solution manual for advance math

8/3/2019 Solution Manual for Advance Math

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Final answers are shaded in green.

SSOOLLUUTTIIOONNSSMMAANNUUAALL IINNAADDVVAANNCCEEDDEENNGGIINNEEEERRIINNGGMMAATTHHEEMMAATTIICCSS

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TECHNOLOGICAL INSTITUTE OF THE PHILIPPPINES

#938 Aurora Blvd., Cubao, Quezon City

College of Engineering and ArchitectureDepartment of Electronics Engineering

FINAL PROJECT

In Partial Fulfillment of the RequirementsNeeded for the completion of the subject

Advanced Engineering Mathematics for ECE(EC353)

Submitted By:Agustin, John Christopher V.Arrobang, Ma. Bernadette T.

Moreno, Kendrick Kent L.Susbilla, Mark Anthony F.

Submitted To:Engr. Armil S. Monsura

Instructor

March 18, 2011


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UNIT 1

COMPLEX

NUMBERS


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Example 1.1

Let z

8 j3 and z

9 j 2Find:(a)The real part of z1 and z2. (b) The imaginary part of z1 and z2.(c) The sum z1 z2. (d) The product z1 z2Solutions:(a)The following are the real part of z and z

Re (z) 8Re (z) 9(b)The following are the imaginary part of z and z

Im (z) 3Im (z) 2

(c)Using the general rule for addition of complex numbers,z z (x x) j ( y y )z z (8 9) j ( 3 2 )

(d)Using the general rule for multiplication of complex numbers,zz (xx yy ) j (xy xy )

zz (8)(9) (3)(2) j (8)(2) (9)(3))zz 7 8 j 1 1

z z 17 j


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Example 1.2

Let z

8 j3 and z

9 j2.Find:(a)The difference z z (b) The quotient Solutions:(a)Using the general rule for subtraction of complex numbers,

z z (x x) j(y y)z z (8 9) j(3 2)

(b)Using the general rule for division of complex numbers,zz xx yyx y j xy xyx y

Drill Problem 1.1

Direction: For items 1 to 9, let z 2 j3 and z 4 j5. Showing the details of your work (inthe rectangular form x jy):1.) (5z 3z)

z z 1 j5

zz 8(9) (2)(3)9 2 j 9(3) 8(2)9 2 zz 6685 j 4385


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Solution:First, substitute the values for z 2 j3 and, z 4 j5 then perform it algebraically

(5z 3z) 5(2 j 3) 3(4j5)(5z 3z) 1 0 j1 5 1 2 j1 5(5z 3z) 22

2.)

z

z

Solution:Using the general rule for complex conjugate defined as z x jy ,For z 2 j3 and z 4 j5

To find zz we can use FOIL methodzz (2 j3) (4 j5)zz 8 j10 j12 j15zz 8 15 j12 j10

3.) Re ZSolution :

Substituting the value of z 2 j3 to , 1z 1(2 j 3)Get the square of z 2 j3 by using FOIL method

(5z 3z) 484

zz 23 j2


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1z 1(2 j 3)(2j3)

1z 14 j 6 j 6 j9 1z 14 9 j 1 2 1z 15j12After getting the simplified form of , we can now multiply it by the complex conjugate ofthe denominator

1z 15j12 . 5j125j12By cross multiplication, 1z 5j122 5 j6 0 j6 0 j2144Since j2 -1, we can now simplify the denominator

1z 5j1225144

1z 5j12169 Therefore,

4.) Re (z) , Re (z)Solution: Re (z)Since z 4 j5 we can get z by FOIL method

z ( 4 j5) (4 j5)z 16 j20 j20 j25

Re 1z 5169


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z 16 25 j40z 9 j40

Re (z)Sincez 4 j5 , thereforeRe (z) 4

Re (z) 4

5.) Solution:zz 4 j 52 j 3

Multiply both the numerator and the denominator by the complex conjugate of thedenominator which is 2 j3zz 4 j 52 j 3 2 j 32 j 3zz 8 j 1 0 j 1 2 j154 j 6 j 6 j9 zz 8 1 5 j1 0 j1 24 9

z

z 7j2213 Therefore,

Re (z) 9

Re (z) 16

zz 713 j 2233


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6.) , Solution:

Since z 2j3 and z 4 j 5Multiplying the numerator and the denominator by the denominators complexconjugate, we have,z

z 2 j 3

4 j 5 4 j 5

4 j 5

zz 8 j 1 2 j 1 0 j151 6 j2 0 j2 0 j25zz 8 1 5 j 2 21 6 2 5 zz 7j2241 Therefore,

7.) (4z z)Solution:

Substitute the values for 1 and 2 in the expression then solve it algebraically

(4z z) 4(2 j 3) (4 j 5)(4z z) (8 j 1 2 4 j 5)Performing binomial expansion,(4z z) (4 j1 7)(4z z) 1 6 j 6 8 j 6 8 j289

zz 741 j 2241


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(4z z) 16289j136Therefore

8.) , Solution: Since

z 2j3 and z 2 j 3Multiplying the numerator and the denominator by the denominators complexconjugate, we havezz 2 j 32 j 3 2 j 32 j 3zz 4 j 6 j 6 j94 j 6 j 6 j9

zz 4 9 j 1 24 9 zz 5j1213 Thereforezz 513 j 1213

Multiplying the numerator and the denominator by the denominators complexconjugate, we have

(4z z)

273j136


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zz 2 j 32 j 3 2 j 32 j 3

z

z 4 j 6 j 6 j

94 j 6 j 6 j9zz 4 9 j 1 24 9 zz 5j1213 Therefore

9.) Solution:

For the numerator: Perform addition of complex numbersz z 2 j 3 4 j 5

z z 6 j 2For the denominator: Perform subtraction of complex numbersz z 2 j 3 4 j 5z z 2 j 8Substituting the values that we have in performing the addition and subtraction ofcomplex number to the expression and multiplying the numerator and thedenominator by the denominators complex conjugate, we havez zz z 6 j 2 2 j8 2 j8 2 j8z zz z 1 2 j 4 j 4 8 j164 j 1 6 j 1 6 j64 z zz z 1 2 1 6 j4 8 j44 6 4

zz 513 j 1213


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z zz z 28j4468 Therefore

Direction: For items 10 to 13, let z x jy. Find the rectangular form.10.)Im(z), Im(z)

Solution:Im(z)

z x j yz (x j y)(x j y)z (x j2 xy jy)(x j y)

z (xy j2xy)(x j y)

z (x y) j2xy(x j y)

z (x y)x j2 xy (x y)j y j2xyGrouping like terms,z x xy 2xy j2xy y(x y)

Therefore

Im(z)since Im(z) jy

z zz z 717 j 1117

Im(z) 3xy y


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Im(z) (jy)Im(z) y

Im(z)

y

11.)Re Solution:Since z x j yMultiply the numerator and the denominator by the denominators complexconjugate, we have 1z 1x j y x j yx j y

1z x j yx j x y j x y jy 1z x j yx jy 1

z x j y

x

y

1z xx y jyx yTherefore

12.)

Im(1 j)z

Solution:Perform first z by binomial expansion

z (x j y)(x j y)z x j x y j x y jy

Re 1z xx y


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z x j2 xy jyz x j2 xy y

Then, Im (z

) 2xyDealing with (1 j) , we can say that it is just equal to (1 j) in termsof laws of exponents, so we have:(1 j) (1 j)

(1 j) (1 j)(1 j)(1 j) 1 j j j

(1 j) 1 1 j 2(1 j) j2(1 j) (j2)

(1 j) (j)(2)(1 j) (1)(16)

(1 j) 16Multiplying the answers that we haveIm(1 j)z 16 (2xy)

13.)

Re Z

Solution: 1z 1(x j y)2

Im(1 j)z 32xy


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Performing the operations, 1z 1(x j y)2

1z 1(x j y)(x j y) 1z 1x2 j y x j y x j2y2 1z 1x2 y2 j2xy

Multiply the numerator and the denominator by the denominators complexconjugate, we have 1z 1x2 y2 j2xy x2 y2 j2xyx2 y2 j2xy 1z x2 y2 j2xyx4 x2y2 j2x3y y4 x2y2 j2x3y j 2 x3y j 2 x3y 4 x2y2

1z x2 y2 j2xyx4 2x2y2 4x2y2 y4

1z x2

y2

j2xyx4 2x2y2 y4 1z x2 y2x4 2x2y2 y4 j2xyx4 2x2y2 y4Well have

14.)Verify the following laws of conjugation(z z) z zIf 5 j3 and 2 j 4

Re 1z x yx 2xy y or Re 1z x y(x y)


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Solution: (z z) z z

(5 j 3) (2 j 4)

(5 j 3) (2 j 4)7 j 7 jTherefore

(zz) zzIf 5 j3 and 2 j 4

(zz) zz(5 j 3)(2 j 4) (5 j 3)(2 j 4)1 0 j 2 0 j 6 j12 10 j20 j6 j1210 12 j14 10 12 j14

22 j14 22 j14Therefore(zz) zz is correct.15.)Show that j2 -1 , j3 -j , j4 1, and -j , -1 , j , 1 . From the results of theseevaluate j2312 and its reciprocal j -2312

