First-Order Circuits cont'd

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Transcript First-Order Circuits cont'd

First-Order Circuits Cont’d

Dr. Holbert April 17, 2006 ECE201 Lect-20 1

Introduction

• In a circuit with energy storage elements, voltages and currents are the solutions to linear, constant coefficient differential equations.

• Real engineers almost never solve the differential equations directly.

• It is important to have a qualitative understanding of the solutions.

ECE201 Lect-20 2

Important Concepts

• • The differential equation for the circuit

Forced

(particular) and

natural

(complementary) solutions •

Transient

and

steady-state

• 1st order circuits: the responses

time constant

(  ) ECE201 Lect-20 3

The Differential Equation

• Every voltage and current is the solution to a differential equation.

• In a circuit of order

n

, these differential equations have order

n

.

• The number and configuration of the energy storage elements determines the order of the circuit.

n

 # of energy storage elements ECE201 Lect-20 4

The Differential Equation

• Equations are linear, constant coefficient:

a n d n x

(

t

)

dt n

a n

 1

d n

 1

x

(

t

)

dt n

 1  ...

a

0

x

(

t

) 

f

(

t

) • The variable

x

(

t

) could be voltage or current.

• The coefficients

a n

through

a

0 depend on the component values of circuit elements.

• The function

f(t)

depends on the circuit elements and on the sources in the circuit.

ECE201 Lect-20 5

Building Intuition

• Even though there are an infinite number of differential equations, they all share common characteristics that allow intuition to be developed: – Particular and complementary solutions – Effects of initial conditions ECE201 Lect-20 6

Differential Equation Solution

• • • The total solution to any differential equation consists of two parts:

x

(

t

)

= x p

(

t

)

+ x c

(

t

)

Particular

(

forced

) solution is

x p

(

t

) – Response particular to a given source

Complementary

(

natural

) solution is

x

– Response common to all sources, that is, due to the “passive” circuit elements

c

(

t

) ECE201 Lect-20 7

The Forced Solution

• The forced (particular) solution is the solution to the non-homogeneous equation:

a n d n x

(

t

)

dt n

a n

 1

d n

 1

x

(

t

)

dt n

 1  ...

a

0

x

(

t

) 

f

(

t

) • The particular solution is usually has the form of a sum of

f(t)

and its derivatives.

– If

f(t)

is constant, then

v p (t)

is constant ECE201 Lect-20 8

The Natural Solution

• The natural (or complementary) solution is the solution to the homogeneous equation:

a n d n x

(

t

)

dt n

a n

 1

d n

 1

x

(

t

)

dt n

 1  ...

a

0

x

(

t

)  0 • Different “look” for 1 st and 2 nd order ODEs ECE201 Lect-20 9

First-Order Natural Solution

• The first-order ODE has a form of

dx c

(

t

)

dt

  1

x c

(

t

)  0 • The natural solution is

x c

• Tau (  ) is the

time constant

(

t

• For an RC circuit,  =

RC

• For an RL circuit,  =

L/R

) 

Ke

t

/  ECE201 Lect-20 10

Initial Conditions

• The particular and complementary solutions have constants that cannot be determined without knowledge of the initial conditions.

• The initial conditions are the initial value of the solution and the initial value of one or more of its derivatives.

• Initial conditions are determined by initial capacitor voltages, initial inductor currents, and initial source values.

ECE201 Lect-20 11

Transients and Steady State

• The steady-state response of a circuit is the waveform after a long time has passed, and depends on the source(s) in the circuit.

– Constant sources give DC steady-state responses • DC SS if response approaches a constant – Sinusoidal sources give AC steady-state responses • AC SS if response approaches a sinusoid • The transient response is the circuit response minus the steady-state response.

ECE201 Lect-20 12

Step-by-Step Approach

1.

2.

3.

4.

Assume solution (only dc sources allowed):

x

(

t

) = K 1 + K 2 e

-t/

 At

t

=0 – , draw circuit with

C L

as open circuit and as short circuit; find I L (0 – ) or V C (0 – ) At

t

=0 + , redraw circuit and replace

C

or

L

with appropriate source of value obtained in step #2, and find

x

(0)=K 1 +K 2 At

t

=  , repeat step #2 to find

x

(  )=K 1 ECE201 Lect-20 13

Step-by-Step Approach

5.

Find time constant (  ) Looking across the terminals of the

C

or

L

element, form Thevenin equivalent circuit;  =R Th C or  =L/R Th 6.

Finish up Simply put the answer together.

ECE201 Lect-20 14

Class Examples

• Learning Extension E7.3

• Learning Extension E7.4

• Learning Extension E7.5

ECE201 Lect-20 15