Engineering Circuit Analysis Prof. Li Chen, Imperial College of London

Engineering Circuit Analysis Prof. Li Chen, Imperial College of London
References： W. H. Hayt, Jr., J. E. Kemmerly and S. M. Durbin, Engineering Circuit Analysis, McGraw-Hill, 6th Edition, ISBN

Teaching Schedule Weeks Chapters References 1, 2
Basis concepts and laws of electronics Hayt: Ch 1 2 5 3, 4 Basis analysis methods to circuits Hayt: Ch 3 4 5 Basis RL & RC circuits Hayt: Ch 6 6, 7, 8 Sinusoidal steady state analysis Hayt: Ch 7 9 Midterm 10, 11 AC circuit power analysis Hayt: Ch 8 12, 13 Polyphase circuit Hayt: Ch 9 14 Magnetically coupled circuits Hayt: Ch 10 15, 16 Fourier circuit analysis Hayt: Ch 12 17 Review (Q & A)

Ch1 Basic Concepts and Laws of Electric Circuits
Engineering Circuit Analysis Ch1 Basic Concepts and Laws of Electric Circuits 1.1 Basic Concepts and Electric Circuits 1.2 Basic Quantities 1.3 Circuit Elements 1.4 Kirchhoff's Current and Voltage Laws References: Hayt: Ch1, 2, 5;

1.1 Basic Concepts and Electric Circuits
Signal processing and transmission Amplifiers Speaker transmitter Circuits Kinescope Antenna Electrical power conversion and transmission Power Supplies Transmission Loads

Electrical power conversion and transmission

Concept of Abstraction Question: What is the current through the bulb? In order to calculate the current, we can replace the bulb with a resistor. R is the only subject of interest, which serves as an abstraction of the bulb. Solution:

Resistance: R = V/I, 1 =1V/A, ohm; Conductance: G = 1/R = 1A/V, siemens (S); 1S = 1A/V, i(t) = G × v(t); Instantaneous current and voltage at time t; A resistor is a circuit element that transforms the electrical energy (e.g. electricity  heat); Commonly used devices that are modeled as resistors include incandescent, heaters, wires and etc; A circuit consists of sources, resistors, capacitors, inductors and conductors; Elements are lumped. Conductors are perfect. Incandescent 白帜 dissipate 消散 Lumped circuit abstraction!

The AM Radio System Transmitter Receiver Understanding the AM radio requires knowledge of several concepts Communications/signal processing (frequency domain analysis) Electromagnetics (antennas, high-frequency circuits) Power (batteries, power supplies) Solid state (miniaturization, low-power electronics)

Example 1: The AM audio system Example 2: The telephone system
1.1 Basic Concepts and Electric Circuits Example 1: The AM audio system Example 2: The telephone system

The AM Radio System A signal is a quantity that may vary with time. * Voltage or current in a circuit * Sound (sinusoidal wave traveling through air) * Light or radio waves (electromagnetic energy traveling through free space) The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis, which allows us to represent signals as combinations of sinusoids (sines and cosines).

The AM Radio System Frequency is the rate at which a signal oscillates. Duration of the signal T, frequency of the signal f = 1/T. High Frequency Low Frequency

The AM Radio System Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz. Our visual system perceives the frequency of the electromagnetic energy as color: is 460THz, is 570THz, and is 630THz. An AM radio signal has a frequency of between 500kHz and 1.8MHz. FM radio and TV uses different frequencies. red green blue Mathematical analysis of signals in terms of frequency Most commonly encountered signals can be represented as a Fourier series or a Fourier transform. A Fourier series is a weighted sum of cosines and sines.

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) + ···
1.1 Basic Concepts and Electric Circuits The AM Radio System Fourier Series: A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines. Given function f(t) with angular frequency ω and period T, its Fourier series can be written as: f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) + ··· =

Example: Given function during a period: t For the example : , k is even. , k is odd.

The AM Radio System Example-Fourier Series 1st series + 3rd series 1st series (k = 1) 3rd series (k = 3) Signals can be represented in terms of their frequency components. The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals.

The AM Radio System Transmitter Block Diagram Signal Source Modulator Power Amplifier Antenna Modulator The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (i.e. 605kHz-615kHz) freq 5kHz Frequency domain representation of input freq 610kHz Frequency domain representation of output

The AM Radio System Modulator: Time Domain Input Signal Output Signal

The AM Radio System Power Amplifier A typical AM station broadcasts several kW Up to 50kW-Class I or Class II stations Up to 5kW-Class III station Up to 1kW-Class IV station Typical modulator circuit can provide at most a few mW Power amplifier takes modulator output and increases its magnitude Antenna The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space.

The AM Radio System Receiver Block Diagram RF Amplifier IF Mixer Envelope Detector Audio Antenna Speaker

The AM Radio System The antenna captures electromagnetic energy and converts it to a small voltage or current. In the frequency domain, the antenna output is Antenna frequency Undesired Signals Desired Signal Carrier Frequency of desired station interferences interferences

The AM Radio System RF (Radio Frequency) Amplifier RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits. RF Amplifier also performs as a Bandpass filter for the signal Bandpass filter attenuates the other components outside the frequency range that contains the desired station frequency Undesired Signals Desired Signal Carrier Frequency of desired station

The AM Radio System IF (Intermediate Frequency) Mixer
The IF Mixer shifts its input in the frequency domain from the carrier frequency to an intermediate frequency of 455kHz frequency Undesired Signals Desired Signal 455 kHz IF Amplifier The IF amplifier bandpass filters the output of the IF mixer, eliminating all of the undesired signals. frequency Desired Signal 455 kHz

The AM Radio System Envelope Detector Computes the envelope of its input signal Output Signal Input Signal

The AM Radio System Audio Amplifier Amplifies signal from envelope detector Provides power to drive the speaker Hierarchical System Models Modelling at different levels of abstraction Higher levels of the model describe overall function of the system Lower levels of the model describe necessary details to implement the system In the AM receiver, the input is the antenna voltage and the output is the sound energy produced by the speaker. In EE, a system is an electrical and/or mechanical device, a process, or a mathematical model that relates one or more inputs to one or more outputs. System Inputs Outputs

The AM Radio System Top Level Model AM Receiver Input Signal Sound Second Level Model RF Amplifier IF Mixer Envelope Detector Audio Antenna Speaker Power Supply

The AM Radio System Half-wave Rectifier Low-pass Filter Low Level Model Envelope Detector. Circuit Level Model Envelope Detector + - R C Vout Vin

1.2 Basic Quantities Units Electric charge (q) in Coulombs (C)
Standard SI Prefixes pico (p) nano (n) micro () milli (m) kilo (k) mega (M) giga (G) tera (T) Electric charge (q) in Coulombs (C) Current (I) in Amperes (A) Voltage (V) in Volts (V) Energy (W) in Joules (J) Power (P) in Watts (W)

1.2 Basic Quantities Current Constant current Time varying current
A mount of electric charges flowing through the surface per unit time. Constant current Time rate of change of charge Time varying current Unit (1 A = 1 C/s) Notation: Current flow represents the flow of positive charge Alternating versus direct current (AC vs DC) i(t) t DC AC Time – varying current Steady current

1.2 Basic Quantities Current Positive versus negative current
Negative charge of -2C/s moving Positive charge of 2C/s moving or Negative charge of -2C/s moving Positive charge of 2C/s moving or P1.1, In the wire electrons moving left to right to create a current of 1 mA. Determine I1 and I2. Ans: I1 = -1 mA; I2 = +1 mA. Current is always associated with arrows (directions)

1.2 Basic Quantities Voltage(Potential) + Energy per unit charge.
It is an electrical force drives an electric current. Voltage Units: 1 V = 1 J/C Positive versus negative voltage + – 2 V -2 V +/- of voltage (V) tell the actual polarity of a certain point . DN Two “Do Not (DN)” +/- of current (I) tell the actual direction of particle’s movement .

1.2 Basic Quantities Voltage (Potential) Example Vab = ?
 a b a、b, which point’s potential is higher？  b a Vab = ? a b +Q from point b to point a get energy ，Point a is Positive? or negative ?  

1.2 Basic Quantities Voltage (Potential) a b c´ c d d´ Example I

1.2 Basic Quantities Voltage (Potential) Example Va=? K Open I K Close

1.2 Basic Quantities Example I I

1.2 Basic Quantities Power +
One joules of energy is expanded per second. P = W/t Rate of change of energy + – v(t) i(t) p(t) = v(t) i(t) v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering. Used to determine the electrical power is being absorbed or supplied if P is positive (+), power is absorbed if P is negative (–), power is supplied

1.2 Basic Quantities Power Example + + Note : + 2A -5V 5V 2A +5V -5V
– -5V Power is supplied. delivered power to external element. + – 5V 2A Power is absorbed. Power delivered to Note : + – +5V -5V 2A -2A Power absorbed .

