E05-1

P1. Unit step function Hu may be defined in three ways, which allows three interpretations for the delayed unit step function vu due to a constant delay c :

!

01015.0,000

)(tfortforortfor

tuH ,

!!

ctctforctctfororctctfor

ctutu Hv01015.0,000

)()(

But here we do not need to specify )()0( cuu VH . Also replace 0 by 0 . Then the Laplace Transform )(sU v of vu is by definition 1

^ ` f 0 )()( dtetuuLsU tsvvv The function is presented separately for two successive time intervals. 0 Split therefore the integration interval into two subintervals: c t

f c tsvc ts

vv dtetudtetusU )()()( 0 Substitute next the appropriate presentations for both subintervals:

f f f c tsc tsc

c

tsc tsv dtedtedtdtedtesU

0

00010)(

Find the integral (antiderivative) function of the integrand (the function to be integrated):

z

f

f

0

0)(

sift

sifs

e

sU

c

ts

cv or with another notation

z f

f

0

0)(

sift

sifs

esU

c

c

ts

v

For 0 s (yielding 100 ee t ) the result is unbounded, which means that no transform exists for

0 s , i.e. in the origin of the complex plane. But for 0zs we obtain

scs

stssU

tv

fo

)exp()exp(lim)(

The existence depends on the asymptotic behavior of stets )exp( for fot . According to the James and lecture notes the limit exists and is bounded if and only if ^ ` ^ ` 0Re0Re ss , i.e. iff (if and only if) ^ ` 0Re !s . For this condition the value of the limit is 0 (zero). Therefore the Laplace transform requested is

se

sscsU csv

1)exp()( for ^ ` 0Re !s but does not exist otherwise. For 0 c the time domain function reduces to the ordinary unit step function and the transform reduces to

s/1 , i.e. to that of the ordinary unit step function.

E05-2

P2. The unit pulse function with width W is defined as

! Wtfor

WtforWtupf 00/1)( 1/W

t W Here the area of the shaded rectangle between the horizontal axis and the unit pulse function is equal to

1)/1( WW (one, unity) for any positive width W of the rectangular pulse. In probability science and statistics this function is the probability density function of an even distribution on the interval W...0 . BTW, rand of Matlab uses 1 W . The integral of pfu over the whole range is 1, the appropriate value for the final value of a cumulation function of a distribution, the ultimate or maximum possible probability. The Laplace transform requested is by definition, since we may again replace 0 by 0 ,

^ ` f 0 )exp()()( dttstuuLsU pfpfpf Splitting this according to the subintervals yields

f WW

pf dttsdttsWsU )exp(0)exp()/1()( 0

Wpf dttsWsU 0 )exp()/1()( (A)

For 0 s the function )exp( ts is equal to 1, which gives 1)/1()( WWsU pf (the area!). For 0zs we have:

s

Ws

WsWs

tsW

sUW

pf)0exp()exp(1)exp(1)(

0

WsWssU pf )exp(1)(

With 0os this transform derived for 0zs will approach the value 1. In fact, this occurs from any direction of the environment of the origin which is necessary for the limit to exist. The unity value was obtained above also in the case with 0 s . These observations show that is )(sU pf continuous even in the origin. To summarize, the Laplace Transform )(sU pf of the Unit Pulse Function is given by

WsWssU pf )exp(1)( for 0zs and 1)0( pfU

Here the formula for 0zs will approach the value for 0 s as 0os . This justifies the usual practice to use the transform formula for 0zs even for 0zs without too many further comments. Important. The case o 0W is interesting: the pulse area is still equal to 1, and the transform will approach the value 1. This occurs since the integral in Equation (A) approaches the value W since

1oste on the integration interval. In the case o 0W the function pfu is called the unit impulse function, which seems to have a unity Laplace Transform. Plot the function integrated for a real positive value of s and interprete the transform as an area

E05-3

P3. Poles of a Laplace transform (or any other function of a complex variable) are points (of the complex plane) such that the absolute value of the transform grows without any finite bound when the free complex variable approaches the point tightly enough. Here each transform is a rational function with a finite order denominator polynomial and a numerator polynomial the degree of which does not exceed that of the denominator polynomial. Therefore the pole condition is satisfied just iff (if and only if) division with zero is approached when Assuming that cancellations can be excluded thanks to the ability to select another row of the table when a cancellation threatens the poles will be precisely the points for which the limit of the denominator polynomial is zero. So the only possible poles here are the roots of the denominator polynomial. The denominator polynomial is also called the charpo or the characteristic polynomial.

