Download 2. Particular Solution

College of Communication Engineering
Undergraduate Course:
Signals and Linear Systems
Lecturer: Kunbao CAI
1
Chapter 2 Time-Domain Analysis of Continuous-Time Systems
Lecture 5
Main Contents:
2.1 Introduction
2.2 Linear Constant-Coefficient Differential
and Their Solutions
Equations
1. Mathematical Models
2. Nth-Order Linear Constant-Coefficient Differential
Equations
2.2.1 Classical Solution of the Differential Equations
1. Homogeneous Solution
2. Particular Solution
3. Complete Solution
4. An Example
2
Chapter 2 Time-Domain Analysis of Continuous-Time Systems
2.1 Introduction
What is the system analysis?
For a given system model with its input, the process to find
its response is called the system analysis.
When we read scientific literatures, we may meet some
terminologies such as the system identification and the system
synthesis that are beyond the scope of our course.
What is the time-domain analysis of the systems?
The so-called time-domain analysis of a continuous-time
system is that the solving procedure for its solution is totally
confined within the time domain.
From this definition, you may ask that if there is the system
analysis in other domains. The answer is yes. In this chapter,
however, we will concentrate our attention to the time-domain
analysis of the systems.
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2.1 Introduction (continued)
Mathematical Models of the Continuous-Time Systems
A large class of continuous-time systems can be
represented by three time-domain mathematical models as
follows.
(1) Linear constant-coefficient differential equations.
(2) Impulse response of LTI systems.
(3) State-variable formulations.
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2.2 Linear Constant-Coefficient Differential Equations
and Their Solutions
1. Mathematical Models
Figure 2.1 Parallel RLC circuit with a current source
The mathematical model relating its input is (t ) and output v(t )
can be determined as follows.
Step 1. Establish the relation between the current and voltage for
every passive circuit element.
iR (t )  v(t ) R
t
Inductor: iL (t )  1   v( )d
L
Capacitor: iC (t )  C dv (t ) dt
Resistor:
(2-1)
(2-2)
(2-3)
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1. Mathematical Models
Step 2. Using KCL (Kirchhoff’s Current Law) , we can obtain a current
equilibrium equation as
iR (t )  iL (t )  iC (t )  is (t )
(2-4)
Step 3. Substituting Eqs.(2-1), (2-2) and (2-3) into Eq.(2-4) and
applying some basic operations, we can obtain a second-order linear
differential equation of the given circuit, which is of the form
d 2 v(t ) 1 dv(t ) 1
1 dis (t )


v
(
t
)

(2-5)
dt 2
RC dt
LC
C dt
In the viewpoint of systems, this is called the single-input single-output
equation of the dynamic system described by the circuit in Figure 2.1 (on
slide 5).
▲ If a known input is applied to the system at time t  0 , then Eq.(2-5)
together with two initial conditions v (0  ) and v(0  ) is sufficient to
determine the system output v(t ) for t  0  uniquely .
▲ Furthermore, if v(0  )  0 and v(0  )  0 , then Eq.(2-5) together
with these two zero-initial conditions determines a causal LTI system with
order two.
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1. Mathematical Models (continued)
In this chapter, we consider a class of continuous-time dynamic
systems whose single-input single-output relation can be characterized by
an nth-order linear constant-coefficient differential equation
where
d n r (t )
d n 1r (t )
dr (t )

