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Preface Introduction Place of Electrical Circuits in Modern Technology
The design of the circuits has 2 main objectives: 1) To gather, store, process, transport, and present information. 2) To distribute and convert energy between various forms. The study of circuits provides a foundation for areas of electrical engineering such as: Communication system Computer system Control system Electronics Electromagnetic Power systems Signal processing SJTU

Motivation for doing this course
About the course Circuit Theory Circuit Analysis Circuit Synthesis Circuits (given) Excitation Response (unknown) Circuit Analysis What we emphasize on, Since it provides the foundation for understanding the interaction of signal solution. Circuits (unknown) Excitation (given) Response Circuit synthesis (design) In contrast to analysis, a design problem may have no solution or several solutions, SJTU

Fundamental Knowledge
Chapter 1 Fundamental Knowledge SJTU

Circuit and circuit model
Actual electrical component: a battery or a light bulb Ideal circuit component: a mathematical model of an actual electric component. Actual electrical component Ideal circuit component Emphasize the main character Neglect the left character SJTU

Circuit model: A commonly used mathematical model for electric system.
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Time invariant----Time variant Passive----Active
Lumped elements Lumped circuit Actual scale of the circuit is much smaller than the wavelength relating to the running frequency of the circuit. i1=i2 V is certain Circuit Type: Linear----Nonlinear Time invariant----Time variant Passive----Active Lumped----Distributive SJTU

Circuit Variables n Electric current is the time rate of change of charge, measured in amperes (A). A direct current (DC) is a current that remains constant with time. (I) Sort An alternating current (AC) is a current that varies sinusoidally with time. SJTU

i >0 means the real direction is same to the reference direction
i <0 means the real direction is opposite to the reference direction SJTU

Circuit Variables Voltage (or potential difference) is the energy required to move a unit charge through an element, measured in volts(V). Reference direction or voltage polarity V>0 means the real polarity is same to the reference polarity V<0 means the real polarity is opposite to the reference polarity SJTU

passive sign convention
Passive sign convention is satisfied when current enters through the positive polarity of the voltage. Unless otherwise stated, we will follow the passive sign convention throughout this course. SJTU

Circuit Variables Power is the time rate of expending or absorbing energy. Measured in watts(W) using passive sign convention P=VI in a DC circuit The algebraic sum of power in a circuit, at any instant of time, must be zero. Power absorbed = - Power supplied SJTU

Reference polarities for power using passive sign convention
P > absorbing power P < releasing or supplying power Examples Law of conservation of energy must be obeyed in any electric circuit. Power absorbed = - Power supplied SJTU

Energy is the capacity to do work, measured in joules(J)
The energy absorbed or supplied by an element from time t0 to time t is SJTU

Circuit Elements Passive elements: resistors, capacitors, and inductors Active elements: source, operational amplifiers SJTU

Voltage and Current Sources
The most important active elements are voltage or current sources that generally deliver power to the circuit connected to them. There are two kinds of sources: independent and dependent sources. An ideal independent source is an active element that provides a specified voltage or current that is completely independent of other circuit variables. Symbols for independent voltage source Symbols for independent voltage source SJTU

Note: 2 or more voltage sources with different value are not permissible to be connected in parallel 2 or more current sources with different value are not permissible to be connected in series Voltage sources connected in series is equivalent to one voltage source Current sources connected in parallel is equivalent to one current source A voltage source connected to any branch in parallel is equivalent to itself A current source connected to any branch in series is equivalent to itself permited SJTU

Symbols for a) dependent voltage sources b) dependent current sources
An ideal dependent (or controlled) source is an active element in which the source quantity is controlled by another voltage or current. Symbols for a) dependent voltage sources b) dependent current sources There are a total of four variations, namely: 1. A voltage –controlled voltage source (VCVS) 2. A current –controlled voltage source (CCVS) 3. A voltage –controlled current source (VCCS) 4. A current –controlled current source (CCCS) A diamond is used to represent a dependent source. Both the dependent current source and the dependent voltage source may be controlled by either a voltage or a current elsewhere in the circuit. So SJTU

What is the difference between independent and dependent sources?
 VCVS I1 CCVS g VCCS  CCCS What is the difference between independent and dependent sources? SJTU

Resistance is the capacity of materials to impede the flow of current.
Resistors The circuit element used to model the current –resisting behavior of a material is the resistor. Resistance is the capacity of materials to impede the flow of current. The resistance R of an element denotes its ability to resist the flow of electric current; it is measured in ohms (Ω) Symbol: SJTU

Nonlinear Time Invariant Nonlinear Time Variant Short Circuit
Open Circuit i u t1 t2 Nonlinear Time Invariant Nonlinear Time Variant Short Circuit SJTU

V=Ri (passive sign convention) -------Ohm’s Law
Linear Resistor: The resistance of the idea resistor is constant and its value does not vary over time. The relation between voltage and current.(VAR) v i V=Ri (passive sign convention) Ohm’s Law Since the value of R can range from zero to infinity, it is important that we consider the two extreme possible value of R: R= is called a short circuit; V=0; R=∞------is called an open circuit, I=0; SJTU

About nonlinear resistor
Conductance G is the reciprocal of the resistance, measured in siemens (s) Power : P=vi (passive sign convention) always absorbs power from the circuit Other methods of expressing : About nonlinear resistor SJTU

Chapter 2 Basic laws SJTU

Some Basic Concepts Branch----- A branch represents a single element such as a voltage source or a resistor. {But usually we think a branch as a path flowing the same current. So maybe includes more than one element.} SJTU

Loop-------A loop is any closed path in a circuit.
Some Basic Concepts Node A node is the point of connection between two or more branches Loop A loop is any closed path in a circuit. Mesh------A mesh is a loop which does not contain any other loops within it. SJTU

KIRCHHOFF’S LAWS Kirchhoff’s Current Law (KCL):
The algebraic sum of currents entering a node (or a closed boundary) is zero. (Based on the law of conservation of charge) The sum of the currents entering a node = the sum of the currents leaving the node. SJTU

KCL also applies to a closed boundary.
Note: 1) KCL is available to every node at anytime. 2) KCL is related only to the currents instead of the elements. 3) Pay attention to the current direction. SJTU

KIRCHHOFF’S LAWS Kirchhoff’s Voltage Law (KVL):
The algebraic sum of all voltages around a closed path (or loop) is zero. (Based on the principle of conservation of energy) To illustrate KVL, consider the circuit : V1-V2-V3+V4-V5=0 or V2+V3+V5=V1+V4 Sum of voltage drops=Sum of voltage rises SJTU

TWO SORTS OF CONSTRAINTS
1. Topological constraints Determined by the way of connection among the elements. (Such as KCL KVL) 2. Element constraints Determined by the elements (VAR) Using two sorts of constraints, we can analysis any lumped circuit (solve out all the voltages and currents). SJTU

That is called 2b analysis.
In a circuit with b branches and n nodes, there are 2b variables should be valued. Then: KCL for n nodes: only n-1 equations are independent. KVL for loops: only b-n+1 equations are independent. (only KVL for meshes) VAR for branches: b equations. So, (n-1)+(b-n+1)+b=2b, 2b equations to value 2b variables. That is called 2b analysis. SJTU

