1 Lecture #1 EGR 272 – Circuit Theory II Reading Assignment: Chapter 8 in Electric Circuits, 6th Edition by Nilsson Welcome to EGR 272 Circuit Theory II

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Lecture #1 EGR 272 – Circuit Theory II

Reading Assignment: Chapter 8 in Electric Circuits, 6th Edition by Nilsson

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EGR 272Circuit Theory II

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EGR 271: Chapters 1 - 7 in Electric Circuits, 6th Edition by Nilsson EGR 272: Chapters 8 - 15 in Electric Circuits, 6th Edition by Nilsson

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Review - Key relationships for capacitors and inductors

Capacitors Inductors

Current

Voltage

Power p(t) = v(t)i(t) p(t) = v(t)i(t)

Energy

Cannot change vC(0+) = vC(0-) iL(0+) = iL(0-)

instantaneously

Steady-state behavior open-circuit short-circuit

dvi(t) C

dt

div(t) L

dt

0

1v(t) i(t)dt v(0)

C

t

0

1i(t) v(t)dt i(0)

L

t

21W CV

2 21

W Li2

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Chapter 8 – Second-Order Circuits

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)

Example: Draw several circuits with C’s and L’s and identify the order of each circuit.

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Lecture #1 EGR 272 – Circuit Theory II• 2nd-order circuits have 2 independent energy storage elements (inductors and/or

capacitors)

• Analysis of a 2nd-order circuit yields a 2nd-order differential equation (DE)

• A 2nd-order differential equation has the form:

• Solution of a 2nd-order differential equation requires two initial conditions: x(0) and x’(0)

• All higher order circuits (3rd, 4th, etc) have the same types of responses as seen in 1st-order and 2nd-order circuits

• Since 2nd-order circuits have two energy-storage types, the circuits can have the following forms:

1) Two capacitors2) Two inductors3) One capacitor and one inductor

A) Series RLC circuit *B) Parallel RLC circuit *C) Others

* The textbook focuses on these two types of 2nd-order circuits

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1 o2

d x dx a a x(t) f(t)

dt dt

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Form of the solution to differential equationsAs seen with 1st-order circuits in Chapter 7, the general solution to a differential equation has two parts:

x(t) = xh + xp = homogeneous solution + particular solution

or x(t) = xn + xf = natural solution + forced solution

where xh or xn is due to the initial conditions in the circuit

and xp or xf is due to the forcing functions (independent voltage and current sources for

t > 0).

The forced response The forced response is due to the independent sources in the circuit for t > 0. Since the natural response will die out once the circuit reaches steady-state, the forced response can be found by analyzing the circuit at t = . In particular,xf = x()

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The natural responseA 2nd-order differential equation has the form:

where x(t) is a voltage v(t) or a current i(t).

To find the natural response, set the forcing function f(t) (the right-hand side of the DE) to zero.

Substituting the general form of the solution Aest yields the characteristic equation:

s2 + a1s + ao = 0

Finding the roots of this quadratic (called the characteristic roots or natural frequencies) yields:

The roots of the quadratic equation above may be real and distinct, repeated, or complex. Thus, the natural response to a 2nd-order circuit has 3 possible forms:

2

1 o2

d x dx a a x(t) f(t)

dt dt

2

1 o2

d x dx a a x(t) 0

dt dt

2

1 1 o1 2

-a a - 4as , s

2

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1) Overdamped response

Roots are real and distinct [ (a1)2 > 4ao ]

Solution has the form:

Sketch the form of the solution.

Discuss the concept of the dominant root.

1 2s sn 1 2x A e A et t

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2) Critically damped response

Roots are repeated [ (a1)2 = 4ao ] so s1 = s2 = s = -a1/2

Solution has the form:


sn 1 2x A t A e t

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3) Underdamped response

Roots are complex [ (a1)2 < 4ao ] so s1 , s2 = j

Show that the solution has the form:


Discuss the concept of the exponential envelope.

Sketch xn if A1 = 0, A2 = 10, =-1, and = .

Sketch xn if A1 = 0, A2 = 10, =-10, and = 100.

n 1 2x e A cos( t) A sin( t) t

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Illustration: The transient response to a 2nd-order circuit must follow one of the forms indicated above (overdamped, critically damped, or underdamped). Consider the circuit shown below. v(t) is 0V for t < 0 and the steady-state value of v(t) is 10V. How does it get from 0 to 10V?

Discuss the possible responses for v(t)

Define the terms damping, rise time, ringing, and % overshoot

+ _ 10 V

t = 0

C v (t)

+

_

R L

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Examples: When is each of the 3 types of responses desired? Discuss the following cases:

• An elevator

• A cruise-control circuit

• The output of a logic gate

• The start up voltage waveform for a DC power supply

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Lecture #1 EGR 272 – Circuit Theory IIProcedure for analyzing 2nd-order circuits

1. Write the differential equation for x(t). It is generally easiest to solve for inductor currents or

capacitor voltages and then to use the result to find other variables. Note: If the circuit is a

series RLC circuit or a parallel RLC circuit, standard formulas for and wo can be used to find

the characteristic equation: s2 + 2s + wo2.

2. Find the initial conditions, x(0) and x’(0).

A) Find x(0) by analyzing the circuit at t = 0- (find all capacitor voltages and inductor currents)

B) Find x’(0) by analyzing the circuit at t = 0+ (use the values for vC and iL found at t = 0- in the circuit).

3. Solve the differential equation.

A) Find the forced response, xf , by analyzing the circuit at t = . xf = x().

B) Find the natural response, xn

1) set the right side of the DE to zero

2) find the characteristic equation and the roots of the equation

3) assume that xn has the standard form of the solution for an overdamped, underdamped, or critically damped circuit. There will be two unknowns in this solution.

C) Find the total response, x(t) = xn + xf . Use the two initial conditions to solve for the two unknowns in the total response.

Next class: detailed circuit examples

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1 Lecture #1 EGR 272 – Circuit Theory II Reading Assignment: Chapter 8 in Electric Circuits, 6th Edition by Nilsson Welcome to EGR 272 Circuit Theory II