connected to a balanced delta-connected load, find the line and phase

currents. Take the load impedance per phase as 60 ‹30°W and line impedance per phase as 1 +

j1W..

Problem 3–25 points — CLO 3

In the circuit of Figure 1, 𝑅 = 2 Ω, 𝐿 = 1 𝑚𝐻, and 𝐶 = 0.4 𝜇𝐹.

(a) Find the resonant frequency and the half power frequencies.

(b) Calculate the quality factor and bandwidth.

(c ) Determine the amplitude of the current at 𝜔 sub 0,𝜔 sub 1, and 𝜔 sub 2

Problem 4–25 points — CLO 3

Obtain the transfer function of the active filter in Figure 2. What kind of filter is it?

Exam 3 Course Content [FINAL]:

Dear Students,
We are scheduling Exam — III on Wednesday , August 05, 2020 @ 9:30 am. This Exam will start at 9:30 am and end at 11:45 am.
Following chapters are included in this exam.
Chapter 15 — Introduction to the Laplace Transform
Chapter 16 — Applications of the Laplace Transform
Chapter 19 — Two-Port Network
Please note that you are allow to use handwritten one piece of A4 size paper on both side. Please also study Homework- 6, Homework -7 and Homework — 8 in addition to example and practice exercise from lecture notes.
I wish you good luck.
Thanks,
Muhammad

This is the email we received about a week before our exam, we are allowed to create a one page cheat.

I will write it up and link it in the summary

The idea of transformation should be familiar by now. When using

phasors for the analysis of circuits, we transform the circuit from the time

domain to the frequency or phasor domain. Once we obtain the phasor

result, we transform it back to the time domain.

The Laplace transform method follows the same process:

We use the Laplace transformation to transform the circuit from the time domain to the frequency domain, obtain the solution, and apply the inverse Laplace transform to the result to transform it back to the time domain.

Chapter 15: Introduction to the Laplace Transform

Topic Overview

Our goal in this chapter 15 — Introduction to the Laplace Transform is to develop techniques for analyzing circuits with a wide variety of inputs and responses. Such circuits are modeled by differential equations whose solutions describe the total response behavior of the circuits. Mathematical methods have been devised systematically determine the solution of differential equations. We will introduce the powerful methods of Laplace transformation, which involves turning differential equations into algebraic equations, thus greatly facilitating the solution process.

The Laplace transform is significant for a number of reasons. First, it can be applied to a wider variety of inputs than phasor analysis. Second, it provides an easy way to solve circuit problems involving initial conditions, because it allow us to work with algebraic equations instead of differential equations. Third, the Laplace transform is capable of providing us, in one single operation, the total response of the circuit comprising both the natural and forced responses.

We will begin with the definition of Laplace transform which gives rise to its most essential properties works. We also consider some properties of the Laplace transform that are very helpful in circuit analysis. We then consider the inverse Laplace transform.

In short, we will focus on on the mechanics of the Laplace transform.

Learning Outcomes

After successful completion of this chapter, you will be able to:

Understand the Laplace transform, its importance in circuit analysis, and how to determine the Laplace transform of functions common to circuit analysis
Understand the properties of Laplace transform
Understand the inverse Laplace transform and how to determine its given functions in the s-domain.

Definition of the Laplace Transform

Given a function f(t), its Laplace transform, denoted by F(s) or [ f(t)], is defined by:

where s is a complex variable given by

The Laplace transform is an integral transformation of a function f (t) from

the time domain into the complex frequency domain, giving F (s).

When the Laplace transform is applied to circuit analysis, the differential equations represent the circuit in the time domain. The terms in the differential equations take the place of f (t). Their Laplace transform, which corresponds to F(s), constitutes algebraic equations representing the circuit in the frequency domain.

Being able to look at circuits and systems in the s-domain can help us to understand how our circuits and systems really function.

The same differential equations can be used to describe any linear circuit, system, or process.

A system is a mathematical model of a physical process relating the input to the output.

It is entirely appropriate to consider circuits as systems. Historically, circuits have been discussed as a separate topic from systems, so we will continue using circuits and systems in this chapter realizing that circuits are nothing more than a class of electrical systems.

Chapter 16: Applications of the Laplace Transform

We are now prepared to employ the Laplace transform to analyze circuits.