SERIES RESISTORS AND VOLTAGE DIVISION
As we know, Series-connected means that the same current flows in them. The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances. See illustration with 2 resistors: R is the equivalent resistance. It can be applied to any number of resistors SJTU

The equivalent power of any number of resistors connected in series is the sum of the individual powers. Concept of Equivalent: Be equivalent to the outside, not the inside. Principle of voltage division: if a voltage divider has N resistors(R1,R2,…RN) in series with the source voltage v, the nth resistor(Rn)will have a voltage drop of SJTU

PARALLEL RESISTORS AND CURRENT DIVISION
As we know, Parallel-connected means that the same voltage covers over them. The equivalent conductance of resistors connected in parallel is the sum of their individual conductances. See illustration with 2 resistors: G G G is the equivalent conductance. It can be applied to any number of resistors SJTU

Principle of current division
The equivalent power of any number of resistors connected in parallel is the sum of the individual powers. Principle of current division If a current divider has N conductors (G1,G2…GN) in parallel with the source current i, the nth conductor (Gn) will have current SJTU

MIXED CONNECTION AND ITS EQUIVALENT RESISTANCE
Examples (we combine resistors in series and in parallel) SJTU

WYE-DELTA TRANSFORMATIONS
Situations often arise in circuit analysis when the resistors are neither in parallel nor in series. e d R12=? SJTU

WYE-DELTA TRANSFORMATIONS
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Can you imagine another way of transformation?
So e d 3 1 2 Can you imagine another way of transformation? SJTU

Chapter 3 Methods of Analysis SJTU

So far, we have analyzed relatively simple circuits by applying Kirchhoff’s laws in combination with Ohm’s law. We can use this approach for all circuits, but as they become structurally more complicated and involve more and more elements, this direct method soon becomes cumbersome. In this chapter we introduce two powerful techniques of circuit analysis: Nodal Analysis and Mesh Analysis. These techniques give us two systematic methods of describing circuits with the minimum number of simultaneous equations. With them we can analyze almost any circuit by to obtain the required values of current or voltage. SJTU

Steps to Determine Node Voltages:
Nodal Analysis Steps to Determine Node Voltages: Select a node as the reference node(ground), define the node voltages 1, 2,… n-1 to the remaining n-1nodes . The voltages are referenced with respect to the reference node. Apply KCL to each of the n-1 independent nodes. Use Ohm’s law to express the branch currents in terms of node voltages. Solve the resulting simultaneous equations to obtain the unknown node voltages. SJTU

Fig. 3.2 Typical circuit for nodal analysis
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So at node 1 and node 2, we can get the following equations.
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In terms of the conductance, equations become
Can also be cast in matrix form as Some examples SJTU

Fig. 3.5 For Example 3.2: (a) original circuit, (b) circuit for analysis
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Nodal Analysis with Voltage Sources(1)
Case 1 If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. As in the figure right: SJTU

Case 2 If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a supernode; we apply both KVL and KCL to determine the node voltages. As in the figure right: A supernode may be regarded as a closed surface enclosing the voltage source and its two nodes. SJTU

Case 3 If a voltage source (dependent or independent) is connected with a resistor in series, we treat them as one branch. As in the figure right: i V11 V22 SJTU

Example P113 SJTU

Steps to Determine Mesh Currents:
Mesh Analysis Steps to Determine Mesh Currents: Assign mesh currents i1, i2,…in to the n meshes. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents. Solve the resulting n simultaneous equations to get the mesh currents. SJTU

Fig. 3.17 A circuit with two meshes
When setting mesh currents,it’s conventional to assume that each mesh current flows clockwise. SJTU

In matrix form: Apply KVL to each mesh. SJTU

Fig For Example 3.5 SJTU

Mesh Analysis with Current Sources(1)
Case 1 When a current source exists only in one mesh: Consider the figure right. SJTU

Mesh Analysis with Current Sources(2)
Case 2 When a current source exists between two meshes:Consider the figure right. 2 solutions: Set v as the voltage across the current source, then add a constraint equation. Use supermesh to solve the problem. SJTU

2, A supermesh has no current of its own.
Solution 1: supermesh Solution 2: Note: 1, The current source in the supermesh is not completely ignored; it provides the constraint equation necessary to solve for the mesh current. 2, A supermesh has no current of its own. 3, A supermesh requires the application of both KVL and KCL. SJTU

Fig For Example 3.7 supermesh SJTU

Fig. 3.32 For Example 3.10; the schematic of the circuit in Fig. 3.31.
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Nodal Versus Mesh Analysis
Both provide a systematic way of analyzing a complex network. When is the nodal method preferred to the mesh method? A circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. Based on the information required. Node voltages required nodal analysis Branch or mesh currents required mesh analysis Examples: E page130 SJTU

Chapter 4 Circuit Theorems SJTU

Linearity Property Linearity is the property of an element describing a linear relationship between cause and effect. A linear circuit is one whose output is linearly ( or directly proportional) to its input. SJTU

Superposition(1) The superposition principle states that voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. SJTU

Steps to Apply Superposition Principle:
Turn off all independent source except one source. Find the output(voltage or current) due to that active source using nodal or mesh analysis. Repeat step 1 for each of the other independent sources. Find the total contribution by adding algebraically all the contributions due to the independent sources. SJTU

Source Transformation(1)
A source transformation is the process of replacing a voltage source Vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa. Vs=isR or is=Vs/R SJTU

Source Transformation(2)
It also applies to dependent sources: SJTU

Fig. 4.17 for Example, find out Vo
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So, we get vo=3.2V SJTU

Example: find out I (use source transformation )
7 2A 6V I SJTU

Substitution Theorem I1=2A, I2=1A, I3=1A, V3=8V
       I1=2A, I2=1A, I3=1A, V3=8V I1=2A, I2=1A, I3=1A, V3=8V I1=2A, I2=1A, I3=1A, V3=8V SJTU

Substitution Theorem If the voltage across and current through any branch of a dc bilateral network are known, this branch can be replaced by any combination of elements that will maintain the same voltage across and current through the chosen branch. SJTU

Substitution Theorem OR SJTU

Thevenin’s Theorem A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth, where Vth is the open-circuit voltage at the terminals and Rth is the input or equivalent resistance at the terminals when the independent source are turned off. SJTU

(a) original circuit, (b) the Thevenin equivalent circuit
d SJTU

Simple Proof by figures
+ V=Voc-RoI SJTU

Consider 2 cases in finding Rth:
Thevenin’s Theorem Consider 2 cases in finding Rth: Case 1 If the network has no dependent sources, just turn off all independent sources, calculate the equivalent resistance of those resistors left. Case 2 If the network has dependent sources, there are two methods to get Rth: SJTU

Thevenin’s Theorem Case 2 If the network has dependent sources, there are two methods to get Rth: Turn off all the independent sources, apply a voltage source v0 (or current source i0) at terminals a and b and determine the resulting current i0 (or resulting voltage v0), then Rth= v0/ i0 SJTU

Thevenin’s Theorem Case 2 If the network has dependent sources, there are two methods to get Rth: 2. Calculate the open-circuit voltage Voc and short-circuit current Isc at the terminal of the original circuit, then Rth=Voc/Isc Rth=Voc/Isc SJTU

