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SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL 03

Ph: 9942099122

DEPARTMENT OF EEE| Transforms and Partial Differential Equations - MA 2211 - IV YEAR

1

Unit I -FOURIER SERIESTwo Mark Question & Answers

1. State Dirichlets conditions for a given function to expand in Fourier series.

i) )(xf is well defined, periodic and single valued.ii) )(xf has finite number of finite discontinuities and no infinite discontinuities.iii) )(xf has finite number of finite maxima and minima.

2. Write the formulas of Fourier constants for )(xf in ).2,( lcc .

Solution:

lc

c

n

lc

c

n

lc

c

dxl

xnxf

lb

dxl

xnxf

la

dxxfl

a

2

2

2

0

sin)(1

cos)(1

)(1

3. Find the constant 0a of the Fourier series for the function 20,)( xkxf .

Solution:

2

0

2

0

0

1)(

1kdxdxxfa

k

x

k 220

ka 20 .

4. Given 20,)( 2 xxxf which one of the following is correct.

(a) an even function (b) an odd function (c) neither even nor odd

Ans: (a)

5. Find nb in expanding )2()( xlxxf as Fourier series in the interval ).2,0( l

Solution:

l

n dxl

xnxlx

lb

2

0

sin)2(1

=

l

l

n

l

xn

l

n

l

xn

xl

l

nl

xn

xlxl

2

0

3

33

2

22

2

cos

)2(

sin

)22(

cos

)2(1

=0.

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6. In the expansion of xxf sinh)( ),( as a Fourier series find the coefficient of na .

Solution: Given xxf sinh)( . )(sinh)sinh()( xfxxxf

)(xf is an odd function. Hence .0na

7.Find an in expanding2)1()( xxf as Fourier series in the interval ).1,0(

Solution:

1

0

2 cos)1(2 xdxnxan

=

1

0

3322

2 sin2cos

)1)(1(2sin

)1(2

n

xn

n

xnx

n

xnx

=2

22

120)000(

n =

22

4

n.

8. Explain periodic function with example.

Solution:

A function )(xf is said to be periodic if there exists a number 0T such that

)()( xfTxf for all x in the domain of the definition of function. The least value of T

satisfying the above condition is called the fundamental period or simply period of the

function )(xf

E.g. xxf sin)(

)(sin)2sin()2( xfxxxf

Hence xsin is a periodic function with period 2 .

9. In the expansion of xxf sin)( ),( as a Fourier series find the coefficient of na .

Solution: Given xxf sin)(

)(sin)sin()( xfxxxf

)(xf is an odd function. Hence .0na

10. Find the Fourier constant na for xxcos in ( , ).

Solution: Let xxxf cos)(

)(cos)cos()( xfxxxxxf )(xf is an odd function.

Hence .0na

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11. Find the constant term 0a and the coefficient na of nxcos in the Fourier series

expansion of3)( xxxf in ).,(

Solution:3)( xxxf

)( xf )( 33 xxxx

)()( xfxf is an odd function.

000 naanda .

12. Find naa ,0 in expanding axxf sin)( as a Fourier series in ),( .

Solution: )(sin)sin()(sin)( xfaxaxxaxf

Hence )(xf is an odd function.

00a and .0

na 13. If xxf )( expanded as a Fourier series in x .Find 0a .

Solution:

dxxdxxfa 1

)(1

0 =

0

2xdx [ x is an even function]

=

0

2

2

2

x= 0a

14. Find the constants nb for xxf )( in x .

Solution: Given xxf )(

)()(

)(

xfxf

xxxf

)(xf is an even function.

Hence nb =0.

15. Find nb in the expansion of2x as a Fourier series in ( ), .

Solution:

Given 2)( xxf

Now )()()( 22 xfxxxf

)(xf is an even function.Hence nb =0.

16. Find the Fourier constant nb for xxsin in ( , ) .

Solution: Let xxxf sin)(

Therefore )(xf is even function of x in ( ),

The Fourier series of )(xf contains cosine terms only.

Which implies bn=0.

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17. Does xxf tan)( possess a Fourier expansion in (0, ).

Solution: xxf tan)( has an infinite discontinuity at x =2

.

Since the Dirichlets conditions on continuity is not satisfied, the

Function xxf tan)( has no Fourier expansion.

18. If xxxf 2)( is expressed as a Fourier series in the interval (-2,2), to which value

this series converges at x =2.

