Laplace Transform, Engineering-Mathematics-3 ,Ch-6 in BME in PDF

[20122013]

Note: I some time mistakely type capital “S” instead of small “s” so be care-full.

BIO-MEDICAL ENGINEERING GUIDE MOHAMMAD SIKANDAR KHAN LODHI

ENGINERING-MATHEMATICS-3, LAPLACETRANSFORM, CH-6, IN BIO-MEDICAL-ENGCOURSES:[ Part-1] My blog -> www.medical-image-processing.blogspot.com

SECTION NO 6.1 , 6.2 , 6.3,6.4,6.5 AND 6.6 ONLY , FROM BOOK : ADVANCED ENGINEERING MATHEMATICS 7TH EDITION by ERWIN-KREYSZIG.

21/FEB/2013

LAPLACE-TRANSFORM [NOTE: I SOME TIME MISTAKELY TYPE CAPITAL “S” INSTEAD OF SMALL “S” SO BE CARE-FULL.] BOOK : ADVANCED ENGINEERING MATHEMATICS 7th EDITION by ERWIN-KREYSZIG.

[ENGINEERING-MATH-3,LAPLACE-TRANSFORM, CH-6,IN BME]BY SIKANDAR-LODHI

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Table of Contents (6.1 START HERE) .......................................................................................................................................... 2 PROBLEM SET 6.1 .................................................................................................................................... 11 [ 6.2 ] [ START HERE ]. ................................................................................................................................ 42 Problem-set 6.2....................................................................................................................................... 61 [ 6.3 ] [ start here ] ...................................................................................................................................... 70 Problem set 6.3 ....................................................................................................................................... 77 [6.4 ] [ start here ] ....................................................................................................................................... 87 Problem-set 6.4 : ..................................................................................................................................... 90 [6.5] start here ............................................................................................................................................ 94 Problem set 6.5 ....................................................................................................................................... 96 [ 6.6 ] start here ........................................................................................................................................ 100 Problem set 6.6 ..................................................................................................................................... 105

CH-6( START-HERE ): ON LAPLACE-TRANSFORM: (6.1 START HERE) LAPLACE-TRANSFORM,INVERSE-TRANSFORM,LINEARITY:  If f(t) is the given function defined for all *t≥0+, then we taken the laplace-transform, of f(t), then if the resulting value is exist [(i-e) has some finite value + , it’s the function of “s”, also called F(s) in frequency domain.  FORMULA OF LAPLACE TRANSFORM[LT]:

Where : t=time and s=frequency Remember that => the given function f(t) depends on “t” and the new function F(s)[ its laplace-transform + depend on “s” .


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So, F(s) is the Laplace-Transform of given function f(t).  FORMULA OF INVERSE LAPLACE TRANSFORM[ILT]:

----------------------EXAMPLE#1  LAPLACE TRANSFORM Let,-> f(t)=1, when *t≥0+ FIND: Note: the interval of integration in (eq-A) below is infinite, therefore we called it as Improper-integral. FORMULA: (A)

Above equation is called as eq-A,

{:.when ( S>0 ) }


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Henced proved _---------------------------------finished _-----------EXAMPLE#2: Let -> f(t)=eat, when *t≥0+ Where [ a=any constant ] Find-> Solution:


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-------------------------finished------------ THEOREM#1:[LINEARITY-OF-LAPLACE-TRANSFORM] The laplace transform is a “linear-operator Or linear-operation”.

-------------finished--------------Proof # 1:


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Solution:


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Henced proved. --------------------------------finished-----------------


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Henced proved ---------------------finished-----------Proof # 3:


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Henced proved -------------------finished----------------{ !=factorial, 4!=4x3x2x1=24}


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-------------------finished-----------


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PROBLEM SET 6.1

------------finished------------Q2) [ 6.1 ] Find laplace transform=?


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--------finished-----------

------------finished-------6.1


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Q4) find laplace transform=?

----------finished--------6.1 Q5 ) find laplace transform=?


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-----------finished----------


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Solution:

Q9) find laplace transform=?


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------------finished--------Q17) find laplace transform=? Given:


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Solution:


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--------------finished-------------Q18 ) find laplace-transform=? GIVEN:


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Solution:


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-------------finished-----------Q19 ) find laplace transform=? Given:

FORMULA-OF-STRAIGHT-LINE-EQUATION’S:


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[

]

Lets put all value in above equations, [

]

[ [

] ]

[

]

Where :[

] so,

[

]

After applying laplace transform on eq-B, we get, [

]

[

].

[ ]


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[ ]. [

].

[

].

[where: s=frequency, and , k=Amplitude ] [

].

[ ]. [

].

--------------------------Rough work: Let consider :[ So, [

] from avove eq-G: ]


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Note: I some time mistakely type capital “S” instead of small “s” so be care-full. –

[

– =0

—

– =0 2}] ] –

[

–

–

[

–

[ [

–

–

–

— —

]

— —

]

] ]

–

[

2] =0 = ] ] ] --------------------Now put the value of eq-k in eq-G, we get, [

2] =0 = ]. [ [

–

]

–


]

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Note: I some time mistakely type capital “S” instead of small “s” so be care-full. –

[

]

–

[

]

---------finished-----------6.1 Q 20) Find laplace transform=? Given:

FORMULA-OF-STRAIGHT-LINE-EQUATION’S: [

]

Solution: [ [

] ] [

]


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[ So, where: [

] , and

]

[

]

We applying laplace transform on eq-A, [

[

]

]

[

]

[ [

] ]

[

]

[

]

……. …. [

]


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-----------finished----------6.1

Given:

.