Examples SJTU

Norton’s Theorem A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN, where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off. SJTU

(a) Original circuit, (b) Norton equivalent circuit
d (c) N SJTU

Examples SJTU

Maximum Power Transfer
Replacing the original network by its Thevenin equivalent, then the power delivered to the load is RL a b SJTU

We can confirm that is the maximum power by showing that
Power delivered to the load as a function of RL We can confirm that is the maximum power by showing that SJTU

Maximum Power Transfer (several questions)
If the load RL is invariable, and RTh is variable, then what should RTh be to make RL get maximum power? If using Norton equivalent to replace the original circuit, under what condition does the maximum transfer occur? Is it true that the efficiency of the power transfer is always 50% when the maximum power transfer occurs? SJTU

Examples SJTU

Tellegen Theorem If there are b branches in a lumped circuit, and the voltage uk, current ik of each branch apply passive sign convention, then we have . SJTU

Inference of Tellegen Theorem
If two lumped circuits and have the same topological graph with b branches, and the voltage, current of each branch apply passive sign convention, then we have not only SJTU

Example SJTU

Reciprocity Theorem 2  3  6  3  6  2  SJTU

Reciprocity Theorem (only applicable to reciprocity networks)
Case 1 The current in any branch of a network, due to a single voltage source E anywhere else in the network, will equal the current through the branch in which the source was originally located if the source is placed in the branch in which the current I was originally measured. SJTU

Case 2 SJTU

Case 3 SJTU

example SJTU

Voltage source transfer
An isolate voltage source can then be transferred to a voltage source in series with a resistor. SJTU

Current source transfer
Examples SJTU

Maximum Power Transfer Tellegen Theorem Inference of Tellegen Theorem
Summary Maximum Power Transfer Tellegen Theorem Inference of Tellegen Theorem Reciprocity Theorem Source Transfer Linearity Property Superposition Source Transformation Substitution Theorem Thevenin’s Theorem Norton’s Theorem SJTU

Operational Amplifier
Chapter 5 Operational Amplifier

A typical op amp: a) pin configuration, b) circuit symbol
Introduction An op amp is an active circuit element designed to perform mathematical operation of addition, subtraction, multiplication, division, differentiation, and integration. A typical op amp: a) pin configuration, b) circuit symbol

Terminal Voltage and Currents
Vd=V2-V1 (differencial input) Vo=AVd=A(V2-V1) A: gain (open-loop voltage gain)

Actual op amp: Ri(very large); Ro(very small); A(very large)
Idealization: Ri=> ; Ro=>0 ; A=>  As Vo is limited, we can get from Vo=A(V2-V1) that: V2=V1 As Ri=> , we can get i1=0, i2=0

Ideal op amp Ideal op amp model i1 i2 i1=0 , i2=0 ; V1=V2

Applications Inverting Amplifier

Noninverting amplifier
Applications Noninverting amplifier If Rf=0; R1=, then=>

Applications Vi Vo + - First stage Second stage The voltage follower

Applications Summing amplifier

Difference Amplifier 2 1 3 4 a b Note: be careful while using nodal analysis, do not set nodal equation at the output terminal.

Example: Find out Vo

P.P.5.10 As a voltage follower, va = v1 = 2V
va = v1 = 2V where va is the voltage at the right end of the 20 k resistor. As an inverter, vb = Where vb is the voltage at the right end of the 50k resistor. As a summer v0 = = [6 - 15] = 9V

Capacitors and Inductors
Chapter 6 Capacitors and Inductors

Capacitors A capacitor consists of two conducting plates separated by an insulator (or dielectric). C: capacitance Unit: farad(F) A typical capacitor

Values: picofarad (pF) -- microfarad (mF)
Polyester Mica Electrolytic Ceramic Types: fixed variable Circuit symbols for capacitors a) fixed capacitor b) variable capacitor

VAR of Capacitor A capacitor is an open circuit to dc
Passive sign convention A capacitor is an open circuit to dc The voltage on the capacitor cannot change abruptly (in condition that i is limited)

Stored Energy in Capacitor
Is capacitor active? equivalent circuit for the parallel capacitors

Equivalent circuit for the series capacitors

Inductors An inductor consists of a coil of conducting wire.
L: inductance Unit: henry (H) Typical form of an inductor

Values: microhenrys(H)--- tens of henrys(H)
Fixed variable Iron, steel Plastic, air Types: Circuit symbols for inductors

VAR of Inductor The current through an inductor cannot change abruptly
What happens when the current is constant? An inductor acts like a short circuit to dc. Integrating form: The current through an inductor cannot change abruptly (in condition that v is limited).

Stored Energy in Inductor
Is inductor active? Equivalent circuit for the series inductors

Equivalent circuit for the parallel inductors

Chapter 7 First-Order Circuit SJTU

First-order Circuit Complete Response Initial and Final Conditions
Items: RC and RL Circuits First-order Circuit Complete Response Initial and Final Conditions First-order Circuit Sinusoidal Response SJTU

1. RC and RL Circuits Two major steps in the analysis of a dynamic circuit use device and connection equations to formulate a differential equation. solve the differential equation to find the circuit response. SJTU

FORMULATING RC AND RL CIRCUIT EQUATIONS
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Eq.(7-1) RC Eq.(7-2) Eq.(7-3) Eq.(7-4) . Eq.(7-5) RL Eq.(7-6) SJTU

RC Circuit: ZERO-INPUT RESPONSE OF FIRST-ORDER CIRCUITS
makes VT=0 in Eq.(7-3) we find the zero-input response Eq.(7-7) Eq.(7-7) is a homogeneous equation because the right side is zero. A solution in the form of an exponential Eq.(7-8) where K and s are constants to be determined SJTU

characteristic equation
Substituting the trial solution into Eq.(7-7) yields OR Eq.(7-9) characteristic equation a single root of the characteristic equation zero -input response of the RC circuit: SJTU

Fig. 7-3: First-order RC circuit zero-input response
Eq.(7-10) time constant TC=RTC Fig. 7-3: First-order RC circuit zero-input response SJTU

Graphical determination of the time constant T from the response curve
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RL Circuit: The root of this equation
The final form of the zero-input response of the RL circuit is Eq.(7-13) SJTU

EXAMPLE 7-1 The switch in Figure 7- 4 is closed at t=0, connecting a capacitor with an initial voltage of 30V to the resistances shown. Find the responses vC(t), i(t), i1(t) and i2(t) for t 0. Fig. 7-4 SJTU

SOLUTION: SJTU

EXAMPLE 7-2 Find the response of the state variable of the RL circuit in Figure 7-5 using L1=10mH, L2=30mH, R1=2k ohm, R2=6k ohm, and iL(0)=100mA Fig. 7-5 SJTU

SOLUTION: SJTU

2. First-order Circuit Complete Response
When the input to the RC circuit is a step function** Eq.(7-15) The response is a function v(t) that satisfies this differential equation for t 0 and meets the initial condition v(0). If v(0)=0, it is Zero-State Response. Since u(t)=1 for t  0 we can write Eq.(7-15) as Eq.(7-16) SJTU

divide solution v(t) into two components:
natural response forced response The natural response is the general solution of Eq.(7-16) when the input is set to zero. SJTU

seek a particular solution of the equation
The forced response is a particular solution of Eq.(7-16) when the input is step function. seek a particular solution of the equation Eq.(7-19) The equation requires that a linear combination of VF(t) and its derivative equal a constant VA for t  0. Setting VF(t)=VA meets this condition since . Substituting VF=VA into Eq.(7-19) reduces it to the identity VA=VA. Now combining the forced and natural responses, we obtain SJTU