a



a
 a0 r (t )
n 1
1
dt n
dt n 1
dt
d m e(t )
d n 1e(t )
de(t )
 bm

b



b
 b0 e(t )
n 1
1
m
n

1
dt
dt
dt
(2-6)
r (t ) : system output or response;
e(t ) : system input or excitation;
ai ' s and bi ' s : constant coefficients (real).
▲ If a known input is applied to the system at t  0 , then Eq.(2-6)
together with n independent initial conditions
v(0  ), v(0  ),, v ( n 1) (0  )
can uniquely determine the system output r (t ) for t  0  .
▲ Furthermore, if all of the n initial conditions are equal to zero, then
Eq.(2-6) together with these zero-initial conditions determines an nth-order
causal LTI system.
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2.2.1 Classical Solution of the Differential Equations
Assume that a dynamic system is described by
n
d k r (t ) m d k e(t )
 ak dt k   bk dt k , (t  0  )
k 0
k 0
(2-7) (back to slide 11 )
where we have assumed, without loss of generality, that an  1 .
With its n initial conditions given by
r (0  ), r (0  ),, r ( n 1) (0  )
the complete solution of the system consists of two parts, that is,
r (t )  rh (t )  rp (t ), for t  0 
(2-8)
where
r (t ) : complete solution;
rh (t ) : homogeneous solution;
rp (t ) : particular solution.
The procedure to solve this mathematical problem is given as
follows.
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1. Homogeneous Solution
The homogeneous equation corresponding to the original
equation in Eq.(2-7) is given by
d n r (t )
d n 1r (t )
dr (t )

a



a
 a0 r (t )  0
n 1
1
dt n
dt n 1
dt
From this equation we can get the following
equation
n  an 1n 1    a1  a0  0
(2-9)
algebraic
(2-10)
This is called the characteristic equation of the system. Its n roots
1 , 2 ,, n
are called the characteristic roots of the systems, or the natural
frequencies of the system.
The function form of the homogeneous solution depends on
the multiplicity of characteristic roots.
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1. Homogeneous Solution (continued)
(1) Simple Characteristic Roots
If all of the characteristic roots are simple, then the
homogeneous solution is of the form
rh (t )  c1
e 1t
 c2
e 2 t
   cn
e n t

n
 ci e i t , t  0 
(2-11)
where ci ' s are n unknown constants which will be determined in a
later step.
(2) Repeated Characteristic Roots
Assuming that 1 is a repeated root with order k and the
remainder n  k roots are all simple, the homogeneous solution is
of the form
k
rh (t )   ci t k i e 1t 
i 1
k 1
n
 ci e i t , t  0 
i  k 1
(2-12)
Also, n unknown constants will be determined in a later step.
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2. Particular Solution
The function form of the particular solution which
satisfies the original differential equation in Eq.(2-7) (on slide 8)
depends on the excitation e(t ) . To find the particular solution, we
generally need to follow the steps:
(1) Substituting the known e(t ) into Eq.(2-7).
(2) By inspecting the function form on the right-hand side of the
equation resulted in step (1) and considering the characteristic roots,
we can select an appropriate particular solution with unknown
constants. This step has been assumed to be familiar with you, while
taking the course of advanced mathematics.
(3) Substituting the selected particular solution into the equation
obtained in the step (1), we can determine the values of the unknown
constants. Thus the particular solution can be completely determined.
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3. Complete Solution
The complete solution with n unknown constants is given by
n
r (t )  rh (t )  rp (t )   ci e i t  rp (t ), t  0 
(2-13)
Here we have assumed that all of the characteristic roots are simple.
Clearly, n independent initial conditions
(2-14)
r (0  ), r (0  ),, r ( n 1) (0  )
are needed to determine n constants in Eq.(2-13). There are several
methods to find the initial conditions at t  0  from the initial
conditions at t  0  , which will be introduced through several
Examples.
From Eqs.(2-13) and (2-14), we can obtain a set of
Simultaneous linear equations as follows.
i 1
r (0  )  c1  c2    cn  rp (0  )
r (0 )  c   c     c   r  (0 )

