Download CLASSICAL SOLUTION OF 1st and 2nd ORDER DEQ`S OBTAINED

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SIGNALS AND SYSTEMS
3. CLASSICAL SOLUTION OF 1st and 2nd ORDER DEQ’S OBTAINED VIA CIRCUIT
ANALYSIS
Introduction
The kinds of DEQ’s considered here are obtained from circuit analysis. They will be a
continuous system that will be linear, causal and time invariant (superposition applies,
changes in excitation signals now can’t change the past and circuit parameters are
constant over time). We consider the classical solution instead of Laplace solutions as
this is part of a sequence that leads to Laplace and other advanced methods. It is assumed
that the student has knowledge of basic circuit analysis via DC or AC techniques.
Obtaining The DEQ From A Circuit
In your Circuits I course you developed proficiency in the analysis of DC and AC
circuits. We will build on that. You were taught to use the circuit laws with impedance as
follows:
L  j L
1
C
j C
Now you will do the same with the following change:
L  sL
1
C
sC
j has been replaced by the letter s . The reason will become clear shortly. With this
change, write the equations and do the algebra to solve symbolically for a desired
response. Consider the following example:
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Figure 1. A simple LRC circuit
We use the voltage divider to obtain an expression for VOUT :
VOUT 
1
sC
V
1 IN
R  sL 
sC
Do some algebra to clean it up:
VOUT
(1)
1
LC

VIN
R
1 
 2
s  s 

L LC 

(2)
Identify the system operator:
s2  s
R
1

L LC
(3)
and multiply both sides by that expression:
R
1 
1

VOUT  s 2  s 
VIN

L LC  LC

(4)
Now let the system operator operate on the unknown, VOUT with: s 

d2
d
and s 2  2
dt
dt

(we will use the dot notation for derivatives: V OUT and V OUT ). Thus we obtain the DEQ
for VOUT :

V OUT 
R 
1
1
V OUT 
VOUT 
VIN
L
LC
LC
(5)
The procedure illustrated by this simple example will obtain the system DEQ for most
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circuits. You may find that the numerator, given by
in this simple example, may also
LC
be an operator that acts upon the forcing function V IN and you could end up with an


overall forcing function that has additional terms of V IN or V IN . Keep in mind that V IN
is a known and one would use superposition to handle these cases.
In general the approach taken to solve, is to obtain the forced response and the natural
response. The total response is the sum of the two. The natural response will be due to
initial conditions and the forced response is from V IN and is essentially what remains
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after all transients from initial conditions die away (assuming the system is stable). The
forced response can be solved complete (no constants TBD). The natural response will
have some constants that are TBD. The sum of these 2 responses is assumed to come
from superposition. For the natural response, the forcing function V IN is set to zero and
the resulting homogeneous DEQ is solved. The resulting natural solution will have
constants (the number is set by the highest power of derivative) and are obtained by
forming the total solution and fitting it to boundary conditions. We look at the 1st order
case first.
VTotal  VNat  VForced
(5a)
Solution of 1st Order Systems
First order systems come from analysis of circuits that have only one energy storage
element ( L or C ). They will result in DEQ’s of the following general form, where the
constants ( A, B, C ) are known parameters.

A X OUT  BX OUT  CVIN
(6)
In general solutions (forced or natural) are obtained by “cut and try”. We guess and plug
the guess in to see if it works. For the forced response the guess is that the response is
simply a scaling of the forcing function. Let’s assume that V IN is a step function. Thus
the forced response is guessed to be a constant.
d
0
dt




A X FORCED  BX FORCED  CVIN  BX FORCED  CVIN  X FORCED 
C
VIN
B
(7)
The homogeneous DEQ is given by Equation 8:

A X NAT  BX NAT  0
(8)
For the natural solution we assume the following guess as a solution:
X NAT  Ke t
(9)
Equation 9. is plugged into Equation 8. and results in:
 AKe t  BKe t  0   A  B  0
B

A
(10)
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We assume that the boundary condition of the output at time=0 is known:
X OUT (t  0)  X O
(11)
Form the total solution as per Equation 5a and plug in the boundary condition:
X TOTAL  Ke
B
 t
A

