Welcome to EGR 272 Circuit Theory II

Welcome to

EGR 272Circuit Theory II

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EGR 271: Chapters 1 - 6 in Electric Circuits, 10th Edition by Nilsson EGR 272: Chapters 7-10, 12 - 15 in Electric Circuits, 10th Edition by Nilsson

1Chapter 7 EGR 272 – Circuit Theory II


Sequence of Electrical/Computer Engineering Courses at TCC

Notes:1. Classes available at the Virginia Beach & Chesapeake Campuses and the Tri-Cities Center2. EGR 271-272 transfers to Virginia Tech as ECE 20043. EGR 270 transfers to Virginia Tech as ECE 25044. EGR 262 does not transfer to Virginia Tech

EGR 271 (3 cr)Circuit Theory I

ODU equiv: ECE 201Offered: F, Sp, Su

EGR 272 (3 cr)Circuit Theory II

ODU equiv: ECE 202Offered: F, Sp

EGR 262 (2 cr)Fund. Circuits Lab


EGR 270 (4 cr)Fund. Of Computer EGR


EGR 125 (4 cr)Into to Engineering

Methods (C++)

MTH 279 (4 cr)DifferentialEquations

EGR 110 (3 cr)Engineering

Graphics

Chapter 6 Review: – A quick review of some key concepts before we begin the course with Chapter 7:


Key relationships:10

C

L

+ v(t) -

i(t)

+ v(t) -i(t)

+ V -

IRi(t) v(t):Resistor

dtdvC i(t) :Capacitor

dtdiL v(t):Inductor

Note: All three components are passive devices, so use passive sign convention. The equation assumes that current enters the positive terminal.

Chapter 6 Review: – Other important relationships:


Resistor Capacitor Inductor

Voltage

Current

Power

Energy

Steady-state behavior

Resistor Open-circuit Short-circuit

Cannot change instantaneously

(not applicable) VoltageV(0+) = v(0-)

Currenti(0+) = i(0-)

dtdiL v(t)

dtdvC i(t)

RI V

RV I

t

0

v(0)i(t)dt C1 v(t)

t

0

i(0) v(t)dt L1 i(t)

RV RI VI P

22 VI P VI P

t

0

w(0)p(t)dt w(t)

(t)Li21 w(t)

w(0)p(t)dt w(t)

2

0

t

(t)Cv21 w(t)

w(0)p(t)dt w(t)

2

0

t

Reading Assignment: Chapter 7 in Electric Circuits, 10th Ed. by Nilsson

Chapter 7 – Circuit Analysis with Capacitors and InductorsKVL and KCL involving circuits with capacitors and inductors result in differential equations (D.E.) rather than algebraic equations.The order of a differential equation is equal to highest derivative.

0

2

1 02

n n-1

n-1 1 0n n-1

dx1st-order DE: + a x(t) = f(t)dtd x dx2nd-order DE: + a + a x(t) = f(t)dt dt

d x d x dxnth-order DE: + a + + a + a x(t) = f(t)dt dt dt


Examples: Draw several circuits with inductors and capacitors and determine the order of each circuit.

Order of a

circuit

Order of the differential equation

(DE) required to describe the circuit

The number of independent* energy

storage elements(C’s and L’s)

= =

* C’s and L’s are independent if they cannot be combined with other C’s and L’s (in series or parallel, for example)

6

Circuit Order Chapter 7 EGR 272 – Circuit Theory II

First-Order Circuits A first-order circuit can only contain one energy storage element (a capacitor or an inductor). The circuit will also contain resistance (discuss). So there are two types of first-order circuits:• RC circuit• RL circuit

Source-Free Circuits A source-free circuit is one where all independent sources have been disconnected from the circuit after some switch action. The voltages and currents in the circuit typically will have some transient response due to initial conditions (initial capacitor voltages and initial inductor currents). We will begin by analyzing source-free circuits as they are the simplest type. Later we will analyze circuits that also contain sources after the initial switch action.