1.2 Basic Quantities Power Power absorbed by a resistor:

1.2 Basic Quantities Power
P1.5 Find the power absorbed by each element in the circuit. 1 2 3 4 5 I1 I2 I3 + - + - - + + - Supply energy : element 1、3、4 . Absorb energy : element 2、5

1.2 Basic Quantities I=0, V=E , P=0 Open Circuit R= Short Circuit R=0

1.2 Basic Quantities Loaded Circuit E R0 R I

1.3 Circuit Elements Key Words: Resistors, Capacitors, Inductors, voltage source, current source

1.3 Circuit Elements Passive elements (cannot generate energy)
e.g., resistors, capacitors, inductors, etc. Active elements (capable of generating energy) batteries, generators, etc. Important active elements Independent voltage source Independent current source Dependent voltage source voltage dependent and current dependent Dependent current source

1.3 Circuit Elements Resistors Dissipation Elements v=iR
P=vi=Ri2=v2/R >0 ， v-i relationship v i Resistors connected in series: Equivalent Resistance is found by Req= R1 + R2 + R3 + … R1 R2 R3 Resistors connected in parallel 1/Req=1/R1 + 1/R2 + 1/R3 + … R1 R2 R3

1.3 Circuit Elements Capacitors
Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator). Charge on the two conductors creates an electric field that stores energy. The voltage difference between the two conductors is proportional to the charge: q = C v The proportionality constant C is called capacitance. Units of Farads (F) - C/V 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device. 1F=106F, 1F=106PF

1.3 Circuit Elements Capacitors store energy in an electric field
+ - v(t) The rest of the circuit v-i relationship vC(t+) = vC(t-) Energy stored Capacitors connected in series: Equivalent capacitance is found by 1/Ceq=1/C1 + 1/C2 + 1/C3 + … series parallel Capacitors connected in parallel Ceq= C1 + C2 + C3 + …

1.3 Circuit Elements Capacitors P1.7 For (1) : i(t) + circuit 0.2F
- v(t) circuit 0.2F t i(t) 1A -1A 1s 2s t v(t) 5V 1s 2s (1)

1.3 Circuit Elements Capacitors P1.7 For (2) : For (1)、(2) : i(t) +
- v(t) circuit 0.2F For (2) : For (1)、(2) : t i(t) 1A -1A 1s 2s t w (t) 2.5J 1s 2s (2)

1.3 Circuit Elements Inductors
store energy in a magnetic field that is created by electric passing through it. v-i relationship i(t) + - v(t) circuit L iL(t+) = iL(t-) Energy stored: Inductors connected in series: Leq= L1 + L2 + L3 + … Inductors connected in parallel: 1/Leq=1/L1 + 1/L2 + 1/L3 + …

Independent voltage source
1.3 Circuit Elements Independent voltage source RS＝0 + VS v i VS Ideal practical

Independent current source
1.3 Circuit Elements Independent current source Ideal v i IS RS＝ ∞ I practical

1.3 Circuit Elements Voltage source connected in series:
Voltage source connected in parallel:

1.3 Circuit Elements Voltage controlled (dependent) voltage source (VCVS) + _ Current controlled (dependent) voltage source (CCVS) + _ Q: What are the units for  and r?

1.3 Circuit Elements Voltage controlled (dependent) current source (VCCS) _ + Current controlled (dependent) current source (CCCS) Q: What are the units for  and g?

1.3 Circuit Elements Independent source dependent source
Can provide power to the circuit; Excitation to circuit ; Output is not controlled by external. dependent source Can provide power to the circuit; No excitation to circuit; Output is controlled by external.

1.3 Circuit Elements Review
So far, we have talked about two kinds of circuit elements: Sources (independent and dependent) active, can provide power to the circuit. Resistors passive, can only dissipate power. The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements!

Nodes, Branches, Loops, KCL, KVL
1.4 Kirchhoff's Current and Voltage Laws Key Words: Nodes, Branches, Loops, KCL, KVL

1.4 Kirchhoff's Current and Voltage Laws
Nodes, Branches, Loops, mesh Node: point where two or more elements are joined (e.g., big node 1) Branch: Component connected between two nodes (e.g., component R4) Loop: A closed path that never goes twice over a node (e.g., the blue line) The red path is NOT a loop Mesh: A loop that does not contain any other loops in it.

Nodes, Branches, Loops, mesh P1.8 A circuit containing three nodes and five branches. Node 1 is redrawn to look like two nodes; it is still one nodes.

KCL sum of all currents entering a node is zero sum of currents entering node is equal to sum of currents leaving node KCL Mathematically i1(t) i2(t) i4(t) i5(t) i3(t)

KCL sum of all currents entering a node is zero sum of currents entering node is equal to sum of currents leaving node P1.9

KCL P1.10 KCL-Christmas Lights + - 120V 50* 1W Bulbs Is Find currents through each light bulb: IB = 1W/120V = 8.3mA Apply KCL to the top node: IS - 50IB = 0 Solve for IS: IS = 50 IB = 417mA

KCL P1.11 We can make supernodes by aggregting node.

KCL In case of parallel : Current divider  N V G1 G2 I + - I1 I2

KVL sum of voltages around any loop in a circuit is zero. KVL Mathematically A voltage encountered + to - is positive. A voltage encountered - to + is negative.

KVL KVL is a conservation of energy principle A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage If the charge comes back to the same Initial point the net energy gain Must be zero.

KVL P1.13 Determine the voltages Vae and Vec. Vec = 0

KVL Voltage divider + + R1 V1 N - V + R2 V2 - - Important voltage Divider equations

KVL Voltage divider P1.14 Example: Vs = 9V, R1 = 90kΩ, R2 = 30kΩ Volume control?

Ch2 Basic Analysis Methods to Circuits
Engineering Circuit Analysis Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits 2.2 Basic Nodal and Mesh Analysis 2.3 Useful Circuit Analysis Techniques References: Hayt-Ch3, 4

2.1 Equivalent Circuits Key Words: Equivalent Circuits Network Equivalent Resistance, Equivalent Independent Sources

Equivalent Circuits Network
Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Circuits Network о Two-terminal Circuits Network о a b c d 6 5 15 о b о N1 I a + _ V о N2 I a b + V

Equivalent Resistance How do we find I1 and I2?
Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance How do we find I1 and I2? I1 + I2 = I R1 R2 V + - I1 I2 I

Equivalent Resistance
Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance i(t) + - v(t) i(t) + - v(t) Req Req is equivalent to the resistor network on the left in the sense that they have the same i-v characteristics.

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance Condition : without knowing V&I . We only know Rs Series and parallel Resistance Method 1 (source-free) Condition : without knowing Rs . We only know V&I Method 2 I V R ab o =  a b source -free  source Voc Method 3 Isc

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance In practice , Vs-source voltage ≠ VL- Local Let The IL-VL curve : 10 100 VL(V) IL(A) , Open Circuit (OC) . , Short Circuit (SC) .

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance P2.1  VS R1 R2 R3 a b How do we find Rab? + - Method 1  V R1 R2 R3 a b I Method 2 Method 3

Source Transformation
Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Source Transformation Ideally: An ideal current source has the voltage necessary to provide its rated current An ideal voltage source supplies the current necessary to provide its rated voltage Practice: A real voltage source cannot supply arbitrarily large amounts of current A real current source cannot have an arbitrarily large terminal voltage

Source Transformation
Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Source Transformation Vs + - Rs Is Rs Note: Consistency between the current source ref. direction and the voltage Source ref. terminals.

Equivalent Source How do we find I1 and I2? Is2 V R1 R2 I1 I2 Is1
Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Source How do we find I1 and I2? Is2 V R1 R2 + - I1 I2 Is1 Ieq

parallel Current Source
Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Source Series Voltage Source  + - VS1 VS2 VSn  + - VS parallel Current Source  IS1 IS2 ISn IS I RS=RS1// RS2//…// RSn

Branch Analysis, Nodal Analysis,
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Key Words: Branch Analysis, Nodal Analysis, Mesh (Loop) Analysis Why? The analysis techniques previously (voltage divider, equivalent resistance, etc.) provide an intuitive approach to analyzing circuits They are not systematic and cannot be easily automated by a computer Comments: Analysis of circuits using node or loop analysis requires solutions of systems of linear equations. These equations can usually be written by inspection of the circuit.

Branch Analysis P2.2 How do we find I1 and I2, I3? KVL Mesh 2:
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Branch Analysis P2.2 How do we find I1 and I2, I3? KVL Mesh 1: I3 I1 I2 Mesh 2: KCL

Suppose m branches, n nodals
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Branch Analysis Suppose m branches, n nodals write KCL equation for each independent node. ——(n-1) KCL equations write KVL equation for each independent mesh/loop ——m-(n-1) KVL equations

We can write the following equations:
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Branch Analysis Here’s a quick example of a circuit that we will see later when we model the operation of transistors. For now, let’s assume ideal independent and dependent sources. Ⅰ We can write the following equations: Ⅰ: Ⅱ: Ⅲ : Ⅱ Ⅲ

1) Choose a reference node
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 1) Choose a reference node + - V 500W 1kW I1 I2 The reference node is called the ground node.