Solution:

(z z) z z is correct

(1) 1For j2

--1Since j 1 (1) 1(1)(1) j

For j3

-jSince j 1 (1) 1(1) 1For j

4

1Since j 1


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To evaluate j and j. We need to divide the exponent by 4 and if the results will be0 then j 11 then j j2 then j 13 then j j

For jSince j 1

Getting the reciprocal of . -j -j

11 1

For 1

Since j2

-1

1j j1j . jj j

1j j

For j

Since j2 -jGetting the reciprocal

jSince j2 -1

11 1

For 1Since j4 1


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For j 23124 578

Since there is no remainder, we can say that

For j By laws of exponents, we can say that

j 1j since j 1Therefore,

Example 1.3

For z 1 j, z 1 j and z 3 j 33 ind(a)The modulus r, the principal argument and express each in polar form.(b)All possible arguments(c)The plot of each one in complex plane

SolutionsFor z(a)The modulus rr (1) (1)

j 1

j 1

r 2


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The principal argument

Expressing in polar form,

(b)All possible arguments

(c)Graph in the complex plane

or


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For z(a)The modulus r

r (1)

(1)


Arg z tan yx

Arg z tan 11Arg z 4

Since is directed angle from the positive axis to the terminal point z. Here, allangles are measured in the counter clockwise sense. Thus,

Arg z 4 Arg z 34 Expressing in polar form,

(b)All the possible arguments

r 2

z 2 cos 34 j sin 34 or z 2 34

argz 34 2n ( n 1, 2, )


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For

(a)The modulus r



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Expressing in polar form,

(b)All the possible arguments


Example 1.4

Given and

(a)Find the product of and the quotient in rectangular form, without converting to polarform.


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(b)Find the product zz and the quotient Z by converting first to polar form, then back torectangular formSolutions:

(a)ProductPerform distributionzz ( 2 j2)(j3)zz j6 j6

Quotient zz 2 j2j3 Multiplying j3 to both numerator and denominator, we havezz 2 j2j3 j3j3zz j6j26j9 Since, j 1 zz 6 j 69 Therefore

(b)Convert z and z into their polar formConvert z firstr x y

r (2) (2)

zz 6 j 6

z

z 2

3 j 2

3


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r 22Arg z tan y

x

Arg z tan 22 since all angles must be in counter clockwise sense,Arg z 4

Arg z 34

Now, convert z r x yr 0 3r 3

Arg z tan yx

Arg z tan 30

Arg z 2Now that we have our values converted in polar form, let us get first theproduct of z and zzz rrcos() jsin()

zz 22 (3) cos34 2 j s i n 34 2zz 62cos 54 jsin 54


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zz 62 22 j 22

Then, we can also get the quotient of z and zzz rr cos() jsin()zz 223 cos34 2 j s i n 34 2zz

223

22

j 22

Example 1.5

Direction: Evaluate the following, expressing the answer in rectangular form(a) (3 j4)Solution:z (3 j4) z (3 j4)

First, get the magnitude and the principal argument

r 3 4r 5 tan 43 0 . 9 3

zz 6 j 6

zz 23 j 23


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Using De Moivres Formula,z r (cosnjsin n) r

We can now substitute the values that we computedz 5 (cos 0.93 j sin 0.93 )z 5cos(3)0.93j sin (3)0.93

(b)(1 j)Solution:

z ( 1 j) z ( 1 j )First, get the magnitude and the principal argumentr 1 1

r 2

tan 11 4Using De Moivres Formula,z r (cosnjsin n) r

z 2 ( cos 4 j sin 4 )

z 117j43

z 64


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Example 1.6

Find all the roots of the following in rectangular form and plot them(a)j(b)5j12Solutions:(a)j

Using De Moivres Formula

z

r

cos 2 k

n r 1 n 2 2 k 0, 1If k 0 we can obtain the 1st root

1st root 1 cos

2 2 (0)

2 j sin

2 2 (0)

2

1st root 1 22 j 22

If k 1 we can obtain the 2nd

root2nd root 1 cos 2 2 (1)2 j sin 2 2 (1)2 2nd root 1 ( 22 j 22 )

1st root 22 j 22


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Graph of the Roots:

(b)

If k 0 we can obtain 1stroot


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If k 1 we can obtain the 2nd root

Graphs of the roots

Drill Problems 1.2

Direction: For items 1 to 8, represent the following numbers in polar form. Show the details of your

work.

1.)Solution:

We must first obtain the magnitude and the principal argument


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r 3 2

t a n

yx

t a n 33 4 Since the formula to convert to polar form isz r(cos js in) r

We can substitute the values that we have in the equation2.) j2,j2

Solution:(a)j2

We must first obtain the magnitude and the principal argumentr (0) (2)r 4r 2 t a n yx

t a n 20 2Since the formula to convert to polar form isz r (cos js in) r

z 3 2 cos 4 jsin 4 or z 32 4


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We can substitute the values that we have in the equation

3.) 5Solution:We must first obtain the magnitude and the principal argument

r (5) 0r 25r 5 tan yx tan 05

Since the formula to convert to polar form is

z r (cos j sin) rWe can substitute the values that we have in the equation

4.) j Solution:We must first obtain the magnitude and the principal argumentr 12 14

r 0 . 9 3

z 2 cos

2 jsin

2 or z 2

2

z 5(cos0jsin0) or 5


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t a n yx

t an 14 12 1.004Since the formula to convert to polar form is

z r(cos js in) rWe can substitute the values that we have in the equation

5.)

Solution: We must first obtain the magnitude and the principal argumentFor the numerator,

32j2r 32 (2)

r 2 2 t a n yx

tan1 232

z 0 . 9 3(cos1.004jsin1.004) or z 0.93 1.004


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0.44For the denominator,

2 j 23r 2 23

r 223

t a n yx t a n 232 2.70

We can now perform the division in polar form,zz rr zz 22223 0.442.70zz 3 2.26Converting to its polar form, we can come up to

zz 3cos(2.26) jsin(2.26)or zz 3 2.26


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6.) Solution:

We must first obtain the magnitude and the principal argumentFor the numerator,32j2

r 32 (2)

r 2 2

t a n yx tan 232 0 . 4 4

For the denominator,

2 j 23r 2 23

r 223

t a n yx t a n 232 2 . 7 0


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Using the general equation for finding the equation for the division ofcomplex numbers in polar formz

z r

r

zz 22223 0.442.70Therefore

7.)

Solution: We must first obtain the magnitude and the principal argumentFor the numerator,

6 j5r (6) (5)r 6 1

t a n yx tan1 56 0.69 2 . 4 5

For the denominator,r (0) (3)

r 9r 3 t a n yx

zz 3cos() jsin() or zz 3


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tan1 30

2

Using the general equation for finding the equation for the division ofcomplex numbers in polar formzz rr zz 613 2.45 2

8.) Solution:

We must first obtain the magnitude and the principal argumentFor the numerator,2 j 3r (2) (3)r 4 9r 1 3

t a n yx t a n 32 0 . 9 8

zz 613 cos (0.87) jsin(0.87) or zz 613 0.87


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For the denominator,5 j 4

r (5)

(4)

r 2 5 1 6r 4 1 t a n yx tan1 45 0.67

Using the general equation for finding the equation for the division ofcomplex numbers in polar formzz rr zz 1341 0.980.67

For items 9 through 11, find all the roots of the given expression in rectangular form. Plot the rootsin the complex plane.9.)1

Let z1,We must first find the values for r and ,r x y Arg z tan yx

r (1)0 A r g z t a n r 1 A r g z

zz cos(0.31) jsin(0.31) or zz 53341 0.31


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r 1 n 4 , k 0, 1, 2, & 3

Using De Moivres formula,z r cos 2 kn jsin 2 kn We can now find the first root,

if k0,1st root 1 cos 2 (0)4 jsin 2 (0)4

1st root cos4 jsin

41st root 22 j 22 For the second root,

if k 1,2nd root 1 cos 2 (1)4 js in 2 (1)4

2nd root cos 34 js in 34

For the third root,if k 2,

3rd root 1 cos 2 (2)4 jsin 2 (2)4 3rd root cos 54 jsin 54

2nd root 22 j 22

3rd root 22 j 22


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For the fourth root,

if k 3,

Graph in the complex plane

10.)Let

We must first find the values for r and ,


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r5 Argz0.927295218r 5 0.927295218n 4 , k 0, 1, 2, & 3Using De Moivres formula,

z r cos 2 kn jsin 2 kn We can now find the first root,

if k 0,1st root 5 cos 0.9272952182(0)3 js in 0.9272952182(0)3

For the second root,

if k 1,2nd root 5 cos 0.9272952182(1)3 js in 0.9272952182(1)3 For the third root,

if k 2,

3rd root 5

cos0.9272952182(2)

3 jsin0.9272952182(2)

3

1st root 1.628937 j0.5201745

2nd root 1.26495 j1.1506

3rd root 0.36398 j1.670788


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Graph in the complex plane

11.)