Fig. 7-12: Step response of first-order RC circuit
using the initial condition:  K=(VO-VA) The complete response of the RC circuit: Eq.(7-20) The zero-state response of the RC circuit: t0 Fig. 7-12: Step response of first-order RC circuit SJTU

the initial and final values of the response are
The RL circuit in Figure 7-2 is the dual of the RC circuit Eq.(7-21) Setting iF=IA SJTU

The constant K is now evaluated from the initial condition:
The initial condition requires that K=IO-IA, so the complete response of the RL circuit is Eq.(7-22) The zero-state response of the RC circuit: t0 SJTU

The complete response of a first-order circuits depends on three quantities:
The amplitude of the step input (VA or IA) The circuit time constant(RTC or GNL)  The value of the state variable at t=0 (VO or IO) SJTU

Find the response of the RC circuit in Figure 7-13
EXAMPLE 7-4 Find the response of the RC circuit in Figure 7-13 SOLUTION: SJTU

EXAMPLE 7-5 Find the complete response of the RL circuit in Figure 7-14(a). The initial condition is i(0)=IO Fig. 7-14 SJTU

(a) What is the circuit time constant?
EXAMPLE 7-6 The state variable response of a first-order RC circuit for a step function input is (a) What is the circuit time constant? (b) What is the initial voltage across the capacitor? (c) What is the amplitude of the forced response? (d) At what time is VC(t)=0? SJTU

(b) The initial (t=0) voltage across the capacitor is
SOLUTION: (a) The natural response of a first-order circuit is of the form . Therefore, the time constant of the given responses is Tc=1/200=5ms (b) The initial (t=0) voltage across the capacitor is (c) The natural response decays to zero, so the forced response is the final value vC(t). (d) The capacitor voltage must pass through zero at some intermediate time, since the initial value is positive and the final value negative. This time is found by setting the step response equal to zero: which yields the condition SJTU

COMPLETE RESPONSE The first parts of the above equations are Zero-input response and the second parts are Zero-state response. What is s step response? SJTU

EXAMPLE 7-7 Find the zero-state response of the RC circuit of Figure 7-15(a) for an input Fig. 7-15 SJTU

The first input causes a zero-state response of
The second input causes a zero-state response of The total response is the superposition of these two responses. Figure 7-15(b) shows how the two responses combine to produce the overall pulse response of the circuit. The first step function causes a response v1(t) that begins at zero and would eventually reach an amplitude of +VA for t>5RC. However, at t=T<5TC the second step function initiates an equal and opposite response v2(t). For t> T+5RC the second response reaches its final state and cancels the first response, so that total pulse response returns to zero. SJTU

3. Initial and Final Conditions
Eq.(7-23) the general form : SJTU

The state variable response in switched dynamic circuits is found using the following steps:
STEP 1: Find the initial value by applying dc analysis to the circuit configuration for t<0 STEP 2: Find the final value by applying dc analysis to the circuit configuration for t>0. STEP 3: Find the time constant TC of the circuit in the configuration for t>0 STEP 4: Write the step response directly using Eq.(7-23) without formulating and solving the circuit differential equation. SJTU

Example: The switch in Figure 7-18(a) has been closed for a long time and is opened at t=0. We want to find the capacitor voltage v(t) for t0 Fig. 7-18: Solving a switched dynamic circuit using the initial and final conditions SJTU

There is another way to find the nonstate variables.
Generally, method of “three quantities” can be applied in step response on any branch of First-order circuit. Get f(0) from initial value of state variable Get f()---use equivalent circuit Get TC---calculate the equivalent resistance Re, TC=ReC or L/ Re Then, SJTU

How to get initial value f(0)?
the capacitor voltage and inductor current are always continuous in some condition. Vc(0+)=Vc(0-); IL(0+)=IL(0-) ---use 0+ equivalent circuit C: substituted by voltage source; L: substituted by current source Find f(0) in the above DC circuit. How to get final value f(∞)? Use ∞ equivalent circuit(stead state) to get f(∞). C: open circuit; L: short circuit How to get time constant TC? The key point is to get the equivalent resistance Re. SJTU

forced response natural response Zero-input response Zero-state response SJTU

EXAMPLE 7-8 The switch in Figure 7-20(a) has been open for a long time and is closed at t=0. Find the inductor current for t>0. SOLUTION: Fig. 7-20 SJTU

EXAMPLE The switch in Figure 7-21(a) has been closed for a long time and is opened at t=0. Find the voltage vo(t) Fig. 7-21 SJTU

another way to solve the problem:
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4. First-Order Circuit Sinusoidal Response
If the input to the RC circuit is a casual sinusoid Eq.(7-24) SJTU

where SJTU

EXAMPLE 7-12 The switch in Figure 7-26 has been open for a long time and is closed at t=0. Find the voltage v(t) for t 0 when vs(t)=[20 sin 1000t]u(t)V. SOLUTION: Fig. 7-26 SJTU

Summary Circuits containing linear resistors and the equivalent of one capacitor or one inductor are described by first-order differential equations in which the unknown is the circuit state variable. The zero-input response in a first-order circuit is an exponential whose time constant depends on circuit parameters. The amplitude of the exponential is equal to the initial value of the state variable. The natural response is the general solution of the homogeneous differential equation obtained by setting the input to zero. The forced response is a particular solution of the differential equation for the given input. For linear circuits the total response is the sum of the forced and natural responses. SJTU

Summary For linear circuits the total response is the sum of the zero-input and zero-state responses. The zero-input response is caused by the initial energy stored in capacitors or inductors. The zero-state response results form the input driving forces. The initial and final values of the step response of a first and second-order circuit can be found by replacing capacitors by open circuits and inductors by short circuits and then using resistance circuit analysis methods. For a sinusoidal input the forced response is called the sinusoidal steady-state response, or the ac response. The ac response is a sinusoid with the same frequency as the input but with a different amplitude and phase angle. The ac response can be found from the circuit differential equation using the method of undetermined coefficients Homework: zero-input: step response: op amp : SJTU

Chapter 8 Second-Order Circuit SJTU

What is second-order circuit?
A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements. Typical examples of second-order circuits: a) series RLC circuit, b) parallel RLC circuit, c) RL circuit, d) RC circuit SJTU

The Parallel RLC Circuit Second-Order Circuit Complete Response
The Series RLC Circuit The Parallel RLC Circuit Second-Order Circuit Complete Response SJTU

FORMULATING SERIES RLC CIRCUIT EQUATIONS
1. The Series RLC Circuit FORMULATING SERIES RLC CIRCUIT EQUATIONS Eq.(7-33) SJTU

The initial conditions
To solve second-order equation, there must be two initial values. SJTU

ZERO-INPUT RESPONSE OF THE SERIES RLC CIRCUIT
With VT=0(zero-input) Eq.(7-33) becomes Eq.(3-37) try a solution of the form then characteristic equation Eq.(7-39) SJTU

