EECE 301 Signals & Systems Prof. Mark Fowler
Transcript of EECE 301 Signals & Systems Prof. Mark Fowler
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EECE 301 Signals & SystemsProf. Mark Fowler
Discussion #3a• Review of Differential Equations
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Differential Equations ReviewDifferential Equations like this are Linear and Time Invariant:
)()(...)()(...)()(0101
1
1 tfbdt
tdfbdt
tfdbtyadt
tydadt
tyda m
m
mn
n
nn
n
n +++=+++ −
−
−
-coefficients are constants ⇒ TI
-No nonlinear terms ⇒ Linear
.,)()(),(,)()(),( etcdt
tyddt
tydtydt
tyddt
tydtf p
p
k
kn
p
p
k
kn
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
Examples of Nonlinear Terms:
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In the following we will BRIEFLY review the basics of solving Linear, Constant Coefficient Differential Equations under the Homogeneous Condition
“Homogeneous” means the “forcing function” is zero
That means we are finding the “zero-input response” that occurs due to the effect of the initial coniditions.
⇒ Write D.E. like this:
( ) ( ) )(...)(...
)(
01
)(
011
1 tfbDbDbtyaDaDaD
DP
mm
DQ
nn
n
444 3444 2144444 344444 21ΔΔ ==
−− +++=++++
⇒.. EqDiff )()()()( tfDPtyDQ =
m is the highest-order derivative on the “input” side
n is the highest-order derivative on the “output” side
We will assume: m ≤ n
)()( tyDdt
tyd kk
k
≡Use “operational notation”:
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Due to linearity: Total Response = Zero-Input Response + Zero-State Response
Z-I Response: found assuming the input f(t) = 0 but with given IC’s
Z-S Response: found assuming IC’s = 0 but with given f(t) applied
( ) 00)(...
0)()(:..
011
1 >∀=++++⇒
=⇒−
− ttyaDaDaD
tyDQED
zin
nn
zi(▲)
numberscomplex possibly areand)(Consider 0
λ
λ
ccety t=
“linear combination” of yzi(t) & its derivatives must be = 0
Can we find c and λ such that y0(t) qualifies as a homogeneous solution?
Finding the Zero-Input Response (Homogeneous Solution)
Assume f(t) = 0
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Put y0(t) into (▲) and use result for the derivative of an exponential:
0)...( 011
1 =++++ −−
tnn
n eaaac λλλλ
must = 0
solutiona is
solutiona is
solutiona is
2
1
2
1
tn
t
t
nec
ec
ec
λ
λ
λ
M
Then, choose c1, c2,…,cn to satisfy the given IC’s
tn
ttzi
necececty λλλ +++= ...)(:Solution I-Z 2121
tnn
tn
edt
ed λλ
λ=
Characteristic polynomial
Q(λ) has at most n unique roots
(can be complex)
))...()(()( 21 nQ λλλλλλλ −−−=⇒
So…any linear combination is also a solution to (▲)
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{ }nitie 1=λ Set of characteristic modes
Real Root: tii
iej σσλ ⇒+= 0
real real t
0>iσ
0=iσ0<iσ
iii jωσλ +=tjtt iii eee ωσλ +=
0<iσ
t
0>iσ
t
0=iσ
t
Complex Root:
Mode:
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To get only real-valued solutions requires the system coefficients to be real-valued.
⇒ Complex roots of C.E. will appear in conjugate pairs:
ωσλωσλjj
k
i
−=+=
Conjugate pair
tjtk
tjti
tk
ti eeceececec ki ωσωσλλ +=+
For some real CθjeC2
θjeC −
2
0)cos( >+ ttCe t θωσUse Euler!
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Repeated Roots
Say there are r repeated roots
))...()(()()( 321 prQ λλλλλλλλλ −−−−= p = n - r
ZI Solution:
( ) :modesother ...)( 111211 ++++= − tr
rziietctccty λ
effect of r-repeated roots
See examples on the next several pages
We “can verify” that: trttt etettee i 111 12 ,...,,, λλλλ − satisfy (▲)
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Find the zero-input response (i.e., homogeneous solution) for these three Differential Equations.
Example (a)
)()(2)(3)(
)()(2)(3)(
2
2
2
tDftytDytyD
dttdfty
dttdy
dttyd
=++
=++
The zero-input form is:
0)(2)(3)(
0)(2)(3)(
2
2
2
=++
=++
tytDytyD
tydt
tdydt
tyd
The Characteristic Equation is:0)2)(1(0232 =++⇒=++ λλλλ
Differential Equation Examples
w/ I.C.’s
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The Characteristic Equation is:0)2)(1(0232 =++⇒=++ λλλλ
2&1 21
The Characteristic Roots are:−=−= λλ
The Characteristic “Modes” are:
tttt eeee 221 & −− == λλ
The zero-input solution is:tt
zi eCeCty 221)( −− +=
The System forces this form through its Char. Eq.