Let


Using De Moivres formula,


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We can now solve for the first root,if k 0

1st root 1

cos0 2 (0)

8 js in0 2 (0)

8 1st root (cos0jsin0)

For the second root,if k 1,

2nd root 1 cos 0 2 (1)8 js in 0 2 (1)8 2nd root cos 4 js in 4

For the third root,if k 2,3rd root 1 cos 0 2 (2)8 jsin 0 2 (2)8

3rd root cos 2 jsin 2

For the fourth root,if k 3,4th root 1 cos 0 2 (3)8 jsin 0 2 (3)8

1st root 1

2nd root 22 j 22

3rd root j


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4th root cos 34 js in 34

For the fifth root,if k 4,5th root 1 cos 0 2 (4)8 jsin 0 2 (4)8

5th root (cosjsin)

For the sixth root,if k 5,

6th root 1 cos 0 2 (5)8 js in 0 2 (5)8

6th root cos 54 js in 54

For the seventh root,if k 6,

7th root 1 cos 0 2 (6)8 jsin 0 2 (6)8 7th root cos 32 jsin 32

4th root 22

j 22

5th root 1

6th root 22 j 22

7th root j


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For the eighth root,

if k 7,

Graph in the

complex plane

For items 12 & 13, evaluate the following, expressing the final answers in rectangular form.

12.)Let



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r 99 A r g z t a n 99

r 9 2 A r g z 4Using De Moivres formula,

z r(cosnjsinn) rn(9 j 9) 92 cos3 4 js in3 4

(9 j 9) 92 22 j 22 (9 j 9) 729 (2) 22 22

13.)( 2 j6)Let z 2 j6

We must first find the values for r and ,r xy A r g z t a n yxr (2)6 A r g z t a n 62r210 Argz1.249045772Using De Moivres formula,

z r(cosnjsinn) rn( 2 j6) 210cos2(1.249045772) js in2(1.249045772)

(9 j 9) 1458j1458

( 2 j6) 32j24


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For items 14 & 15, prove the following trigonometric identities using De Moivres formula.14.)cos2cos s i n 15.)sin22cossin

Solution for numbers 14 and 15Using the De Moivres formula,(c os js in) cosnjsinnwhere in n 2 (c os js in) cos2jsin2

cos j 2 c o s s i n j sin c os 2 js in2

(cos

s in

) j2cossincos2jsin2Since in trigonometry, cosine is located at the x-axis where your x-axis (real axis) isconsidering only the real part of the equation, we can already say that:

Also in trigonometry, sine is located at the y-axis where it is called imaginary axisbecause it only considering the imaginary part of the equation, therefore we can saythat:

Example 1.7

Evaluate the following expressions, expressing answers in rectangular form.a.)cos(1 j)

Solution:Using the trigonometric identity of sum of two angles of cosine,cos(A B) cosAcosBsinAsinB

cos(1 j) cos1cosjsin1sinj

cos2cos s in

sin22cossin


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Since,cosjcosh1

sinjsinhjcos (1 j) cos 1cosh 1 j sin1 sinh1b.) sinh(4 j 3)

Solution:

Using the trigonometric identity of difference of two angles of hyperbolic sine,sinh (A B) sinhAcoshBcoshAsinhBsinh(4 j 3) sinh4 cosh(j3) cosh4sinh(j3)Since, cosjcosh1

sinjsinhjsinh(4 j 3) sinh(4) cos(3) jcosh(4) sin(3)

Example 1.8

Evaluate the following logarithms, expressing the answers in rectangular form.

a.)

ln1,Ln1

Solution: ln 1Since it doesnt consider the principal argument, we will use the equation

cos (1 j) 0.8337 j 0.9889

sinh(4 j 3) 27.0168j 3.8537


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l n z l n r j j 2 nLet z 1

r x

y

A r g z 0r 10 r 1l n 1 l n 1 j(0) j2n

Ln 1Since it consider the principal argument, we will use the equationL n z l n|z| jA rg z

Let z 1r xy A r g z 0r 10

r 1 L n 1 l n 1 j(0)

ln 1 0 j2n ; n 0,1,2,

L n 1 l n 1 j(0)


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b.) ln(3 j 4) , Ln (3 j4)Solution:

ln(3 j 4)Since it doesnt consider the principal argument, we will use the equationl n z l n r j j 2 n

Let z 3 j4r xy A r g z t a n yxr 3(4) A r g z t a n

r5 Argz0.927295

ln(3 j 4) l n 5 j(0.927295) j2n

Ln (3 j4)Since it consider the principal argument, we will use the equation

L n z l n|z| jA rg zLet z 3 j4r xy A r g z t a n r 3(4) A r g z t a n 43r5 Argz0.927295Ln(3 j 4) l n 5 j(0.927295)

Example 1.9

Evaluate the following, expressing the answers in rectangular form.a.) j

ln(3 j 4) 1.609 j 0.927 j2n ; n 0,1,2,

Ln(3 j 4) 1.609 j 0.927


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Solution:The general power of a complex number z x jy are defined by theformula,

z

e

Let zj and cjWe must first solve for the value of r and r xy Arg z tan yxr 01 Arg z tan 10

r 1 2Then solve for lnz l n z l n r j j 2 nl n j l n 1 j 2 j2nl n j j 2 j2nTherefore,

z

e

z ez eb.) (1 j)()

Using the general power for a complex number,

Let z 1 j and c 2 j

r xy Arg z tan r 11 Arg z tan 11

r 2 4

j e


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Solve for the value of lnz,l n z l n r j j 2 n

ln(1 j) l n 2 j 4 j2nThen, z e (1 j)() e() (1 j)() e (1 j)() e ee

Since e c o s j s i n e c o s j s i n (1 j)() 2e cos 2 jsin 2 cosln2jsinln2

(1 j)() 2ejcosln2jsinln2

(1 j)() 2ejcosln2j sinln2(1 j)() 2e sinln2 jcosln2(1 j)() 2e sin12 ln2 jc os 12 ln2


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Drill Problem 1.3

For number 1 through 4, find the principal value of lnz, rectangular form, when z equals,1.) -10Solution:|z| r x y Arg z tan

|z| r (10) (0) Arg z tan 010|z

| r 10 Using the formula: l n z l n|z| j Arg z

2.) 2 j 2

Solution:|z| r x y Arg z tan yx|z| r 2 2 Arg z tan 22|z| r 2 2 4

Using the formula: l n z l n|z| j Arg zln(2 j 2) l n 2 2 j 4

ln(10) l n 1 0 j

ln(2 j 2) 12 l n 8 j 4


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3.) 2 j 2Solution:

|z| r x y Arg z tan yx|z| r 2 (2) Arg z tan 22|z| r 2 2 4

Using the formula:

l n z l n|z| j Argzln(2 j 2) l n 2 2 j 4

4.)

je

Solution:|z| r x y Arg z tan yx

|z| r 0 (e) Arg z tan e0|z| r e 2

Using the formula: l n z l n|z| j Arg zln(je) l n e j 2

ln(2 j 2) 12 l n 8 j 4


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For number 5 through 8, evaluate the following answers in rectangular form.5.) ln e Solution:

Let z eThen, find the values for r and

r x y Arg z tan yxr e 0 Arg z tan 0er e 0

Since it doesnt consider the principal value, we will use the formulal n z l n r j j 2 n

l n e l n e j 0 j 2 n

6.) lne Solution:

Considering the value of z e and c j, we use the formula

z

e

Let z eSolve for the values of r and r x y Arg z tan yxr e 0 Arg z tan 01

ln(je) 1 j 2

ln(e) 1 j2n ; n (0,1,2, )


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r e Since it doesnt consider the principal value, we will use the formula,

l n z l n r j j 2 nln(e) l n e 1 j j 2 nz e

e e() lne ln e()

lne

j(1 j j2 n)

lne ( j j) j2nlne ( j j2 n)7.) ln(4 j 3)

Solution:Since it doesnt consider the principal value, we will use the formulal n z l n r j j 2 nLet z 4 j3, find the values for r and r x y Arg z tan yxr 4 3 Arg z tan 3

4

r5 0.6435011ln(4 j 3) ln5j0.6435011j2n

lne ( 1 2 n)j ; n (0,1,2, )

ln(4 j 3) 1.6094 j0.6435 j2n ;n (0,1,2, )


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8.) ln eSolution:

Considering the value of z e and c j3, we use the formulaz e Let us solve for the value of r and r x y Arg z tan yxr e 0 Arg z tan 0e

r e 0l n z l n r j j 2 nl n e l n e j(0) j2n

l n e 1 j 2 ne e()

ln e l n e()

For numbers 9 through 12, find the principal value in rectangular form.9.) j , j2

Solution: jConsidering the value of z j and c j2, we use the formulaz e

ln e j3 j2n ; n (0,1,2, )


