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Math Review with Matlab:
Differential
Equations
First Order Constant
Coefficient Linear
Differential Equations
S. Awad, Ph.D.
M. Corless, M.S.E.E.
E.C.E. Department
University of Michigan-Dearborn
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
First Order Constant
Coefficient Linear
Differential Equations


First Order Differential Equations
General Solution of a First Order
Constant Coefficient Differential Equation

Electrical Applications

RC Application Example
2
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
First Order D.E.

A General First Order Linear Constant
Coefficient Differential Equation of x(t) has
the form:
dx(t )
 x(t )  f (t )
dt

Where  is a constant and the function f(t) is
given
3
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Properties



A General First Order Linear
Constant Coefficient DE of
x(t) has the properties:
dx(t )
 x(t )  f (t )
dt
The DE is a linear combination of x(t) and its derivative
x(t) and its derivative are multiplied by constants
 dx(t ) 


 dt 
2

There are no cross products

In general the coefficient of dx/dt is normalized to 1
4
Differential Equations: First Order Systems
dx (t )
dt
Math Review with Matlab
U of M-Dearborn ECE Department
Fundamental Theorem

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:
dx(t )
 x(t )  f (t ) SOLUTION
dt

and x(t)=xc(t) is any solution to the homogenous equation
dx(t )
 x(t )  0
dt

x(t )  x p (t )
SOLUTION
x(t )  xc (t )
Then x(t) = xp(t)+xc(t) is also a solution to the original DE
dx(t )
 x(t )  f (t ) SOLUTION
dt
x(t )  x p (t )  xc (t )
5
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
f(t) = Constant Solution

If f(t) = b (some constant) the general solution to the
differential equation consists of two parts that are
obtained by solving the two equations:
dx p (t )
dt
 x p (t )  b
dxc (t )
 xc (t )  0
dt
xp(t) = Particular Integral
Solution
xc(t) = Complementary
Solution
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dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Particular Integral Solution
dx p (t )
dt

 x p (t )  b
Since the right-hand side is a constant, it is reasonable
to assume that xp(t) must also be a constant
x p (t )  K1

Substituting yields:
b
K1 

7
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Complementary Solution

To solve for xc(t) rearrange terms
dt
c (t )  0


x
dxc (t )

Which is equivalent to:
dt
ln xc (t )  

d

Taking the exponential
of both sides:
xc (t )  e t  c  e t e c

 xc (t )  dt


 
c
 1  dx (t )
Integrating both sides:
ln xc (t )  t  c

Resulting in:
xc (t )  K 2e t
8
Differential Equations: First Order Systems
dx (t )
dt
Math Review with Matlab
U of M-Dearborn ECE Department
First Order Solution Summary

A General First-Order Constant Coefficient
Differential Equation of the form:
dx(t )
 x(t )  b
dt

 and b are constants
Has a General Solution of the form
x(t )  K1  K 2e
t
K1 and K2 are constants
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dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Particular and Complementary
Solutions
x(t )  K1  K 2e
t
x(t )  x p (t )  xc (t )
x p (t )  K1
xc (t )  K 2e
Particular Integral Solution
Complementary Solution
t
10
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Determining K1 and K2

In certain applications it may be possible to directly
determine the constants K1 and K2
x(t )  K1  K 2e

The first relationship can be seen by evaluating for t=0
x(0)  K1  K 2e

t
 0
 K1  K 2
The second by taking the limit as t approaches infinity
x()  Lim x(t )  K1  K 2e
t 

 K1  K 2 0  K1
11
Differential Equations: First Order Systems
dx (t )
dt
Math Review with Matlab
U of M-Dearborn ECE Department
Solution Summary

By rearranging terms, we see that given particular
conditions, the solution to:
dx(t )
 x(t )  b
dt

 and b are constants
Takes the form:
x(t )  K1  K 2e
t
K1  x()
K 2  x(0)  x()
12
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Electrical Applications


Basic electrical elements such as resistors (R),
capacitors (C), and inductors (L) are defined by
their voltage and current relationships
A Resistor has a linear relationship between
voltage and current governed by Ohm’s Law
vR (t )  iR (t ) R
 vR (t ) 
iR (t )
R
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dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Capacitors and Inductors



A first-order differential equation is used to describe
electrical circuits containing a single memory storage
elements like a capacitors or inductor
The current and voltage
relationship for a
capacitor C is given by:
The current and voltage
relationship for an
inductor L is given by:
d vc (t ) 
ic (t )  C
dt
d iL (t ) 
vL (t )  L
dt
iC (t )
C

vC (t )

iL (t )
L

vL (t )

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dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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
VDC
t 0
 vR

R
iC
C

vC

15
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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 vc(0)
 Determining vc()
 Determining

Finally graph the resulting vc(t)
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dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Equation for t > 0



Kirchhoff’s Voltage Law (KVL) states that the sum of the
voltages around a closed loop must equal zero
Ohm’s Law states that the voltage across a resistor is
directly proportional to the current through it, V=IR
Use KVL and Ohm’s Law to write an equation describing
the circuit after the switch closes
vR (t )  vC (t )  (VDC )  0
RiC (t )  vC (t )  VDC  0
RiC (t )  vC (t )  VDC
 vR

R
iC
VDC
C

vC

t 0
17
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Differential Equation



Since we want to solve for vc(t), write the
differential equation for the circuit in terms of vc(t)
Replace i = Cdv/dt
for capacitor current
voltage relationship
Rearrange terms to
put DE in Standard
Form
RiC (t )  vC (t )  VDC
 dvc (t ) 
R C
 vC (t )  VDC

dt 

dvc (t ) vC (t ) VDC


dt
RC
RC
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Differential Equations: First Order Systems
dx (t )
dt
Math Review with Matlab
U of M-Dearborn ECE Department
General Solution
dvc (t ) vC (t ) VDC


dt
RC
RC

The solution will now take the standard form:
dx(t )
 x(t )  b
dt
x(t )  K1  K 2e

 can be directly determined

K1 and K2 depend on vc(0) and vc()
t
1

RC
19
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Initial Condition




A physical property of a capacitor is that voltage cannot
change instantaneously across it
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+)
Before the switch closes, the capacitor
was initially discharged, therefore:
Substituting gives:
vc (0  )  vc (0  )

vc (0 )  0V

vc (0 )  0V
20
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Steady State Condition

As t approaches infinity, the capacitor will fully charge to
the source VDC voltage
 vR
vc ()  VDC

iC ()  0
R
VDC
C

vC ()  VDC

t 

No current will flow in the circuit because there will be no
potential difference across the resistor, vR() = 0 V
21
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Solve Differential Equation
1

RC



vc (0 )  0V
vc ()  VDC
Now solve for K1 and K2
K1  vc ()
K1  VDC
K 2  vc (0)  vc ()
K 2  VDC
Replace to solve differential equation for vc(t)
vc (t )  K1  K 2 e
t
vc (t )  VDC  VDC e
t
RC
22
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Time Constant

When analyzing electrical circuits the constant 1/ is
called the Time Constant t
vc (t )  K1  K 2e
t
t
K1 = Steady State Solution


t
1

t = Time Constant
The time constant determines the rate at which the
decaying exponential goes to zero
Hence the time constant determines how long it takes to
reach the steady state constant value of K1
23
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Plot Capacitor Voltage

For First-order RC circuits the Time Constant t = 1/RC
vc (t )  VDC  VDC e
t
RC
24
dx (t )
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
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
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