. . . Nodal Analysis 1) Choose a reference node 500W 1kW I1 I2 I4 I5
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 1) Choose a reference node + - V 500W 1kW I1 I2 . . . Ⅰ I4 I5 I6 I7 I8 Ⅱ Ⅲ KCL Ⅰ: Ⅱ: Ⅲ :

V1, V2, and V3 are unknowns for which we solve using KCL.
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 2) Assign node voltages to the other nodes 500W 1kW I1 I2 1 2 3 V1 V2 V3 I4 I5 I6 I7 I8 V1, V2, and V3 are unknowns for which we solve using KCL.

2.2 Basic Nodal and Mesh Analysis Nodal Analysis 3) Apply KCL to each node other than the reference-express currents in terms of node voltages. 500W 1kW I1 I2 1 2 3 V1 V2 V3 I4 I5 I6 I7 I8 Node ① : Node ② : Node ③ : KCL Node ① : Node ② : Node ③ : ,

4) Solve the resulting system of linear equations. Node 1: Node 2:
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 4) Solve the resulting system of linear equations. Node 1: Node 2: Node 3: 500W 1kW I1 I2 1 2 3 V1 V2 V3 The left hand side of the equation: The node voltage is multiplied by the sum of conductances of all resistors connected to the node. The neighbourly node voltages are multiplied by the conductance of the resistor(s) connecting to the two nodes and to be subtracted. The right hand side of the equation: The right side of the equation is the sum of currents from sources entering the node.

4) Solve the resulting system of linear equations. Node 1: Node 2:
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 4) Solve the resulting system of linear equations. Node 1: Node 2: Node 3: 500W 1kW I1 I2 1 2 3 V1 V2 V3 Matrix Notation(Symmetric)

4) Solve the resulting system of linear equations.
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 4) Solve the resulting system of linear equations. Node 1: Node 2: Node 3: 500W 1kW I1 I2 1 2 3 V1 V2 V3 G11V1+G12V2 +G13V3 =I11 G21V2+G22V2 +G23V3 =I22 G31V1+G32V2+G33V3=I33

What if there are dependent sources?
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis What if there are dependent sources? 1kW 5mA 100Ib + - Vo 50W Ib 2 1 V2 V1 Example: Node ①： Node ② ： Matrix is not symmetric due to the dependent source.

What if there are voltage sources?
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis What if there are voltage sources? 0.7V R1 Ib V2 V3 R3 V4 + - 1 2 3 4 + 1kW 50W + Vo 100Ib R4 3kW V1 - R2 1kW - Difficulty: We do not know Ib – the current through the voltage source? Equations: KCL at node 2, node 3, node 4, and Unknowns: Ib, V1, V2 (V3),V4 Node 2: Node 3: Independent Voltage Source: Node 4:

What if there are voltage sources?
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis What if there are voltage sources? CURRENT CONTROLLED VOLTAGE SOURCE Io=? KCL AT SUPERNODE

Advantages of Nodal Analysis
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis Advantages of Nodal Analysis Solves directly for node voltages. Current sources are easy. Voltage sources are either very easy or somewhat difficult. Works best for circuits with few nodes. Works for any circuit.

1) Identifying the Meshes
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh(Loop) Analysis 1) Identifying the Meshes 1kW 1kW V1 + - + - Mesh 1 Mesh 2 V2 1kW Mesh: A special kind of loop that doesn’t contain any loops within it.

2) Assigning Mesh Currents
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh(Loop) Analysis 2) Assigning Mesh Currents 1kW + - V2 I1 I2 V1 3) Apply KVL around each loop to get an equation in terms of the loop currents. For Mesh 1: I1 ( 1kW + 1kW) - I2 1kW = V1 - I1 1kW + I2 ( 1kW + 1kW) = -V2 -V1 + I1 1kW + (I1 - I2) 1kW = 0 For Mesh 2: (I2 - I1) 1kW + I2 1kW + V2 = 0

4) Solve the resulting system of linear equations.
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh(Loop) Analysis 3) Apply KVL around each loop to get an equation in terms of the loop currents. 1kW + - V2 I1 I2 V1 I1 ( 1kW + 1kW) - I2 1kW = V1 - I1 1kW + I2 ( 1kW + 1kW) = -V2 4) Solve the resulting system of linear equations.

Mesh (Loop) Analysis Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis I1 I2 I3 I5 I6 I4 Im1 Mesh 1 Mesh 1: Mesh 2: Mesh 3: Im2 Mesh 2 Im3 Mesh 3

Mesh (Loop) Analysis P2.4(P2.2) -12
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis P2.4(P2.2) Mesh 1: Im1 Im2 Mesh 2: -12

What if there are current sources?
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? The current sources in this circuit will have whatever voltage is necessary to make the current correct. We can’t use KVL around the loop because we don’t know the voltage. P2.6 Im1 Super Mesh Im2 Mesh 1 P I = ? + - V Super Mesh: Im1 Im2 Im3 Mesh 2 Mesh 2: Mesh 1:

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? The Supermesh does not include this source! 2kW The Supermesh surrounds this source! 2mA I3 1kW + 2kW I1 12V 4mA I2 - I0

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? 1kW 2kW 12V + - 4mA 2mA I0 I1 I2 I3 The 4mA current source sets I2: I2 = -4mA The 2mA current source sets a constraint on I1 and I3: I1 - I3 = 2mA We have two equations and three unknowns. Where is the third equation? -

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? Mesh 2 Mesh 1 P2.6 Im1 Im2 Mesh 1: Mesh 2: Node 3: + - V Im3 Node 3

Mesh(Loop) Analysis Ch2 Basic Analysis Methods to Circuits
2.2 Useful Circuit Analysis Techniques Mesh(Loop) Analysis Dependent current source. Current sources not shared by meshes. We treat the dependent source as a conventional source. Equations for meshes with current sources We are asked for Vo. We only need to solve for I3 . Replace and rearrange Then KVL on the remaining loop(s) And express the controlling variable, Vx, in terms of loop currents

Advantages of Loop Analysis
Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis Advantages of Loop Analysis Solves directly for some currents. Voltage sources are easy. Current sources are either very easy or somewhat difficult. Works best for circuits with few loops. Disadvantages of Loop Analysis Some currents must be computed from loop currents. Choosing the supermesh may be difficult.

Linearity Superposition Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques Key Words: Linearity Superposition Thevenin’s and Norton’s theorems

Linearity leads to many useful properties of circuits:
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Linearity Linearity is a mathematical property of circuits that makes very powerful analysis techniques possible. Linearity leads to many useful properties of circuits: Superposition: the effect of each source can be considered separately. Equivalent circuits: Any linear network can be represented by an equivalent source and resistance (Thevenin’s and Norton’s theorems)

Linearity leads to simple solutions:
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Linearity Linearity leads to simple solutions: Nodal analysis for linear circuits results in systems of linear equations that can be solved by matrices 500W 1kW I1 I2 1 2 3 V1 V2 V3

2.3 Useful Circuit Analysis Techniques Linearity The relationship between current and voltage for a linear element satisfies two properties: Homogeneity Additivity *Real circuit elements are not linear, but can be approximated as linear

Example: Resistor: V = R I If current is KI, then voltage is R KI = KV
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Linearity Homogeneity: Let v(t) be the voltage across an element with current i(t) flowing through it. In an element satisfying homogeneity, if the current is increased by a factor of K, the voltage increases by a factor of K. Additivity Let v1(t) be the voltage across an element with current i1(t) flowing through it, and let v2(t) be the voltage across an element with current i2 (t) flowing through it In an element satisfying additivity, if the current is the sum of i1 (t) and i2 (t), then the voltage is the sum of v1 (t) and v2 (t). Example: Resistor: V = R I If current is KI, then voltage is R KI = KV If current is I1 + I2, then voltage is R(I1 + I2) = RI1 + RI2 = V1 + V2

Superposition Superposition is a direct consequence of linearity
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition I2 I Superposition is a direct consequence of linearity It states that “in any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”

How to Apply Superposition?
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition How to Apply Superposition? To find the contribution due to an individual independent source, zero out the other independent sources in the circuit. Voltage source  short circuit. Current source  open circuit. Solve the resulting circuit using your favorite techniques. Nodal analysis Loop analysis

Superposition Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques Superposition For the above case: Zero out Vs, we have : Zero out E2, we have : R1 R3 E2 R2 I2’ I + _ Vs R1 R3 E2 R2 I2’’

Superposition P2.7 Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques Superposition 2kW 1kW 12V + - I0 2mA 4mA P2.7

Superposition P2.7 KVL for mesh 2:
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition P2.7 2kW 1kW Io’ 2mA KVL for mesh 2: I1 I2 Mesh 2