Solution:

The value of the Fourier series of )(xf at x =2is

4]2424[

2

1)]2()2([

2

1 ff

19. If f(x) is an odd function defined in (-l,l) , what are the values of a0and an ?

Solution:Since f(x) is an odd function of x in (-l,l) , its Fourier expansion contains

sine terms only. 5a =0 and an=0.

20. If f(x) is discontinuous at x =a, what does its Fourier series represent at the point?

[or] Define the value of the fourier series of )(xf at a point of discontinuity.

Solution:

The value of the Fourier series at x=a is

)(af = )](lim)(lim[2

1

00hafhaf

hh

)].()([2

1 afaf

21. If

50

cos)(

xxf

2,

0,

x

xand f(x+2 )() xf for all x, find the sum of the

Fourier series of f(x) at x=?

Solution:

2

49]50[cos

2

1)]()([

2

1)( fff .

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22. If the Fourier series for the function f(x)=

2,sin

0,0

xx

x is

f(x)=

........

7.5

6cos

5.3

4cos

3.1

2cos21

2

sin xxxx

deduce that

.4

2..........

7.5

1

5.3

1

3.1

1

Solution:

Put x =2

in the Fourier expansion of f(x),

.....

7.5

1

5.3

1

3.1

121

2

1

2

f

1

2

1......

7.5

1

5.3

1

3.1

12

Since 0

2

f

.4

2

22

2.......

7.5

1

5.3

1

3.1

1

23. Suppose the function xxcos has the series expansion

1

sinnxbn in ),,( find the

value of 1b .

Solution:

01 sincos

2xdxxxb

0

2sin1

xdxx

0

4

2sin

2

2cos1

xxx

2

1

2

1

.

24. Find the constant term in the Fourier expansion of xxf 2cos)( in ),( .

Solution:

x2cos is an even function of x in ),( .

x2cos

1

0 cos2 n

n nxaa

1

0 cos22

2cos1

n

n nxaax

The constant term is 0a =1

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25. Find the Fourier constants nb for xxsin in ),( .

Solution:

Given xxxf sin)(

Now )(sin)sin()( xfxxxxxf

)(xf is an even function. .0nb

26. Find ,n

a in expandingx

e

as Fourier series in ),( .

Solution:

eenn

en

e

nxnnxn

enxdxenxdxxfa

n

nn

xx

n

)1()1()1(

1)1(

11

sincos1

1cos

1cos)(

1

222

2

27. Define root mean square value of a function )(xf in .bxa

Solution:

R.M.S value

b

a

dxxfab

y .)(1 2

28. Find the root mean square value of the function xxf )( in the interval (0, l).

Solution:

R.M.S value33

11 3

0

2 ll

ldxx

ly

l

29. Find the R.M.S value of the function xxf )( in ),0( .

Solution:

R.M.S =

0

2dxx

=

0

3

3

x

=

3

3

=3

.

30. Find the root mean square value of the function2)( xxf in the interval (-1,1)

Solution:

51

52

21

521

21

1

1

5

4

xdxxl

y

l

l

.

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31. Find the R.M.S value of the function ,0)( 2 inxxf .

Sol:55

11 2

0

5

0

4

x

dxxy

32. State parsevals identity (Theorem) of Fourier series.

Solution:

If f(x) has a Fourier series of the form

)sincos(2

)(1

0 nxbnxaa

xf nn

in (0,2 ), then

).(

2

1

4

)]([

2

1 2

1

22

02

0

2

n

n

n baa

dxxf

33. State parsevals identity for the half range cosine expansion of f(x) in ( l,0 ).

Solution:

.2

)]([2

1

22

0

0

2

n

l

aa

dxxfl

34. If the Fourier series of the function f(x) =x+x2

in the interval ( ), is

),sin2

cos4

()1(3 1

2

2

nxn

nxn

n

then find the value of the infinite series

.......3

1

2

1

1

1222

Solution: Given f(x) = )sin2

cos4

()1(3 21

2

nxn

nxn

n

Put x = , f(

1

2

2 4

3)(

n

f )( = 222 ][2

1)]()([

2

1 ff

3

214

143/

2

12

12

22

nn

22 2

1

1

1

643

2.....

3

1 22

2

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35. State Parsevals identity for full range expansion of )(xf as Fourier series in ).2,0( l

Solution:

Let f(x) be a periodic function with period 2l defined in the interval ).2,0( l

then )(

2

1

4)(

2

1 2

1

2

2

0

22

0

n

n

n

l

baa

dxxfl

.