[

]

Time domain Frequency domain

Solutions: . . . . ---------------------------------. [ [

] ]


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--------------Now put the value of eq-S into eq-P, we get. . . . Answer. --------------finished----[ 6.1 ] Q 29) find f(t)=? Given

.

Solution: . . . --------------Formula:(eq-1). (eq-2). 


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----------Now by using the formula of eq-1 then eq-P becomes, . . Answer. ------------finished-----------6.1 Q 31) find f(t)=? Given-> F(s)=[1/S4 ]; Solution: . . . ------------Formula: [ 

. ] .

-----------------Now just applying above formula on above eq-P , we get,


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. . . . Answer . ----------------finished-----------[ 6.1 ] Q 33 ) find f(t)=? Given->

.

Solution:->

. . .

Using “Partial Fractional Method” we consider

from above eq-

P: . . . .


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Let,-> [A+B]=0; … …. We get,  A=1, B=-1 . Answer. --------finished--------[ 6.1 ] Q 30 ) find f(t)=? Given:-> F(s)= [4/(S+1)(S+2)] Solution: . . Consider [4/(S+1)(S+2)] : So, . . . .


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. . . Let:->

.

So, by placing the value of { eq-i } in [ eq-

] we get,

. . .

. .

Now, ------.

…

.

---------For B:->

.

When

. . . . . .


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For A:-> . .

[:.

. ]

. . . When

[

] [

]

So, just placing the value of A and B in

. . .

Now by placing the value of [ eq-G ] in [eq-j ] then [ eq-j ] becomes. . . Now simply solve it, . ------------Formulas:-> .


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. . . ----------. . . answer. -----------finished-----------[ 6.1 ] Q 34 ) find f(t)=? Given:->

.

Solution:.

.

.


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-------------------. . --------------------Where: -> [

]

Now by applying the formula of [eq-i ] on [ eq-P ] we get, .

.

. Answer ---------finished----------[ 6.1 ] Q 35 ) find f(t)=? Given:->

.

Solution:->

. .


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 Consider

:-

 For value of “A” & “B” Values:. . . . . . . . . Let:->

.

So,  For A:. [:.

]] . .

.


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 For B:. . . Now, by placing the value of [eq-1 & 2] in [ eq-N ] we get. . ---------. . -----------. Now by placing the value of [ eq-N ] in [ eq-B ] we get the f(t). . .

. .


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. Answer. -------------finished-----------[ 6.1 ] Q 36 ) find f(t)=? Given->

.

Solution:->

. . . . .

-------------Formula’s: . . ------------. Answer. --------finished----------


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[ 6.1 ] Q 38 ) find f(t)=? Given:->

.

Solution:->

. . . . . . . Answer.

-------finished-------------[ 6.1 ] Q 32 ) find f(t)=? Given:->

.

Solution:->

. .


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. . . . . Answer. ----------finished-----------[ 6.1 ] new-topic.  EXISTENCE OF LAPLACE-TRANSFORM’S:Its suffices to require that f(t) be Piece-wise-continuous on every finite interval in the range [ ] That is :

By the definitions, a function f(t) is a piece wise continuous on a finite interval [ ], if f(t) is defined on that interval and is such


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subdivided into finitely many sub-intervals, in each of which f(t) is continuous and has finite limits as time(t) approaches either, endpoint of interval of subdivision from the interior, it follows from this definitions that finite Jump’s are the only dis-continuities that a piecewise continuous function may have, these are known as Ordinary discontinuities. ------------------[ 6.1 ] THEOREM-2 [EXISTENCE-THEOREM-FOR-LAPLACE-TRANSFORM’S+:Let, f(t) be a function that is piece-wise continuous on every finite interval in the range [ ], and satisfies below [ eq-2 ]. .

And for some constant [ ] and [ ]. Then, the laplace-transform of f(t) exists for all [

].


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Proof:->since f(t) is piece-wise continuous, [ over any finite interval on t-axis. From [eq-2], Let, we have a conditions [

], is integrable

], we get, . . . . . . . . . . . . OR .


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Where : [

] is need for last integral.

------------FINISHED----------[ 6.2 ] [ START HERE ].  TRANSFORM’S-OF-DERIVATIVE’S-AND-INTEGRAL’S:THEOREM # 1:  LAPLACE-TRANSFORM-OF-THE-DERIVATIVE-OF-f(t):Suppose that f(t) is continuous for all [ ], where the derivatives of f(t) is [f(t)’] which is pice-wise continuous, on every finite interval in the range [ ], then laplace transform of derivatives of function [f(t)’] is exist when there is a conditions is [ ]. So, taking laplace transform on its derivatives [f(t)’], That is: o PROOF: . [

] .

Let consider

. .

….. …..