2.3 Useful Circuit Analysis Techniques Superposition P2.7 2kW 1kW I’’0 4mA I1 KVL for mesh 2: I2 Mesh 2

2.3 Useful Circuit Analysis Techniques Superposition P2.7 2kW 1kW 12V + - I’’’0 KVL for mesh 2: I2 Mesh 2

I0 = I’0 +I’’0+ I’’’0 = -16/3 mA Superposition P2.7
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition 2kW 1kW 12V + - I0 2mA 4mA 2kW 1kW 12V + - I0 2mA 4mA P2.7 I0 = I’0 +I’’0+ I’’’0 = -16/3 mA

2.3 Useful Circuit Analysis Techniques Thevenin’s theorem Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis

Circuit with independent sources Thevenin equivalent circuit
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Thevenin’s theorem Independent Sources RTh Voc + - Circuit with independent sources Thevenin equivalent circuit

No Independent Sources
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Thevenin’s theorem No Independent Sources RTh Circuit without independent sources Thevenin equivalent circuit

Very similar to Thevenin’s theorem
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Norton’s theorem Very similar to Thevenin’s theorem It simply states that any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single current source and a single resistor

Norton Equivalent: Independent Sources
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Norton’s theorem Norton Equivalent: Independent Sources Isc RTh Circuit with one or more independent sources Norton equivalent circuit

Norton Equivalent: No Independent Sources
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Norton’s theorem Norton Equivalent: No Independent Sources RTh Circuit without independent sources Norton equivalent circuit

The Thevenin’s theorem:
Ch2 Basic Analysis Methods to Circuits Motivation of applying the Thevenin’s theorem and Norton’s theorem: Sometimes, in a complex circuit, we are only interested in working out the voltage /current or power being consumed by a single load (resistor); We can then treat the rest of the circuit (excluding the interested load) as a voltage (current) source concatenated with a source resistor; Simplify our analysis. The Thevenin’s theorem: Given a linear circuit, rearrange it in the form of two networks of A and B connected by two wires. Define Voc as the open-circuit voltage which appears across the terminals of A when B is disconnected. Then all currents and voltage in B will remain unchanged, if we replace all the independent current or voltage source in A by an independent voltage source which is in series with a resistor (RTh). The Norton’s Theorem: Given a linear circuit, rearrange it in the form of two networks of A and B connected by two wires. Define isc as the short-circuit current which appears across the terminals of A when B is disconnected. Then all currents and voltage in B will remain unchanged, if we replace all the independent current or voltage source in A by an independent current source isc which is in parallel with a resistor (RN).

Equivalent transform between a Thevenin equivalent circuit and a Norton equivalent circuit Thevenin equivalent Norton equivalent VL N If and VL VL =

Therefore, for a source transform:
Ch2 Basic Analysis Methods to Circuits Therefore, for a source transform: Thevenin Norton: Norton Thevenin :

Example 45 (P88) Find a Thevenin and Norton equivalent circuit for the following circuit excluding A B

Thevenin equivalent Norton equivalent

Application of Thevenin’s theorem when there are only independent sources. step : 1) Determine of two connection points between network A and network B. 2) Determine of two connection points by replacing the voltage source by a short- circuit or the current source by a open-circuit. Similarly, for Norton’s Theorem steps : 1) Determine between the two connection points between network A and B. 2) Determine by two connection points by replacing the voltage source by a short- circuit or the current source by an open circuit. Test the above case?

Practice 4.6 (P90)

Thevenin equivalent :

When there are multiple independent source, we shall use “superposition” . Example 4.7 (P91) Find the Thevenin and Norton equivalent for the network excluding the resistor. To determine , when only 4v voltage source is functioning.

When only 2mA current source is functioning :
Ch2 Basic Analysis Methods to Circuits When only 2mA current source is functioning : Therefore,

To determine Thevenin equivalent:
Ch2 Basic Analysis Methods to Circuits To determine Thevenin equivalent:

Try to look into this problem from the Norton approach.
Ch2 Basic Analysis Methods to Circuits Norton equivalent Try to look into this problem from the Norton approach. (Figure out the Norton equivalent circuit first)

When there are both independent source and dependent source. —Dependent source cannot be “zero out” as far as its controlling variable is not zero. —Similar as before, —But we cannot determine directly, however, we can use Example 4.8 (P92) — Determine the Thevenin equivalent of the following circuit To determine since ,applying KVL to the supermesh:

Therefore its Thevenin equivalent is:
Ch2 Basic Analysis Methods to Circuits To determine Therefore its Thevenin equivalent is:

When there are only dependent sources:
Ch2 Basic Analysis Methods to Circuits When there are only dependent sources: VOC = 0 RTH can be determined by implying a test (imaginary) voltage across the two terminals. 3 Ω 2Ω i 1.5i Example 4.9 (p93) Open circuit : To determine RTh, imagine an independent current source xA. as : i = -xA, Apply KCL: Giving : 0.6Ω 3 Ω 2Ω i 1.5i v xA - - - - v = 0.6x V RTh = 0.6Ω

Circuits with independent sources
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Thevenin’s theorem Circuits with independent sources Compute the open circuit voltage, this is Voc Compute the Thevenin resistance (set the sources to zero – short circuit the voltage sources, open circuit the current sources), and find the equivalent resistance, this is RTh Circuits with independent and dependent sources: Compute the open circuit voltage Compute the short circuit current The ratio of the two is RTh Circuits with dependent sources only* Voc is simply 0 RTh is found by applying an independent voltage source (V volts) to the terminals and finding voltage/current ratio * Not required by this course.

2.3 Useful Circuit Analysis Techniques Norton’s theorem Circuits with independent sources, w/o dependent sources Compute the short circuit current, this is Isc Compute the Thevenin resistance (set the sources to zero – short circuit the voltage sources, open circuit the current sources), and find the equivalent resistance, this is RN Circuits with both independent and dependent sources Find Voc and Isc Compute RN= Voc/ Isc Circuits w/o independent sources* Apply a test voltage (current) source Find resulting current (voltage) Compute RN

Maximum power transfer
Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Maximum power transfer For every choice of RL we have a different power. How do we find the maximum value? Consider PL as a function of RL and find the maximum of such function The maximum power transfer theorem The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent resistance of the circuit. The value of the maximum power that can be transferred is

KVL, KCL, I — V Combination rules Node method Mesh method
Ch2 Basic Analysis Methods to Circuits Analysis methods Review KVL, KCL, I — V Combination rules Node method Mesh method Superposition Thévenin Norton Any circuits linear circuits

Ch3 Basic RL and RC Circuits
Engineering Circuit Analysis Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits 3.2 First-Order RL Circuits 3.3 Examples References: Hayt-Ch5, 6

Transient Response of RC Circuits, Time constant
Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Key Words: Transient Response of RC Circuits, Time constant

3.1 First-Order RC Circuits Used for filtering signal by blocking certain frequencies and passing others. e.g. low-pass filter Any circuit with a single energy storage element, an arbitrary number of sources and an arbitrary number of resistors is a circuit of order 1. Any voltage or current in such a circuit is the solution to a 1st order differential equation. Ideal Linear Capacitor Energy stored A capacitor is an energy storage device  memory device.

3.1 First-Order RC Circuits R + - C vs(t) vc(t) vr(t) One capacitor and one resistor The source and resistor may be equivalent to a circuit with many resistors and sources.

3.1 First-Order RC Circuits Transient Response of RC Circuits Switch is thrown to 1 KVL around the loop: Initial condition Called time constant

3.1 First-Order RC Circuits Time Constant RC R=2k C=0.1F

3.1 First-Order RC Circuits Transient Response of RC Circuits Switch to 2 Initial condition

3.1 First-Order RC Circuits Time Constant R=2k C=0.1F

3.1 First-Order RC Circuits

3.2 First-Order RL Circuits Key Words: Transient Response of RL Circuits, Time constant

3.2 First-Order RL Circuits Ideal Linear Inductor i(t) + - v(t) The rest of the circuit L Energy stored: One inductor and one resistor The source and resistor may be equivalent to a circuit with many resistors and sources.

3.2 First-Order RL Circuits Transient Response of RL Circuits Switch to 1 KVL around the loop: Initial condition Called time constant

3.2 First-Order RL Circuits Time constant t i (t) . Indicate how fast i (t) will drop to zero. It is the amount of time for i (t) to drop to zero if it is dropping at the initial rate

3.2 First-Order RL Circuits Transient Response of RL Circuits Switch to 2 Initial condition

3.2 First-Order RL Circuits Transient Response of RL Circuits Input energy to L L export its energy , dissipated by R

Summary Initial Value （t = 0） Steady Value (t  ) time constant  RL Circuits Source (0 state) Source-free (0 input) RC Circuits

Summary The Time Constant For an RC circuit,  = RC For an RL circuit,  = L/R -1/ is the initial slope of an exponential with an initial value of 1 Also,  is the amount of time necessary for an exponential to decay to 36.7% of its initial value

Summary How to determine initial conditions for a transient circuit. When a sudden change occurs, only two types of quantities will remain the same as before the change. IL(t), inductor current Vc(t), capacitor voltage Find these two types of the values before the change and use them as the initial conditions of the circuit after change.