36. If the Fourier series corresponding to f(x) = xin the interval (0,2 ) is

)sincos(2 1

0 nxbnxaa

nn

without finding the value of a0,an,bn.Find the values of

)(2

2

1

22

0nn ba

a

Solution: By parsevals identity

3

8

3

1

2

1.2)(

2

22

0

32

0

22

1

22

0

xdxxba

ann .

37. If )sincos(2

cos1

03 ntbntaa

t nn

n

in 20 t , find the sum of the series

).(21

42

1

2

2

0n

n

n baa

Solution: .cos2

1)(

2

1

4

2

0

62

1

22

0

tdtba

an

n

n

2

0

6cos2

4

tdt

16

5

22

1

4

3.

6

52

.

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41. Find the value of anin the cosine series expansion of f(x)=5 in the interval(0,8).

Solution:Here l=8 and f(x) =5

an= dxxn

8

0 4cos5

4

2

= 00sin2sin5

4

4sin

4

5

8

0

nnn

xn

.

42. Find the sine series for the function f(x) =1, 0

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2cos1

21

2cos

2cos

22

0

n

n

n

n

l

nlxn

l

l

l

xnn

nxf

n

sin

2cos1

2)(

1

2sin2cos1sin

4sin

4 2

1

2

l

xnn

nn

44. Define Harmonic Analysis.

Ans: The process of finding the fourier series for a function given by numerical value isknown as Harmonic Analysis.

45. If 4,02)( inxxf then find the value of 2a in the fourier series expansion.Sol:

Given 4,02)( inxxf

l

n dxl

xnxf

la

2

0

cos)(1

4

0

22

2cos2

2

1dx

xxa

0

11cossin22

4

0

2

xxx.

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Unit II -Fourier Transforms

Two Mark Question & Answers

1) State Fourier Integral Theorem

Sol:

If )(xf is piecewise continuous, has piecewise continuous derivatives in

every finite interval in ),( and absolutely integrable in ),( ,then

dtdsetfxf txis )()(2

1)(

(or) equivalently

0

)}(cos{)(1

)( dtdstxstfxf

.

This is known as Fourier Integral of )(xf .

2) Define Fourier transform and its inverse transform. (or)Write the Fourier

transform pair.

Sol: The Fourier transform of a function )(xf is

.)(2

1)]([ dxexfxfF isx

The function

dsexfFxf isx)]([2

1)(

is called the inverse formula for the

Fourier transform of )]([ xfF .

3) Define Fourier sine transform and its inverse.

Sol: Fourier sine transform of )(xf is defined as

0

sin)(2

)]([ sxdxxfxfFs

.

Its inverse is defined by .sin)]([2

)(0

sxdsxfFxf s

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4) Find the Fourier Transform of )(xf 1 1x

0 1x

Sol:

We know that

.)(2

1)]([ dxexfxfF isx

1

12

1dxeisx

1

12

1

is

eisx

is

ee isis

22

2

si

issin2

2

1

2

2

s

ssin2

.

5) Define Fourier cosine transform and its inverse.

Sol: Fourier cosine transform of )(xf is defined as

0

cos)(2

)]([ sxdxxfxfFc

.

Its inverse is defined by .cos)]([2

)(0

sxdsxfFxf c

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8) Find the Fourier Cosine Transform of )(xf x 10 x

x2 21 x

0 2x

Sol: We know that

0

.cos)(2

)]([ sxdxxfxfFc

1

0

2

1

cos)2(cos2

sxdxxsxdxx

2

1

2

1

0

2

cossin)2(

cossin2

s

sx

s

sxx

s

sx

s

sxx

s

s

s

s

s

s

ss

s

s

s sin2coscos1cossin22222

12coscos2122

sss

.

9) Find the Fourier cosine transform of )(xf 0

,cosx

ax

ax

0

Sol:

aa

c dxxsxssxdxxsF

00

])1cos()1[cos(2

12coscos

2)(

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1)1sin(

1)1sin(

21

1)1sin(

1)1sin(

21

0 sas

sas

sxs

sxs

a

10) Find the Fourier cosine transform of )(xf 1 ax 0

0 ax

Sol:

aa

cssxsxdxsF

00

sin2cos2)(

ssasin2

11) Find the Fourier cosine Transform ofaxe ,a>0.

Sol: We know that

0

.cos)()]([ sxdxxfxfFc .

0

.cos2][ sxdxeeF axaxc

22

2

bs

a

.

12) Find the Fourier cosine transform of xxf )( .