The IC’s determine the specific values of the Ci’s
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The zero-input solution is:tt
zi eCeCty 221)( −− +=
and it must satisfy the ICs so:0)0(0 21
02
01 =+⇒+== −− CCeCeCyzi
The derivative of the z-s soln. must also satisfy the ICs so:
522)0(5 210
20
1 =+⇒−−=′=− −− CCeCeCyzi
Two Equations in Two Unknowns leads to:
5&5 21 =−= CC
The “particular” zero-input solution is:321321mode second
2
modefirst 55)( tt
zi eety −− +−=
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Because the characteristic roots are real and negative…the modes and the Z-I response all decay to zero w/o oscillations
0 1 2 3 4 5 6 7 8 9 10-6
-4
-2
0Fi
rst M
ode
Plots for Example 2.1 (a)
0 1 2 3 4 5 6 7 8 9 100
2
4
6
Seco
nd M
ode
0 1 2 3 4 5 6 7 8 9 10-1.5
-1
-0.5
0
Zer
o-In
put R
espo
nse
t (sec)
Plots for Example (a)
slower decay for the e-t term
faster decay for the e-2t term
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Example (b):
)(5)(3)(9)(6)(
)(5)(3)(9)(6)(
2
2
2
tftDftytDytyD
tfdt
tdftydt
tdydt
tyd
+=++
+=++
The zero-input form is:
0)(9)(6)(
0)(9)(6)(
2
2
2
=++
=++
tytDytyD
tydt
tdydt
tyd
The Characteristic Equation is:0)3(096 22 =+⇒=++ λλλ
w/ I.C.’s
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The Characteristic Equation is:
The Characteristic Roots are:3&3 21 −=−= λλ
The Characteristic “Modes” are:
tttt teteee 33 21 & −− == λλ
The zero-input solution is:tt
zi teCeCty 32
31)( −− +=
The System forces this form through its Char. Eq.
The IC’s determine the specific values of the Ci’s
0)3(096 22 =+⇒=++ λλλ
Using the “rule” to handle repeated roots
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Following the same procedure (do it for yourself!!) you get…
The “particular” zero-input solution is:
tttzi etteety 3
mode second
3
modefirst
3 )23(23)( −−− +=+= 321321
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0 1 2 3 4 5 6 7 8 9 100
1
2
3F
irst M
ode
Plots for Example 2.1 (b)
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
Sec
ond
Mod
e
0 1 2 3 4 5 6 7 8 9 100
1
2
3
t (sec)Zer
o-In
put R
espo
nse
Plots for Example (b)
Same Decay Rates
Effect of “t out in front”
Because the characteristic roots are real and negative…the modes and the Z-I response all .
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Example (c):
)(2)()(40)(4)(
)(2)()(40)(4)(
2
2
2
tftDftytDytyD
tfdt
tdftydt
tdydt
tyd
+=++
+=++
The zero-input form is:
0)(40)(4)(
0)(40)(4)(
2
2
2
=++
=++
tytDytyD
tydt
tdydt
tyd
The Characteristic Equation is:0)62)(62(04042 =++−+⇒=++ jj λλλλ
w/ I.C.’s
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The Characteristic Equation is:
The Characteristic Roots are:62&62 21 jj −−=+−= λλ
The Characteristic “Modes” are:
tjtttjtt eeeeee 6262 21 & −−+− == λλ
The zero-input solution is:tjttjt
zi eeCeeCty 622
621)( −−+− +=
The System forces this form through its Char. Eq.
The IC’s determine the specific values of the Ci’s
0)62)(62(04042 =++−+⇒=++ jj λλλλ
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Following the same procedure with some manipulation of complex exponentials into a cosine…
The “particular” zero-input solution is:
)3/6cos(4)( 2 π+= − tety tzi
Set by the ICs
Imag. part of root controls oscillation
Real part of root controls Decay
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1F
irst M
ode
Plots for Example 2.1 (c)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
Sec
ond
Mod
e
0 0.5 1 1.5 2 2.5 3 3.5 4-4-2024
t (sec)Zero
-Inp
ut R
espo
nse
Can’t Easily Plot the Modes Because They Are ComplexPlots for Example (c)
Because the characteristic roots are complex… have oscillations!Because real part of root is negative… !!!
4e-2t
-4e-2t
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Big Picture…
The structure of the D.E. determinesthe char. roots, which determine the “character” of the response:
• Decaying vs. Exploding (controlled by real part of root)• Oscillating or Not (controlled by imag part of root)
The D.E. structure is determined by the physical system’s structure and component values.