EECE 301 Signals & SystemsProf. Mark Fowler

Discussion #3a• Review of Differential Equations

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:

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”:

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

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 (▲)

{ }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:

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!

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 (▲)

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

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

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 −− +−=

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

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

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

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

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 .

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

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

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

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

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.