Sol: We know that

0

.cos)(2)]([ sxdxxfxfFc

0

cos2 sxdxx

0

.2

dxxePR isx

0

2)()(.

2

is

e

is

exPR

isxisx

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.12

1.

2

2

2

s

sPR

13) Find the Fourier Sine transform of .)( xexf

Sol: We know that

0

sin)(2

)]([ sxdxxfxfFs

0

2

1

2sin

2

s

ssxdxe x

14) Find the Fourier cosine transform of xx ee 23 32 .

Sol:

0

2323 cos)32(2

]32[ sxdxeeeeF xxxxc

0 0

23 cos3cos22

sxdxesxdxe xx

4

1

9

126

4

23

9

32

22222 ssss

15) Find the Fourier cosine transform of xx ee 52 25 .

Sol:

0

5252cos)25(]25[ sxdxeeeeF

xxxx

c

0 0

52

cos2cos5

2

sxdxesxdxe

xx

25

1

4

1210

25

52

4

25

22222 ssss

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16) Find the Fourier cosine transform of axe ax cos

Sol:

0

coscos2

]cos[ sxdxaxeaxeF axaxc

0

])cos()[cos(2

2dxxasxas

e ax

00

)cos()cos(2

1dxasedxase axax

2222 )()(2

1

asa

a

asa

a

17) Find the Fourier cosine transform of axe ax sin

Sol:

0

cossin2

]sin[ sxdxaxeaxeF axaxc

0

])sin()[sin(2

2 dxxasxase ax

00

)sin()sin(2

1xdxasexdxase axax

2222 )()(2

1

asa

as

asa

as

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18) Find the Fourier Sine Transform of )(xf x 10 x

x2 21 x

0 2x :

Sol:We know that

0

sin)(2

)]([ sxdxxfxfFs

1

0

2

1

sin)2(sin2

sxdxxsxdxx

2

1

2

1

0

2

sinsin)2(

sincos2

s

sx

s

sxx

s

sx

s

sxx

222

sincos2sinsincos2

s

s

s

s

s

s

s

s

s

s

2

cos1sin2

2

s

ss

.

19) Find the Fourier sine transform ofx

1.

Sol:22

.2

.sin21

0

dxx

sx

xFS

0 2

sin dx

x

mx

20) Find the Fourier sine transform ofxx ee 25 53

Sol: [SF xx ee 25 53 ]

0

25 sin)53(2

sxdxee xx

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45

253

2

sin5sin32

22

0

2

0

5

s

s

s

s

sxdxesxdxe xx

21) Find the Fourier sine transform ofxx

ee 7447

Sol: [SF xx

ee 7447

]

0

74 sin)47(2

sxdxee xx

494

167

2

sin4sin72

22

0

7

0

4

s

s

s

s

sxdxesxdxe xx

22) Find the Fourier sine transform of )(xf

0

,sinx

ax

ax

0

Sol:

aa

S dxxsxssxdxxsF00

])1cos()1[cos(2

12sinsin

2)(

1

)1sin(

1

)1sin(

2

1

1

)1sin(

1

)1sin(

2

1

0 s

as

s

as

s

xs

s

xs a

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23) Solve the integral equation

0

cos)( exdxxf

Sol: Given

0

cos)( exdxxf .

0

2cos)(

2

exdxxf

exfFc

2

)]([

0

1cos

222)(

xdeeFxf c

21

12)(

xxf

.

24) State and prove change of scale property.

Sol:

.)(2

1)]([ dxeaxfaxfF

isx

a

dtetf

ta

si

)(2

1

( by putting axt )

a

sF

aaxfF 1)]([ if 0a

Similarly

a

dtetfaxfF

ta

si

)(2

1)]([

if 0a

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a

sF

aaxfF

1)]([ if 0a

Hence

a

sF

aaxfF

1)]([

25) If ][sFc

is the Fourier cosine transform of )(xf ,prove that the Fourier cosine

transform of )(axf is .1

a

sF

a c

Sol:

0

cos)(2

)]([ sxdxaxfaxfFc

Put .0:,, ta

dtdxtax

0

cos)(2

a

dtt

a

stf

0

cos)(21

dxxa

sxf

a

= .1

a

sF

a c

.

26) If ][sFS is the Fourier cosine transform of )(xf , prove that the Fourier cosine

transform of )(axf is .1

a

sF

a S

Sol:

0

sin)(2

)]([ sxdxaxfaxfFS

Put .0:,, ta

dtdxtax to

0

sin)(2

a

dtt

a

stf

0

sin)(21

tdta

stf

a

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= .1

a

sFa S

27) If )(sF is the Fourier transform of )(xf , find the Fourier transform of )( axf .

