Transcript Differential Equations - University of Michigan
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
2
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
First Order D.E.
U of M-Dearborn ECE Department
A
General First Order Linear Constant Coefficient Differential Equation
of
x(t)
has the form:
dx
(
t
)
x
(
t
)
dt f
(
t
) Where is a
constant given
and the function
f(t)
is 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
)
dt f
(
t
) The DE is a
linear combination
of x(t) and its derivative x(t) and its derivative are multiplied by
constants
There are
no cross products
dx
(
t
)
dt
2 In general the coefficient of
dx/dt
is
normalized
to 1 4
dx
(
t
)
dt
Differential Equations: First Order Systems
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)=x p (t)
is
any
solution to:
dx
(
t
)
x
(
t
)
dt f
(
t
) SOLUTION
x
(
t
)
x p
(
t
) and
x(t)=x c (t)
is any solution to the homogenous equation
dx
(
t
)
x
(
t
)
dt
0 SOLUTION
x
(
t
)
x c
(
t
) Then
x(t) = x p (t)+x c (t)
is also a solution to the original DE
dx
(
t
)
x
(
t
)
dt f
(
t
) SOLUTION
x
(
t
)
x p
(
t
)
x c
(
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
)
x p
(
t
)
dt
b
dx c
(
t
)
x c
(
t
) 0
dt
x p (t)
= Particular Integral Solution
x c (t)
= Complementary Solution 6
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Particular Integral Solution
dx p
(
t
)
x p
(
t
)
dt
b Since the right-hand side is a constant, it is reasonable to assume that
x p (t) must also be a constant
x p
(
t
)
K
1 Substituting yields:
K
1 b 7
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Complementary Solution
To
solve for x c (t)
dt dx c
(
t
)
x c
(
t
) rearrange terms 0
x c
1 (
t
)
dt dx c
(
t
) Which is equivalent to:
dt d
ln
x c
(
t
) Taking the exponential of both sides:
x c
(
t
)
e
t
c
e
t e c
Integrating both sides: ln
x c
(
t
)
t
c
Resulting in:
x c
(
t
)
K
2
e
t
8
dx
(
t
)
dt
Differential Equations: First Order Systems
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
)
dt
b and b are constants Has a
General Solution
of the form
x
(
t
)
K
1
K
2
e
t
K 1 and K 2 are constants 9
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Particular and Complementary Solutions
x
(
t
)
x
(
t
)
K
1
x p
(
t
)
K
2
e
t
x c
(
t
)
x p
(
t
)
K
1
Particular Integral Solution
x c
(
t
)
K
2
e
t
Complementary Solution
10
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
U of M-Dearborn ECE Department
Determining K
1
and K
2
In certain applications it may be possible to
directly determine
the constants K 1 and K 2
x
(
t
)
K
1
K
2
e
t
The first relationship can be seen by evaluating for t=0
x
( 0 )
K
1
K
2
e
0
K
1
K
2 The second by taking the limit as t approaches infinity
x
( )
t Lim
x
(
t
)
K
1
K
2
e
K
1
K
2
K
1 11
dx
(
t
)
dt
Differential Equations: First Order Systems
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
Takes the form:
x
(
t
)
K
1
K
2
e
t
and b are constants
K
1
K
2
x
( )
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 voltage
has a
linear relationship
and
current
between governed by Ohm’s Law
v R
(
t
)
v R
(
t
)
i R
(
t
)
R i R
(
t
)
R
13
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:
i c
(
t
)
C d
v c
(
t
)
dt C
i C
(
t
)
v C
(
t
) The current and voltage relationship for an
inductor L
is given by:
v L
(
t
)
L d
i L
(
t
)
dt L
i L
(
t
)
v L
(
t
) 14
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
v R
V DC t
0
R i C C
v C
15
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
Plan of Attack
U of M-Dearborn ECE Department
Write a
first-order differential equation
circuit
for time t>0
The solution will be of the form
K 1 +K 2 e -
t
for the These constants can be found by: Determining Determining
v c (0)
Determining
v c (
)
Finally
graph
the resulting
v c (t)
16
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
v R Ri C Ri C
(
t
) (
t
)
v C v C
(
t
) (
t
) (
t
)
v C
(
t
) (
V DC
V DC
V DC
) 0 0
V DC
v R R t
0
i C C
v C
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 v c (t), write the differential equation for the circuit in terms of v c (t) Replace
i = Cdv/dt
for capacitor current voltage relationship
Ri C
(
t
)
v C
(
t
)
V DC R
C dv c dt
(
t
)
v C
(
t
)
V DC
Rearrange terms to put DE in
Standard Form
dv c
(
t
)
dt
v C
(
t
)
RC
V DC RC
18
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
General Solution
dv c
(
t
)
dt v C
(
t
)
RC
V DC RC
U of M-Dearborn ECE Department
The solution will now take the standard form:
dx
(
t
)
x
(
t
) b
dt x
(
t
)
K
1
K
2
e
t
can be directly determined
K 1
and
K 2
depend on
v c (0)
and
v c (
)
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 v c (0 ) is the same as t approaching from the left v c (0 + )
v c
( 0 )
v c
( 0 ) Before the switch closes, the capacitor was
initially discharged
, therefore:
v c
( 0 ) 0
V
Substituting gives:
v c
( 0 ) 0
V
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 V DC voltage
v c
( )
V DC V DC
v R
R C i C
( ) 0
v C
( )
V DC t
No current will flow in the circuit because there will be no potential difference across the resistor,
v R (
) = 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 v c
( 0 ) 0
V v c
( )
V DC
Now solve for
K 1
K K
1 2
v v c c
( ( ) 0 )
v c
and
K 2
( )
K K
2 1
V DC
V DC
Replace to solve differential equation for v c (t)
v c
(
t
)
K
1
K
2
e
t v c
(
t
)
V DC
V DC 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
called the
Time Constant
t the constant 1/ is
v c
(
t
)
K
1
K
2
e
t
t t 1 t
= Time Constant K 1 = Steady State Solution
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 K 1 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
v c
(
t
)
V DC
V DC e
t RC
24
dx
(
t
)
dt
Differential Equations: First Order Systems
Math Review with Matlab
Summary
U of M-Dearborn ECE Department
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
capacitor
electrical circuit
a resistor and 25