1 1
2 2
n n
p





r ( n 1) (0  )  c11n 1  c2 n2 1    cn nn 1  rp( n 1) (0  )
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Solving this set of equations yields n constant values. Thus, we
totally determine the complete solution.
Example 2.1 Consider a linear dynamic circuit shown in
Figure 2.2. The circuit has reached steady state at t  0  . If the
switch is moves from position 1 to position 2 at t  0 , calculate
(Back to
i (t ) for t  0  .
slide 16)
(Back to
slide 17)
Figure 2.2 The linear dynamic circuit for Example 2.1
Solution:
Step 1. Modeling the circuit for t  0  .
Applying two basic laws for the circuit analysis, we can
obtain following three equations.
13
Continue Example 2.1
KVL:
KCL:
R1i (t )  vC (t )  e2 (t )
di (t )
vC (t )  L L  R2 iL (t )
dt
dv (t )
i (t )  C C  iL (t )
dt
(2-15a)
(2-15b)
(2-15c)
From these three equations, we can get a differential equation as
follows.
d 2 i (t )
1  R2 ) di(t )  ( 1  R2 )i (t )

(
dt 2
R1C L dt
LC R1 LC
d 2 e2 (t ) R2 de2 (t )
1
1 e (t )



R1 dt 2
R1 L dt
R1 LC 2
Substituting the circuit-element parameters into the above equation,
we can obtain
d 2 i (t )
di(t )
d 2 e2 (t )
de2 (t )

7

10
i
(
t
)


6
 4e2 (t )
dt 2
dt
dt 2
dt
(2-16)
(t  0  )
14
Continue Example 2.1
Step 2. Find the homogeneous solution with undetermined
constants.
Characteristic equation: 2  7  10  0
Characteristic roots: 1  2 and 2  3
Homogeneous solution:
ih (t )  c1e  2t  c2 e 3t , t  0 
(2-17)
Step 3. Determine the particular solution.
Substituting e2 (t )  4V for t  0  into Eq.(2-16), we can get
d 2 i (t )
di(t )
(2-18)

7
 10i (t )  16, (t  0  )
dt 2
dt
According to the knowledge obtained in the course of advanced
calculus, the particular solution is of the form
i p (t )  A
Substituting this into Eq.(2-18), the particular solution is given by
i p (t )  8 / 5
15
Continue Example 2.1
Step 4. Find the complete solution .
The complete solution:
i (t )  ih (t )  i p (t )  c1e  2t  c2 e 3t  8 , t  0 
5
We need i (0  ) and i (0  ) to determine its two constants.
(2-19)
Since the circuit in Figure 2.2 has reached steady state
at t  0  , we know that:
the inductor: a short circuit;
(See slide 13.)
the capacitor: an open circuit.
Therefore, we have
i (0  )  i L (0  ) 
e1 (0  )
2

4A ,
R1  R2 1  (3 / 2) 5
i (0  )  0
and
vC ( 0  ) 
R2
e1 (0  )  3 / 2  2  6 V
R1  R2
1  (3 / 2)
5
16
Continue Example 2.1
After switching, the designed current satisfies
i (t )  [e2 (t )  vC (t )] / R1 for t  0 
(See slide 13.)
Since the inductor current and the capacitor voltage for the given
circuit can not abruptly change at t  0  , we have
i L (0  )  i L ( 0  )  4 A
5
Thus,
and
vC (0  )  vC (0  )  6 V
5
i (0  )  [e2 (0  )  vC (0  )] / R1
 [e2 (0  )  vC (0  )] / R1  (4  6 ) / 1  14 A
5
5
and
i (0  )  [e2 (0  )  vC (0  )] / R1  [e2 (0  )  1 iC (0  )] / R1
C
 {e2 (0  )  1 [i (0  )  iL (0  )]} / R1
C
 [0  1 (14  4 ) / 1  2 A/s
1 5 5
17
Continue Example 2.1
From these two initial conditions and Eq.(2-19)
i (t )  ih (t )  i p (t )  c1e  2t  c2 e 3t  8 , t  0 
5
(2-19)
we get a set of simultaneous linear equations
i (0  )  c1  c2  8  14
5 5

i (0  )  2c1  5c2  2
Solving this set of equations yields
c1  4
and c2   2
3
15
Substituting them into Eq.(2-19), we obtain the designed complete
Solution
i (t )  ( 4 e  2t  2 e 3t  8 )A for t  0 
3
15
5
(2-20)
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Thank you for
your attention!
The End!
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