Sol:

dxaxfeaxfF isx )(2

1)]([

Put :,, tdtdxtax

dttfe tais )(.

2

1 )(

)()]([ sFeaxfF ias

28) If )(sF is the Fourier transform of )(xf ,find the Fourier transform of )(xfeiax .

Sol:

dxexfexfeF isxiaxiax )(

2

1)]([

)()(2

1 )( asFdxexf xasi

29) If )(sF is the Fourier transform of )(xf , find the Fourier transform of )( axf .

Sol:

dxaxfeaxfF isx )(

2

1)]([

Put :,, tdtdxtax

dttfe atis )(.2

1 )(

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)()]([ sFeaxfF ias

30) If )(sF is the Fourier transform of )(xf ,find the Fourier transform

of )(xfe iax

.

Sol:

dxexfexfeF isxiaxiax )(2

1)]([

)()(

2

1 )( asFdxexf xasi

31) If )(sF is the Fourier transform of )(xf , derive the formula for the Fourier

transform of axxf cos)( in terms of F.

Sol:

dxaxexfaxxfF isxcos)(2

1]cos)([

=

dxe

eexf

isxiaxiax

2)(

2

1

=

dxeexf xasixasi )()()(2

121

=

dxexfdxexf xasixasi )()( )(2

1)(

2

1

2

1

= )]()([2

1asFasF

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32) If )(sFs is the Fourier sine transform of )(xf ,show that

)]()([2

1]cos)([ asFasFaxxfF SSS

Sol:

]cos)([ axxfFS

0

sincos)(2

sxdxaxxf

=

0 ])sin())[sin((2

2

1

dxxasxasxf

0 0

)sin()(2

2

1)sin()(

2

2

1xdxasxfxdxasxf

.

)]()([2

1asFasF SS

33) If )(sFc is the Fourier cosine transform of )(xf ,show that

)]()([2

1]cos)([ asFasFaxxfF

ccc

Sol:

]cos)([ axxfFc

0

coscos)(2

sxdxaxxf

=

0

])cos())[cos((2

2

1dxxasxasxf

0 0

)cos()(2

)cos()(2

2

1xdxasxfxdxasxf

.

)]()([2

1asFasF cc

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34) If )(sFS is the Fourier sine transform of )(xf ,show that

)]()([2

1]sin)([ asFsaFaxxfF ccS

Sol:

]sin)([ axxfFS

0

sinsin)(2

sxdxaxxf

=

0

])cos())[cos((2

2

1dxxsaxsaxf

0 0

)cos()(2

)cos()(2

2

1xdxasxfxdxsaxf

.

)]()([2

1asFsaF cc

35) If )(sFs is the Fourier sine transform of )(xf ,show that

)]()([2

1]sin)([ asFasFaxxfF SSc

Sol: ]sin)([ axxfFc

0

cossin)(2

sxdxaxxf

=

0

])sin())[sin((2

2

1dxxsaxsaxf

0 0

)sin()(2

)sin()(2

2

1xdxsaxfxdxsaxf

.

)]()([2

1asFsaF SS

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36) Find )]([ xfxF n and

n

n

dxxfdF )( in terms of Fourier transform of )(xf .

Sol: )]([)()]([ sFds

dixfxF

n

n

nn

)()()( sFisxfdx

dF n

n

n

where )]([)( xfFsF .

37) Prove that )]([)]([ sFds

d

xxfF cS

Sol:We know that

0

.cos)(2

)]([ sxdxxfxfFc

Diff both sides w.r.t s

0 0

)sin)((2

)(cos)(2

)]([ dxsxxxfdxsxs

xfxfFds

dc

=

0

)]([sin)(2

xxfFsxdxxxf S

(i.e) )]([)]([ sFds

dxxfF cS

38) Show that ][)]([ sFds

dxxfF Sc

Sol:

0 sin)(

2

][ sxdxxfsFS

Diff both sides w.r.t s,

0

)]([)cos)((2

)( xxfFdxsxxxfsFds

dcS

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39) Find the Fourier cosine transform of axxe

Sol:

222

22

22)(

22][][

as

sa

as

s

ds

deF

ds

dxeF axS

ax

c

since

0

22

2sin

2][

as

ssxdxeeF

axax

S

40) Find the Fourier sine transform of

ax

xe

Sol:

22222 )(

222][][

as

as

as

s

ds

deF

ds

dxeF

ax

S

ax

S

since

0

22

2cos

2][

as

asxdxeeF axax

c

41) State the Parsevals identity for Fourier Transform.

