leo lam © 2010-2011 signals and systems ee235. leo lam © 2010-2011 breeding what do you get when...
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Leo Lam © 2010-2011
Signals and Systems
EE235
Leo Lam © 2010-2011
Breeding
• What do you get when you cross an elephant and a zebra?
• Elephant zebra sin theta
Leo Lam © 2010-2011
Today’s menu
• Friday: LCCDE Zero-Input Response• Today: LCCDE Zero-State Response (forced)
Leo Lam © 2010-2011
Zero input response (example)
4
• 4 steps to solving Differential Equations:
• Step 1. Find the zero-input response = natural response yn(t)
• Step 2. Find the Particular Solution yp(t)• Step 3. Combine the two• Step 4. Determine the unknown constants using initial
conditions
)()()( tytyty pn
Leo Lam © 2010-2011
From Friday (example)
5
• Solve
• Guess solution:
• Substitute:
• We found:
• Solution:
Characteristic roots = natural frequencies/
eigenvalues
Unknown constants:Need initial conditions
Leo Lam © 2010-2011
Zero-state output of LTI system
6
• Response to our input x(t)• LTI system: characterize the zero-state with h(t)
• Initial conditions are zero (characterizing zero-state output)
• Zero-state output:
Total response(t)=Zero-input response (t)+Zero-state output(t)
Td(t) h(t)
)()()( thtxty
Leo Lam © 2010-2011
Zero-state output of LTI system
7
• Zero-input response:
• Zero-state output:
• Total response:
Total response(t)=Zero-input response (t)+Zero-state output(t)
)()()( thtxty
“Zero-state”: d(t) is an input only at t=0
Also called: Particular Solution (PS) or Forced
Solution
Leo Lam © 2010-2011
Zero-state output of LTI system
8
• Finding zero-state output (Particular Solution)
• Solve: • Or: Guess and check• Guess based on x(t)
Leo Lam © 2010-2011
Trial solutions for Particular Solutions
9
• Guess based on x(t)
tbtb cossin 10
ta cos1 tbtb cossin 10
tata cossin 10 tbtb cossin 10
Input signal for time t> 0 x(t)
Guess for the particular function yP
nta0 n
nn btbtb ...110
nnn atata ...1
10 nnn btbtb ...1
10
tea 0
teb 0
ta sin1
Leo Lam © 2010-2011
Particular Solution (example)
10
• Find the PS (All initial conditions = 0):• Looking at the table:• Guess:
• Its derivatives:
)()(sin452
2
tutydt
dy
dt
yd
0 1sin cospy b t b t tbtb cossin 10 ta sin1
)sin()cos( 10 tbtbdt
dy p
)cos()sin( 102
2
tbtbdt
yd p
Leo Lam © 2010-2011
Particular Solution (example)
11
• Substitute with its derivatives:
• Compare:
)()(sin452
2
tutydt
dy
dt
yd
)sin()cos( 10 tbtbdt
dy p )cos()sin( 102
2
tbtbdt
yd p
t
tbtbtbtbtbtb
sin
cossin4sincos5cossin 101010
ttbbtbb sincos)35(sin)53( 1010
tttbbtbb cos0sin1cos)35(sin)53( 1010
Leo Lam © 2010-2011
Particular Solution (example)
12
)()(sin452
2
tutydt
dy
dt
yd
tttbbtbb cos0sin1cos)35(sin)53( 1010 • From• We get:
• And so:
153 10 bb 035 10 bb
34
5 ,
34
310 bb
tty
tbtby
P
P
cos34
5sin
34
3
cossin 10
Leo Lam © 2010-2011
Particular Solution (example)
13
)()(sin452
2
tutydt
dy
dt
yd
• Note this PS does not satisfy the initial conditions!
ttyP cos34
5sin
34
3
34
5)0cos(
34
5)0sin(
34
3)0( Py
34
3)0sin(
34
5)0cos(
34
3)0(
dt
dyPNot 0!
Leo Lam © 2010-2011
Natural Response (doing it backwards)
14
)()(sin452
2
tutydt
dy
dt
yd
• Guess:• Characteristic equation:
• Therefore: 4,1
0452
ttn eCeCty 4
21)(
Leo Lam © 2010-2011
Complete solution (example)
15
)()(sin452
2
tutydt
dy
dt
yd
• We have
• Complete Soln:
• Derivative:
ttn eCeCty 4
21)( ttyP cos34
5sin
34
3
tteCeCty tt cos34
5sin
34
3)( 4
21
tteCeCdt
tdy tt sin34
5cos
34
34
)( 421
Leo Lam © 2010-2011
Complete solution (example)
16
• Last step: Find C1 and C2
• Complete Soln:
• Derivative:
• Subtituting:
tteCeCty tt cos34
5sin
34
3)( 4
21
tteCeCdt
tdy tt sin34
5cos
34
34
)( 421
034
5
)0cos(34
5)0sin(
34
30)0(
21
)0(42
01
CC
eCeCy
034
34
)0sin(34
5)0cos(
34
340
)0(
21
)0(42
01
CC
eCeCdt
dyTwo equationsTwo unknowns
Leo Lam © 2010-2011
Complete solution (example)
17
• Last step: Find C1 and C2
• Solving:
• Subtitute back:
• Then add u(t):
034
521 CC 0
34
34 21 CC
51
1,
6
121 CC
tteety tt cos34
5sin
34
3
51
1
6
1)( 4
)(cos34
5sin
34
3
51
1
6
1)( 4 tutteety tt
yn(t)
yp(t)
y(t)
Leo Lam © 2010-2011
Another example
18
• Solve:
• Homogeneous equation for natural response:
• Characteristic Equation:
• Therefore:
)()( 2 tuedt
dyty t
0)( dt
dyty
1
01
tn Cety )(
Input x(t)
Leo Lam © 2010-2011
Another example
19
• Solve:
• Particular Solution for • Table lookup:
• Subtituting:
• Solving: b=-1, a=-2
)()( 2 tuedt
dyty t
tp bety )(
ttt ebebe 2
No change in frequency!
)(2 tue t
tp ety 2)(
tbtb cossin 10
ta cos1 tbtb cossin 10
tata cossin 10 tbtb cossin 10
Input signal for time t> 0 x(t)
Guess for the particular function yP
nta0 n
nn btbtb ...110
nnn atata ...1
10 nnn btbtb ...1
10
tea 0
teb 0
ta sin1
Leo Lam © 2010-2011
Another example
20
• Solve:
• Total response:
• Solve for C with given initial condition y(0)=3
• Tada!
)()( 2 tuedt
dyty t
tt eCety 2)(
4
3)0( )0(2)0(
C
eCey
)(4)( 2 tueety tt
Leo Lam © 2010-2011
Stability for LCCDE
21
• Stable with all Re(lj)<0• Given:
• A negative l means decaying exponentials
n
j
tjn
jeCty1
)( ttn eCeCty 3
21)(
Characteristic modes
Leo Lam © 2010-2011
Stability for LCCDE
22
• Graphically• Stable with all Re(lj)<0
• “Marginally Stable” if Re(lj)=0• IAOW: BIBO Stable iff Re(eigenvalues)<0
Im
ReRoots overhere are stable
Leo Lam © 2010-2011
Summary
• Differential equation as LTI system• Stability of LCCDE