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[

];

Henced proved. There fore: [

];

Key THEOREM # 2 :- [ LAPLACE-TRANSFORM-OF-THE-DERIVATIVE-OF-ANYORDER-“n”+: Let, -> [

];

And [

];

Then [

];

THE GENERAL-FORMULA OF THEOREM-2:Where:- [

= No of derivatives on f(t) ]

[

];

Or


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[

1.

1 (0)];

Or

Where the eq-z is the laplace transform of nth derivative of functions f(t). --------------finished-----------[ 6.2 ]  DIFFERENTIAL-EQUATIONS, INITIAL-VALUE-PROBLEM[ IVP ]:Let,

system

. Where a & b are constant.  IVP:GIVEN:-> [

] and [

].

Solution:-> . .


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Now by applying laplace transform on both side we get, . .

Above [eq-T+ is called the “subsidiary-equation”. . . . . Answer. --------------. Answer. WHERE:->[

].

Transfer-function is often denoted by H(s), but we need H(s) much more frequentely for other purposes. . Answer. . .


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-- IVP:GIVEN:-> [

] and [

].

--If [-> y(0)=y’(0)=0], . . . . . Than, this is simply [-> Y(s)=R(s).Q(s)]. Thus the transfer-function is Q(s) THE GENERAL-FORMULA-OF-TRANSFER-FUNCTION:

NOTE: Q(s) depends only on a and b, but neither on r(t) nor on the initial-condition’s. Last step: we reduce [ eq-7 + ,usually by partial fraction’s, as in integral calculus} to a sum of terms whose inverse can be found from the table, so that the solution [ ] of [eq-A] is obtained.


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--------------------finished---------[ 6.2 ] Example # 5 : [Initial-value-Problem[I.V.P]]: GIVEN: [

].

Initial-value-Problem[I.V.P]: [

].

REQUIRED: [

].

And solve it by I.V.P , So, STEP-1: To make eq-A as y(t) [ or to find the solution of eq-A ] by using Laplace-Transform, [

].

By applying the laplace transform on eq-A we get, [

].

[

].

[ [

]. ].


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[

].

[

].

[

].

---------By using Initial-value-Problem[I.V.P]: [

].

----------Than we get. [

].

[

].

[ [

]. ].

[

].

[

].

[

].


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[

]. ].

[ let

CONSIDER [1/s2[s2-1] ]: A=? , B=? . Using Partial-fraction method: [

] ]

[

]

[

]

[

]

[ [

]

[

]

[

]

[

]

{:.[eq-i] -> 0=A+B }


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So,

]

[ [

]

[

].

So, {:.[eq-i] -> 0=A+B; [eq-e] -> A=-B ; } For B: When [

].

[eq-e] -> A=-B ;  -1=-B ;  B=1 ; So, When [

] and B=1 :-

Than, [ [

] ].

So, by placing the value of eq-Z in eq-D than eq-D becomes, [ENGINEERING-MATH-3,LAPLACE-TRANSFORM, CH-6,IN BME]BY SIKANDAR-LODHI

Page 50-OF-109


].

[

].

[

].

[ Above eq-S is the solution of “subsidiary-equations”.

Now by taking the inverse-laplace transform on eq-S, we get the solution of given problem.

].

[

].

[

].

[ ].

[ [

].

[

]. Answer, It’s

the solution of given problem. ----------finished-----------


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In practice, instead of justifying the use of formulas and theorems in this method, one simply checks at the end whether y(t) satisfies the given equations and initial conditions. “Gains in the method”, as compared to that in Ch # 2, and illustrated by the example, are as follows: 1. No determination of a general solution of the homogeneous equation. 2. No determination of values for arbitrary constants in a general solution. “Shifted data problems” is a short name for “initial value problems” in which the initial conditions refer to some later instant instead of [ t=0 ]. We explain the idea of solving such a problem by the laplace-transform in terms of a simple example. ---------Section # 6.2 : Example # 6 :- [ shifted-data-problem=Initial-value-Problem]:Solve the I.V.P Given:-


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[

];

Initial-Condition: [

];

Given-General-Solution-Of-Eq-A :[

];

Required:* y(0)=? ; y’(0)=? +; Solution:  FOR-SUBSIDIARY-EQUATION:[

];

By apply laplace transform [

];

[

];

[

]; its subsidiary

equation. FOR-SOLUTION-OF-SUBSIDIARY-EQUATION:[

]; its subsidiary

equation. [

];


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[

];

[

];

by multiplying [ 1/(S2+1)] on each sides , [

];

[

];

[

];

CONSIDER [ 1/S2[S2+1] ]:Using Partial-fraction-method. [

];

[

];

[

];

[

];

[

];

[

];

Let, -> [ A+B=0]; So, [

];


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[

]; [

]; [

For B:- [

];

];

[ A+B=0]; [ 1+B=0]; [ So, [

B= 1 ]; ];

[

];

[

];

[

];

By placing value of eq-z in eq-E we get; [

];

[

];

For y(t) :By apply Inverse laplace transform on eq-k; [

];

[

];

[

];

[ [

]; ];


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Note: I some time mistakely type capital “S” instead of small “s” so be care-full. [

];

Let, [->

]; so the eq-T becomes

[

];