This usually involves three steps:

1. Transform the circuit from the time domain to the s-domain.

2. Solve the circuit using nodal analysis, mesh analysis, source

transformation, superposition, or any circuit analysis technique

with which we are familiar.

3. Take the inverse transform of the solution and thus obtain the

solution in the time domain.

As we did in phasor analysis, we transform a circuit in the time domain to the frequency or s-domain by Laplace transforming each term in the circuit.

For a resistor:

The voltage-current relationship in the time domain is:

v(t) = R•i(t)

Taking the Laplace transform, we get:

V(s) = R•I(s)

For an inductor:

v(t) = L •di(t)/dt

Taking the Laplace transform of both sides gives

V(s) = L[s•I(s) − i(0−)] = [s•L•I(s)] − [L•i(0−)]

I(s) =[1/sL ]•V(s) + [i(0−)/s]

For a capacitor:

i(t) = C •dv(t)/dt

which transforms into the s-domain as

I(s) = C[s•V(s) − v(0−)] = [s•C•V(s)] − [C•v(0−)]

V(s) =1/sC •I(s) +[v(0−)/s]

Why use Laplace Transforms in Circuit Analysis?

Representation of a capacitor: (a) time-domain (b, c) s-domain equivalents.

Another advantage is that a complete response — transient and steady state — of a network is obtained

If we assume zero initial conditions for the inductor and the capacitor, the above equations reduce to:

Resistor: V(s) = R • I(s)

Inductor: V(s) = sL • I(s)

Capacitor: V(s) = 1/sC •I(s)

The s-domain equivalents are shown in Fig. 16.3.

We define the impedance in the s-domain as the ratio of the voltage transform to the current transform under zero initial conditions:

Z(s) = V(s)/I(s)

Thus, the impedances of the three circuit elements are

Resistor: Z(s) = R

Inductor: Z(s) = s•L

Capacitor: Z(s) = 1/sC

The admittance in the s-domain is the reciprocal of the impedance, or

Y(s) = 1/Z(s)= I(s)/V(s)

Examples:

Find v sub o(t) in the circuit of Fig. 16.4, assuming zero initial conditions.

Solution:

We first transform the circuit from the time domain to the s-domain.

u(t) ⇒ 1/s

1 H ⇒ sL = s

1/3F ⇒ 1/sC= 3/s

The resulting s-domain circuit is

We now apply mesh analysis.

For mesh 1,

1/s = ( 1 + 3/s ) I sub 1 − 3/s • I sub 2

For mesh 2,

0 = − 3/s • I sub 1 + ( s + 5 + 3/s) • I sub 2

I sub 1 = 1/3(s^2 + 5s + 3)• I sub 2

Substituting this in to 1/s = ( 1 + 3/s ) I sub 1 − 3/s • I sub 2

1/s= ( 1 + 3/s)• 1/3(s^2 + 5s + 3)• I sub 2 − 3/s • I sub 2

Multiplying through by 3s gives

3 = (s^3 + 8s^2 + 18s)I sub2 ⇒ I sub 2 = 3/s^3 + 8s^2 + 18s

V sub o(s) = s(I sub 2) = 3/(s^2 + 8s + 18)

= 3/√2 • √2 /[(s + 4)2 + ( √2)^2]

Taking the inverse transform yields

v sub o(t) = 3/√2•e^−4t • sin √2t V

Where t ≥ 0

Chapter 19: Two Port Networks

Learning Objectives

By using the information and exercises in this chapter you will be

able to:

1. Understand the variety of two-port parameters that make analyzing circuits easier.

2. Understand impedance parameters and how to use them effectively in analyzing certain classes of circuit analysis problems.

3. Understand admittance parameters and how to use them effectively in analyzing certain classes of circuit analysis problems.

4. Understand hybrid parameters and how to use them effectively in analyzing certain classes of circuit analysis problems.

5. Understand transmission parameters and how to use them effectively in analyzing certain classes of circuit analysis problems.

6. Understand the relationships between all two-port parameters.

7. Understand how to interconnect networks using the characteristics of the variety of parametric relationships.

A two-port network is an electrical network with two separate ports for input and output.

Summary

1. A two-port network is one with two ports (or two pairs of access terminals), known as input and output ports.

2. The six parameters used to model a two-port network are the impedance[z], admittance [y], hybrid [h], inverse hybrid [g], transmission[T], and inverse transmission [t] parameters.

3. The parameters relate the input and output port variables as