In general, a quadratic characteristic equation has two roots:
three distinct possibilities: Case A: If , there are two real, unequal roots Case B: If , there are two real, equal roots Case C: If , there are two complex conjugate roots SJTU

A source-free series RLC circuit
Special case: Vc(0)=V0, IL(0)=0 V0 V(t) tM I(t) SJTU

t > tM tM>t>0 What happens when R=0? SJTU

Second Order Circuit with no Forcing Function
vc(0) = Vo , iL(0) = Io. I. OVER DAMPED: R=2 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) = -0.7 e t +2.7 e t A vc(t) = e t e t V SJTU

II. CRITICALLY DAMPED: R= , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) = 2e t -5.83t e t A vc(t) = e t t e t V SJTU

III. UNDER DAMPED: R=0.5 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) =4.25 e -0.75t cos(1.2t ) A vc(t) = 2 e -0.75t cos(1.2t ) V SJTU

IV. UNDAMPED: R=0 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) =2.915 cos(1.414t ) A vc(t) =1.374 cos(1.414t ) V SJTU

For RT=8.5kohm, the characteristic equation is
EXAMPLE 7-14 A series RLC circuit has C=0.25uF and L=1H. Find the roots of the characteristic equation for RT=8.5kohm, 4kohm and 1kohm SOLUTION: For RT=8.5kohm, the characteristic equation is whose roots are * These roots illustrate case A. The quantity under the radical is positive, and there are two real, unequal roots at S1=-500 and S2=-8000. SJTU

For RT=4kohm, the characteristic equation is whose roots are
whose roots are This is an example of case B. The quantity under the radical is zero, and there are two real, equal roots at S1=S2=-2000. * For RT=1kohm the characteristic equation is whose roots are The quantity under the radical is negative, illustrating case C. In case C the two roots are complex conjugates. * SJTU

In case A the two roots are real and unequal and the zero-input response is the sum of two exponentials of the form Eq.(7-48a) In case B the two roots are real and equal and the zero-input response is the sum of an exponential and a damped ramp. Eq.(7-48b) In case C the two roots are complex conjugates and the zero-input response is the sum of a damped cosine and a damped sine. Eq.(7-48c) SJTU

EXAMPLE 7-15 The circuit of Figure 7-31 has C=0.25uF and L=1H. The switch has been open for a long time and is closed at t=0. Find the capacitor voltage for t  0 for (a) R=8.5k ohm, (b) R=4k ohm, and (c) R=1k ohm. The initial conditions are Io=0 and Vo=15V. SOLUTION: (a) In Example 7-14 the value R=8.5kohm yields case A with roots S1=-500 and S2= The corresponding zero-input solution takes the form in Eq.(7-48a). Fig. 7-31 SJTU

The initial conditions yield two equations in the constants K1 and K2:
Solving these equations yields K1=16 and K2 =-1, so that the zero-input response is SJTU

(b) In Example 7-14 the value R=4kohm yields case B with roots S1=S2= The corresponding zero-input response takes the form in Eq.(7-48b): The initial conditions yield two equations in the constants K1 and K2: Solving these equations yields K1=15 and K2= 2000 x 15, so the zero-input response is SJTU

c) In Example 7-14 the value R=1k ohm yields case C with roots . The corresponding zero-input response takes the form in Eq.(7-48c): The initial conditions yield two equations in the constants K1 and K2: Solving these equations yields K1=15 and K2=( ) , so the zero-input response is SJTU

Fig. 7-32 SJTU

Ciritically damped situation Underdamped situation
In general, a quadratic characteristic equation has two roots: Eq.(7-40) three distinct possibilities: Case A: If , there are two real, unequal roots Case B: If , there are two real, equal roots Case C: If , there are two complex conjugate roots Overdamped situation Ciritically damped situation Underdamped situation SJTU

2. The Parallel RLC Circuit
FORMULATING PARALLEL RLC CIRCUIT EQUATIONS Eq. 7-55 SJTU

The initial conditions
Equation(7-55) is second-order linear differential equation of the same form as the series RLC circuit equation in Eq.(7-33). In fact, if we interchange the following quantities: we change one equation into the other. The two circuits are duals, which means that the results developed for the series case apply to the parallel circuit with the preceding duality interchanges. The initial conditions iL(0)=Io and SJTU

set iN=0 in Eq.(7-55) and obtain a homogeneous equation in the inductor current:
A trial solution of the form IL=Kest leads to the characteristic equation Eq. 7-56 SJTU

There are three distinct cases:
Case A: If (GNL)2-4LC>0, there are two unequal real roots and the zero-input response is the overdamped form Case B: (GNL)2-4LC=0, there are two real equal roots and the zero-input response is the critically damped form Case C:(GNL)2-4LC<0, there are two complex, conjugate roots and the zero-input response is the underdamped form SJTU

From Eq.(7-56) the circuit characteristic equation is
EXAMPLE 7-16 In a parallel RLC circuit RT=1/GN=500ohm, C=1uF, L=0.2H. The initial conditions are Io=50 mA and Vo=0. Find the zero-input response of inductor current, resistor current, and capacitor voltage SOLUTION: From Eq.(7-56) the circuit characteristic equation is The roots of the characteristic equation are SJTU

Evaluating this expression at t=0 yields
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(a) Find the initial conditions at t=0
EXAMPLE 7-17 The switch in Figure 7-34 has been open for a long time and is closed at t=0 (a) Find the initial conditions at t=0 (b) Find the inductor current for t0 (c) Find the capacitor voltage and current through the switch for t  0 Fig. 7-34 SOLUTION: (a) For t<0 the circuit is in the dc steady state SJTU

The roots of this equation are
(b) For t  0 the circuit is a zero-input parallel RLC circuit with initial conditions found in (a). The circuit characteristic equation is The roots of this equation are The circuit is overdamped (case A), The general form of the inductor current zero-input response is using the initial conditions SJTU

The derivative of the inductor response at t=0 is
The initial capacitor voltage establishes an initial condition on the derivative of the inductor current since The derivative of the inductor response at t=0 is The initial conditions on inductor current and capacitor voltage produce two equations in the unknown constants K1 and K2: SJTU

(c) Given the inductor current in (b), the capacitor voltage is
Solving these equations yields K1=30.3 mA and K2= ma The zero-input response of the inductor current is (c) Given the inductor current in (b), the capacitor voltage is For t 0 the current isw(t) is the current through the 50 ohm resistor plus the current through the 250 ohm resistor SJTU

3. Second-order Circuit Complete Response
The general second-order linear differential equation with a step function input has the form Eq. 7-60 The complete response can be found by partitioning y(t) into forced and natural components: Eq. 7-61 yN(t) --- general solution of the homogeneous equation (input set to zero), yF(t) is a particular solution of the equation ∴ yF=A/ao SJTU

Combining the forced and natural responses
Eq. 7-67 EXAMPLE 7-18 The series RLC circuit in Figure 7-35 is driven by a step function and is in the zero state at t=0. Find the capacitor voltage for t  0. SOLUTION: Fig. 7-35 SJTU

the natural response is underdamped (case C)
By inspection, the forced response is vCF=10V. In standard format the homogeneous equation is the natural response is underdamped (case C) SJTU