7

Source-free RC circuitConsider the RC circuit shown below. Note that it is source-free because no sources are connected to the circuit for t > 0. Use KCL to find the differential equation:

+ _ VX

t = 0

R C v (t)

+

_

dv 1 v(t) 0 for t 0dt RC

-tRC

Xv(t) V e for t 0

and solve the differential equation to show that:


Checks on the solution• Verify that the initial condition is satisfied.• Find dv(t)/dt and regenerate the D.E. from the solution.• Show that the energy dissipated over all time by the resistor equals the initial

energy stored in the capacitor.


General form of the D.E. and the response for a 1st-order source-free circuitIn general, a first-order D.E. has the form:

dx 1 x(t) 0 for t 0dt

-t

x(t) x(0)e for t 0

Solving this differential equation (as we did with the RC circuit) yields:

where = (Greek letter “Tau”) = time constant (in seconds)

Notes concerning :1) for the previous RC circuit the DE was:

so

(for an RC circuit)

dv 1 v(t) 0 for t 0dt RC

= RC


2) is related to the rate of exponential decay in a circuit as shown below

3) It is typically easier to sketch a response in terms of multiples of than to be concerning with scaling of the graph (otherwise choosing an appropriate scale can be difficult). This is illustrated on the following page.

-t

x(t) x(0)e

Smaller (faster decay)

Larger (slower decay)

t

x(0)

0

0


Graphing functions in terms of :Illustration:A) Calculate values for x(t) = x(0)e-t/ for t = , 2, 3, 4, and 5.

t x(t) = x(0)e-t/

B) Graph x(t) versus t for t = 0, , 2, 3, 4, and 5.x(t)

2 3 4 5 0t


Time for a circuit to completely decayFrom the last page note that x(5) = x(0)e-5/ = x(0)e-5 = 0.007x(0)This means that magnitude of x(t) is only 0.7% of its initial value by time 5 (or the function has lost 99.3% of its original value). Technically a decaying exponential function never reaches zero, but we see that by time t = 5 it is very close. So we generally use the approximation that:

5 = time for a circuit to decay

Example: Some circuits connect a “bleeder resistor” in parallel with a capacitor when the circuit is turned off in order to safely discharge the capacitor (which might otherwise have a significant voltage across it for a long time). For the circuit shown below, what value of Rbleeder should be used in order to discharge the capacitor in 10 seconds (the circuit is turned off at time t = tx)?

Rbleeder

t = tx

Circuit470uF

t = tx


Example: The switch in the circuit shown had been closed for a long time and then opened at time t = 0. A) Determine an expression for v(t).B) Graph v(t) versus t.

+ _ 50V

t = 0

15k 10 uF v (t)

+

_


Example: (continued) C) How long will it take for the capacitor to completely discharge?D) Determine the capacitor voltage at time t = 100 ms.E) Determine the time at which the capacitor voltage is 10V.


Equivalent Resistance seen by a CapacitorFor the RC circuit in the previous example, it was determined that = RC. But what value of R should be used in circuits with multiple resistors?In general, a first-order RC circuit has the following time constant:

= REQ C

where Req is the Thevenin resistance seen by the capacitor.More specifically,

seen from the terminals of the capacitor for t > 0EQwith independent sources killed

R = R

CCircuitCircuit

t > 0independent

sources killed

REQ


Example: Determine an expression for v(t). Graph v(t) versus t.

+ _ 50V t = 0

30 10 F v (t) +

_

40 20 90


Source-free RL circuitConsider the RL circuit shown below. Use KVL to find the differential equation:

and use the general form of the solution to a first-order D.E. to show that:

IX

t = 0

R L

i(t) di R i(t) 0 for t 0dt L

-tRL

Xi(t) I e for t 0

= L/R


Equivalent Resistance seen by an InductorFor the RL circuit in the previous example, it was determined that = L/R. As with the RC circuit, the value of R should actually be the equivalent (or Thevenin) resistance seen by the inductor.In general, a first-order RL circuit has the following time constant:

where

seen from the terminals of the inductor for t > 0EQwith independent sources killed

R = R

EQ

L = R

LCircuitCircuit

t > 0independent

sources killed

REQ


Example: Determine an expression for i(t). Sketch i(t) versus t.

t = 0

150

10 mH

i(t) 100 V + _

50

75


First-order circuits with DC forcing functions:In the last class we consider source-free circuits (circuits with no independent sources for t > ). Now we will consider circuits having DC forcing functions for t > 0 (i.e., circuits that do have independent DC sources for t > 0).