[

];

I.V.P

[

]; [

];

[

];

[

];

[ [-> 0=[A+B]/

]; ];

[A+B=0]; [B=- A ]; its not suitable value for A and B; So, we try different I.V.P [

];

So, [

];

By differentiating w.r.t ” t”. [

[

];

];

Let,


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Note: I some time mistakely type capital “S” instead of small “s” so be care-full. [A=y(0) , B=

];

When: I.V.P -> [

];

[

];

[

];

[

];

Let -> [B-A=0 ]; [

];

[

];

[

];

[

];

By placing the value of B in eq-PP’, we get , [

];

[

];

…. ….. [ A = +1 ]; For B:- When [ A = +1 ]; [

];

[

];

[

];


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Note: I some time mistakely type capital “S” instead of small “s” so be care-full. For y(t) :[

];

[

];

[ A = +1 ]; [

];

Placing value of A abd B in eq-SS we get ; [

];

[

];

[

]; Answer

-------------finished-here------------

----------- LAPLACE-TRANSFORM-OF-THE-INTEGRAL-OF-AFUNCTION:  Since differentiation and integration are inverse processes, and since, roughly speaking, differentiation of a function corresponds to the multiplications of its transform by “s”, we expect integration of a function to correspond to division of its transform by “s”, because division is the inverse operation of multiplications: [

]

[

]

Theorem # 3 : [ Integration of f(t) ]: If f(t) is piecewise continuous and satisfies an inequality of the form eq-2 from sec.6.1, that is,


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[

]

[ {:.

]; }.

Proof: suppose that f(t) is piecewise continuous and satisfies [eq-2 from sec # 6.1+, for some “ ” and “ ”. Clearly, if *eq-2 from sec # 6.1] holds for some negative [-ve+ “ ”, it also holds for positive *+ve+ “ ”, and we may assume that “ ” is positive [+ve]. Then the integral [

] is continuous, and by using [eq-2

from sec # 6.1] we get the following equations, [ [ [

] ] ]


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[

] , [:.

[

] {:. (

]

)}

This shows that g(t) also satisfies an inequality of the form [eq-2 from sec # 6.1]. Also, ->[

->[

], and the graph of f(t) and g’(t) are same,

];

->[

];

->[

];

{:.

, ->

->[ ->[ ->[

} ];

]; ];

For g(t):


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->[

];

->[ Or

]; {:. M and

are constant }.

->[

];

----------------finished ------------ LAPLACE-TRANSFORM [ ]:- [ define ] The laplace-transform replaces operations of calculus by operations of algebra on transforms . ----------------Problem-set 6.2 Q-1 ) using equation-1 and 2 from book , then find Laplace of f(t) ? Given: [

];

Required: [F(s)=?]; Formula: [

.];


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[

1 (0) ;];

For f(0) :[

];

[

];

[

];

For f’(t) :[

];

[ ]; {:.

};

By placing the value of eq-J & eq-K in eq-1 we get, [

.];

[

.];

[

.];

[

.];

[

.];

[

.]; its a correct Answer


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--------------finished ----------[ 6.2 ] Q 11 ) using theorems 1 and 2 , derive the following transforms that occur in applications ? Given: [

];

Solution: Let, [

];

[

];

[

];

Formula: [

.];

For f(0) :-

[

];

[

];

[ [

]; ];

By placing all required value in eq-1, [ENGINEERING-MATH-3,LAPLACE-TRANSFORM, CH-6,IN BME]BY SIKANDAR-LODHI

Page 63-OF-109


[

.];

[

.];

[

.];

[

.];

Let [

];

[

];

[

];

[

];

[

];

[

];

[ ]; By placing the value of eq-B in eq-Z, then eq-Z become, [

.];

[ .]; [


.];

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[

.];

…. ….. …. [

]; hence –proved,

--------------------finished-----------[ 6.2 ] Q 13 ) prove that Given: [

];

Solution: [ {:.

]; };

[ [

]; ];

……. ……


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.. [

]; hence proved

L.H.S = R.H.S --------------Finished ---------[ 6.2 ] Q 19 ) Application of theorem-3 , find f(t) or inverse-Laplace-transform ? Given: [

];

Solution:

[

];

Let [

];

[

];

Now, [ [

]; ];


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[

];

…. .. [

];

Let,  A+B = 0 ; C=0 ;  B= 2 ;  8 = - A.4 ; -> A= - 2 ; [

];

[ [

]; ];

[

];

[

];

[

];

[ [

]; ];

So,


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[

];

[

];

… … [

]; correct answer

----------------finished --------------

Initial value problem [I.V.P ] using laplace transform , solve: Q 27 ) INITIAL-VALUE-PROBLEM [

];

Initial condition : [

];

Solution: [

];

I.V.P [

];

[

];

Apply laplace transform [

];

[

];


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[

];

[

];

[

];

[

];

[

];

[ [

]; ];

[

]; in frequency-domain;

For y(t) :[

];

[ [ [

]; ]; ];

[

];

[

];

[

];


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[

+; it’s the right answer * good +

--------------finished --------------------------- [ exercise no 6.2 finished here ] -------------