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Let us solve for the value of r and r x y Arg z tan yx

r 0 1 Arg z tan 10r 1 2Using the general equation,lnz ln r j Argz

l n j l n 1 j 2

l n j j 2j e j e

j2Considering the value of z j2 & c j, we use the formula

z e Let us solve for the value of r and

r x y Arg z tan yxr 0 2 Arg z tan 20r 2 2

j e


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Using the general equation,l n z l n r j

l n j 2 l n 2 j 2

j2 ej2 e using logarithmic exponent rulee ee

j2 e esince,e cos j sin , therefore

10.) 4() Solution:Considering the value of z 4 & c (3 j), we use the formulaz e Let us solve for the value of r and

r x y Arg z tan yxr 4 0 Arg z tan 04r 4 0Using the general equation for natural logarithm of a complex number,

l n z l n|z| jln(4) l n 4 j(0)ln(4) l n 4

4() e() using the logarithmic exponent rule e ee

j2 e(cosln2jsinln2)


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4() e e

Since e

c o s j s i n e 4(cosln4jsinln4)

11.)(1 j)() Solution:

Considering the value of z 1 j and c 1 j, we use the formulaz e r x y Arg z tan yx

r 1 (1) Arg z tan 11

r 2 4

Using the formula for natural logarithm of z,l n z l n r j

l n z l n 2 j 4

(1 j)() e() (1 j)() e Using the logarithmic exponent rule, e ee

z 6 4 cos(ln 4) jsin(ln 4)


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(1 j)() e e Since e c o s j s i n e e e cos ln2

4 jsin ln2

4

12.)(1)()Solution:

Considering the value of z 1 and c (1 j 2), we use the formula

z e r x y Arg z tan yxr (1) 0 Arg z tan 01r 1

Using the general formula of natural logarithm of z,l n z l n r j l n z l n 1 j

l n z j (1)() e()()

(1)() ee(1)() eesince e c o s j s i n

z 2 e cos ln2 4 jsin ln2 4


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e e(cos2jsin2)

For numbers 13 through 15, solve for z in rectangular form13.) l n z 2 j Solution:

To eliminate the natural logarithm, we need to equate on both sides,e ez ez eesince,e c o s j s i n

z e cos

2 js in

2

14.) lnz0.3j0.7Solution:

To eliminate the natural logarithm, we need to equate on both sides,

e e(..)

z e(..)z e.e.

z e

z e


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since,e c o s j s i n z e.(cos0.7jsin0.7)

15.) l n z e j Solution:To eliminate the natural logarithm, we need to equate on both sides,

e e()

z ee since,e c o s j s i n z e(c os js in)

z e

z1.032j0.870

z15.154


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UNIT 2

LAPLACE ANDINVERSE LAPLACE

TRANSFORM


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Example 2.1

Direction: Using the Laplace integral, find the Laplace transform of the following:a.) f(t) 1Solution:Using the general equation of the function defined for s ,F(s) ef(t)dt

F(s) l i m ef(t)dt F(s) l i m e(1)dt

F(s) l i m edt F(s) l i m 1s e 1 T0F(s) l i m

1

se 1

F(s) 1s e() 1F(s) 1s 0 1 Therefore,

b.) f(t) eSolution:using the general equation of the function defined for s ,

F(s) 1s


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F(s) e f(t)dt

F(s) l i m e

f(t)dt

F(s) l i m e(e)dt F(s) l i m edt F(s) l i m e()dt

F(s) l i m 1s a e() 1 T0F(s) 1s a lime() 1 T0F(s) 1s a lime() 1

F(s) 1s a e() 1

F(s) 1s a 0 1

c.) f(t) t

Solution:Using the general equation of the function defined for s ,F(s) ef(t)dt F(s) l i m ef(t)dt

F(s) 1s a


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F(s) l i m e(t)dt Integrating by parts we let: ()

1 Then,

F(s)lim t 1s e 1s edt T0F(s)lim ts e 1s (e)dt T0F(s)lim ts e 1s 1s e T0

F(s)lim ts e 1s (e) T0F(s)lim ts e 1s (e) T0

F(s)lim Ts e 1s e 1sF(s)lim Ts e l i m 1s e l i m 1sF( s ) 1s lime 0 1s

F( s ) 1s e() 1s

F ( s ) 0 1sF(s) 1s


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d.) f(t) cos t , is a constantSolution:Using the general equation of the function defined for s ,F(s) ef(t)dt F(s) l i m ef(t)dt

F(s) l i m e(cost)dt Integrating by parts we let

1 Then,

F(s)costs e 1s esintdt

F(s) 1s e cos t T0 s e sintdt

Integrating by parts for e sintdt , we let:

1

Then, e costdt 1s e sint T0 ( 1s e) cost dt e costdt 1s e sint T0 s e costdt


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e costdt 1s e cos t T0 s 1s e sint T0 s e costdt

e

costdt

1s e

cos t T0

s

e

sint T0

s

e

costdt

e costdt 1s e cos t T0 s e sint T0 s e costdt e costdt s e costdt 1s e cos t T0 s e sint T0

1 s e costdt 1s e cost T0 s e sint T0 11 s

e costdt 1s e cos t 1 s s e sint 1 s Then, applying the limits,

F(s) 1s e cosT1 s s e sinT 1 s

F(s) ss e cosTs

s e sinTs Sincecos 1 sin 0 e 0

Applying the limits: lim

F(s) ss e cos()

e

sin()s F(s) ss


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e.) f(t) sint , is a constantSolution:Using the general equation of the function defined for s

F(s) ef(t)dt F(s) l i m ef(t)dt

F(s) l i m e(sint)dt Integrating by parts we let:

1 Then,

e sintdt 1s e sint T0 s e costdt

Integrating by parts for e costdt , we let: du

e costdt 1s e cos t T0 s e sintdt e sintdt 1s e sint T0 s 1s e cost T0 s e sintdt

e sintdt 1s e sint T0 s 1s e cost T0 s e sintdt 1 s e sintdt 1s e sint T0 s e cost T0 s e sintdt


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e sintdt 1s e sint T0 s e cost T0 ( 1s s )

e sintdt ( 1s ) ss e sint T0 ss e sint s Applying the limits:

F(s) se sinTs s e cosT

F(s) s

s e cosTs

s e

sinTs Sincecos 1 sin 0 e 0

Example 2.2

Direction: Using linearity theorem and the previously obtained Laplace transform pairs, find theLaplace transform pairs, find the Laplace transform ofa.) coshat

Solution:Since coshat e e2 We can now use the linearity theorem,coshat e e2

F(s) s


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sinhat 12 2as a

c.) costSolution:Since

cost e e2

We can now apply the Linearity theorem cos t e e2 cos t 12 e eSince

e 1s j and e 1s j

cos t 12 1s j 1s j cos t 12 s j s j s j

cos t 12 2ss

F(s) as a

F(s) ss


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d.) sintSolution:Sincesint e ej2 Using the Linearity theorem

sint e ej2 sin t 1j2 e e

Since e 1s j and e 1s j sin t 1j2 1s j 1s j sin t 1j2s j s j s j

sin t 1j2

2js

Example 2.3

Direction: Apply the shifting theorem and the previously obtained Laplace transform of the

following:a.) e costSolution: cost ss2 2 .s s a

F(s) s


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b.)

e

sintSolution: sint s2 2 .s s a

Example 2.4

Direction: Find the Laplace transform of the following functions using the table (variables otherthan t are considered constant).a.) t 2tSolution:

By linearity, t 2t t 2 t

Since t 2!s and t 1sThen,t 2t 2!s 1s

b.) costSolution:Since

cost s a( s a )2 2

sint ( s a )2 2

t 2t 2 1s 1s


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cost ss Then,

c.) e coshtSolution:Using shifting theorem,

cosh t ss aThen, e cosht ss 1 .s s 2

e cosht s 2( s 2 ) 1

e cosht s 2s 2 s 4 1d.) e Solution:

Using logarithmic propertiese

e

e

Then, e e e

Since e is a constant, and e

cos t ss

e cosht s 2s 2 s 3


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e.) cos(t )Solution:Using a trigonometric identity (sum of two angles of cosine)

cos()coscossinsinSo,

cos(t)costcossintsinSince cos and sin are constant, and cost ss and sin t s Then,

cos(t ) scoss sins

Simplifying further, we get

Example 2.5

Direction: Find the inverse Laplace transform of the following functions using he table (variablesother than s are constants).a.) Solution:

By Linearity theorem,

e es 2 b

cos(t ) s c os s ins


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4s3s 4ss 3s

Since ss cost and s sintThen,

b.) Solution:Expanding the term,s 3s 12s 3 12

s 3s 12s 3 12s Sincen!s tThen

c.)

Solution:Completing the square of the denominator well have15s 4 s 2 9 15( s 2 ) 25Since e sint ( s a )

4 s 3 s 4 cos t 3 s int

s 3s 12s 1 32 t 12 t


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Then 15(s 2) 25 155 5( s 2 ) 5Therefore,

d.)

Solution:Performing partial fraction expansion 8s(s4) As Bs 4 s(s4)

8 A(s 4) B(s)s : 0 A Bs: 8 4 A

Well have the value of A and B as,A 2; B -2 8s(s 4) 2s 2s 4Therefore,

e.) Solution:Performing partial fraction expansion 1s 2s 5 As 2 Bs 5 s 2s 5

15(s 2) 25 3e sin5t

8s(s 4) 2 2 e


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1 A s 5 Bs 2s: 0 A Bs: 1 A5 B2By elimination method,

A B Then,

1s 2s 5 5 32 s 5 32s 4 Therefore,

Drill Problem 2.1Direction: Find the Laplace transform of the following functions. Variables other than t areconstants.