Sol:

dxxfdssF 22

)()( where )]([)( xfFsF

42) State the Parsevals identity for Fourier sine and cosine transform.

Sol:

0 0

22

)()( dxxfdssFc where )]([)( xfFsF cc .

0 0

22 )()( dxxfdssFS where )]([)( xfFsF SS

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43) State the Convolution theorem for Fourier transforms

Sol: If )()]([ sFxfF and )()]([ sGxgF

Then )().()](*)([ sGsFxgxfF where

dttxgtfgf )()(2

1*

44) Find the Fourier transform ofxa

e

Sol: We know that

.)(2

1)]([ dxexfxfF

isx

[F xae

] =

dxsxisxedxee

xaisxxa)sin(cos

2

1

2

1

=

22

0

2cos

2

2

as

asxdxe ax

45) If

0

2 sin1

2sxds

s

se x

then show that

medxm

mxx

21sin

0

2

.

Sol: Given

0

2 sin1

2sxds

s

se x

that is

0

2 sin12

sxdss

se x

.

Put mx ,we get

0

2

0

2 sin

1sin

12mxdx

x

xsmds

s

se m

.

46) Prove that )()]([ sFxfF

Sol:

dxexfxfF

isx)(21)]([

Put yx where yx ,

dydx where yx ,

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=

dyeyfdyeyf isyisy )(

2

1)()(

2

1

= )()(2

1)(

2

1 )( sFdxexfdxexf xsiisx

47) Prove thatn

nnn

ds

FdixfxF )()]([ .

Sol:

dxexfsF isx)(2

1)(

.)()(2

1)(

2

1)(

dxixexfdxexfds

dsF

ds

d isxisx

.)()(2

1)()(

2

1)( 2

2

2

dxixexfdxixexfds

dsF

ds

d isxisx

In general )].([)()(2

1)()( xfxFidxexfxisF

ds

d nnisxnnn

n

Hence ).()()]([ sFds

dixfxF

n

nnn

48) Prove that ))(( xfFS )(ssF

Sol:

00

))((sin2

sin)(2

))(( xfsxdsxdxxfxfFS

0

0 cos)()(sin

2sxdxsxfxsxf

)(cos)(2

0

ssFsxdxxfs

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49) Give the function which is self reciprocal under fourier sine and cosine transform.

Ans:x

xf 1)(

50) State modulation theorem in Fourier transforms.

Ans:

If )]([)( xfFsF , then )]()([2

1]cos)([ asFasFaxxfF

51) Find the fourier sine transform ofx

1.

Sol:

0

sin)(2

)]([ sxdxxfxfFS

=

0

sin12

sxdxx

From complex analysis,2

sin10

sxdxx

(contour integration over semi circle)

22

2]

1[

xFS

52) State Parsevals identity on complex fourier transform.

Sol:

dxxfdssF

22

)()( where )]([)( xfFsF

53) Find the finite fourier sine transform of 402)( xxxf .

Sol:

0

sin)(2

)]([ sxdxxfxfFS

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=

4

0

sin22

dxsxx

=

4

0

2

sincos22

s

sx

s

sxx

)]([ xfFS =

2

4sin4cos422

s

s

s

s

54) If )()}({ sFxfF , prove that )()}({2

22

sFds

d

xfxF

Sol: Since,

dxexfsF isx)(2

1)(

.)()(2

1)(

2

1)(

dxixexfdxexfds

dsF

ds

d isxisx

dxixexf

ds

dsF

ds

d isx )()(

2

1)(

2

2

dxexfxdxixexf isxisx )(2

1)()(

2

1 22

)}({ 2 xfxF

Hence )()}({2

22 sF

ds

dxfxF .

55) Find the Fourier cosine transform of

x

xxxf

,0

0,)(

Sol:

00

cos2

cos)(2

)]([ sxdxxsxdxxfxfFc

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22

0

2 1cos2cossin2 sss

ssx

ssxx

UNIT-IV Applications of Partial Differential Equations

Two Marks Questions And Answers

1. Classify : 022

22

2

2

y

u

x

u

y

u

yx

u

x

u

Solution:

Here, 1,2,1 CBA

04442 ACB

The given P.D.E is parabolic.

2. Classify: 032 yxyyxyxx fffff

Solution:

Here, 1,1,2 CBA

0981)1)(2(4142 ACB

The given P.D.E is hyperbolic.