[ 6.3 ] [ start here ] s-SHIFTING, t-SHIFTING, UNIT-STEP-FUNCTION:Two very important properties concern the shifting on s-axis and the shifting on the t-axis, as expressed in two shifting theorems [ theorems 1 and 2 of this section ]. s-Shifting: Replacing s by (s-a) in F(s) : Theorem-1 (first-shifting theorem: s-shifting ): {THEOREM-1 (1st shifting theorem) :[ define] -> it concerns shifting on the s-axis: the replacement of s in F(s) by (s-a) corresponds to the multiplication of the original function f(t) by [ e+a.t ] . } If f(t) has the laplace-transform of F(s), where [ s > ] , then [ ] has the laplace-transform F(s-a) where [ ], thus , if [ ];, then [

];

Key-> hence if we know the laplace-transform F(s) of f(t) , we get the laplace-transform of [ + by “shifting on the s-axis” *i.e by replacing s with [ ] , to get F(s-a) ].


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Fig : first shifting theorem, shifting on s-axis [ in this figure , a > 0 ].

[

]; By taking inverse laplace transform

[

];

Proof of theorem 1 .by definition, [

]; and

therefore , [

[

];

]; its theorem 1 .

--------------finished----


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Formula: replacing { s[s-a] } f(t)

-----------------[new-topic] -> t-Shifting: Replacing t by [ t-a ] in f(t) : we shell now state the 2nd – shifting theorem [ theorem-2 ], which concerns shifting on the t-axis: the replacements of t in f(t) by [ corresponds roughly to the multiplication of the transform F(s) by [ ], the theorem-2 given as follows.

]

 THEOREM-2 [Second (2nd ) Shifting Theorem; t-shifting ] : If f(t) has the transform F(s), then the function [

];

With arbitrary [

] has the transform [

].

 Hence if we know that transform F(s) of f(t), we get the transform of the function in eq-2 ,  Whose variable has been shifted * “shifting on the t-axis” ), by multiplying F(s) by [ ]. [

];


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…… If [

], then

[

];

By taking the inverse laplace transform on both side of eq-4, and interchanging side we obtain the companion formula. [

];

------------Proof-of-theorem-2: Show that : -> [

];

from the definition we have [

];

[

];

[

];

Let, [

[

];

]; {:. a=constant }  For new lower limit , when [

[

]: ];


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 For new upper limit, when [ [

]: ];

Put these values in above integral in [

];

[

];

[

];

Fig-92-a): f(t)=cos [t] and fig-92-b): f(t-a).u(t-a)=cos(t-2).u(t-2).where a=2.

We can write this as an intergral from 0 to infinity if we make sure that the integrand is zero for all t from 0 to a . We may easily accomplish this by multiplying the present integrand by the step function u(t-a), thereby obtaining eq-4 and completing the proof: [

];

Its fair to say that we are already approaching the stage where we can attack problems fro which the laplace transform method is preferable to the usual


Page 74-OF-109


method, as the examples in the next section will illustrate. In this connection we need the transform of the unit step function u(t-a), [

]; this formula follows directly from

the definition because. [unit-step function:

];

TABLE OF VALUE FOR ABOVE FIGURE-92-b : Any value of

0

1

AT

AT

[

];

Further application follow in the problem set and in the next sections. --------------finished—  UNIT-STEP-FUNCTION [or HEAVISIDE-FUNCTION ]

:

Define : Unit-Step-Function u(t-a) : [

];

, :. a = 0, 1,2,3,4,………..,Nth ; a=+ ve = any arbitrary positive number including zero };


Page 75-OF-109


Fig-90 [at a

0]

fig-91 [at a

0]

 The “unit-step-function u(t-a)” is a basic building block of various function, as well shell see, and it greatly increase the usefulness of laplace transform methods .  At present we can use it to write

in eq-2 in the form f(t-a).u(t-a),

that is , [

];

Figure 92 is the graph of f(t) for t>0, but shifted a unit to the right. ------------Some concept beyond this text book advance engineering mathematics 7th edition are given below .


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 JUMP-DISCONTINEOUS:- Any function is discontinuous for at any single point where any one limit is exist . this “Unit-step function u(t) ” is applied on a jump-discontineous functions only . also by shifting algument u(t) the function translate u(t-a) at * “a” greater then or equal to zero+ . By multiplying by a constant “M” we can change the height *or Amplitude+ of the “unit-step function”.-> [

];

Therefore:

---------------Problem set 6.3 Q 9) find laplace transform ? Given : -> [

];

Solution: Let , [ [

]; ];


Page 77-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. Let, -> [

];

[

];

[

];

[

];

So, ]; answer check this

[

answer from book ------------finished----Q 13 ) f(t)=? Given :-> [

];

Solution: Let, [

];

[ Let, -> [

]; ];

[ Let , -> [

]; ]; , -> [

So, -> [ So , [

]; ];

];Answer

------------finished-------Proof:


Page 78-OF-109


[unit-step function:

];

Solution: Consider L.H.S : [

];

Apply laplace transform [

];

[

];

[

];

[

];

[

];

[

];

[

];

[

];

[

];

[ [ [

]; ]; ]; it’s a formula to memorized or learned .