3.3 Examples About Calculation for The Initial Value iC iL i t=0 + _ 1A  - vL(0+) vC(0+)=4V i(0+) iC(0+) iL(0+)

3.3 Examples (Analyzing an RC circuit or RL circuit) Method 1 1) Thévenin Equivalent.(Draw out C or L) Simplify the circuit Veq , Req 2) Find Leq(Ceq), and  = Leq/Req ( = CeqReq) 3) Substituting Leq(Ceq) and  to the previous solution of differential equation for RC (RL) circuit .

3.3 Examples (Analyzing an RC circuit or RL circuit) Method 2 1) KVL around the loop  the differential equation 2) Find the homogeneous solution. 3) Find the particular solution. 4) The total solution is the sum of the particular and homogeneous solutions.

3.3 Examples (Analyzing an RC circuit or RL circuit) Method 3 (step-by-step) In general, Given f(0+)，thus A = f(0+) – f(∞) Initial Steady 1) Draw the circuit for t = 0- and find v(0-) or i(0-) 2) Use the continuity of the capacitor voltage, or inductor current, draw the circuit for t = 0+ to find v(0+) or i(0+) 3) Find v(), or i() at steady state 4) Find the time constant t For an RC circuit, t = RC For an RL circuit, t = L/R 5) The solution is:

3.3 Examples P3.1 vC (0)= 0, Find vC (t) for t  0. Method 3: Apply Thevenin theorem : s

P3.2 vC (0)= 0, Find vC (t) for t  0.
Ch3 Basic RL and RC Circuits 3.3 Examples P3.2 vC (0)= 0, Find vC (t) for t  0. Apply Thevenin’s theorem : s

Ch4 Sinusoidal Steady State Analysis
Engineering Circuit Analysis Ch4 Sinusoidal Steady State Analysis 4.1 Characteristics of Sinusoidal 4.2 Phasors 4.3 Phasor Relationships for R, L and C 4.4 Impedance 4.5 Parallel and Series Resonance 4.6 Examples for Sinusoidal Circuits Analysis References: Hayt-Ch7

Any steady state voltage or current in a linear circuit with a sinusoidal source is a sinusoid All steady state voltages and currents have the same frequency as the source In order to find a steady state voltage or current, all we need to know is its magnitude and its phase relative to the source (we already know its frequency) We do not have to find this differential equation from the circuit, nor do we have to solve it Instead, we use the concepts of phasors and complex impedances Phasors and complex impedances convert problems involving differential equations into circuit analysis problems  Focus on steady state; 􀂄 Focus on sinusoids.

Period: T , Ch4 Sinusoidal Steady State Analysis
4.1 Characteristics of Sinusoidal Key Words: Period: T , Frequency: f , Radian frequency  Phase angle Amplitude: Vm Im

v、i t t1 t2 Ch4 Sinusoidal Steady State Analysis
4.1 Characteristics of Sinusoidal v、i t t1 t2 Both the polarity and magnitude of voltage are changing.

Period: T — Time necessary to go through one cycle. (s)
Ch4 Sinusoidal Steady State Analysis 4.1 Characteristics of Sinusoidal Period: T — Time necessary to go through one cycle. (s) Frequency: f — Cycles per second. (Hz) f = 1/T Radian frequency(Angular frequency):  = 2f = 2/T (rad/s） Amplitude: Vm Im i = Imsint， v =Vmsint v、i t  2 Vm、Im

Effective Value of a Periodic Waveform
Ch4 Sinusoidal Steady State Analysis 4.1 Characteristics of Sinusoidal Effective Roof Mean Square (RMS) Value of a Periodic Waveform — is equal to the value of the direct current which is flowing through an R-ohm resistor. It delivers the same average power to the resistor as the periodic current does. Effective Value of a Periodic Waveform

Phase (angle) Ch4 Sinusoidal Steady State Analysis
4.1 Characteristics of Sinusoidal Phase (angle) Phase angle 0 ①如果正弦波的起始最小值发生在时间起点之前，则为正值。 ②如果正弦波的起始最小值发生在时间起点之后，则为负值。 <0

— v(t) leads i(t) by (1 - 2), or i(t) lags v(t) by (1 - 2)
Ch4 Sinusoidal Steady State Analysis 4.1 Characteristics of Sinusoidal Phase difference — v(t) leads i(t) by (1 - 2), or i(t) lags v(t) by (1 - 2) — v(t) lags i(t) by (2 - 1), or i(t) leads v(t) by (2 - 1) v、i t v i Out of phase。 t v、i v i v、i t v i In phase.

The sinusoidal waves whose phases are compared must:
Ch4 Sinusoidal Steady State Analysis 4.1 Characteristics of Sinusoidal Review The sinusoidal waves whose phases are compared must: ① Be written as sine waves or cosine waves. ② With positive amplitudes. ③ Have the same frequency. 360°—— does not change anything. 90° —— change between sin & cos. 180°—— change between + & -

Phase difference P4.1, Find Ch4 Sinusoidal Steady State Analysis
4.1 Characteristics of Sinusoidal Phase difference P4.1, Find If

Phase difference P4.2, v、i v i t Ch4 Sinusoidal Steady State Analysis
4.1 Characteristics of Sinusoidal Phase difference P4.2, v、i t v i -/3 /3 •  P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？

Complex Numbers Ch4 Sinusoidal Steady State Analysis 4.2 Phasors
A sinusoidal voltage/current at a given frequency , is characterized by only two parameters :amplitude an phase Key Words: Complex Numbers Rotating Vector Phasors

Angular frequency ω is known in the circuit.
Ch4 Sinusoidal Steady State Analysis 4.2 Phasors E.g. voltage response Time domain Frequency domain Complex form: Phasor form: Angular frequency ω is known in the circuit. A sinusoidal v/i Complex transform Phasor transform By knowing angular frequency ω rads/s.

Rotating Vector i Im i Im  Ch4 Sinusoidal Steady State Analysis
4.2 Phasors Rotating Vector Im  t x y i t  Im i t1 P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？ i(t1) A complex coordinates number: Real value: Imag 194

Rotating Vector y   x Ch4 Sinusoidal Steady State Analysis
4.2 Phasors Rotating Vector Vm x y   P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？

— Rectangular Coordinates
Ch4 Sinusoidal Steady State Analysis 4.2 Phasors Complex Numbers — Rectangular Coordinates  |A| a b real axis imaginary axis — Polar Coordinates conversion： P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？ j——旋转90的算子

Arithmetic With Complex Numbers
Ch4 Sinusoidal Steady State Analysis 4.2 Phasors Complex Numbers Arithmetic With Complex Numbers Addition: A = a + jb, B = c + jd, A + B = (a + c) + j(b + d) Real Axis Imaginary Axis A B A + B P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？

Ch4 Sinusoidal Steady State Analysis 4.2 Phasors Complex Numbers Arithmetic With Complex Numbers Subtraction : A = a + jb, B = c + jd, A - B = (a - c) + j(b - d) Real Axis Imaginary Axis A B A - B P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？

Ch4 Sinusoidal Steady State Analysis 4.2 Phasors Complex Numbers Arithmetic With Complex Numbers Multiplication : A = Am  A, B = Bm  B A  B = (Am  Bm)  (A + B) Division: A = Am  A , B = Bm  B A / B = (Am / Bm)  (A - B) P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？

4.2 Phasors Phasors A phasor is a complex number that represents the magnitude and phase of a sinusoid: Phasor Diagrams A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes). A phasor diagram helps to visualize the relationships between currents and voltages. P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？

A real-valued sinusoid is the real part of a complex exponential.
Ch4 Sinusoidal Steady State Analysis 4.2 Phasors Complex Exponentials A real-valued sinusoid is the real part of a complex exponential. Complex exponentials make solving for AC steady state an algebraic problem. P4.2, v、i波形如图，问，v、i初相各为多少？若将时间起点右移/3，则v、i初相有何改变？改变否？若时间起点右移，则v、i初相有何改变？ 改变否？若将时间起点左移/3 ，则v、i初相有何改变？ 改变否？

I-V Relationship for R, L and C,
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Key Words: I-V Relationship for R, L and C, Power conversion

v~i relationship for a resistor
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Resistor v~i relationship for a resistor Suppose Relationship between RMS: v、i t v i Wave and Phasor diagrams：

Time domain frequency domain
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Time domain frequency domain Resistor With a resistor θ﹦φ, v(t) and i(t) are in phase .