3. Classify: 0343 yxyyxyxx UUUUU

Solution:

Here, 4,3,1 CBA

07169)4)(1(4942 ACB

The given P.D.E is elliptic.

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4. Classify: 11,,0)1( 2

2

22

2

2

yxy

uyx

ux

Solution:

}011110{

,0)1(44

)1(4)1(4)1(404

)1(,0,

222

222

2222222

22

yyyandx

yxACB

yxyxyxACB

yCBxA


5 Classify: 0)4()25()1(2

2

2

2

2

2

2

2

t

ux

tx

ux

x

ux

Solution:

094201642025

)4)(1(4)25(4

4,25,1

4242

22222

222

xxxx

xxxACB

xCxBxA


6. Classify: 04

UUUx

yUUxyyxyxx .

Solution:

hyperbolicxyifxyACB

xy

x

yACB

xCyBA

222

222

,04

4)1(44

4,,1

parabolicxyif 2,0

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ellipticxyif 2,0

7. Classify: 0,0,0 yxyfxf yyxx

Solution:

0004404

,0,

2

andyxxyxyACB

yCBxA

Given P.D.E is elliptic.

8. Classify the following P.D.E 0322 22 UUUxxyUUyxyyxyxx

Solution:

0444

,2,

22222

22

xyyxACB

xCxyByA

The given P.D.E is parabolic.

9. Classify: 07222

yxyyxx UUUUy

Solution:

04404

1,0,

222

2

yyACB

cByA


10. (a)Classify: 0 yyxx

fxf

Solution:

ellipticifxxxACB

CBxA

0,04404

1,0,

2

parabolicifx 0,0

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.0,0 hyperbolicifx

(b). Classify: 026432

22

2

2

u

y

u

y

u

yx

u

x

u

11. State onedimensional wave equation.

The one-dimensional wave equation is

2

2

22

2

x

ya

t

y

12. What is the constant2a in the wave equation ?2 xxtt uau

Solution:

m

Ta 2 Where T is the tension caused by stretching the string before fixing it at the end

points, and m is the mass per unit length of the string.

13. State the possible solutions of one dimensional wave equation.

Solution:

The possible solutions of one dimensional wave equation are

DCxBAxtxy

patDpatCpxBpxAtxy

DeCeBeAetxy patpatpxpx

),(

sincossincos),(

),(

14. State the assumptions made in the derivation of one dimensional wave equation.

Solution:

(1)The mass of the string per unit length is constant.

(2)The string is perfectly elastic and does not offer any resistance to bending

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(3)The tension T caused by stretching the string before fixing it at the end points is constant at all

points of the deflected string and at all times.

(4) T is so large that other external forces such as weight of the string and friction may be

considered negligible.

(5)Deflection y and the slopex

y

at every point of the string are small, so that their higher powers

may be neglected.

15. Write the boundary conditions and initial conditions for solving the vibration of string equation, if

the string (of length l) is subjected initial displacement )(xf

Solution:

Boundary conditions:

0),(

0),0(

tly

ty

Initial conditions:

00

tty

)()0,( xfxy =initial development.


the string (of length l) is subjected to the initial velocity )(xg

Solution:

Boundary conditions: Initial conditions:

0),(

0),0(

tly

ty

)(

0)0,(

0

xgt

y

xy

t

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the string is subjected to initial displacement )(xf and initial velocity )(xg

Solution:


0),(

0),0(

tly

ty

)()0,(

)(0

xfxy

xgt

y

t

18. A tightly stretched string of length 2l is fastened at both ends. The mid point of the string isdisplaced by a distance b transversely and the string is released from rest in this position. Write the

boundary and initial conditions.

Solution:

l

bxy

l

x

b

y

l

x

b

y

xx

xx

yy

yy

lineOM

lAblMO

0

0

0

0

:

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

12

1

12

1

l

xlb

l

blbxbly

l

lxbby

l

lx

b

byll

lx

ba

by

lineMA

)2(

)(

2

:

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0),2(

0),0(

tly

ty 0

0

tt

y

lxl

bx0,

)0,(xy =

lxll

xlb2,

)2(

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19. A uniform string of length l is fastened at both ends. The string is at rest, with the point bx drawn aside through a small distance d and released at time t=0.Write the initial conditions,

Solution:

0

)(

)(

0

0

0

0

0

)0,(),0,0(

0

12

1

12

1

12

1

12

1

tt

y

ditionsInitialcon

blxldy

bl

dbdxdbdly

bl

bxddy

bl

bx

d

dy

xx

xx

yy

yy

lineBA

xb

dy

b

x

d

y

b

x

d

y

xx

xx

yy

yy

lineOB

lAo

)0,(xy =

lxbbl

xld

bxxbd

,)(

0,

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20. The points of trisection of a tightly stretched string of length 30 cm with fixed ends pulled aside

through a distance of 1cm on opposite sides of the position of equilibrium and the string is released

from rest. Write the boundary and initial conditions.