--------------finished---------------


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Q 21 ) representing the hyperbolic function in treams of exponential function and applying the first shifting theorem? Show that : Given: [

];

Solution: -----------------Formula’s: [ [

]; ];

---------------[

];

Consider L.H.S : [

];

…… .. [

];

[

];

……. ……….


Page 80-OF-109


…… .. [

]; henced proved , its correct answer.

-------------finished-here--Q 26 ) represent the given function into Unit-step function and find laplacetransform ? Given :

Solution:  FOR REPRESENTING GIVEN FUNCTION INTO UNIT STEP FUNCTION: [

];

So, applying laplace ransform : [

{:. [

];

] };


Page 81-OF-109


];

[

];

….. …. … [

];

[

];it’s the correct answer ,

but reader must verify it by your self . ------------finished-here----Q 23) Given:

Solution: Let, [

];

Apply laplace transform


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[

];

[

]; Answer , its correct answer.

-----------finished here----------Q 33 ) sketch the given function and find laplace transform ? Given: [

];

Solution: [

];

Let, [

];

Let , [

]; [

];

So, [

];

[ {:. [

]; }; ];

Apply laplace [

]; correct Answer

--finished------Q 41) in each case sketch the given function , [which is assume to be zero outside the given interval and find the laplace transform ?


Page 83-OF-109


Given: -> [

];

Solution: Sketch : [

];

{:.Unit-step-function: =>[

];}

[

];

Graph-sketching:

So, [

];

[

];

Formula: [

];

So, let [

];


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[

];

[

];

For V(s) :- When [ t – 1 ] : [

];

For V(s) :- When [ t – 2 ] : [

];

Apply laplace on eq-A [

];

[

];

[

];

We take L.H.S [

];

[

+; answer it’s the

correct answer . --------------finished -----------Q 47) sketch and find inverse laplace transform ? Given : -> [

];

Solution; [ [ [

]; ]; ]; correct-Answer .


Page 85-OF-109


---------finished------Q 51) [Initial-Value-problem(I.V.P)] , use laplace and solve ? Given:-> [

];

Initial-condition [I.V.P]: [

];

Solution: [

];

Apply laplace [

];

[

];

…… [

];

[

];

[

];

[ [

]; ];

[ [

]; ]; its correct answer

---------------finished---------------------CH-6 , LAPLACE TRANSFORM EX # 6.3 FINISHED HERE ------


Page 86-OF-109


[6.4 ] [ start here ] 

FURTHER-APPLICATIONS, “DIRAC’S-DELTA-FUNCTION [

]” :-

Introduction: In this section we consider some further applications and then introduce “Dirac’s-delta-Function ” , which greatly increase the usefulness of Laplace-transform in connection with “Impulsiveinputs”. * note : I do not cover any further example on 6.4 over the application but this example cover by the question 1 below on 6.4 ]. 

SHORT-IMPULSES.”DIRAC’S-DELTA-FUNCTION [

{“DIRAC’S-DELTA-FUNCTION [

+ “ :-

]”=”UNIT-IMPULSE-FUNCTION [

]” };

 Phenomena of an impulsive nature, such as the action of very large force [or voltage ] over very short intervals of time , are of great practical-interest, since they arise in various applications.  This situation occurs for instance, when a tennis ball is hit , e.t.c and so on .  Our present goal is to show how to solve problems involving short impulses by laplacetransform’s.  In mechanics, the impulse of a force f(t) over a time interval, say , [ ] is defined to be the integral of f(t) from “a” to “*a+k+”.  Of particular practical-interest is the case of a very short “k” * and its limit k0 (k approaches to zero[0] ) ], that is , the impulse of a force acting only for an instant . To handle the case, we consider the function that is : [ Fig # 102 [ the function

]; in [Eq-1] :-


Page 87-OF-109


Where, Ik is the rectangular area of the “Impulse fk (t) ”. ]; it’s the area of

[

the Impulse [or rectangular]; We can represent in terms of two unit-step function’s * section # 6.3] namely, [

];

From eq-5 in section# 6.3 , we get laplace-transform. [

];

[

];

[

];

……. …. .. [

];  The limit of as [k0], is denoted by “ its called the Diracdelta-function”.  The quotient in [Eq-3] has the limit equals to one as [ k0], as follows by “l’H pital’s-rule”.

Differentiate the numerator and denominator w.r.t “k”. [

];


Page 88-OF-109


 We note that is not a function in the ordinary sense as used in Calculus, but a so-called “Generalized-function” because *Eq-1] and [Eq-2] with [k0] {k --approaches to-> 0 } imply. [

]; infinitely long magnitude of impulse at point

“a” . And [

];

Graph:

Note: the reach infinity at [ t=a ] , infinite magnitude. 

is a very short impulse with

it’s an ordinary function which is every where zero [0] except at a single-point [ t=a ] on above graph , must have the integral is zero [0] .

Let, ,:. a = + ve = 0,1,2,3,….,+N -;


Page 89-OF-109


Graph-The –Following function: 1. 2. 3.

. . .