Power Transient Power Note: I and V are RMS values. Average Power v、i
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Resistor Power Transient Power p0 Note: I and V are RMS values. v、i t v i Average Power P=IV

4.3 Phasor Relationships for R, L and C Resistor P4.4 ， , R=10，Find i and P。

v~i relationship Suppose Ch4 Sinusoidal Steady State Analysis
4.3 Phasor Relationships for R, L and C Inductor v~i relationship Suppose

v~i relationship Relationship between RMS: For DC，f = 0，XL = 0.
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Inductor v~i relationship Relationship between RMS: For DC，f = 0，XL = 0. v(t) leads i(t) by 90º, or i(t) lags v(t) by 90º

v ~ i relationship Represent v(t) and i(t) as phasors:
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Inductor v ~ i relationship i(t) = Im ejwt Represent v(t) and i(t) as phasors: The derivative in the relationship between v(t) and i(t) becomes a multiplication by j in the relationship between and The time-domain differential equation has become the algebraic equation in the frequency-domain. Phasors allow us to express current-voltage relationships for inductors and capacitors in a way such as we express the current-voltage relationship for a resistor.

v ~ i relationship Wave and Phasor diagrams： v、i v eL i t
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Inductor v ~ i relationship Wave and Phasor diagrams： v、i t v i eL

Power P t Energy stored: v、i t v i Average Power Reactive Power
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Inductor Power P t Energy stored: v、i t v i + - Average Power Reactive Power （Var）

P4.5，L = 10mH，v = 100sint，Find iL when f = 50Hz and 50kHz.
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Inductor P4.5，L = 10mH，v = 100sint，Find iL when f = 50Hz and 50kHz.

v ~ i relationship Suppose: Relationship between RMS:
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Capacitor v ~ i relationship Suppose: Relationship between RMS: For DC，f = 0， XC   i(t) leads v(t) by 90º, or v(t) lags i(t) by 90º

v ~ i relationship Represent v(t) and i(t) as phasors:
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Capacitor v ~ i relationship v(t) = Vm ejt Represent v(t) and i(t) as phasors: The derivative in the relationship between v(t) and i(t) becomes a multiplication by j in the relationship between and The time-domain differential equation has become the algebraic equation in the frequency-domain. Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor.

v ~ i relationship Wave and Phasor diagrams： v、i i v t
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Capacitor v ~ i relationship Wave and Phasor diagrams： v、i t v i

Power P t Energy stored: v、i t v i Average Power: P=0 Reactive Power
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Capacitor Power P t Energy stored: v、i t v i + - Average Power: P=0 Reactive Power （Var）

4.3 Phasor Relationships for R, L and C Capacitor P4.7，Suppose C=20F，AC source v=100sint，Find XC and I for f = 50Hz, 50kHz。

4.3 Phasor Relationships for R, L and C Review (v-I relationship) Time domain Frequency domain , , v and i are in phase. , v leads i by 90°. , v lags i by 90°. R C L

Frequency characteristics of an Ideal Inductor and Capacitor:
Ch4 Sinusoidal Steady State Analysis 4.3 Phasor Relationships for R, L and C Summary R： L： C： Frequency characteristics of an Ideal Inductor and Capacitor: A capacitor is an open circuit to DC currents; A Inducter is a short circuit to DC currents.

complex currents and voltages.
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Key Words: complex currents and voltages. Impedance Phasor Diagrams

Complex voltage， Complex current， Complex Impedance
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Complex voltage， Complex current， Complex Impedance AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law: Z is called impedance. measured in ohms ()

Impedance is a complex number and is not a phasor (why?).
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Complex Impedance Complex impedance describes the relationship between the voltage across an element (expressed as a phasor) and the current through the element (expressed as a phasor) Impedance is a complex number and is not a phasor (why?). Impedance depends on frequency

Resistor——The impedance is R
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Complex Impedance ZR = R  ＝ 0; or ZR = R  0 Resistor——The impedance is R or Capacitor——The impedance is 1/jwC or Inductor——The impedance is jwL

Complex Impedance Voltage divider: Current divider:
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Complex Impedance Impedance in series/parallel can be combined as resistors. Voltage divider: Current divider:

Complex Impedance P4.8, Ch4 Sinusoidal Steady State Analysis

First compute impedances for resistor and capacitor:
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Complex Impedance Phasors and complex impedance allow us to use Ohm’s law with complex numbers to compute current from voltage and voltage from current 20kW + - 1mF 10V  0 VC w = 377 Find VC P4.9 How do we find VC? First compute impedances for resistor and capacitor: ZR = 20kW = 20kW  0 ZC = 1/j (377 *1mF) = 2.65kW  -90

Complex Impedance 20kW 1mF 10V  0 VC w = 377 Find VC P4.9 20kW  0
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Complex Impedance 20kW + - 1mF 10V  0 VC w = 377 Find VC P4.9 20kW  0 + - 2.65kW  -90 10V  0 VC Now use the voltage divider to find VC:

4.4 Impedance Complex Impedance Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state. All the analysis techniques we have learned for the linear circuits are applicable to compute phasors KCL & KVL node analysis / loop analysis superposition Thevenin equivalents / Norton equivalents source exchange The only difference is that now complex numbers are used.

ik- Transient current of the #k branch
Ch4 Sinusoidal Steady State Analysis 4.4 Impedance Kirchhoff’s Laws KCL and KVL hold as well in phasor domain. KCL: ik- Transient current of the #k branch KVL： vk- Transient voltage of the #k branch

4.4 Impedance Admittance I = YV, Y is called admittance, the reciprocal of impedance, measured in siemens (S) Resistor: The admittance is 1/R Inductor: The admittance is 1/jL Capacitor: The admittance is j  C

4.4 Impedance Phasor Diagrams A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes). A phasor diagram helps to visualize the relationships between currents and voltages. I = 2mA  40, VR = 2V  40 VC = 5.31V  -50, V = 5.67V   2mA  40 – 1mF VC + 1kW VR V Real Axis Imaginary Axis VR VC V

RLC Circuit, Ch4 Sinusoidal Steady State Analysis
4.5 Parallel and Series Resonance Key Words: RLC Circuit, Series Resonance Parallel Resonance

Series RLC Circuit (2nd Order RLC Circuit ) Phasor
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Series RLC Circuit (2nd Order RLC Circuit ) v vR vL vC Phasor 

Series RLC Circuit (2nd Order RLC Circuit ) Phase difference:  
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Series RLC Circuit (2nd Order RLC Circuit ) Z X = XL-XC R  Phase difference:  XL>XC   >0，v leads i by ——Inductance Circuit XL<XC   <0，v lags i by ——Capacitance Circuit XL=XC   =0，v and i in phase——Resistors Circuit

Series RLC Circuit (2nd Order RLC Circuit )
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Series RLC Circuit (2nd Order RLC Circuit ) v vR vL vC

P4.9, R. L. C Series Circuit，R = 30，L = 127mH，C = 40F，Source
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Series RLC Circuit (2nd Order RLC Circuit ) P4.9, R. L. C Series Circuit，R = 30，L = 127mH，C = 40F，Source . Find 1) XL、XC、Z；2) and i; 3) and vR; and vL; and vC; 4) Phasor diagrams; v vR vL vC P4.10，Computing by (complex numbers) Phasors

Series Resonance (2nd Order RLC Circuit ) Resonance frequency and f0 f
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Series Resonance (2nd Order RLC Circuit ) Resonance condition Resonance frequency and ——Series Resonance f0 f X

Zmin；when V=constant, I=Imax=I0。
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Series Resonance (2nd Order RLC Circuit ) Resonance condition: Zmin；when V=constant, I=Imax=I0。 When ,  Quality factor Q,

Series Resonance (2nd Order RLC Circuit )
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Series Resonance (2nd Order RLC Circuit )

Parallel Resonance frequency
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Parallel RLC Circuit   When , In phase with Parallel Resonance Parallel Resonance frequency In generally Zmax Imin:

Parallel RLC Circuit Z  . Quality factor Q,
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Parallel RLC Circuit   Z  . Quality factor Q,

Parallel RLC Circuit P4.10, Find i1、 i2、 i
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Parallel RLC Circuit P4.10, Find i1、 i2、 i v  i i1   i2

？ Parallel RLC Circuit Phasor Diagrams and Complex Numbers
Ch4 Sinusoidal Steady State Analysis 4.5 Parallel and Series Resonance Parallel RLC Circuit Review For sinusoidal circuit， Series ：？ Parallel : Two Simple Methods: Phasor Diagrams and Complex Numbers

Bypass Capacitor Ch4 Sinusoidal Steady State Analysis
4.6 Examples for Sinusoidal Circuits Analysis Key Words: Bypass Capacitor RC Phase Difference Low-Pass and High-Pass Filter

4.6 Examples for Sinusoidal Circuits Analysis Bypass Capacitor P4.11, Let f ＝ 500Hz，Determine VAB before the C is connected . And VAB after parallel C = 30F Before C is connected After C is connected v i

f = 300Hz, R = 100。 If vo - vi= /4，C =？
Ch4 Sinusoidal Steady State Analysis 4.6 Examples for Sinusoidal Circuits Analysis RC Phase Difference P4.12, f = 300Hz, R = 100。 If vo - vi= /4，C =？

