s. awad, ph.d. m. corless, m.s.e.e. e.c.e. department university of michigan-dearborn differential...

S. Awad, Ph.D.

M. Corless, M.S.E.E.

E.C.E. Department

University of Michigan-Dearborn

Differential Equations

Math Review with Matlab:

First Order Constant Coefficient Linear


U of M-Dearborn ECE DepartmentMath Review with Matlab

2

Differential Equations: First Order Systemsdt

tdx )(

First Order Constant Coefficient Linear


First Order Differential Equations

General Solution of a First Order Constant Coefficient Differential Equation

Electrical Applications

RC Application Example


3


tdx )(

First Order D.E. A General First Order Linear Constant Coefficient

Differential Equation of x(t) has the form:

)()()(

tftxdt

tdx

Where is a constant and the function f(t) is given


4


tdx )(

In general the coefficient of dx/dt is normalized to 1

Properties

The DE is a linear combination of x(t) and its derivative x(t) and its derivative are multiplied by constants

)()()(

tftxdt

tdx

A General First Order Linear Constant Coefficient DE of x(t) has the properties:

2)(

dt

tdx There are no cross products


5


tdx )(

SOLUTION

SOLUTION

SOLUTION

Fundamental Theorem

)()()(

tftxdt

tdx

A fundamental theorem of differential equations states that given a differential equation of the form below where x(t)=xp(t) is any solution to:

0)()(

txdt

tdx

and x(t)=xc(t) is any solution to the homogenous equation

Then x(t) = xp(t)+xc(t) is also a solution to the original DE

)()()(

tftxdt

tdx

)()( txtx p

)()( txtx c

)()()( txtxtx cp


6


tdx )(

f(t) = Constant Solution If f(t) = (some constant) the general solution to the

differential equation consists of two parts that are obtained by solving the two equations:

)()(

txdt

tdxp

p

0)()(

txdt

tdxc

c

xp(t) = Particular Integral

Solution

xc(t) = Complementary

Solution


7


tdx )(

Particular Integral Solution

Since the right-hand side is a constant, it is reasonable to assume that xp(t) must also be a constant

)()(

txdt

tdxp

p

1)( Ktxp

Substituting yields:

1K


8


tdx )(

Complementary Solution To solve for xc(t) rearrange terms

0 ) () (

t xdt

t dxc

c

dt

t dx

t xc

c

) (

) (

1 Which is equivalent to:

) ( lnt xdt

dc c t t xc ) ( ln

Integrating both sides:

c t c tce e e t x

) (

tce K t x

2 ) (

Taking the exponential of both sides:

Resulting in:


9


tdx )(

First Order Solution Summary A General First-Order Constant Coefficient

Differential Equation of the form:

)()(

txdt

tdx and are constants

teKKtx 21)(

Has a General Solution of the form

and are constants


10


tdx )(

Particular and Complementary Solutions

1)( Ktxp tc eKtx 2)(

Particular Integral Solution Complementary Solution

)()()(

)( 21

txtxtx

eKKtx

cp

t


11


tdx )(

Determining K1 and K2 In certain applications it may be possible to directly

determine the constants K1 and K2

teKKtx 21)(

210

21)0( KKeKKx

The second by taking the limit as t approaches infinity

12121 0)()( KKKeKKtxLimxt

The first relationship can be seen by evaluating for t=0


12


tdx )(

Solution Summary By rearranging terms, we see that given particular conditions, the

solution to:

)()(

txdt

tdx and are constants

Takes the form:

teKKtx 21)(

)()0(

)(

2

1

xxK

xK


13


tdx )(

A Resistor has a linear relationship between voltage and current governed by Ohm’s Law