So,

 Nevertheless, in impulse problem’s its convenient to operate on as though it were an ordinary-function. -----------------Problem-set 6.4 : Q 1 ) Find the current in the RC-series-circuit, with Given:  R=100 , Ohm’s.  C=0.1 F, farad.  v(t)=100 volts if [ 1 i(t)=? ; Solution:


Page 90-OF-109


[

];

We can re-presented the v(t) in terms of unit-step-function: that is , [

];

Formula: eq of RC series circuit, see section # 1.8 ; [

];

So, placing the given value of R and C in eq-A , we get. [

];

[ [

]; ];

[

];

[

];

applying laplace transform on both side of eq-B , so then , [

];

[

];

[

];

[

];

----------Formulas: [

];


Page 91-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. ------------[

];

[

];

……. … .. [

];

[

];

Let, [

];

[ {:.

]; };

[

];

[

];

So, For f(t-1) :[ [ [

]; ]; ];

For f(t-2) :[

];

So,


Page 92-OF-109


];

[

];

[

];

[

];

[ unit step function .

]; Answer , we can represented i(t) in terms of

Graph this answer i(t) if possible. ---------------finished here--------6.4 , Q 7 ) Find y(t)=? Given: [

];

I.V.P ; [

];

Solution: [

];

By applying laplace transform. [

];

………. [

];

…………….. [

];

[ [

]; ];


Page 93-OF-109


];

--------I.V.P ; [

];

--------[

];

[

];

[

];

[

];

[

];

[

];

[

];

[

]; correct Answer.

-----------finished-here-----------------Ex # 6.4 finished here------------

[6.5] start here    

DIFFERENTIATION-AND-INTEGRATION-OF-TRANSFORM’S:DIFFERENTIATION-OF-TRANSFORM’S:It can be shown that if f(t) satisfies the conditions of existence theorem in section No 6.1, Then the derivative of its transform with respect to “s” can be obtained by differentiating under the integral sign wwith respect to “s”, thus , [

];

Apply derivative w.r.t “s”. [ [

]; ];

………. …….


Page 94-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. … [

Consequently, if F(s)=

];

,then,

[

];

--------Some concept beyond this text book->eq-1, can also re-write in this way as follows, [

];

{:. -> n = any real-No };

----------Differentiation of transform of a function corresponds to multiplication of function by “-t” . Equivalently, [ ]; This property enables us to get new transforms from given ones, as we show next. Serial No for formula’s [Eq-2]-> [eq-3]-> [eq-4]-> [eq-5]-> [eq-B]beyond this book->

Learn the above all 6 or 7 formulas: ----------------

INTEGRATION-OF-TRANSFORM’S:-

Similarly, if f(t) satisfies the condition’s of the existence theorem in sec # 6.1, and limit of f(t)/t , as “t” approaches to zero [0] from the right, exists, then, [

];

In this manner, integration of transform F( ) of a function f(t) corresponds to the division of f(t) by t . Equivalently


Page 95-OF-109


];

[

];

--------------Derivations [ un-important]: In fact: [

];

And it can be shown that under the above assumptions we may reverse the order of integration, that is [

];

[

];

{:.

};

[

];

[

];

Hence proved. ------------------------

Problem set 6.5 Q 19 ) find f(t) by using eq-6 or eq-1, ? Given: [

];

Solution: [

];

Applying log properties. [

];


Page 96-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. --------------------Re-view the Formula’s : [

];

[

];

[

];

[

];

[

];

[

];

--------------------So, FOR F’ (s) :[

];

Apply derivatives w.r.t s. [ [

]; ];

[

];

[

];

[

];

We required f(t). [

];

For f(t) :-


Page 97-OF-109


];

[

];

[

];

{:. [

] };

So, let, [

];

So, [

];

So, formula: [

];

So, [ {:. [

]; ] };

[

];

[

];

[

];

[ [ [ [

]; ]; ]; ];


Page 98-OF-109


];

[

];

[

];

By placing the value of [eq-GG] in [Eq-G] we get, [

]; };

[ [

]; +; it’s the correct answer .

-----finished here------[ 6.5 ] Q 5) find F(s)=? Given -> [

];

Solution: Let, [ [

]; ];

[ n=1 ];

Formula: [

];

So,

[

];

For F(s) :[ [

]; ];


Page 99-OF-109


];

Formula: [

];

[

];

{:. [

], n=1 };

[

];

[

];

[

];

…………. ………. … [

]; might be correct Answer, check it out.

--------------finished------------------Ex # 6.5 finished here-----

[ 6.6 ] start here



“CONVOLUTION”. “INTEGRAL-EQUATION’S” :-

INTRODUCTION: Another important general-property of Laplace-transform has to do with is products of transforms.  It often happens that we are given two transform’s F(s) and G(s) whose inverses f(t) and g(t) we know, and we would like to calculate the inverse of the product “H(s)=F(s).G(s)” , from those known inverses f(t) and g(t) .  This inverse h(t) is written “(f*g)(t)“, which is a standard-notation, and is called the “Convolution” of f(t) and g(t).  How can we find h(t) from f(t) and g(t) ? this is stated in the following theorem.  Since the situation and task just described arise quite often in application’s, this theorem is of considerable practical-importance.


Page 100-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. *Convolution of f(t) and g(t) = h(t) = (f*g)(t)+; it’s a standard-Notation’s. 