RC---- High-Pass Filter
Ch4 Sinusoidal Steady State Analysis 4.6 Examples for Sinusoidal Circuits Analysis Low-Pass and High-Pass Filter RC---- High-Pass Filter P4.13, The voltage sources are vi= sin2100t(V), R＝200， C＝50F， Determine VAC and VDC in output voltage vo. VDC = 240V

4.6 Examples for Sinusoidal Circuits Analysis Low-Pass and High-Pass Filter

4.6 Examples for Sinusoidal Circuits Analysis

4.6 Examples for Sinusoidal Circuits Analysis Low-Pass and High-Pass Filter

4.6 Examples for Sinusoidal Circuits Analysis

in the circuit of the following fig. P4.14, Find
Ch4 Sinusoidal Steady State Analysis 4.6 Examples for Sinusoidal Circuits Analysis Complex Numbers Analysis in the circuit of the following fig. P4.14, Find v1=120sint v2  i3  i1  i2

P4.15, Let . Use Thevenin’s theorem to find
Ch4 Sinusoidal Steady State Analysis 4.6 Examples for Sinusoidal Circuits Analysis Complex Numbers Analysis P4.15, Let Use Thevenin’s theorem to find v v

Engineering Circuit Analysis
CH5 AC circuit power analysis 5.1 Instantaneous Power 5.2 Average Power 5.3 Effectives values of Current ＆ Voltage 5.4 Apparent Power and Power Factor

Ch5 AC circuit power analysis
5.1 Instantaneous Power Given instantaneous power current ＆ voltage, the instantaneous power delivered to any devices is : If it is a resistor If it is a inductor If it is a capacitor

5.1 Instantaneous Power The total amount of power supplied by the source = total amount of power delivered to the circuit elements Known Power supplied by the source Power delivered to R: Power delivered to L: PL(t) = VL(t) ∙ iL(t) R + - V0 - +

5.1 Instantaneous Power Since

5.1 Instantaneous Power How if both ＆ sinusoidal signals? It is known , leading to possible to be the averaged power? twice of Time invariant

) Ch5 AC circuit power analysis 5.1 Instantaneous Power
Example 8.1 Given a voltage source of μ(t) V, a 5 μF capacitor and a 200 Ω resistor being connected in series. Find at t = 1.2ms, the powers being absorbed by the resistor and capacitor respectively. If PS = PC + PR? When t<0 ,Vs=40V When t=0 , Vs=100V , and + ) - ~ Hence In this 1st order RC circuit Hence ,

5.1 Instantaneous Power The average power (over time) for the instantaneous power The average power over time intervals to --If is periodic as , where is the period Validate the previous suggestion !?

5.1 Instantaneous Power Example : the averaged power for It is a periodic signal with a period of Phase different between ＆ , or Validate the previous suggestion !

5.2 Average Power Two observations: -- power absorbed by an ideal resistor, , -- power absorbed by a purely reactive element, Example 8.2 (p.213) Given and Determine P and p(t).

5.2 Average Power P

) Ch5 AC circuit power analysis 5.2 Average Power
Example 8.4 Determine the average power absorbed by each of the passive elements and supplied by each of the sources. + ) - ~ Apply KVL, we can determine the currents are: The current that passes through the resistor is: Hence , Both L & C are reactive elements , and For the source on the left mesh For the source on the right mesh The power of the left supplies 50W power and it is absorbed by the resistor and the source on the right

5.2 Average Power Remark : superposition is applicable for evaluating the averaged power of a non periodic function which can be decomposed into two periodic functions with different periods,

i1 = 2 cos10t – 3 cos20t A Ch5 AC circuit power analysis
5.2 Average Power Example 8.6 (p218) Determine the average power delivered to the 4 Ω resistor by the current i1 = 2 cos10t – 3 cos20t A. The two different frequency components of the current can be considered separately. i1 = 2 cos10t – 3 cos20t A

5.3 Effectives values of Current ＆ Voltage -- represent the actual effect of a periodic function (current / voltage ) in term of delivering / absorbing power -- transforms an AC circuit into an equivalent DC circuit -- square root of a mean square (RMS)

5.3 Effectives values of Current ＆ RMS of the Sinusoidal wave. Given , For a sinusoidal function , its RMS is of its magnitude

5.3 Effectives values of Current ＆ Effective value of multiple-frequency circuit Power can be determined using super position of 50Hz of 100Hz In general, given a circuit with frequencies,

5.3 Effectives values of Current ＆ Example …

5.4 Apparent Power and Power Factor Given One can determine the averaged power delivered to the network is or However, the effective power delivered to the network would be is defined as the apparent power – The maximal value of the average power can be. Note that the unit of the Papp is VA.

5.4 Apparent Power and Power Factor Power factor ( ) In the above case, If passive elements are all resistive, If passive elements are all reactive, If passive elements are neither purely resistive, nor purely reactive , , capacitive load, since lags behind , inductive load, since leads , a leading PF of “—current is leading” a lagging PF of “—current is lagging”

5.4 Apparent Power and Power Factor Example 8.8 (P222) Determine the apparent power supplied by the source, and the power factor of the combined loads. + - + - Note that for the impedance 3+j4Ω,only the resistive component will absorb power delivered by the source and

CH6 Polyphase Circuits 6.1 Polyphase System 6.2 Notations 6.3 Single-phase Three-wire Systems 6.4 Three Phase connection 6.5 The Delta ( ) Connection 6.6 Power measurement

Ch6 Polyphsae Circuits 6.1 Polyphase System
Polyphase system : system with polyphase sources Single source (Vs) Notice the instantaneous voltage maybe zero  The instantaneous power will be zero They all have 120o phase differences  The instantaneous power will never be zero. Poly sources ( )

Ch6 Polyphsae Circuits 6.1 Polyphase System
The incident with a zero voltage has been exempted. The source power can be delivered more stably. The polyphase systems can provide multiple output voltage levels. Polyphase systems in practice certain sources which maybe approximated by ideal voltage sources, or ideal voltage sources in series with small internal impedances.

Ch6 Polyphsae Circuits 6.1 Polyphase System a b c d e f g h i j k l
For note c : For note f : For note j :

The voltage of a point with respect to b point
Ch6 Polyphsae Circuits 6.2 Notations The voltage of a point with respect to b point a +; b -; Similarly, Iab denotes the current from point a to b. Test with graphical analysis ? (Using the phasor diagram)

6.3 Single-phase Three-wire Systems
Ch6 Polyphsae Circuits 6.3 Single-phase Three-wire Systems Function: allowing household electronics operating at two levels of voltages to be applied. Voltage characteristics 1-phase 3-wire Source Household electronics may either operate with or with Phase characteristics

Ch6 Polyphsae Circuits 6.3 Single-phase Three-wire Systems Current characteristics This is no current in the neutral wire. How if the two are NOT equal, and all the wires have impedances ? This is a more practical scenario.

Ch6 Polyphsae Circuits 6.3 Single-phase Three-wire Systems Example 9.1 (P242) Determine the power delivered to the and the Loads. ② Determine the power lost in the three lines represented by respectively. rms and ③ Determine the transmission efficiency? rms total power absorbed by the loads η = total power generated by the sources Hints: observe a structure with regular meshes and know impedances, we can determine the currents I1, I2 and I3 in order to find out the power being lost and delivered!

Ch6 Polyphsae Circuits 6.3 Single-phase Three-wire Systems Apply KVL for the three meshes. Rearranging them in a matrix form as

Ch6 Polyphsae Circuits 6.3 Single-phase Three-wire Systems If can be calculated: Hence, the average power delivered to each of the loads are: Total loaded power

Ch6 Polyphsae Circuits 6.3 Single-phase Three-wire Systems Power lost in three wires are: Total lost power Transmission efficiencyη Total power generated by the two voltage sources is: Transmission efficiency

6.4 Three Phase connection
Ch6 Polyphsae Circuits 6.4 Three Phase connection Voltage characteristics Balanced three-phase sources (phasor voltages)

Ch6 Polyphsae Circuits 6.4 Three Phase connection Positive phase sequence (abc) (clockwise rotation) Negative phase sequence (cba) (Anti-clockwise rotation)

Ch6 Polyphsae Circuits 6.4 Three Phase connection Line-to-line voltages (take the abc sequence as an example) Hence verifies KVL.

Ch6 Polyphsae Circuits 6.4 Three Phase connection Voltage types magnitude Phasor difference Phase voltages ( ) Line-to-line voltages ( )

Ch6 Polyphsae Circuits 6.4 Three Phase connection Current characteristics

Ch6 Polyphsae Circuits 6.4 Three Phase connection Consider three impedances are connected between each line and the neutral line. Hence When balanced impedances are applied to each of the three lines and the neutral line carries no current.

Ch6 Polyphsae Circuits 6.4 Three Phase connection Example 9.2 (P247) Phase voltages: line-to-line voltage: Line currents: Power absorbed by the three loads

Ch6 Polyphsae Circuits 6.4 Three Phase connection Example 9.2 (P247) How about the instantaneous power? Note: Van = 200V rms Similarly , the instantaneous total power absorbed by the loads are : The total instantaneous power is NEVER ZERO.