The general solution to a differential equation has two parts:

x(t) = xh + xp = homogeneous solution + particular solutionor x(t) = xn + xf = natural solution + forced solution

where xh or xn is due to the initial conditions in the circuitand xp or xf is due to the forcing functions (independent voltage and current sources for t > 0).xp or xf in general take on the “form” of the forcing functions, so DC sources imply that the forced response function will be a constant (DC), sinusoidal sources imply that the forced response will be sinusoidal, etc.Since we are only considering DC forcing functions in this chapter, we assume that


xf = B (a constant)

Recall that a 1st-order source-free circuit had the form Ae-t/ . Note that there was a natural response only since there were no forcing functions (sources) for t > 0. So the natural response was

xn = Ae-t/ The complete response for 1st-order circuit with DC forcing functions therefore will have the form x(t) = xf + xn or

x(t) = B + Ae-t/ The “Shortcut Method”An easy way to find the constants B and A is to evaluate x(t) at 2 points. Two convenient points at t = 0- and t = since the circuit is in steady-state at these two points. This approach is sometimes called the “shortcut method.”So, x(0) = B + Ae0 = B + A And x() = B + Ae- = BShow how this yields the following expression found in the text:

x(t) = x() +[x(0) - x()]e-t/


“Shortcut Method” - ProcedureThe shortcut method will be the key method used in this chapter to analyze 1st-order circuit with DC forcing functions:

Notes:• The “shortcut method” also works for source-free circuits, but x() = B = 0

since the circuit is dead at t = .• If variables other than vC or iL are needed, it is generally easiest to solve for

vC or iL first and then use the result to find the desired variable.

1) Analyze the circuit at t = 0-: Find x(0-) = x(0+), where x = vC or iL .2) Analyze the circuit at t = : Find x().3) Find = REQC or = L/REQ .4) Assume that x(t) has the form x(t) = B + Ae-t/ and solve for B and

A using x(0) and x().


Example: Find v(t) and i(t) for t > 0.

+ _ 240V

t = 0

10 uF v(t)

+

_

15

60

20

5

i(t)


Example: Find v(t) and i(t) for t > 0.

t = 0

3 mH

i(t)

40 V 75 100 V + _

+ _ v(t)

+

_

300 20


Example: Capacitor charging circuit.Find v(t) for t > 0.

+ _ VX

t = 0

C v (t)

+

_

R + _ VX

t = 0

R C v (t)

+

_

Example: Capacitor discharging circuit. Find v(t) for t > 0.

26Chapter 7 EGR 272 – Circuit Theory IICharging and discharging capacitors – the forms of these two simple circuits are commonly referred to in later courses.

1st-order Circuits with Dependent Sources

Dependent sources affect the resistance seen by the inductor or capacitor and

therefore affect the value of for the circuit.

Two approaches can be used to find :

1) When REQ seen by an inductor or a capacitor, remove the L or C, kill any independent sources, and place any value independent source at the terminals. Then

2) Write a DE for the circuit (any variable) and can be easily determined from the DE since it has the form:

EQterminal voltageR terminal current

dx 1 x(t) f(t)dt


Example: Find v(t) for t > 0 if v(0) = 2 V.

A) Use method 1: Find = REQC using EQterminal voltageR terminal current

+ _ 18 V 0.1 F v (t)

+

_

20 + -

3v(t) 40


Example: (continued) Find v(t) for t > 0 if v(0) = 2 V.

B) Use method 2: Write a DE for v(t).

+ _ 18 V 0.1 F v (t)

+

_

20 + -

3v(t) 40


Unit Step FunctionsUnit step functions have several important uses in electrical engineering,

including:• representing piecewise-continuous signals• representing switches• defining functions for use with one-sided Laplace transforms (in EGR 261)

Definition: u(t) = unit step function where

and u(t) is represented by the graph shown below.