Electrical Applications Basic electrical elements such as resistors (R), capacitors (C), and

inductors (L) are defined by their voltage and current relationships

Rtitv RR )()(

)(tiR R

)(tvR


14


tdx )(

Capacitors and Inductors

The current and voltage relationship for a capacitor C is given by:

dt

tvdCti c

c

)()(

The current and voltage relationship for an inductor L is given by:

dt

tidLtv L

L

)()(

A first-order differential equation is used to describe electrical circuits containing a single memory storage elements like a capacitors or inductor

)(tiC

)(tvCC

)(tiL

)(tvL

L


15


tdx )(

RC Application Example Example: For the circuit below, determine an

equation for the voltage across the capacitor for t>0. Assume that the capacitor is initially discharged and the switch closes at time t=0

R

C

CvDCVCi

Rv

0t


16


tdx )(

Plan of Attack

Write a first-order differential equation for the circuit for time t>0

The solution will be of the form K1+K2e-t

These constants can be found by: Determining Determining vc(0)

Determining vc()

Finally graph the resulting vc(t)


17


tdx )(

Equation for t > 0

Use KVL and Ohm’s Law to write an equation describing the circuit after the switch closes

Kirchhoff’s Voltage Law (KVL) states that the sum of the voltages around a closed loop must equal zero

R

C

CvDCVCi

Rv

0tDCCC

DCCC

DCCR

VtvtRi

VtvtRi

Vtvtv

)()(

0)()(

0)()()(

Ohm’s Law states that the voltage across a resistor is directly proportional to the current through it, V=IR


18


tdx )(

Differential Equation Since we want to solve for vc(t), write the

differential equation for the circuit in terms of vc(t)

DCCc

DCCC

Vtvdt

tdvCR

VtvtRi

)()(

)()( Replace i = Cdv/dt for capacitor current voltage relationship

Rearrange terms to put DE in Standard Form

RC

V

RC

tv

dt

tdv DCCc )()(


19


tdx )(

General Solution

The solution will now take the standard form:

RC

V

RC

tv

dt

tdv DCCc )()(

)()(

txdt

tdxteKKtx 21)(

RC

1

can be directly determined

K1 and K2 depend on vc(0) and vc()


20


tdx )(

Initial Condition A physical property of a capacitor is that voltage cannot

change instantaneously across it

Before the switch closes, the capacitor was initially discharged, therefore:

Vvc 0)0(

Therefore voltage is a continuous function of time and the limit as t approaches 0 from the right vc(0-) is the same as t approaching from the left vc(0+)

)0()0( cc vv

Substituting gives: Vvc 0)0(


21


tdx )(

Steady State Condition As t approaches infinity, the capacitor will fully charge to

the source VDC voltage

No current will flow in the circuit because there will be no potential difference across the resistor, vR() = 0 V

DCc Vv )( R

C

DCC Vv )(DCV

0)( Ci

Rv

t


22


tdx )(

Solve Differential Equation

Now solve for K1 and K2

Vvc 0)0( DCc Vv )(

RC

1

)()0(

)(

2

1

cc

c

vvK

vK

DC

DC

VK

VK

2

1

Replace to solve differential equation for vc(t)

tc eKKtv 21)( RC

t

DCDCc eVVtv

)(


23


tdx )(

Time Constant When analyzing electrical circuits the constant 1/ is

called the Time Constant

The time constant determines the rate at which the decaying exponential goes to zero

t

c eKKtv

21)( 1

K1 = Steady State Solution = Time Constant

Hence the time constant determines how long it takes to reach the steady state constant value of K1


24


tdx )(

Plot Capacitor Voltage For First-order RC circuits the Time Constant = 1/RC

RCt

DCDCc eVVtv

)(


25


tdx )(

Summary Discussed general form of a first order

constant coefficient differential equation

Proved general solution to a first order constant coefficient differential equation

Applied general solution to analyze a resistor and capacitor electrical circuit

s. awad, ph.d. m. corless, m.s.e.e. e.c.e. department university of michigan-dearborn differential...

Documents

constant n

order systems

order solution summary

general solution n

constants n

terms n

complementary solution

inductors n