Continuous Time Convolution Properties

• Commutative: )(*)()(*)( txththtx

• Associative:)(*)()(*)( txththtx =

• Associative:))(*)((*)()(*))(*)(( tztytxtztytx =

• Distributive:))()(()()())()(( yy

)(*)()(*)())()((*)( tztxtytxtztytx +=+

Commutativity)()()(*)( dthxthtx −= ∫

∞

τττ

let

)()()()(

2t =−

∫∞−

ττ

)()(

let 2

dhtx

t

−= ∫∞

τττ

ττ

)()( 222 dhtx= ∫∞

∞−

τττ

)()( 222 dhtx −= ∫∞−

τττ

)(*)( txth=

Connecting SystemsConnecting Systems( )

h2(t)h1(t)x(t) y(t)z(t)

h(t)=h1(t)*h2(t)x(t) y(t)

(t)*)(t)*(t)((t)*(t)(t) 2112 hhxhzy ==

(t)*(t))(t)*(t)(*(t) 21 hxhhx == (t)(t))(t)(t)((t) 21 hxhhx

Connecting SystemsConnecting Systemsh1(t)

x(t)

h2(t)

h1(t)

y(t)+

h2(t)

h(t)=h1(t)+h2(t)x(t) y(t)

(t)*(t)(t)*)(t)(t)((t)*(t)(t)*(t)(t) 21

xhxhhxhxhy

=+=+=

(t)(t)(t))(t)(t)( 21 xhxhh =+=

Inverse SystemsInverse Systems:nconvolutioforidentityis][nδ

][][*][:nconvolutiofor identity is ][

nxnnxn

=δδ

• Let’s look for a system such that:][][*][ iff

][][*][*][][*][][

21

122

nnhnhnxnxnhnhnynhnz

δ====

• The system with impulse response h2 is said to be the inverse system of h1

][][][ 21

y 1

Inverse SystemsInverse Systems• Inverse system can recover input x(t) from output y(t)

• Not always possible to find inverse systemh2(t)h1(t)x(t) x(t)y(t)

• Generally solve equation:)()(*)( tthth δ= )()()( 21 tthth δ=

Linear Constant‐Coefficient Differential EquationsEquations• It will often be useful to describe systems using equations involving the rate of change in some q g gquantity, e.g. SHM

xd 2

• This is a differential equationcxdtxd

−=2

• This is a differential equation• This is an ordinary differential equation (ODE) y q ( )because there is only one independent variable and derivatives with respect to that variable

• The general form of the ODE is given as:The general form of the ODE is given as:∑∑m kn k txdbtyd )()( ∑∑==

=k

kkk

kk dttxdb

dttyda

00

)()(

• The equation above is linear with real co‐efficients ak and bkk k

Example• Consider an RLC circuit• The voltage across the battery is V

1• The voltage across the capacitor is • The voltage across the resistor is q

C1

dqRThe voltage across the resistor is• The voltage across the inductor is dt

R

2

2

dtqdL

dt

• Using Kirchoff ’s voltage law we have:Using Kirchoff s voltage law we have:Vq

CdtdqR

dtqdL =++

12

2

• Which describes the operation of the given RLCcircuitcircuit• Many systems can be described using ODEsMany systems can be described using ODEs

Solving LCCDEsSolving LCCDEs• The approach is to find the general form of all pp gpossible solutions to the equation and then apply a number of conditions to find the appropriate solutionsolution.• The two main types of problems are:The two main types of problems are:

– initial value problems, which involve constraints on the solution and its derivatives at a single point, and b d l bl h h l– boundary value problems, which involve constraints on the solution or its derivatives at several points.

• The number of initial conditions needed for anThe number of initial conditions needed for an Nth order differential equation, which is the order of the highest order derivative, is N, and a unique solution is always guaranteed if these are supplied.

• Boundary value problems can be slightly more complicated and will not necessarily have acomplicated and will not necessarily have a unique solution or even a solution at all for a given set of conditions.g

• Solution is given by:)()()( tytyty +

• yh(t) is the homogeneous (or complementary)

)()()( tytyty ph +=

solution for input x(t)=0. It represents “natural modes” of unforced oscillation• yp(t) is the particular solution due to input x(t), represents the forced responserepresents the forced response

SolvingSolving …• LCCDEs are solved in two parts. • First solve the complementary equation, set to zero. • Then solve the particular solution.

The Homogenous SolutionThe Homogenous Solution• Since the input is equal to zero, we have:

0)(0

==∑ dt

tydan

kk

k

k

)()()( 1− tydtydtyd nn

(1) ... 0)()(...)()(0111 =+++ −− tya

dttyda

dttyda

dttyda nnnn

• Assume the solution is of the form:,)( Cety st=

constants are s and where,)(

Cy

The Homogenous SolutionThe Homogenous Solution• Applying the solution to equation (1)Applying the solution to equation (1)

( )( ) 0... 011

1 =++++ −−

stnn

nn Ceasasasa

• Assuming C≠0, then44444 344444 21 01

11 0... =++++ −− asasasa nn

nn 2

equation sticcharacteri

The Homogenous SolutionThe Homogenous Solution• Factorizing characteristic polynomial we get:g p y g( )( ) ( ) 0...- :such that , 21 =−− ni sssssss• Each si with a corresponding Ci form a solution

tsieCty =)(• Owing to linearity, the homogeneous solution is:ii eCty =)(

Wh C f t t h i iti l diti (IC)

tsn

tsn

tstsh

nn eCeCeCeCty ++++= −−

121121 ...)(• Where Ci are free to match initial conditions (IC)

Repeated RootsRepeated Roots• Repeated roots in characteristic polynomial:

( ) ( )( ).....321 −−− L ssssss

( ) ......)( 3211

11

2321 +++++++= +

−−

tsL

tsL

tsLLh eCeCetCtCtCCty

P i l l iParticular solution• Represents the forced response of the system• Solution depends on the form of the input

W h f f (t) di th f f (t)• We choose form of yp(t) according the form of x(t)

• Substitute y (t) into non‐homogeneous DE andSubstitute yp(t) into non homogeneous DE and compare coefficients on both sides of the equation

Particular SolutionParticular Solution• IfIf

5)( 7etx t= −

)( 7Pety tp = −

• If )377cos(170)( ttx =)377cos()377cos()( 21 tPtPtyp +=

ExampleExample2)(2)(

+ tytdy

( ) 2)(2 =+ tydt

• For t ≥ 0, y(0) = 4• The homogenous solution is assumed to be

sth eCty =)(

• Applying this to the differential equation andApplying this to the differential equation and setting x(t)=02)(2)(

+ ttdyh

02 =+ eCesC stst

2)(2)(=+ ty

dty

hh

0)2(

02

=+

=+

eCs

eCesCst

• The homogenous solution is2−=⇒ st

h eCty 2)( −=The homogenous solution is hy )(

• For the particular solution consider theFor the particular solution, consider the input, which is a constant, thus the particular solution takes a similar formparticular solution takes a similar formPtyp =)(

2)(2)(

=+ tydttdy

pp

22 =⇒ Pdt p

1 =P

• The solution is thus:• The solution is thus:1)()()( 2 +=+= − t

ph eCtytyty

• To find C we take into account the initial )()()( ph yyy

condition, y(0)=4]1[)0()0()0( 2C t

41

]1[)0()0()0()0(2

0

2

=+

+=+=−

=

−

Ce

Ceyyyt

tph

341

==+

CC

3=C

• The total solution is then:The total solution is then:tety 231)( −+=