1 for t 0u(t)

0 for t 0

u(t)

t 0

1


A good way to think of a unit step function is as follows:• u(argument) = 1 for argument > 0• u(argument) = 0 for argument < 0• the transition in u(argument) occurs where argument = 0

Example: Graph 4u(t - 2) The function can also be described as follows:4 for t 2

4u(t - 2) 0 for t 2

u(t)

t 0

4

2

The transition occurs when the argument = 0.Note that when t = 2, u(t - 2) = u(2 - 2) = u(0).

u(argument) = 1 for argument > 0.Note that for t > 2, u(t - 2) has a positive argument.For example, when t = 3, u(t - 2) = u(3 - 2) = u(+1) = 1.

u(argument) = 0 for argument < 0.Note that for t < 2, u(t - 2) has a negative argument.For example, when t = 1, u(t - 2) = u(1 - 2) = u(-1) = 0.


Example: Graph the following functions:1) -2u(t - 10)

2) 3u(t + 2)

3) 4u(-t)

4) 4u(2-t)


Example: Graph the following functions (continued):5) 6u(-4-t)

6) u(-t)

7) Show that 1 - u(t) = u(-t)

8) sin(t) and sin(t)u(t)


Example: Graph the following functions (continued):9) u(t) - u(t - 2)

10) f(t) = 2t

11) 2t[u(t) - u(t - 2)] - discuss the concept of a “window”


Example: Graph the following functions (continued):12) f(t)[u(t - 2) - u(t - 4)] for any f(t)

13) (2t + 6)[u(t + 2) - u(t - 2)]


There are two common types of problems in representing functions using unit step functions:1) Determining the function that represents a given graph

Approach: Represent each unique portion of the function using unit step “windows”2) Graphing a function specified by unit stepsApproach: As each unit step function “turns on”, graph the cumulative function.


Determining the function that represents a given graph:Approach: Represent each unique portion of the function using unit step “windows”

f(t)

t

6

1) Represent f(t) shown below using unit step functions.

2 4 0

12

-6


Example:

f(t)

t

20

2) Represent f(t) shown below using unit step functions.

8 16 0

60

24

38Chapter 7 EGR 272 – Circuit Theory IIExample:

Graphing a function specified using unit steps functions:Approach: As each unit step function “turns on”, graph the cumulative function (i.e., add each part as it turns on).


1) Graph f(t) = 2u(t) + 4u(t – 2) – 8u(t – 4)

Solution: f(t) = 2u(t) + 4u(t – 2) – 8u(t – 4)

Example:

turns ONat t = 0

turns ONat t = 2

turns ONat t = 4

t4-28-64t26422t02

0t0

f(t) so

f(t)

t2 4

2

6

-2

2) Graph f(t) = (t + 2)u(t + 1) + (3 – t)u(t – 2) - (2t-5)u(t – 4)

3) Graph f(t) = 4sin(4t)[u(t) – u(t – 1)]


Using unit step functions to replace switches in circuitsUnit step functions are commonly used to represent switches in circuits. Consider the following examples.

+ _ 10 V

t = 0

C v (t)

+

_

R

v(t) = 10(1 - e-t/RC ) V, t > 0 0, t < 0

+ _ 10u(t) V C v (t)

+

_

R

v(t) = 10(1 - e-t/RC )u(t) V

VX

+

_

Example: A unit step function is used below to replace a switch connecting a voltage source. • Discuss the value of VX. • Discuss the forms of the solution.


2A

t = 0

C v (t)

+

_

R

IX

Example: A unit step function can be used to replace a switch disconnecting a current source. • Discuss the concept of a “make before break” switch• Discuss the value of IX. • Draw two possible circuits using unit step functions that are equivalent to

the circuit shown below.


2u(t) A 5F v (t)

+

_ 20

+

+

_

_

20u(-t) V

40 15

10 10 V

Example: Determine an expression for v(t) in the circuit below. Use the “shortcut method.”


Unit step response to a circuitSuppose that you wished to compare the outputs of two circuits. It would be misleading to compare them if the circuits had different inputs or different initial conditions. A common way to compare them is to use a unit step input [1u(t)V or 1u(t)A] and zero initial conditions (or zero initial stored energy).

Unit step response – the output of a circuit where the input is a unit step and there are zero initial conditions.

Circuit with zero initial conditions

Input is1u(t) V

or1u(t) A

Output (a voltage or a current) is the

“unit step response”


Example: Find the unit step response for VC(t) in the circuit shown below.

+ _ v(t) 1F VC(t)

+

_

1k


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Welcome to EGR 272 Circuit Theory II