The constants K1 and K2 are determined by the initial conditions.
These equations yield K1= -10 and K2= The complete response of the capacitor voltage step response is SJTU

General second-order circuit
Steps: Set a second-order differential equation Find the natural response yN(t) from the homogeneous equation (input set to zero) Find a particular solution yF(t) of the equation Determine K1 and K2 by the initial conditions Yield the required response SJTU

Summary Circuits containing linear resistors and the equivalent of two energy storage elements are described by second-order differential equations in which the dependent variable is one of the state variables. The initial conditions are the values of the two state variables at t=0. The zero-input response of a second-order circuit takes different forms depending on the roots of the characteristic equation. Unequal real roots produce the overdamped response, equal real roots produce the critically damped response, and complex conjugate roots produce underdamped responses. Computer-aided circuit analysis programs can generate numerical solutions for circuit transient responses. Some knowledge of analytical methods and an estimate of the general form of the expected response are necessary to use these analysis tools. SJTU

Chapter 9 Sinusoids and Phasors SJTU

Sinusoids A sinusoid is a signal that has the form of the sine or cosine function. SJTU

t  radians/second (rad/s) f is in hertz(Hz) SJTU

Phase difference: SJTU

Complex Number  SJTU

Phasor a phasor is a complex number representing the amplitude and phase angle of a sinusoidal voltage or current. Eq.(8-1) Eq. (8-2) and Eq.(8-3) SJTU

When Eq.(8-2) is applied to the general sinusoid we obtain
The phasor V is written as Eq.(8-5) SJTU

Two features of the phasor concept need emphasis:
Fig. 8-1 shows a graphical representation commonly called a phasor diagram. Two features of the phasor concept need emphasis: Phasors are written in boldface type like V or I1 to distinguish them from signal waveforms such as v(t) and i1(t). A phasor is determined by amplitude and phase angle and does not contain any information about the frequency of the sinusoid. Fig. 8-1: Phasor diagram SJTU

Phase-domain representation Time domain representation
In summary, given a sinusoidal signal , the corresponding phasor representation is . Conversely, given the phasor , the corresponding sinusoid is found by multiplying the phasor by and reversing the steps in Eq.(8-4) as follows: Eq.(8-6) Phase-domain representation Time domain representation SJTU

Properties of Phasors additive property
Eq.(8-7) Eq.(8-8) Eq.(8-9) SJTU

Time domain representation Phase-domain representation
derivative property Eq.(8-10) Time domain representation Phase-domain representation SJTU

Time domain representation Phase-domain representation
Integral property Time domain representation Phase-domain representation The differences between v(t) and V: V(t) is the instantaneous or time-domain representation, while V is the frequency or phasor-domain representation. V(t) is a real signal which is time dependent, while V is just a supposed value to simplify the analysis SJTU

Fig. 8-2: Complex exponential
The complex exponential is sometimes called a rotating phasor, and the phasor V is viewed as a snapshot of the situation at t=0. Fig. 8-2: Complex exponential SJTU

Construct the phasors for the following signals:
EXAMPLE 8-1 Construct the phasors for the following signals: (b) Use the additive property of phasors and the phasors found in (a) to find v(t)=v1(t)+v2(t). SOLUTION (a) The phasor representations of v(t)=v1(t)+ v2(t) are SJTU

The waveform corresponding to this phasor sum is
(b) The two sinusoids have the same frequent so the additive property of phasors can be used to obtain their sum: The waveform corresponding to this phasor sum is j V2 1 V V1 SJTU

Construct the phasors representing the following signals:
EXAMPLE 8-2 Construct the phasors representing the following signals: (b) Use the additive property of phasors and the phasors found in (a) to find the sum of these waveforms. SOLUTION: (a) The phasor representation of the three sinusoidal currents are SJTU

(b) The currents have the same frequency, so the additive property of phasors applies. The phasor representing the sum of these current is Fig. 8-4 SJTU

The sinusoid corresponding to the phasor jV is
EXAMPLE 8-3 Use the derivative property of phasors to find the time derivative of v(t)=15 cos(200t-30° ). SOLUTION: The phasor for the sinusoid is V=15∠-30 ° . According to the derivative property, the phasor representing the dv/dt is found by multiplying V by j . The sinusoid corresponding to the phasor jV is SJTU

Device Constraints in Phasor Form
Resistor: Re jIm I V Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain. SJTU

Inductor:  Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain. SJTU

Capacitor:  Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain. SJTU

Connection Constraints in Phasor Form
KVL in time domain Kirchhoff's laws in phasor form (in frequency domain) KVL: The algebraic sum of phasor voltages around a loop is zero. KCL: The algebraic sum of phasor currents at a node is zero. SJTU

The Impedance Concept V=ZI or Z= V/I Eq.(8-16)
The IV constraints are all of the form V=ZI or Z= V/I Eq.(8-16) where Z is called the impedance of the element The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms() The impedance is inductive when X is positive is capacitive when X is negative SJTU

The Impedance Concept SJTU

EXAMPLE 8-5 Fig. 8-5 The circuit in Fig. 8-5 is operating in the sinusoidal steady state with and . Find the impedance of the elements in the rectangular box. SOLUTION: SJTU

The Admittance Concept
The admittance Y is the reciprocal of impedance, measured in siemens (S) Y=G+jB Where G=Re Y is called conductance and B=Im Y is called the susceptance How get Y=G+jB from Z=R+jX ? SJTU

Basic Circuit Analysis with Phasors
Step 1: The circuit is transformed into the phasor domain by representing the input and response sinusoids as phasor and the passive circuit elements by their impedances. Step 2: Standard algebraic circuit techniques are applied to solve the phasor domain circuit for the desired unknown phasor responses. Step 3: The phasor responses are inverse transformed back into time-domain sinusoids to obtain the response waveforms. SJTU

Series Equivalence And Voltage Division
where R is the real part and X is the imaginary part SJTU

(a) Transform the circuit into the phasor domain.
EXAMPLE 8-6 Fig. 8-8 The circuit in Fig is operating in the sinusoidal steady state with (a) Transform the circuit into the phasor domain. (b) Solve for the phasor current I. (c) Solve for the phasor voltage across each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c) SJTU

SOLUTION: SJTU

PARALLEL EQUIVALENCE AND CURRENT DIVISION
Rest of the circuit Y1 Y2 YN I V I1 I2 I3 phasor version of the current division principle SJTU

(a) Transform the circuit into the phasor domain.
EXAMPLE 8-9 Fig. 8-13 The circuit in Fig is operating in the sinusoidal steady state with iS(t)=50cos2000t mA. (a) Transform the circuit into the phasor domain. (b) Solve for the phasor voltage V. (c) Solve for the phasor current through each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c). SJTU

SOLUTION: (a) The phasor representing the input source current is Is=0.05∠0° A. The impedances of the three passive elements are Fig. 8-14 SJTU

And the voltage across the parallel circuit is
The current through each parallel branch is The sinusoidal steady-state waveforms corresponding to the phasors in (b) and (c) are SJTU

EXAMPLE 8-10 Fig. 8-15 Find the steady-state currents i(t), and iC(t) in the circuit of Fig (for Vs=100cos2000t V, L=250mH, C=0.5  F, and R=3k ). SOLUTION: Vs=100∠0° SJTU