1.) (t 3)Performing binomial expansion,

(t

3)

t

6t

9(t 3) (t 6t 9)Then, applying the distributive property of Laplace,(t 3) t 6t 9

1s 2s 5 e e5 3


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(t 3) 4!s 6 (2!)s 9s

2.) sin 4tUsing the trigonometric identity for double angle formula of sine,sin A 12 (1 cos 2A)

(sin 4t) 1

2(1cos8t)

(sin 4t) 12( 1 cos8t)(sin 4t) 12 1s ss 64

(sin 4t) 12 s 64 ss (s 64)

3.) e sinh5tUsing the shifting theorem, (sinh5t) 5s 25 .s s 1

(e sinh5t) 5( s 1 ) 254.) sin 3t Using trigonometric identity for the sum of two angles of sine,sin(A B) sinAcosB cosAsinB

(t

3)

24s

12s

9s

(sin 4t) 32s (s 64)

3


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sin3t 12sin3tcos 12 cos3tsin 12sin3t 1

2 c o s 1

2sin3t sin 1

2cos3t

5.) 8sin0.2t (8 sin 0.2t ) 8 sin0.2t

since, sint

s

(8sin0.2t) 8 0.2s 0.04

6.)

sintcostUsing the trigonometric identity for product formula of the angles sine and cosine,sinAcosB 12 sin(A B) 12 sin(AB)sintcost 12 sin(tt) 12 sin(tt)

sintcost 12 sin2t 12 sin0(sintcost) 12 sin2t(sintcost) 12 2s 4

sin3t 12 3 cos 0.5 s sin 0.5s 9

(8 sin 0.2t) 1.6s 0.04 4


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7.)( t 1 )

( t 1 ) t 3t 3 t 1By applying the linearity theorem,(t1) t 3t 1(t1) 3!s 3 (2!)s 2s 1s

(t1) 6s 6s 2s 1s8.)3.8te.

By shifting theorem,(3.8t) 3.8

s .

s ( s 2 . 4 )

(3.8te.) 3.8(s2.4)9.)3te.By shifting theorem,

(3t) 3(4!)s .s (s0.5)

10.)5e sintBy shifting theorem,

(5sint) 5s .s ( s a )

(sintcost) 1s 4

(3te.) 72(s0.5)


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(5e sint) 5(sa)

II. Direction: Find the Inverse Laplace transform of the following functions. Variables other than sare constant.11.)

1s 5 1s 5 1s 5 1s 5

Multiplying in in order to satisfy (sin t) 1s 5 1s 5 15 sin 5 t e

12.)2s16s 16 2ss 16 16s 16

2 s 1 6s 16 25s 16 16s 16Since, ss cosht and s sinht

13.) 102 2 102 1 22

2 s 1 6s 16 2cosh4tsinh4t


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1(s a)( s b) 1a bs a

1a bs b

18.) Performing completing the square in the denominator, we will arrive at:

4 s 2s 6 s 1 8 4 s 2s 6 s 9 1 8 94 s 2s 6 s 1 8 4 s 2(s 3) 9In order to satisfy the form of the numerator to obtain its inverse Laplace transform,we must get the value of x. We will able to form the equation,

4 s 2 4(s 3) x4 s 2 4 s 1 2 xx 2 12x 1 0 4 s 2s 6 s 1 8 4 (s 3)( s 3 ) 9 10(s3) 9

Since, s a(s a) e cosht and (s a) e sinhtTherefore, 4 s 2s 6 s 1 8 4e cosh3t 103 e sinh3t

1(s a)( s b) 1b a (e e)


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19.)

Performing completing the square in the denominator of the equation,s 10s24 s 10s25 24 25s 10s24 (s 5 ) s 10s24 (s 5 )

20.) Performing completing the square in the denominator of the equation,2 s 5 6s 4 s 1 2 2 s 5 6s 4 s 4 1 2 4

2 s 5 6

s

4 s 1 2 2 s 5 6

( s 2 )

16

in order to satisfy the form of the numerator to obtain its inverse Laplace transformwe must get the value of x. We will be able to form the equation:2 s 5 6 2(s 2) x2 s 5 6 2 s 4 x

x 4 5 6x 52

2 s 5 6s 4 s 1 2 2 (s 2)( s 2 ) 16 13 4(s 2) 16 2 s 5 6s 4 s 1 2 2e cosh4t 13e sinh4t

s

10s24 e sinht

2 s 5 6s 4 s 1 2 e(2cosh4t 13e sinh4t)


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Example 2.6Direction: Find the Laplace transform of the following using the differentiation property.

a.) teSolution:f(t) te F(s)

Whenf(0) 0

f(t) kte e

f(t) k f(t) ef(t) k f(s) 1s kDP:f(t) s f(s) f(0)

k f(s) 1s k s f(s) f(0)

k f(s) 1s k s f(s)(s k)f(s) 1s k

b.)

tsintSolution: f(t) t s i n t F(s)f(0) 0f(t) tcostsint

f(0) 0

f(t) 1(s k)


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f(t) ts intcostcost

f

(t)

tsint2cost

f(t) F(s) 2ss DP:f(t) sF(s) s f(0) f(0)

F(s) 2ss sF(s)

(s

)F(s) 2ss

c.) sin tSolution:f(t) sin t F(s)

f(0) 0f(t) 2sin tcos tf(t) 2 sin t2cos t

f(0) 0f(t) 2 F(s) 2cos t

f(t) 0

f(t) 2 F(s) 2 12 1s ss

DP:f(t) s F(s) s f(0) f(0)

f(t) 2s(s )


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f(t) s F(s) 0 02 F(s) 1s ss 4 s f(s)

(s 2)F(s) 1s ss 4F(s) 1s ss 4(s 2) F(s)

s 4 ss(s 4)(s 2)

F(s)

s

4

s

s 6s 8s

F(s) 2(s 2)(s 4s)(s 2)F(s) 2(s 4s)

Example 2.7Direction: Find the inverse Laplace transform of the following using the integration property.

a.) ( )Solution: F(s) 1s(s ) G(s)

G(s) s 1F(s) G(s)s

F(s) 22

s(s2

42

)


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f(t) g()d

g() 1 sin

f(t) g()d f(t) 1 sind f(t) 1 cos f(t) 1

(cost1)

b.) ( )Solution:

F(s) 1

s(s )

F(s) 1s s(s )G(s) 1s(s )

F(s) G(s)s f(t) g()d

g() 1 (1cost)f(t) g()d f(t) 1 (1cost)d

f(t) 1 (1cost)


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f(t) 1 1 sin

f(t) 1

(t 0) 1 (sin0)

Example 2.8

Direction: Find the general solution of the differential equation.a.) y 2y 2y 0 for y (0) 1 and y(0) 3Solution: sY(s) sY(0) Y(0) 2sY(s) y(0) 2Y(s) 0

sY(s) s 3 2 s Y(s) 2 2 Y(s) 0

(s

2 s 2)Y(s) s 1

Y(s) s 1s 2 s 2Y(s) s 1s 2 s 1 1

Y(s) s 1 1 1(s 1) 1 Y(s) s 1

(s 1)

1 2

(s 1)

1

f(t) sint

y(t) e cost2e sint


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b.) y y t for y(0) y(0) 1Solution:

sY(s) sY(0) Y(0) Y(s) 1s

sY(s) s ( 1 ) 1 Y(s) 1s(s 1)Y(s) 1s s 1

Y(s) 1s(s 1) s 1s 1

By Partial Fraction Expansion,1s(s 1) As Bs Cs 1 Ds 11 A s(s 1) B(s 1) C(s)(s 1) D(s)( s 1 )

1 A(s s) B(s 1) C(s s) D(s s)s: 0 A C D

s

: 0 B C Ds: 0 As: 1 BA 0 B 1 C 1 2 D 1 2

1s(s 1) 1s 1 2s 1 1 2s 1 1s 1

y(t) e s i n h t t


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Drill Problem 2.2Direction: Use the differentiation property to find the Laplace transform of the following:

1.) tcos5tSolution:f(0) 0

f(t) t(5sin5t) cos5tf(t) 5tsin5tcos5tf(0) 1f(t) 5t(5cos5t) sin5t5sin5tf(t) 25tcos5t5sin5t5sin5t f(t) 25tcos5t10sin5t f(t) 25f(t) 10sin5t

f(t) 25F(s) 50s 25

DP: s

F(s) sF(s) f

(0)

sF(s) sF(s) f(0) 25F(s) 50s 25sF(s) 25F(s) 1 50s 25sF(s) 25 F(s) s 2 5 5 0s 25 (s 25) F(s) s 25s 25s 25

2.) cos tF(s) s 25(s 25)


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Solution:f(0) 1

f(t) 2cost(sint)f(t) 2costsintf(0) 0f(t) 2cost costsintsintf(t) 2cos tsin tf(t) 2f(t) sin tf(t) 2 F(s) 12 1s ss 4

f(t) 2 F(s) 2s(s 4)f(t) 2F(s) 4s(s 4)DP: sF(s) sF(s) f(0)

sF(s) sF(s) f(0) 2F(s) 4s(s 4)