Ch6 Polyphsae Circuits 6.4 Three Phase connection Example 9.3 (P249) A balanced three-phase system with a line voltage of 300V is supplying a balanced Y-connected load with 1200W at a leading power factor (PF) of 0.8. Determine line cuurent IL and per-phase load impedance Zp. The phase voltage is: Vp = 300/ V. The per-phase power is: 1200W/3 = 400W. Therefore , and IL = 2.89A The phase impedance is: A leading PF of 0.8 implies the current leads the voltage, and the impedance angle is: -argcos(0.8) = -36.9o and Zp = o Ω Note: the apparent power of a Y-Y connected load is P = Van × IAN (phase voltage × line current)

6.5 The Delta ( ) Connection
Ch6 Polyphsae Circuits 6.5 The Delta ( ) Connection The neural line dose not exist. Balanced impedances are connected between each pair of lines.

﹠ Ch6 Polyphsae Circuits 6.5 The Delta ( ) Connection
Voltage characteristics Phase voltages Line voltages ﹠ Current characteristics Phase currents Line currents

Ch6 Polyphsae Circuits 6.5 The Delta ( ) Connection connections connections Phase voltages Line voltages Phase currents Line currents

Ch6 Polyphsae Circuits 6.5 The Delta ( ) Connection Example 9.5 (p251) Determine the amplitude of line current in a three-phase system with a line voltage of 300V that supplies 1200W to a Δ-connected load at a lagging PF of 0.8. The per-phase average power is: 1200W/3 = 400W Therefore, 400W = VL ∙ IP ∙ 0.8 = 300V ∙ IP ∙ 0.8, and IP = 1.667A The line current is: IL = IP = A = 2.89A Moreover, a lagging PF implies the voltage leads the current by argcos(0.8) = 36.9o The impedance is: Note: the apparent power of a Δ connected load is P = Vab × IAB (line voltage × phase current)

Ch6 Polyphsae Circuits 6.6 Power measurement Wattmeter measured by
current coil measured by potential coil current coil Passive Network potential coil E.g.

Ch6 Polyphsae Circuits 6.6 Power measurement 1 2

Ch6 Polyphsae Circuits 6.6 Power measurement reactive (PF=0)
capacitive / inductive (0<PF<1) resistive (PF=1)

. Ch6 Polyphsae Circuits 6.6 Power measurement Example 9.7 (p256) 1
with positive phase sequence. (1) Find the reading of each wattmeter. (2) The total power absorbed by the loads. With positive phase sequence , we know : Wattmeter 1 reads and :

. Ch6 Polyphsae Circuits 6.6 Power measurement Example 9.7 (p256) 1
Wattmeter 1 reads : Wattmeter 2 reads and : Hence ,

CH7 Magnetically Coupled Circuits 7.1 Self-inductance and mutual inductance 7.2 Dot convention 7.3 Transformer

Ch7 Magnetically Coupled Circuits
Physical phenomenon: By having two closely located coils, the magnetic flux created by one coil will affect the other – leading to the mutual inductance. Voltage across a coil is introduced by the self inductance and mutual inductance. Application: The transformer – coverts the ac voltage to a higher or lower value required by the load.

7.1 Self-inductance and mutual inductance Coupled Circuits and v ~ i relationship Magnetic flux: 1 ＝ f(i1) (1 = N11) The flux is proportional to the current in linear inductor: 1(t) = L1i1(t) L is a lumped element abstraction for the coil.  i1 v1  + -

7.1 Self-inductance and mutual inductance Coupled Circuits and v ~ i relationship  v1 + -  i1 v2  i2 ——Ideal Coupled Circuits’ v ~ i relationship L1、L2、M represent Ideal Coupled Inductor, Self inductance vol. Mutual inductance vol.

7.1 Self-inductance and mutual inductance Coupled Circuits and v ~ i relationship  v1 + -  i1  i2  v1 + -  i1 v2  i2 v2

 i1  i2 v1 v2 • Ch7 Magnetically Coupled Circuits 7.2 Dot Convention  v1 + -  i1 v2  i2  v1 + -  i1 v2  i2  i1  i2 v1 v2 • A current entering the dotted terminal of one coil produces an open circuit voltage with a positive voltage reference at the dotted terminal of the second coil.

7.2 Dot Convention Example Apply dot convention , M creates a negative potential at the dot position of the primary mesh Apply dot convention , M creates a negative potential at the dot position of the secondary mesh

7.2 Dot Convention Question：The terminal is dotted，how can we get v ~ i equations to coupled inductor？ u2  i1  i2 v1 • v2  i1  i2 v1 v2 • Suppose direction of the i and is consistent with Dot convention! Steps to determine the coupled circuit voltage : 2.For mutual-inductance voltage + - 1.For self-inductance voltage

7.2 Dot Convention P4.16，For the circuit shown in following figures, determine v1and v2. u2  i2 v1 • v2 i1 L1 L2 M ＋－

7.2 Dot Convention Coupled Circuits and v ~ i relationship  v1 + -  i1 v2  i2 ——Ideal Coupled Circuits’s v ~ i relationship For sinusoidal circuit,

7.2 Dot Convention + - Example (Practice 10.2, p267) Primary mesh : Secondary mesh :

7.2 Dot Convention Example10.2 (KVLS for the three meshes) ~ ( Mesh 1 : M= 2H Mesh 2 : Mesh 3 :

7.3 Transformer Equivalent networks equivalent network In the equivalent network, mutual inductance no longer exists. And the dot convention has been removed , and are also treated as self-inductance.

7.3 Transformer A B C D Example10.4 (P274) One the equivalence : Let Let Apply the equivalent network: Apply the original transformer:

7.3 Transformer equivalent network (※)

7.3 Transformer -- ideal transformer Turn ratio Number of turns of wire forming the coil

7.3 Transformer KVL for both primary and secondary meshes V1 = jwL1I1 – jwMI2 0 = -jwMI1 + (ZL + jwL2)I2 With

7.3 Transformer Impedance Matching Condition of maximizing the power that is being transferred One can apply the ideal transformer to have the above matching to be achieved. if if Current and voltage adjustment

7.3 Transformer Thevenin equivalent for mesh 1 : Example 10.6 (P281) Determine Thevenin equivalent for mesh 2 : Knowing , the equivalent impedance of the resistor in the primary mesh is : Hence ,

CH8 Fourier Circuit Analysis 8.1 Fourier Series 8.2 Use of Symmetry

Ch8 Fourier Circuit Analysis
8.1 Fourier Series Most of the functions of a circuit are periodic functions They can be decomposed into infinite number of sine and cosine functions that are harmonically related. A complete responds of a forcing function = Partial response to each harmonics.

8.1 Fourier Series Harmonies: Give a cosine function : fundamental frequency ( is the fundamental wave form) Harmonics have frequencies Freq. of the 1st harmonics (=fund. freq) Freq. of the 3rd harmonics Freq. of the 4th harmonics Freq. of the 2nd harmonics Amplitude of the nth harmonics (amplitude of the fundamental wave form) Freq. of the nth harmonics

8.1 Fourier Series Example Fundamental: v1 = 2cosw0t v3a = cos3w0t v3b = 1.5cos3w0t v3c = sin3w0t

8.1 Fourier Series - Fourier series of a periodic function Given a periodic function can be represented by the infinite series as

8.1 Fourier Series Example 12.1 Given a periodic function It is known It can be seen , we can evaluate

8.1 Fourier Series Review of some trigonometry integral observations (a) (c) (d) (b) (e)

8.1 Fourier Series Evaluations of Based on (a) (b) ( is also called the DC component of )

8.1 Fourier Series Based on (b) Based on (c) Based on (e) When k=n

8.1 Fourier Series Based on (a) Based on (c) Based on (d) When k=n

8.1 Fourier Series Harmonic amplitude Phase spectrum - -

8.2 Use of Symmetry - Depending on the symmetry (odd or even), the Fourier series can be further simplified. Even Symmetry Observation: rotate the function curve along axis, the curve will overlap with the curve on the other half of Example : Odd Symmetry Observation: rotate the function curve along the axis, then along the axis, the curve will overlap with the curve on the other half .

8.2 Use of Symmetry Symmetry Algebra odd func. =odd func. × even func. Example: (b) even func. =odd func. × odd func. (c) even func. =even func. × even func.

8.2 Use of Symmetry (d) even func. =const. +∑ even func. (No odd func.) Example: (e) odd func. =∑odd func. Note that odd func. + even func.  No symmetry. odd func. odd func.

8.2 Use of Symmetry Apply the symmetry algebra to analyze the Fourier series. If is an even function If is an odd function

8.2 Use of Symmetry Half-wave symmetry f(t) = -f(t ) or f(t) = -f(t )

8.2 Use of Symmetry Fourier series:

Engineering Circuit Analysis Prof. Li Chen, Imperial College of London

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