THEOREM # 1 :- [CONVOLUTION-THEOREM]:-

Let f(t) and g(t) satisfy the hypothesis of existence-theorem [sec # 6.1 ], then the product of their transform’s F(s)= , and G(s)= is the transform H(s)= , of the convolution h(t) of f(t) and g(t) , written [ ] and defined by [Eq-1] below. [

];

Given-> f(t),g(t),G(s)=

, F(s)=

So, -> [

, ];



= convolution of f(t) and g(t) =

;

[

]; 

PROOF-OF-THEOREM # 1 :- by the definition of G(s) and the second[2nd ] shifting theorem, for each fixed “ * tau +” * ] we have

{:. From 6.3, 2nd shifting theorem [eq-4], -> [

]; ];

So, [

];

[

];

[

];

[

]; {:.

,

};

From this and the definition of F(s) we obtain . [

];

[

];

[

Where {:.

];

,

};


Page 101-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. Here we integrate over “t” from * ] to , and then over from 0 to ; this corresponds to the “ “colored wedge—shaped-region” extending to infinity in the -plane, shown in fig-105, Fig # 105 :- Region of integration in the

-plane, in the proof of theorem-1,

Shaded region is the “region of –Integrations”. 

Our assumption’s on f(t) and g(t) are such that the order of integration can be reversed.

Thus, [

];

Interchanging the integral we get; [

];

[

];

{:.

};

This complete the proof . And

[ {:.

]; = it’s the standard notation for convolution -;

-----------finished--------


Page 102-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full.  1. 2. 3. 4.   

PROPERTIES-OF-CONVOLUTION :f*g=g*f ; -> [ Commutative-Law ]; f*(g1+g2 )=f*g1+f*g2 ; -> [ distributive-Law ]; (f*g)*v = f*(g*v); -> [ Associative-Law]; f*0 = 0*f = 0; but [ f*1 f ] in general, as Ex # 2 shows Another un-usal property is that, [(f*f)(t) 0 ] may not hold, as we can see from Ex # 1, very useful applications of convolution occur in a natural-way in the solution of differential-equation’s, as we shall now discuss. DIFFERENTIAL-EQUATION’S:-

From sec # 6.2 we re-call that the subsidiary-equation of the differential-equation [i-e] [

];

Has the solution. [

];

Where, [

];

[

];

Hence for solution y(t) of [ EQ-2], satisfying the initial condition, [

];

[

];

[

];

[

];

[

];

[

];

[ [

]; ];

Convolution-Theorem: {:.[

]};

[

+; it’s the solution of Eq-2 ;

---------------finished----------------

 INTEGRAL-EQUATION’S : Convolution also helps in solving certain integral-equations, that is equations in which the unknown function y(t) appears under the integral [ and perhaps also outside of it].


Page 103-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full.  This concerns only very special ones [those whose integral is of the form of a convolution], so, that it suffices to consider a typical example and a handful of problems,  But we do this because integral-equations are practically important and often difficult to solve. Ex # 5 [ INTEGRAL-EQUATION ]:Solve the integral-equation Given : -> [

];

Solution:1st –Step : [ Equation in terms of convolution] We see that the given equation can be written [

];

2nd –step: [ Application of Convolution-Theorem ] :[

];

Apply laplace-transform [

];

[

];

For Y(s):[

];

…….. …… ….. [

];

3rd –step: [ take inverse-Laplace-transform] or For y(t) :[ [ [

]; ]; ];


Page 104-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. [ [

]; +; it’s a correct answer , but check your self.

{:.

};

-------------finished-here-----------

Problem set 6.6 Q 3 ) find convolution=? Given-> [

];

Solution: Formula: [

];

[

];

Let,

By placing the above value in eq-D, so, [ [

]; ];

[

];

[ [

]; ];

[ [

]; ]; answer .

-----------finished here--------------


Page 105-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. 6.6 Q 17 ) f(t)=? By using convolution Theorem? Given-> [

];

Solution: [

];

[

];

[ [

]; ];

-----------

Formula: [

];

[

];

-------------------Apply convolution [ [

]; ];

--[

];

[

];

[ [

]; ];

[

];

[

];


Page 106-OF-109


];

[

];

[

];

…….. ……simple integrations used here …….. [

]; Answer

--------------finished here-----------[ 6.6 ] Q 31) using convolution theorem on I.V.P , solve: Given-> [

];

I.V.P : [

];

Solution: [

];

Apply laplace –transform [

];

[

];

[

];

[

];

[

];

[

];

[

];

[

];


Page 107-OF-109


];

[

];

[ [

]; ];

[

];

Apply convolution theorem on eq-B : [ [

]; ];

-----------------

Formula: [

];

[

];

-----------[

];

[

];

{:.

};

[

];

[

];

[

];

[ [

]; ];

…….. ………


Page 108-OF-109

Note: I some time mistakely type capital “S” instead of small “s” so be care-full. ---[

]; Answer

------------finished--------------------Ch-6->6.1—to—>6.6-------Finished-here---

X MUHAMMAD-SIKANDAR-KHAN-LODHI OWNER OF MY PERSONAL NOTES


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Laplace Transform, Engineering-Mathematics-3 ,Ch-6 in BME in PDF

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