(s s)F(s) 2F(s) 4s(s 4)(s 2)F(s) s 4s(s 4)(s 2)F(s) s(s 4) 4s(s 4) (s 2)F(s) s 4s 4s(s 4)(s 2)

F(s) (s 2)(s 2)s(s 4)(s 2)

F(s) (s 2)s(s 4)


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3.) sinh at

Solution: f(0) 0f(t) 2 sinhat a cosh atf(t) 2asinhatcoshatf(0) 0f(t) 2asinhat asinhatcoshat acoshat

f(t) 2asinh atcosh at

f(t) 2af(t) cosh atf(t) 2a F(s) 12 1s ss 4af(t) 2a F(s) 12 2s 4as(s 4a)f(t) 2aF(s) a 2s 4as(s 4a)

DP: sF(s) sF(s) f(0)sF(s) sF(s) f(0) 2aF(s) a 2s 4as(s 4a)(s 2a)F(s) 2a s 2as(s 4a)s 2a

4.) cosh tSolution:

f(0) 1

F(s) 2a

s(s

4a

)


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f(t) 2cosh 12 t 12 sinh 12 tf(t) cosh 12 tsinh 12 t

f(0) 0

f(t) 12 cosh 12 t 12 sinh 12 tf(t) 12 f(t) 12 sinh 12 tf(t) 12 F(s) 12 1s ss 1

f(t) 12 F(s) 12s(s 1)f(t) 12 F(s)

14s(s 1)DP:sF(s) sF(s) f(0)

sF(s) sF(s) f(0) 12 F(s) 14s(s 1)

sF(s) 12 F(s) s 14s(s 1) s 12 F(s) s(s 1)

14s(s 1) s 12 F(s) s s

14s(s 1) s 12 F(s) s 12s(s 1)s 12

F(s) s 12s(s 1)


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5.)

sin

t

Solution: f(0) 0f(t) 4 s in tcostf(0) 0

f

(t) 4sin

tcost

f(t) 4sin t (sint) cost (3sin t)(cost)f(t) 4sin t 3 s i n tcos tf(t) 4sin t12sin tcos tf(t) 4f(t) 12sin tcos tf(t) 4F(s) 12sin tcos tf(t) 4F(s) 3sin 2tf(t) 4F(s) 3

21s

ss

16

f(t) 4F(s) 32 s 1 6 ss(s 16) f(t) 4F(s) 24s(s 16)DP:sF(s) sF(s) f(0)sF(s) sF(s) f(0) 4F(s) 24s(s 16)

(s

4)F(s) 24

s(s 16)s 4

6.) F(s) 24s(s 4)(s 16)


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Solution:F(s) G(s)s

f(t) g()d F(s) 10s sF(s) 10s(s )F(s) 10s(s)(s )

G(s) 10s(s )

g(t) 10ef(t) g()d

f(t) 10ed

f(t) 10e

t0

g(t) 10e 10 f(t) g()d

f(t) 10 (e 1)d

f(t) 10 (e 1)dt

f(t) 10 e t f(t) 10e 10t 10


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f(t) 10e 10t10

7.) Solution:

F(s) G(s)s

f(t) g()d

F(s) 1s(s 1)G(s) 1s(s 1)g(t) s intf(t) g()d

f(t) sind f(t) cos t0f(t) c o s t c o s 0g(t) c o s t 1f(t) g()d

f(t) ( c o s 1)d f(t) s i n t0

f(t) 10(e t 1)

f(t) s i n t t


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8.) Solution:

F(s) G(s)s f(t) g()d F(s) 5s(s 5)G(s) 5s 5g(t) 55 5

s

5

g(t) 55 sinh 5tf(t) g()d f(t) 55 sinh 5d f(t) 5

5 sinh 5d

f(t) 55 15 cosh 5 f(t) 55 cosh 5 t0f(t) cosh 5 t0f(t) cosh 5tcosh0

9.)

Solution:F(s) G(s)s f(t) g()d

f(t) cosh 5 t 1


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F(s) 1s(s 4)F(s) 1s(s)(s 4)

G(s) 1s(s 4)

G(s) 12 (sinh2t)f(t) g()d f(t) 12 (sinh2)d

f(t) 12 12 cosh2

f(t) cosh24 t0f(t) cosh2t4 14f(t) g()d f(t) 14 (cosh21)d

f(t) 14 sinh22

f(t) 14 sinh2t2 t

10.)

Solution: F(s) G(s)s f(t) g()d F(s) 2s(s 3)

f(t) 18 sinh2t 14 t


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G(s) 2(3)(s 3)

G(s) 23 sin3t

f(t) g()d f(t) 23 sin3d f(t) 23 sin3d f(t) 23 cos33 f(t) 2cos3t9 2cos09

f(t) 29 (cos3t1)

11.)y 4y 0; y(0) 2.8

Solution: sY(s) y(0) 4 Y(s) 0sY(s) 2 . 8 4 Y(s) 0(s 4)Y(s) 2.8Y(s) 2.8s 4Y(s) 2 . 8 1s 4

12.)y y 17 sin 2t ; y(0) 1Solution:13.) y y 6 y 0 ; y(0) 6 , y(0) 13

f(t) 29 (1cos3t)

y(t) 2.8e


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Solution: sY(s) sy(0) y(0) sY(s) y(0) 6Y(s) 0

s

Y(s)

6 s 1 3 s Y(s

) 6 6 Y

(s

) 0

(s s 6)Y(s) 6 s 7Y(s) 6 s 7s s 6Y(s) 6 s 7(s 3)(s 2)Y(s) 6 s 7(s 3)(s 2) As 3 Bs 2 (s 3)(s 2)Y(s) 6 s 7 A(s 2) B(s 3)

if s 3

2 5 A(s 2)

2 5 5 AA 5if s 2 5 B(s 3) 5 B( 2 3)B 1Y(s) 5 1s 3 1s 2

14.)y 4y 4y 0; y(0) 2.1,y(0) 3.9Solution:

sy(s) sy(0) y(0) 4sy(s) y(0) 4y(s) 0s

y(s) 2.1s3.94sy(s) 4(2.1) 4y(s) 0

(s 4 s 4)y(s) 2.1s4.5Y(s) 2.1s4.5s 4 s 4Y(s) 2.1s4.5(s 2)(s 2)

y(t) 5e e


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Y(s) 2.1s4.5(s 2)(s 2) As 2 B(s 2) (s 2)2 . 1 s 4 . 5 A(s 2) Bs: 0 0s : 2 . 1 A

A 2 . 1s: 4 . 5 A(2) B4.52.1(2) B 4 . 5 4 . 2 BB 0 . 3Y(s) 2.1s 2 0.3(s 2)

Y(s) 2 . 1 1s 2 0 . 3 1(s 2)

Y(s) 2.1e 0.3te

15.)y ky 2ky 0; y(0) 2 , y(0) 2kSolution: sy(s) sy(0) y(0) ksy(s) y(0) 2ky(s) 0

s

y(s)

2 s 2 k k s y(s

) 2 k 2 k

y(s

) 0

(s k s 2 k)y(s) 2 s 4 k

Y(s) 2 s 4 ks k s 2 kY(s) 2 s 4 k(s 2 k)(s k)Y(s) 2(s 2 k)(s 2 k)(s k)Y(s) 2s k

Y(s) 2 1s k

Example 2.8

y(t) e(2.10.3t)

y(t) 2e


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Directions: Write the following function using unit step functions and find its transform.

a.) f(t)

2 0 1. t

1

.cost t

Solution:f(t) 2 0 11

2t 1 2

cost t 2

since f(t)u(t a) ef(ta)for 2 2s 2es t

for 2 2s 2e

s

for; 12 t 12 tu(t 1) tu t 2for; 12 t 12 e(t 1) e t 2


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for ; 12 t 12 et 2 t 1 e t t 4

for;12 t

1s

1s

1

2s e

1s

2s

8s e

for ; cos t cos t u t 2for; cos t e cos t 2

for ;c os t e costcos 2 sintsin 2

for;coste sint

for; cost e s 1Therefore,

Example 2.9Direction: Find the inverse Laplace Transform of

F(s) e es e(s2)

Solution:We will first solve on the first term, by expanding the term, we can get two terms,For es

f(t) 2s

2e

s 1

s 1

s 1

2s e 1

s

2s

8s e/ 1

s

1e/


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F(s) 1s f(t) sint t (t -1)f(t) 1 sin(t)f(t) 1 (sintcoscostsin)

f(t) 1 (sint)Then,

For e

s

F(s) 1s f(t) sint t (t -2)f(t) 1 sin(t2)

f(t) 1

(sintcos2costsin2)f(t) 1 (sint)Now, for the second term,

e(s2)

f

(t) te

t

(t -3)

f(t) (t3)e()

f(t)