The equations for the △ to Y transformation are
Y←→△ TRANSFORMATIONS The equations for the △ to Y transformation are SJTU

The equations for a Y-to- △ transformation are
when Z1=Z2=Z3=ZY or ZA=ZB=ZC=ZN. ZY=ZN /3 and ZN =3ZY balanced conditions SJTU

EXAMPLE 8-12 Use a △ to Y transformation to solve for the phasor current IX in Fig SOLUTION: ABC △ to Y Fig. 8-18 SJTU

Sinusoidal steady-state analysis
Chapter 10 Sinusoidal steady-state analysis SJTU

Steps to analyze ac circuit
Transform the circuit to the phasor or frequency domain Solve the problem using circuit techniques(nodal analysis, mesh analysis, superposition,etc) Transform the resulting phasor to the time domain SJTU

Nodal analysis Fig. 8-28: An example node SJTU

Mesh analysis Fig. 8-29: An example mesh planar circuits:
Circuits that can be drawn on a flat surface with no crossovers the sum of voltages around mesh A is Fig. 8-29: An example mesh SJTU

Use node analysis to find the current IX in Fig. 8-31.
EXAMPLE 8-21 Use node analysis to find the current IX in Fig SOLUTION: or Fig. 8-31 SJTU

DS:example on F page 394, notebook p105
SJTU

EXAMPLE 8-24 The circuit in Fig is an equivalent circuit of an ac induction motor. The current IS is called the stator current, IR the rotor current, and IM the magnetizing current. Use the mesh-current method to solve for the branch currents IS, IR and IM. SJTU

EXAMPLE 8-25 Use the mesh-current method to solve for output voltage V2 and input impedance ZIN of the circuit below. SOLUTION: SJTU

Frequency domain equivalent of the circuit
Example Frequency domain equivalent of the circuit SJTU

Example SJTU

Find Vo/Vi, Zi See F page417 SJTU

Circuit Theorems with Phasors
PROPORTIONALITY The proportionality property states that phasor output responses are proportional to the input phasor where X is the input phasor, Y is the output phasor, and K is the proportionality constant. SJTU

Assume a unit output voltage .
EXAMPLE 8-13 Use the unit output method to find the input impedance, current I1, output voltage VC, and current I3 of the circuit in Fig for Vs= 10∠0° Assume a unit output voltage . By Ohm's law, . By KVL, By Ohm's law, By KCL, By KCL, SOLUTION: SJTU

Given K and ZIN, we can now calculate the required responses for an input
SJTU

SUPERPOSITION Two cases: With same frequency sources.
With different frequency sources EXAMPLE 8-14 Use superposition to find the steady - state voltage vR (t) in Fig for R=20 , L1 = 2mH, L2 = 6mH, C = 20 F, V s1= 100cos 5000t V , and Vs2=120cos (5000t +30 )V. SJTU

SOLUTION: Fig. 8-22 SJTU

With source no. 2 off and no.1 on
EXAMPLE 8-15 Fig. 8-23 Use superposition to find the steady-state current i(t) in Fig for R=10k , L=200mH, vS1=24cos20000t V, and vS2=8cos(60000t+30 ° ). SOLUTION: With source no. 2 off and no.1 on SJTU

With source no.1 off and no.2 on
The two input sources operate at different frequencies, so that phasors responses I1 and I2 cannot be added to obtain the overall response. In this case the overall response is obtained by adding the corresponding time-domain functions. SJTU

More examples See F page403 SJTU

THEVENIN AND NORTON EQUIVALENT CIRCUITS
The thevenin and Norton circuits are equivalent to each other, so their circuit parameters are related as follows: SJTU

Source transformation
SJTU

EXAMPLE 8-17 Both sources in Fig. 8-25(a) operate at a frequency of =5000 rad/s. Find the steady-state voltage vR(t) using source transformations. SOLUTION: + SJTU

EXAMPLE 8-18 Use Thevenin's theorem to find the current Ix in the bridge circuit shown in Fig Fig. 8-26 SJTU

SOLUTION: SJTU

Chapter 11 AC power analysis SJTU

effective value or DC-equivalent value
rms value The RMS value is the effective value of a varying voltage or current. It is the equivalent steady DC (constant) value which gives the same effect. effective value or DC-equivalent value The rms value of a periodic function is defined as the square root of the mean value of the squared function. SJTU

If the periodic function is a sinusoid, then
What do AC meters show, is it the RMS or peak voltage? AC voltmeters and ammeters show the RMS value of the voltage or current. What does '6V AC' really mean, is it the RMS or peak voltage? If the peak value is meant it should be clearly stated, otherwise assume it is the RMS value. SJTU

AC power analysis Instantaneous Power Suppose: i(t) v(t) N
Invariable part Sinusoidal part SJTU

E page415 figure 10.2 SJTU

Stored energy WLav In the sinusoidal steady state an inductor operates with a current iL(t)=IAcos(wt). The corresponding energy stored in the element is Average stored energy SJTU

Stored energy In the sinusoidal steady state the voltage across a capacitor is vc(t)=VAcos(wt). The energy stored in the element is Average stored energy WCav SJTU

There are other methods to calculate P.
Average power The average power is the average of the instantaneous power over one period real power Note : There are other methods to calculate P. 1) 1) 2) SJTU

Instantaneous power, real power
Instantaneous power waveforms for a voltage of 2V peak and a current of 1.5A peak Flowing separately in a resistor, a capacitor and an inductor Inductor case Pav = 0 Resistor case Average power Pav=0.5Vm*Im Pav=vrms*irms SJTU Capacitor case Pav = 0

current leads voltage or current lags voltage
Apparent power S=VrmsIrms (VA) Power factor current leads voltage or current lags voltage < or  >0 SJTU

Capacitor: Q=-VrmsIrms
Reactive power (VAR) Resistor: Q=0 Inductor: Q=VrmsIrms Capacitor: Q=-VrmsIrms To any passive single port network SJTU

The power triangle S Q  P SJTU

EXAMPLE Find the average power delivered to the load to the right of the interface in Figure 8-64. Fig. 8-64 SOLUTION: SJTU

Complex power Complex power is the complex sum of real power and reactive power =P+jQ So =VI* Where V is the voltage phsor across the system and I* is the complex conjugate of the current phasor. The magnitude of complex power is just apparent power SJTU

Are these equations right?
SJTU

Maximum power transfer
Fig. 8-66: A source-load interface in the sinusoidal steady state. SJTU

we know P is maximized when RL=RT
Let XL=-XT then we know P is maximized when RL=RT the maximum average power where |VT| is the peak amplitude of the Thevenin equivalent voltage SJTU

EXAMPLE (a) Calculate the average power delivered to the load in the circuit shown in Figure 8-67 for Vs(t)=5cos106t, R=200 ohm, and RL=200 ohm. (b) Calculate the maximum average power available at the interface and specify the load required to draw the maximum power. SOLUTION: (a) SJTU

If the load must be a resistor, how get the maximum power on it?
Question: If the load must be a resistor, how get the maximum power on it? SJTU