The answers are collected to the table below assuming that a and b are real numbers. In some cases a condition of the form 0zb is assumed not to have a reduction the consequence of which may be dealt (as was proposed) in another row of the table or is trivial. Two examples will be explained in more details. Note that at a pole the transform expression is in fact not the transform of the time domain function because the Laplace integral is not convergent for that point:

0,0,,)(

1)exp(

)(0,)cos()exp(

)(0,)sin()exp(

00,)cos(

00,)sin(

01)exp(

0,0,00,0,00,0,02

0,00,00,01

00011

ImRe)()(

2

22

22

22

22

32

2

aaaaas

tta

bajbabas

asbtbta

bajbabas

bbtbta

bjbbs

sbtb

bjbbs

bbtb

aaas

tas

ts

ts

partspartspolessFtf

rrz

rrz

rrz

rrz

For example, a time domain function f defined with

)cos()exp()( btattf , 0zb , 0tt has a irreducible transform )(sF the charpo of which is

22)()( bass M For 0 b both f and F could be simplified so that another row of the table becomes actual. The roots of M are the solutions of the equation

0)( 22 bas This so called characteristic equation may be replaced by

22222 )()()( bjbjbas

E05-4

bjas r Hence the so called characteristic roots are

bjas r Assuming that a and b are real numbers we see, that the two roots of charpo, here the poles of the transform, are complex conjugates to each other. Another time domain function f defined with

ttf )( , 0tt seems to have the characteristic polynomial 2s . Its characteristic equation is

02 sss This is satisfied iff

0 s or 0 s (first s ) (second s ) So we have the poles 0 and 0 , i.e. two poles at 0 s or, as we often say, a double pole in the origin. The figure given shows the plots for

)cos()exp()(1 btattf (curve rows 1-2)

)exp()(2 attf (curve row 3) for a few parametric cases. The shape of a missing function represented by

)sin()exp()(3 btattf is that of 1f but with a suitable time shift so that the function starts with zero value at zero time. Note that the real part of the pole multiplies the time inside exp function, while the imaginary coordinate of the pole multiplies the time inside sin and/or cos function! A second order twin-pole-function not illustrated A unilateral function 4f defined with

)exp()(4 tattf , 0tt has a transform which has the poles a and a , i.e. a double pole at as . Assuming that the pole is real we can illustrate the plots of 4f as below where a special value of a is adopted vs. a normalized time is used:

E05-5

0 5 10 150

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4Impulse Response

Time (sec)

Ampl

itude

0 0.5 10

0.2

0.4

0.6

0.8

1

1.2

1.4Impulse Response

Time (sec)Am

plitu

de

0 0.5 1 1.5 2 2.50

5

10

15

20

25

30

35Impulse Response

Time (sec)

Ampl

itude

a < 0 a = 0 a > 0 0)(lim 4 fo tft f fo )(lim 4 tft f fo )(lim 4 tft Find answers where you refer to features of poles: a) When will the functions discussed converge to a finite value, perhaps even to zero? b) When a function discussed is not oscillating? c) When is an oscillating function discussed above convergent non-convergent (divergent) & bounded non-convergent & unbounded ? d) How is the exponential decay changed if the pole is moved leftwards? e) How is the temporary oscillation changed if the pole is moved leftwards? f) How is the temporary oscillation changed if the pole is moved upwards? An excellent source for further Laplace studies (also available as an ebook in TUT library net service): James: Advanced Modern Engineering Mathematics, Pearson. P4. b) )sin()cos()( tbtatu ZZ , 0tt , with real parameters Both sine and cosine have a Laplace Transform for ^ ` 0Re !s . Assuming this lets go:

^ `)sin()cos()( tbtaLsU ZZ , ^ ` 0Re !s Sum Rule gives

^ ` ^ `)sin()cos()( tbLtaLsU ZZ , ^ ` 0Re !s

E05-6

Gain Rule implies for constants a , b

^ ` ^ `)sin()cos()( tLbtLasU ZZ , ^ ` 0Re !s Pick the transforms of sine and cosine from the appropriate table of Laplace Transforms,

2222)( ZZ

Z sbssasU , ^ ` 0Re !s

Collect this into a standard ratio of two polynomials:

2222 1)( Z

ZZZ

s

bsas

bsasU , ^ ` 0Re !s Here Z is assumed to be real. Here numerator degree does not exceed denominator degree. Hence each pole must here belong to the set of denominator roots,

022 Zs 22 Z s

jjs r r ZZ 0 But a denominator root is a pole iff it is not also a numerator root. So lets study carefully: 1) For 0za the numerator has a root given by ab /)( Z . Then we have the poles Zjr except in the case where this root is also a den root which occurs iff 0 Z . This means

