EE 313 Linear Systems and Signals
Fall 2010
Differential Equations
Prof. Brian L. Evans
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
Initial conversion of content to PowerPoint
by Dr. Wade C. Schwartzkopf
Time-Domain Analysis
• For a system governed by a
linear constant coefficient
differential equation,
Total response 
zero - input response

when x ( t )  0
results from internal systemconditionsonly
independent of x ( t )
x(t)
y(t)
 zero - state response


responseto non- zero x ( t )
all initialconditionsare zero
dependent on x ( t )
Each component can be computed independently of other
System satisfies linearity property if zero-input response is
zero (i.e. all initial conditions are zero)
Zero-state response is convolution of impulse response and
input signal
5-2
Time-Domain Analysis
• Zero-input response
Response when x(t) = 0
Results from internal
system conditions only
Independent of x(t)
For most filtering
applications (e.g. your
stereo system), we
want a zero-valued
zero-input response
• Zero-state response
Response to non-zero x(t)
when system is relaxed
A system in zero state
cannot generate any
response for zero input
Zero state corresponds to
initial conditions being
zero
5-3
Zero-Input Response
• Simplest case (first-order equation)
d
y0 t   a1 y0 t   0
dt
 a1 t


y
t

C
e
• Solution: 0
d
y0 t    a1 y0 t 
dt
• For arbitrary constant C
How is C determined?
Could C be complex-valued?
• How about the following Nth-order equation?
dN
d N 1
d
yt   a1 N 1 yt     aN 1 yt   aN y t   x(t )
N
dt
dt
dt
5-4
Zero-Input Response
• For the Nth-order equation
Guess solution has form y0(t) = C e  t
Substitute form into differential equation
Factor common terms to obtain

dy0
 C e  t
dt
2
d y0
2 t

C

e
2
dt

k
d y0
k t

C

e
k
dt

N
N 1
t
C


a




a


a
e
1
N
 1N 
 0



non zero
non zero
Q  
y0(t) = C e t is a solution provided that Q() = 0
Factor Q() to obtain N characteristic roots
Q( )    1   2   N   0
5-5
Zero-Input Response
• Assuming that no two i terms are equal
y0 t   C1e 1t  C2 e 2t    C N e N t
• For repeated roots, solution changes
Simplest case of root  repeated twice:
y0 t   C1  C2 t  e t
With r repeated roots


y0 t   C1  C2t    Cr t r 1 e t
• Characteristic modes e  t
Determine zero-input response
Influence zero-state response
5-6
Zero-Input Response
• Could i be complex?
If complex, we can write it in Cartesian form
i   i  ji
Exponential solution e t becomes product of two terms
e it  e i  ji t  e it e jit 
 it
e
dampening term

cosi t   j sin i t 

oscillating term
• For conjugate symmetric roots, and conjugate
symmetric constants,
i t
 i t
1
C1e  C e
 2 C1 e it cosi t  C1 



 
dampening term
oscillating term
5-7
Zero-Input Solution Example
L
• Component values
L = 1 H, R = 4 , C = 1/40 F
Realistic breadboard components?
x(t)
R

C
y(t)
• Loop equations
(D2 + 4 D + 40) [y0(t)] = 0
• Characteristic polynomial
Envelope
2 + 4  + 40 =
( + 2 - j 6)( + 2 + j 6)
• Initial conditions
y(0) = 2 A
ý(0) = 16.78 A/s
y0(t) = 4 e-2t cos(6t - p/3) A
5-8
Impulse Response
• Response to unit impulse
x(t)
y(t)
Set x(t) = d(t) and solve for y(t)
• Linear constant coefficient differential equation
dN
d N 1
d
y t   a1 N 1 y t     a N 1 y t   a N y t  
N
dt
dt
bN  M
dt
dM
d M 1
d




x
t

b
x
t



b
xt   bN xt 
N  M 1
N 1
M
M 1
dt
dt
dt
• With zero initial conditions, impulse response is
h(t )  b0 d (t )  v(t ) u(t )
v(t )  bN  M
dM
d M 1
d
y0 t   bN  M 1 M 1 y0 t     bN 1 y0 t   bN y0 t 
M
dt
dt
dt
b0 is coefficient of dNx(t)/dtN term and could be 0
5-9
Impulse Response
• Where did b0 come from?
• In solving these differential equations for t  0,
x(t )  g (t ) u (t )
y (t )  m(t ) u (t )
• Funny things happen to y’(t) and y”(t)
y ' (t )  m' (t ) u (t )  m(t ) d (t )
y" (t )  m" (t ) u (t )  2 m' (t ) d (t )  m(t ) d ' (t )
• In differential equations class, solved for m(t)
Likely ignored d(t) and d’(t) terms
Solution for m(t) is really valid for t  0+
5 - 10