Maximum power transfer when ZL is restricted
RL and XL may be restricted to a limited range of values. In this situation, the optimum condition for RL and XL is to adjust XL as near to –XT as possible and then adjust RL as close to as possible the magnitude of ZL can be varied but its phase angle cannot. Under this restriction, the greatest amount of power is transferred to the load when the magnitude of ZL is set equal to the magnitude of ZT SJTU

Note: If the load is a resistor, then what value of R results in maximum average-power transfer to R? what is the maximum power then? If ZL cannot be varied but ZT can, what value of ZT results in maximum average-power transfer to ZL? SJTU

Magnetically coupled circuits
Chapter 13 Magnetically coupled circuits SJTU

Mutual inductance A single inductor:  SJTU

Mutual inductance of M21 of coil 2 with respect to coil 1
 SJTU

(for nonmagnetic cores)
21 22 i2(t) v1 v2 N1 N2 (for nonmagnetic cores) SJTU

Dot convention When the reference direction for a current enters the dotted terminal of a coil, the reference polarity of the voltage that it induces in the other coil is positive at its dotted terminal. SJTU

How could we determine dot markings if we don’t know?
Examples How could we determine dot markings if we don’t know? SJTU

Series connection 1 2 M 1 2 M (a)mutually coupled coils in series-aiding connection (b)mutually coupled coils in series–opposing connection Total inductance LT=L1+L2+2M LT=L1+L2-2M SJTU

Parallel connection L1 L2 I + V M L1 L2 I + V M
(a)mutually coupled coils in parallel-aiding connection (b)mutually coupled coils in parallel–opposing connection Equivalent inductance SJTU

Coefficient of coupling
The coupling coefficient k is a measure of the magnetic coupling between two coils k < loosely coupled; k > tightly coupled. SJTU

Tee model SJTU

TEE MODEL SJTU

Examples of the mutual coupled circuits
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Model in frequency field
Linear transformers V R1 R2 ZL L1 L2 M I1 I2 jwM Primary winding Secondary winding R1 R2 V jwL1 jwL2 RL+jXL I1 I2 Model in frequency field SJTU

Total self-impedance of the mesh containing the primary winding
Total self-impedance of the mesh containing the secodary winding SJTU

reflected impedance V R1 jwL1 Zr (reflected impedance) I1 Zr
Equivalent primary winding circuit (reflected resistance) (reflected reactance) SJTU

Equivalent secondary winding circuit
Z22 Equivalent secondary winding circuit SJTU

Ideal transformer + - three properties:
The coefficient of coupling is unity (k=1) The self- and mutual inductance of each coil is infinite (L1=L2=M=∞), but is definite. Primary and secondary coils are lossless. + - 1: n SJTU

+ + - - 1: n + - 1: n + - 1: n SJTU

Transformer as a matching device
+ - 1: n RL + - 1: n RL/n2 - + - 1: n R R + + n2R - - 1: n SJTU

Transformer as a matching device
+ + RL Thevenin equivalent - - 1: n Zin Vs2 Vs1 Z1 Z2 1: n I1 I2 Vs1 Z1 Z2/n2 Vs2/n SJTU

Vs2 Vs1 Z1 Z2 1: n I1 I2 nVs1 n2 Z1 Z2 Vs2 SJTU

Solving Ideal Transformer Problem
Method 1: Write out equations first Loop equations or Nodal equations Two more transformer equations Method 2 : Form equivalent circuit first Reflecting into secondary Reflecting into primary Vs1 Vs2 Z1 Z2 SJTU

The Ideal Transformer SJTU

General transformer model
Lossless, k=1, but L1,L2,M are not infinite + - L1 L2 M + - 1: n L1 SJTU

2. Lossless, k≠1, L1,L2,M are not infinite + - 1: n LM LS1 LS2 + - L1 L2 M SJTU

3. No restriction + - L1 L2 M + + LS1 R1 LS2/n2 R2/n2 LM - - 1: n SJTU

SUMMARY Mutual inductance, M, is the circuit parameter relating the voltage induced in one circuit to a time-varying current in another circuit. The coefficient of coupling, k, is the measure of the degree of magnetic coupling. By definition, 0≤k≤1 The relationship between the self-inductance of each winding and the mutual inductance between the windings is The dot convention establishes the polarity of mutually induced voltage Reflected impedance is the impedance of the secondary circuit as seen from the terminals of the primary circuit, or vise versa. SJTU

SUMMARY The two-winding linear transformer is a coupling device made up of two coils wound on the same nonmagnetic core. An ideal transformer is a lossless transformer with unity coupling coefficient(k=1) and infinite inductance. An ideal transformer can be used to match the magnitude of the load impedance, ZL, to the magnitude of the source impedance, ZS, thus maximizing the amount of average power transferred. SJTU

Chapter 18 Two-port Networks

the four terminals have been paired into ports
four-terminal network two-port network KCL At all times, the instantaneous current flowing into one terminal is equal to the instantaneous current flowing out the other. i1 i2 i3 i4 i1 i4 i1=-i2 ; i3=-i4 i1+i2+i3+i4=0(KCL)

The network is linear(without independent sources).

Impedance Parameters impedance matrix Z Open-circuit input impedance.
Impedance or z parameters are defined by impedance matrix Z Open-circuit input impedance. Open-circuit transfer impedance from port 1 to port 2 Open-circuit transfer impedance from port 2 to port 1 Open-circuit output impedance.

Determining of the z parameters: (a) finding z11 and z21, (b) finding z12 and z22
Examples

(a) T equivalent circuit (for reciprocal case only), (b) general equivalent circuit

Admittance Parameters
Admittance or y parameters are defined by admittance matrix Y Short-circuit input admittance. Short-circuit transfer admittance from port 1 to port 2 Short-circuit transfer admittance from port 2 to port 1 Short-circuit output admittance.

Determination of the y parameters: (a) finding y11 and y21, (b) finding y12 and y22.

(a) -equivalent circuit (for reciprocal case only), (b) general equivalent circuit.

Hybrid Parameters hybrid matrix Z Short-circuit input impedance.
Hybrid or h parameters are defined by hybrid matrix Z Short-circuit input impedance. Open-circuit reverse voltage gain Short-circuit forward current gain Open-circuit output admittance.

The h-parameter equivalent network of a two-port network

Inverse hybrid parameters (g parameters)
The g-parameter model of a two-port network

Transmission Parameters
Transmission or T parameters are defined by Transmission matrix T Open-circuit voltage ratio Negative short-circuit transfer impedance Open-circuit transfer admittance Negative short-circuit current ratio

Inverse transmission parameters

Reciprocal Two-Port Circuits
linear and has no dependent source If a two-port circuit is reciprocal, the following relationships exist among the port parameters:

Symmetric Two-Port Circuit
A reciprocal two-port circuit is symmetric if its ports can be interchanged without disturbing the values of the terminal currents and voltages. If a two-port circuit is symmetric, the following relationships exist among the port parameters: (besides those exist in reciprocal)

For a general two-port with sources: 6
Question: How many calculations or measurements are needed to determine a set of parameters of a two-port circuit? For a general two-port with sources: 6 For a general linear two-port: 4 For a reciprocal two-port: 3 For a symmetric two-port: 2

Relationships between parameters
Example: z parameters y parameters

Interconnection of networks
Series connection of two two-port networks

Parallel connection of two two-port networks

Cascade connection of two two-port networks

Transistor amplifier with source and load resistance

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