C
VIN
B
X TOTAL (t  0)  Ke
K  XO 
B
 ( t 0 )
A
(12)

C
VIN  X O
B
C
VIN
B
(13)
Thus the final solution is:
C

  t C
  X O  VIN e A  VIN
B
B


B
X OUT
(14)
In general for the first order system the natural response will always be of the form of
Equation 9. For the forced response the following guess table will help:
Table 1. Forced Response guess
V IN
X GUESS
Comments
VDC
K
Plug in DEQ and solve for K
VIN cos  t
Ke j t
ditto above but K  K e j

X FORCED  real K e j ( t  )
VIN sin  t
Ke j t
ditto above but


X FORCED  imag K e j ( t  )

VIN e t
Ke t
Plug in and solve for K
VIN t
K O  K1t
Plug in and equate coefficients of like
powers of t to solve for K O , K1
For a first order system obtained from electrical circuit equations where the forcing
function is of the form of a step the solution will be obtained from the following:
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X OUT  X SS   X IN  X SS e

t

(15)
Where: X OUT : is the desired output
X SS : is the output in the steady state
X IN : is the initial output at t  0 
 : is the circuit time constant (for a capacitor:   REQ C , for an
inductor:  
L
)
REQ
REQ : is the equivalent resistance seen by the L or C at t  0 
To find X SS , where t   , replace L with a short circuit and C with an open circuit and
solve the circuit for X OUT . To solve for X IN note that the equivalent resistance of a
battery is a short and the equivalent resistance of a current source is an open. Analyze the
circuit at time zero with L replaced by a current source with the current through it at
t  0  and C replaced by a voltage source with the voltage it had at t  0  . Solve the
resulting circuit for X IN . If there is no initial inductor current or capacitor voltage then at
t  0  the capacitor is a short and the inductor is an open.
Solution of 2nd Order DEQ’s
2nd order DEQ’s come about from analyzing electrical circuits with 2 energy storage
elements. It is possible that a circuit with either 2 L ’s or 2 C ’s, that those elements could
combine into a single equivalent L or C and result in a first order system. The form of the
2nd order DEQ is just like Equation 5. The characteristic polynomial for Equation 5 is
given by Equation 3. For the first order system the form of the natural response is limited
to one option. For the 2nd order system there are 3 choices and which one fits is
determined from the characteristic polynomial. The general form of the characteristic
polynomial is given by Equation 16:
s 2  2 O s   O2
(16)
The general form for the homogeneous DEQ from which that characteristic polynomial
came from is given by Equation 17.


V  2 O V   O2 V  0
Where V : is the desired unknown response
 O : is a constant called the natural frequency, rads/sec
 : is a constant called the dampening coefficient, unit less
(17)
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The solution to Equation 17 is obtained exactly as with the 1st order DEQ case: guess the
form and plug into the homogeneous equation. If the guess of Equation 9 were used it can
be shown that the roots of Equation 16 determine the form of the natural solution.
Equation 16 is a quadratic, so the roots will be real and distinct, real and repeated or
complex conjugates. The value of  will determine which of the 3 cases it will be.
Case 1:   1 Roots are real and distinct. Thus the characteristic polynomial factors via
it’s roots as given by Equation 18.
s 2  2 O s   O2  (s  s1 )(s  s2 )
(18)
For this case the natural solution of Equation 17 is given by Equation 19.
VN  K1e  s1t  K 2 e  s2t
(19)
Where: K 1 and K 2 are constants TBD via boundary conditions
Case 2:   1 Roots are real and repeated. Thus the characteristic polynomial factors via
it’s roots as given by Equation 20.
s 2  2 O s   O2  (s  sO ) 2
(20)
For this case the natural solution of Equation 17 is given by Equation 21.
VN  K1e  sOt  K 2 te  sOt
(21)
Where: K 1 and K 2 are constants TBD via boundary conditions
Case 3:   1 Roots are complex conjugates. Thus the characteristic polynomial factors
via it’s roots as given by Equation 22.
s 2  2 O s   O2  (s    j )(s    j )
(22)
For this case the natural solution of Equation 17 is given by Equation 23.
VN  e t ( K1 cos t  K 2 sin t )
Where: K 1 and K 2 are constants TBD via boundary conditions
(23)
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To solve the complete DEQ add the force response to the natural response as per
Equation 5a and then use the boundary conditions to solve for constants K 1 and K 2 . The
forced response is obtained using the same technique as with 1st order systems. Pick the
form of the forced response from Table 1, plug the form into the original DEQ and then
solve for the constant K . Let’s work some examples with the forcing function always a
unit step.
Example 1: Real distinct roots
Given the following characteristic polynomial that factors into 2 distinct real roots:
(s  3)(s  2)
(24)
The DEQ that this characteristic polynomial might come from is given by:


V  5V  6V  6u (t )
(25)
Boundary conditions are given by:

V (0)  1, V (0)  2
(26)
First we find the forced response. Using Table 1:
VFORCED  K , a constant
Now plug into the original DEQ:
d2
d
( K )  5 (K )  6K  6  K  1
2
dt
dt




 
0
(27)
0
Now form the total response as per Equation 5a and use the boundary conditions to find
K 1 and K 2 .
VTOTAL  K1e 3t  K 2 e 2t  1
(28)
Plug in time =0:
VTOTAL (0)  1  K1  K 2  1  0  K1  K 2
Now take a derivative of Equation 28 and plug in the other boundary condition:
(29)
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
V TOTAL  3K1e 3t  2K 2 e 2t

V TOTAL (0)  2  3K1  2K 2
(30)
Equations 29 and 30 are solved simultaneously to yield:
1   K1 
0   1
 2    3  2  K 
  
 2 
(31)
The MATLAB syntax is:
c=[0;2];
a=[1,1;-3,-2];
k=a^-1*c
k = -2.0000
2.0000
The final result is:
VTOTAL  2e 3t  2e 2t  1
(32)
Example 2: Real repeated roots
Given the following characteristic polynomial that factors into 2 repeated real roots:
( s  2) 2
(33)
The DEQ that this characteristic polynomial might come from is given by:


V  4V  4V  4u (t )
(34)
Boundary conditions are given by:

V (0)  1, V (0)  2
First we find the forced response. Using Table 1:
VFORCED  K , a constant
Now plug into the original DEQ:
(35)
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d2
d
( K )  4 ( K )  4K  4  K  1
2
dt
dt




 
(36)
0
0
Now form the total response as per Equation 5a and use the boundary conditions to find
K 1 and K 2 .
VTOTAL  K1e 2t  K 2 te 2t  1
(37)
Plug in time =0:
VTOTAL (0)  1  K1  1  K1  0
(38)
Now take a derivative of Equation 28 and plug in the other boundary condition:

V TOTAL  K 2 e 2t  2 K 2 te 2t

V TOTAL (0)  2  K 2
(39)
The final result is:
VTOTAL  2te 2t  1
(40)
Example 3: complex conjugate roots
Given the following characteristic polynomial that factors into 2 complex conjugate
roots:
(s  2  j 2)(s  2  j 2)
(41)
The DEQ that this characteristic polynomial might come from is given by:


V  4V  8V  8u (t )
(42)
Boundary conditions are given by:

V (0)  1, V (0)  2
First we find the forced response. Using Table 1:
(43)
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VFORCED  K , a constant
Now plug into the original DEQ:
d2
d
( K )  4 ( K )  8K  8  K  1
2
dt
dt




 
0
(44)
0
Now form the total response as per Equation 5a and use the boundary conditions to find
K 1 and K 2 .
VTOTAL  e 2t ( K1 cos 2t  K 2 sin 2t )  1
(45)
Plug in time =0:
VTOTAL (0)  1  K1  1  K1  0
(46)
Now take a derivative of Equation 45 and plug in the other boundary condition:

V TOTAL  2e 2t ( K 2 sin 2t )  e 2t (2K 2 cos 2t )

V TOTAL (0)  2  2 K 2  K 2  1
(47)
The final result is:
VTOTAL  e 2t (sin 2t )  1
(48)
In the future a similar write-up for the classical solution of difference equations obtained
from DEQ’s will be done.