0 0 1sint 1 20 1 3(t 3)e() t 3


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Drill Problems 2.3

Direction: Sketch or graph the given function (assumed to be zero outside the given interval).Represent it using unit step function. Find its transform.1.) t (0 1)Solution:

f(t) u(t a) e f(t a) t u(t 0) t u(t 1) e t e t 1

t u(t 0) t u(t 1) 1s e 1s 1s

2.) s i n 3 t ( 0 )

Solution: f(t) u(t a) e f(t a) sin3t u(t 0) sin 3t u(t ) e sin3t e sin3(t) sin 3t u(t 0) sin 3t u(t ) sin3t e sin3tcos3cos3tsin3

sin 3t u(t 0) sin 3t u(t ) sin3t e sin3t sin 3t u(t 0) sin 3t u(t ) 3

9 3

9

sin 3t u(t 0) sin3t u(t ) 3s 9 3es 9

t u(t 0) t u(t 1) 1 es es

sin 3t u(t 0) sin3t u(t ) 3s 9 ( 1 e


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3.) t(t 3)Solution:

t

u(t 3) e

(t 3)

t u(t 3) e t 6 t 9 t u(t 3) e 2!s 6s 9s

4.) 1 e ( 0 )Solution: (1 e) u(t 0) (1 e) u(t u) e(1 e) e1 e() (1 e) u(t 0) (1 e) u(t u) e(1 e) e(1 ee)

(1 e) u(t 0) (1 e) u(t u) (1 e) e(1 ee)

(1 e) u(t 0) (1 e) u(t u) 1s 1s 1 e

s e

e

s 1

5.) sin t (t )Solution: sint u t 6 e sin(t 6 )

sint u t 6 e sintcos6costsin6

t u(t 3) e 2s 6s 9s

(1 e) u(t 0) (1 e) u(t u) 1 es (ee) 1s 1


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sin t u t 6 e sint sin t u t 6 e sint

Direction : Find and sketch or graph f(t) is F(s) equals6.)

Solution: se s

ss ss cost t (t 1)

7.) e

Solution: F(s) 1s 1sF(s) t 1 . ( )F(s) t (t 1) 1 u(t 1)F(s) t (t 1 1) u(t 1)

F(s) t(t) u(t1)8.)

sin t u t 6 e s

f(t) 0 t 1c os ( t 1 ) t 1

f(t) t 0 10 t 1


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Solution:F(s) 1

s

2 s 2

By completing the square,F(s) 1(s 1) 1F(s) e sint . ( )

F(s) e() sin(t ) u(t)F(s) e()sintcoscostsin u(t)

F(s) e()sint u(t)

9.) ()

Solution:Expanding the term, we get F(s) 1s k es kF(s) 1s k ees kF(s) e ee . ( )

F(s) e ee()u(t1)

10.)2.5 ..

f(t) 0 t e() s i n t t

f(t) e 0 10 t 1


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Solution:F(s)2.5 e.

s e.

s

F(s)2.5 t . ( .) 1 . ( .)F(s)2.5 u(t 3 . 8) u(t 2.6)

Example 2.10

Direction: Determine the response of a system described by the differential equationy 3y 2 y r( t )For y (0) y(0) 0 and inputs r(t).

a.) r(t) u(t 1) u ( t 2 )may GRAPH DITO

Solution: y(0) y(0) 0r(t) u(t 1) u ( t 2 )sY(s) sy(0) y(0) 3sY(s) 3y(0) 2Y(s) 1s (e e)

f(t) 2.5 2 .63.80 elsewhere


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sY(s) 3sY(s) 2Y(s) 1s (e e)Y(s)s 3 s 2 1

s(e e)

Y(s) (e e)ss 3 s 2By partial fraction expansionFor ess 3 s 21s(s 1)( s 2 ) As 1 Bs 2 Cs

1 A(s 4 s 5) B(2 s 4)(s 1) C ( s 1 )

1 A s(s 2) Bs(s1)C(s 3 s 2 )1 A(s 2s) B(s 1s)C(s 3 s 2 )s: 0 A B C A 1

s: 0 2A B 3C B 1

2

s: 1 2C C 12

F(s) 1s 1 12s 2

12s f(t)e() 12 e() 12

For ess 3 s 21s(s 1)( s 2 ) As 1 Bs 2 Cs


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1 A(s 4 s 5) B(2 s 4)(s 1) C ( s 1 )1 A s(s 2) Bs(s1)C(s 3 s 2 )

1 A(s

2s)

B(s

1s)C(s

3 s 2 )

s: 0 A B C A 1s: 0 2A B 3C B 12s: 1 2C C 12

F(s) 1s 1 12s 2

12s

f(t)e() 12 e() 12

b.)r(t) ( t 1 )Drill Problem 2.4

Direction: Showing the details, find and graph the solution.1.) y y (t 2 ) y(0) 1, y(0) 0

Solution:

s

Y(s) sy(0) y

(0) Y(s) e

sY(s) 10s Y(s) eY(s)s 1 e 10sY(s) es 1 10ss 1

y(t)

0 0 112 e() 12 e() 1 2e() e() 12 e() 12 e() t 2


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For :

F(s) 1s 1f(t) sint t (t 2)f(t) sin(t 2 ) u(t2)Then,

Y(s) 10costsin(t 2 ) u(t2)Y(s) 10 cos t sint cos 2 cos tsin 2 u(t 2)

2.) y 2y 2 y e 5 (t 2) y(0) 0, y(0) 1Solution:sY(s) sy(0) y(0) 2sY(s) 2y(0) 2Y(s) 1s 1 5e

s

Y(s) 1 2sY(s) 2Y(s) 1

s 1 5e

Y(s)s 2 s 2 1s 1 5e 1Y(s) 1( s 1 )s 2 s 2 5es 2 s 2 1s 2 s 2

For () : 1( s 1 )s 2 s 2 A ( 2 s 2 ) Bs 2 s 2 Cs 11 A(2 s 2)(s 1) B(s 1) C(s 2 s 2 )

y(t) 10 cos t sint u(t 2)


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1 A ( 2 s 4 s 2 ) B(s 1) C(s 2 s 2 )

s

: 0 2A C

s: 0 4A B 2Cs: 1 2A B 2CA 12 , B 0 , C 1Then,

Y(s) ( 12 )(2s2)s

2 s 2 1

s 1 5e

s

2 s 2 1

s

2 s 2

Completing the square of the denominators:Y(s) (s1)( s 1 ) 1 1s 1 5e( s 1 ) 1 1( s 1 ) 1

3.)

y

y 10 t 100 (t 1) y(0) 10, y(0) 1

Solution:sY(s) sy(0) y(0) Y(s) 10e 100e

Y(s)s 1 10e 100e 1 0 s 1

Y(s) 10e

s 1 100e

s 1 10ss 1 1s 1

For , 10sinht . ( )

y(t) e s i n t e c o s t e 5e() sin(t 2) u (t 2)


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10es 1 10sinh(t 12) u t 12

For , 100sinht . ( )100es 1 100 sinh(t 1) u t 12

4.)

y 3y 2 y 1 0 s int 1 0 (t 1) y(0) 1, y(0) 1

Solution:sY(s) sy(0) y(0) 3sY(s) 3y(0) 2Y(s) 10s 1 10esY(s) s 1 3sY(s) 3 2 Y(s) 10s 1 10e

Y(s)s 3 s 2 10s 1 10e s 2

Y(s)s 3 s 2 10s 1 10e s 25.) y 4y 5 y 1 u(t 1 0)e e(t 1 0) y(0) 0, y(0) 1Solution:sY(s) sy(0) y(0) 4sY(s) 4y(0) 5Y(s) e e u(t 1 0)

sY(s) 1 4sY(s) 5Y(s) 1s 1 ee 1s 1 1Y(s)s 4 s 5 1s 1 1 ee 1s 1 1

y(t)10sinh(t 12) u t 12 100 sinh(t 1) ut 1210coshtsinht


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Y(s)s 4 s 5 1 s 1s 1 ee 1 s 1s 1 Y(s)s 4 s 5 ss 1 ee ss 1

Y(s) s(s10)s 4 s 5 (ee)(s)(s1)s 4 s 5By partial fraction expansion:

s As 1 B ( 2 s 4 ) Cs 4 s 5 s A(s 4 s 5) B(2 s 4)(s 1) C ( s 1 )s A(s 4 s 5) B(2s 2 s 4 ) C ( s 1 )s: 0 A 2Bs : 1 4 A 2 B C

s: 0 5A 4B CA 110 , B 120 , C 710

s 110s 1 120 (2s4)s 4 s 5 710s 4 s 5 6.) y 2y 3y 100 (t 2) 100 (t 3) y(0) 1, y(0) 0

Solution:

s

Y(s) sy(0) y

(0) 2sY(s) 2y(0) 3Y(s) 100e

100e

sY(s) s 2sY(s) 2 3 Y(s) 100e 100eY(s)s 2 s 3 100e 100e s 2Y(s) 100e 100es 2 s 3 s 2s 2 s 3

Completing the square of the denominator:

y(t) e e cost e sint e() e() cos (t 1 0) e() e u(t 1 0)