0Zs

as

sasU 1)( 2

with a single pole at the origin and making u the step function of amplitude a . 2) For 0 a the numerator has no roots. Then we have the poles Zjr except in the pathological case

where 0 Zb which occurs for both 0 b and 0 Z both of which actually imply 0)( sU with no poles and making u to be the zero function. So for true oscillation the poles are Zjr on the Im axis while the other possibilities are associated with some degenerate cases where the function is not really oscillating but has almost everywhere zero derivative. Now lets have some repetition for improved learning and linking topics: Let the cos and sin functions be cosx and

sinx , respectively. Our function u can be generated using a function generator described below. Derive an elementary block diagram as well as a state space model with matrix objects, and study observability of the model to find something concerning reduction of the time domain function and its Laplace Transform:

)()()( sincos txbtxatu , )()( sincos txtx Z , )()( cossin txtx Z , 1)0(cos x , 0)0(sin x P2 Revisited. The time domain function studied in P2 may be rewritten in terms of Heaviside unit step function Hu as in Equation (1) below. Sum Rule and Gain Rule (or the Linearity Rule joining them) can then be used to get the Laplace Transform. But this approach assumes ^ ` 0Re !s . However, it is not a necessary assumption in the derivation of P2. This illustrates that sometimes a procedure used introduces unnecessary assumptions.

)(1)(1)( WtuW

tuW

tu HHpf (1)

E05-7

P5. Let )()( 2 sUsF pf . For o 0W this has the limit

111)(lim)(lim)()(lim)(lim0000

oooo

sUsUsUsUsF pfW

pfW

pfpfWW

Hence )(tf to be found may also serve as a Dirac impulse by letting o 0W . Lets derive )(tf :

22

22

2

22 )(exp)exp(21)(

))exp(1()exp(1)(sW

WsWsWs

WsWs

WssF

22

)2exp()exp(21)(sW

WsWssF

222222

1)2exp(11)exp(211)(s

WsWs

WsWsW

sF Using the Linearity Rule of Inverse Laplace Transform, the Laplace Transform of the Unit Ramp Function reversely and the Delay Rule reversely (for the Unit Ramp Function) we obtain

t

t

t

WtWtWt

WWtWtWt

Wttt

Wtf

2,22,01

,,02

0,0,01)( 222

tdd

WtWtWWtWtWWtWWtWtW

WttWt

tf

2,)2()(22,0)(2

0,000,000

)(

222

22

2

tdd

WtWtWtWW

WttWt

tf

2,02,2

0,0,0

)( 212

We obtain a triangle symmetric with respect to Wt , with the height equal to W/1 and with unity area Here pfu can serve as the PDF (Probability Density Function) of a scalar random variable distributed evenly on the range W...0 while f could represent the PDF of 21 xx if 1x and 2x are mutually independent random variables but have the same PDF pfu . Illustrate this with rand (which applies 1 W ), hist and +. Solve f also using syms, laplace, * or ^ , ilaplace, ezplot. Make the same also for )(4 sU pf to see a curve shape close to the famous Gaussian bell curve. In fact, each PDF with zero mean and extremely small positive variance is Dirac.

E05-8

P6. Inertia 0!J , coefficient of viscous damping 0!b .

0)()( tbtJ TT equivalence of the functions ^ ` ^ `0)()( LtbtJL TT equivalence of the LTs ^ ` ^ ` 0)()( tbLtJL TT Sum Rule ^ ` ^ ` 0)()( tLbtLJ TT Gain Rule

^ `> @ ^ `> @ 0)0()()0()0()(2 TTTTT tLsbstLsJ Derivative Rules

^ ` ))0()0(()0()()( 2 TTTT bJsJtLsbsJ

^ `sbsJ

bJsJtL 2

))0()0(()0()( TTTT

LT of the natural response

Denominator roots 0 , )( 1bJ , here distinct. Use them to factor the denominator polynomial:

^ `))(()0(

))0()0(()0()( 1bJssJbJsJtL

TTTT

Partial Fraction Expansion due to Distinct Roots:

^ `))(()0)((

))0()0(()()0())(0()0(

))0()0((0)0()( 111

1 bJsbJJbJbJJ

bJsJbJJtL

TTTTTTT

Represent it as a Linear Combination of a few elementary LTs available in Exam Tables:

^ `)(

1)0(1)0()0()( 1bJsbJ

sbbJtL

TTTT

Exploit the LT formula of Step Function and exp function and keep the Linear Combination structure:

0,)0(1)0()0()( )(1 t te

bJ

bbJt tbJTTTT

This has a finite final value,

bbJt

t

)0()0()(lim TTT fo

Check: Solve first )(tZ using LT, then use

)()( tt ZT t dt 0 )()0()( WWZTT