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Y(s) 100e 100e(s 3)( s 1 ) s 2(s 3)( s 1 )By partial fraction expansion:

s 2 A(s 3) B( s 1 )s 2 A ( s 1 ) B(s 3)1 A B , 2 a 3 B

A 14 , B 341 0 0 A ( s 1 ) B(s 3)0 A B , 1 0 0 a 3 B

A 25 , B 25Y(s) 14(s 3)

34(s 1) 100 25(s 3) 25(s 1) e 100 25(s 3) 25(s 1) e

y(t) 14 e 34 e 25e() 25e()u(t 2) 25e() 25e()u(t 3)

7.) y 2y 1 0 y 1 01 u(t 4) 10(t 5) y(0) 1, y(0) 1Solution:sY(s) sy(0) y(0) 2sY(s) 2y(0) 10Y(s) 1010(t 4) 10(t5)sY(s) s 1 2sY(s) 2 1 0 Y(s) 10s 10e 1s 4s

Y(s)s 2 s 1 0 10e 10s 10es 40es

y(t) 14 e 3e 2 5ee()u(t 2) 25 e e(t 3)


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Y(s) 10s(s 2 s 1 0 ) 10es(s 2 s 1 0 ) 40es(s 2 s 1 0 ) 10es 2 s 1 0For ()

10s(s 2 s 1 0 ) As Bs C s Ds 2 s 1 01 0 A s(s 2 s 1 0) B(s 2 s 1 0) Cs(s) Ds1 0 A(s 2s 10s) B(s 2 s 1 0) C(s) Ds

s: 0 A C s: 0 2A B D s: 0 10A 2B s: 1010BA 2 ; B 1 ; C 2 ; D 3

10s(s 2 s 1 0 ) 2s 1s 2ss 2 s 1 0 3s 2 s 1 0For ()

10es(s 2 s 1 0 ) 2s 1s 2ss 2 s 1 0 3s 2 s 1 0

For () 40es(s 2 s 1 0 ) As B s Cs 2 s 1 0 4 0 A(s 2 s 1 0) Bs Cs

s: 0 A B s: 0 2A C s : 4 0 1 0 AA 4 ; B 4 ; C 8

40es(s 2 s 1 0 ) 4s 1s 4ss 2 s 1 0 8s 2 s 1 0Y(s) 2s 1s 2ss 2 s 1 0 3s 2 s 1 0 2s 1s 2ss 2 s 1 0 3s 2 s 1 0 4s 1s 4ss 2 s 1 0 8s 2 s 1 0 10s 2 s 1 0


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Competing the square of the denominatorY(s) 2s 1s 2(s 1)(s 1) 9 3(s 1) 9 2s 1s 2(s 1)(s 1) 9 3(s 1) 9

4s 1s 4(s 1)(s 1) 9 2s 2 s 1 0 10(s 1) 9

8.) y 5y 6 y t u(t ) cost y(0) y(0) 0

Solution: y 5y 6 y t 12 u(t ) cost y(0) y(0) 0sY(s) sy(0) y(0) 5sY(s) 5y(0) 6Y(s) e e( ss 1)Y(s)s 5 s 6 e e( ss 1)

Y(s) e s 5 s 6 se(s 1)s 5 s 6

ecos(t)ecostcossintsinecostCompleting the square of the denominator

Y(s) e (s 2)(s 3) se(s 1)(s 2)(s 3) 1(s 2) 1(s 3) e

By partial fraction expansion:s A(2s) B(s 1) C(s 2) D(s 3)s A(2s)(s 5 s 6) B(s 5 s 6) C(s 1)(s 3) D(s 1)(s 2)

s A(2s 10s 12s) B(s 5 s 6) C(s 3s s 3) D(s 2s s 2)

y(t) 2 t 2e cos3te sin3t 2 (t 4) 2e() cos3(t 4)u(t 4) 4 4 e cos3(t 4) sin3(t 4) u(t 4) sin3(t 5) u(t5)


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A 120 , B 110 , C 25 , D 310

Y(s) 1

(s 2) 1

(s 3) e

120 (2s)(s 2)

110(s 2)

25(s 2)

310(s 3) e

y(t)(e e ) . ( ) ( cost sint e e ) . ( )y(t)e e u(t cos(t ) sin(t ) e () e () u(t)

9.) y 2y 5 y 2 5 t 1 0 0 (t ) y(0) 2, y(0) 5Solution:

sY(s) sy(0) y(0) 2sY(s) 2y(0) 5Y(s) 25s 100esY(s) 2s 5 2sY(s) 4 5 Y(s) 25

s 100e

Y(s)s 2 s 5 25s 100e 2 s 1Y(s) 25ss 2 s 5 100es 2 s 5 (2s1)s 2 s 5By partial fraction expansion:

2 s 1 A(2 s 2) Bs 2 s 5

2s 1 A(2 s 2) B

A 1 , B 325 As Bs C(2 s 2) Ds 2 s 5

25 A(s)(s 2 s 5) B(s 2 s 5) C(2 s 2)(s) D ( s)

y(t)e e u(t cos (t ) sin(t ) e () e () u(t)


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25 A(s 2s 5s) B(s 2 s 5) C(2s 2s) D(s)s: 0 A 2C

s

: 0 2A B 2C D

s: 0 5A2Bs: 25 5 BA 2 , B 5, C 1 , D 3

2(2 s 2)s 2 s 5 3s 2 s 5 2s 5s (2 s 2)s 2 s 5 3s 2 s 5

Y(s) 2s 5s 100es 2 s 5

10.) y 5y 25t 100(t ) y(0) 2, y(0) 5Solution: sY(s) sy(0) y(0) 5Y(s) 25s 100e

sY(s) 2s 5 5Y(s) 25s 100eY(s)s 5 25s 100e 2 s 5

Y(s) 25

ss 5 100es 5

(2s5)s 5

By partial fraction expansion: 2 s 5 A(2s) Bs 5 2s 5 A(2s) B

y(t) 2 5 t 5 0 e() sin2(t ) u(t )


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A 1 , B 5

25 As

Bs

C(2s) Ds 5

25 A(s)(s 5) B(s 5) C(2s)(s) D ( s)25 A(s 5s) B(s 5) C(2s) D(s)s: 0 A 2Cs: 0 B D

s: 0 A

s: 2 5 5 B

A 0 , B 5, C 0 , D 5(2s)s 5 5s 5 5 1s 1(s 5) 100es 5

Example 2.11

Direction: Find the response of a mass-spring system without damping to the following inputsa.) Hammerblow input (t)at t 0,zero initial conditions

Solution:

We are given the conditions,x(0) x(0) 02 0

y(t) cos 5 t 5 t 2 05sin 5(t ) u(t)


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Then,x(t) 2x(t) x(t) F (t)

x(

t)

x(t)

F

(t)

sX(s) sx(0) x(0) X(s) F (s)sX(s) X(s) F (s)X(s)s F (s)X(s) F (s)s

b.) no input, but with non-zero initial conditions (x(0) x( 0 ) 0 )Solution:We are given the condition,

x(0) x(0) 0Then,

x

(t) 2x(t)

x(t) F (t)

sX(s) sx(0) x(0) 2sX(s) 2x( 0 ) X(s) F (s)X(s)s 2 s sx(0) x(0) 2x(0) F (s)X(s) sx(0) x(0) 2x(0)s 2 s F (s)s 2 s

X(s) sx(0) x(0) 2x(0)(s ) ( )

X(s) sx(0)(s ) x(0) 2x(0)(s ) x(t) x(0)e cost x(0)2x(0) e sint

x(t) sint


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let C x(0) and C x(0)2x(0) Therefore

x(t) e (CcostC sint)c.) sinusoidal driving force Asint with ,x(0) x( 0 ) 0 .Solution:We are given the conditions,

x(0) x(0) 0

F(s) A s in tThen,x(t) 2x(t) x(t) F (t)sX(s) sx(0) x(0) 2sX(s) 2x( 0 ) X(s) F (s)X(s)s 2 s sx(0) x(0) 2x(0) F (s)

X(s) sx(0) x(0) 2x(0)s 2 s F (s)s 2 s

X(s) sx(0) x(0) 2x(0)(s ) ( ) As 2 s (s )

By partial fraction expansion:As 2 s (s ) M s Ns O s Ps 2 s A M s(s 2 s ) N(s 2 s )Os(s ) P( s )

s: 0 M O


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s: 0 2M N Ps: 0 M 2 N O s: A N P

M 0 N A

O 0 P A x(t) x(0)e cost x(0)sint A 1s A 1s 2 s

Therefore

d.) sinusoidal driving force Asint with ,x(0) x( 0 ) 0 .Solution:

We are given the conditions,x(0) x(0) 0

F(s) A s in tThen,x(t) 2x(t) x(t) F (t)sX(s) sx(0) x(0) 2sX(s) 2x( 0 ) X(s) F (s)X(s)s 2 s sx(0) x(0) 2x(0) F (s)

X(s) sx(0) x

(0) 2x

(0)s 2 s F (s)s 2 s X(s) sx(0) x(0) 2x(0)(s ) () A(s ) (s )

x(t) x(0)e costx(0)sint A sint A sint

8/3/2019 Solution Manual for Advance Ma

solution manual for advance math

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