Solution:

)0,30(),1,20(),1.10(),0,0( CBAO

,0

:

10

30

10

201

10

201

1030

20

10

1

5

15

5

105

5

101

5

101

10

10

2

1

1020

10

11

1

0

tt

y

ditionsInitialcon

xxy

xy

xy

LineBC

xxy

xy

xy

xyxy

LineAB

),0( ty =

3020,10

30

2010,5

15

100,

10

xx

xx

xx

10101010

0

01

0

12

1

12

1

xy

xyxy

xx

xx

yy

yy

LineOA

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21. State one dimensional heat flow equation.

2

22

x

u

t

u

22. What does "" 2 denote in one dimensional heat flow equation.

c

k2 Where k- Thermal conductivity

- Density

c - Specific heat

23. State three possible solutions of one dimensional heat flow equation.

BAxtxu

epxBpxAtxu

eBeAetxu

tp

tppxpx

),(

sincos),(

),(22

22

24. State two laws used in the derivation of one dimensional heat flow equation.

Laws of thermodynamics:

1. Increase in heat in the element in t time= (specific heat) (mass of the element) (Increase in temperature)

2. The rate of flow of heat across any area A is proportional to A and the temperature

gradient normal to the area, (i.e.)x

u

. Where the constant of the

Proportionality is the thermal conductivity.

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25. A uniform rod of length 50 cm with insulated sides is initially at a uniform temperature c100 .Itsends are kept at c0 write the boundary and initially conditions.

Solution:


0),50(

0),0(

tu

tu

Initial conditions:

cxu 100)0,(

26. Write the boundary and initial conditions in a homogeneous bar of length which is insulated

laterally, if the ends are kept at zero temperature and if, initially, the temperature is k at the centre

of the bar and fully uniformly to zero at its ends.

Solution:

kxy

x

k

yx

k

y

xx

xx

yy

yyThelineOA

BkAo

2

20

2

0

0

0

)0,(:,2

:)0,0(

12

1

12

1

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2

2

2

2

2

2

012

1

12

1

xk

ky

x

k

ky

x

k

ky

xx

xx

yy

yy

ThelineAB

)(2

2

xky

kkxk

y


0),(

0),0(

tu

tu

Initial conditions:

20,

2

x

kx

)0,(xu

x

xk

2,

)(2

27. A rod of length 20 cm has its ends A and B kept at c30 and c90 respectively, until steady stateconditions prevail. Find the steady state solution.

Solution:


)2(90),20(

)1(30),0(

tu

tu

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The solution of steady state is

)3(),( baxtxu

Applying (1) in (3),

30

30)0(

b

bu


3

6020

903020

9020

9020)20(

a

a

a

ba

bau

Substitute a & b in (3),

303)( xxu

28. An insulated rod of length 60 cm has its ends at A and B maintained at c20 and c80 respectively. Find the steady state solution of the rod.

Solution:

One dimensional heat flow equation is

)1(2

22

x

u

t

u

Boundary condition:

)3(80),60(

)2(20),0(

tu

tu

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Applying (3) in (1), we get

2

1

20

10

1020

201020

2020.)20(

a

a

a

a

bau

Substitute a & b in (1)

102

x

u

30. What is the basic difference between the solutions of one dimensional wave equation and one

dimensional heat equation?

Solution:

The suitable solution )sincos)(sincos(),( patDpatCpxBpxAtxy of one

dimensional wave equation is periodic in nature. But the solution

tpepxBpxAtxu 22

)sincos(),( of one dimensional heat flow equation is not

periodic in nature.

31. Write the steady state two dimensional heat flow equation in Cartesian co-ordinates.

Solution:

The two dimensional heat flow equation is

)1(2

2

2

2

y

u

x

u

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32. Write the possible solutions of two dimensional heat flow equation.

The possible solutions are

))((),(

sincos),(

sincos),(

DCxBAxyxu

DeCepxBpxAyxu

pyDpyCBeAeyxu

pypy

pxpx

33. A rectangular plate with insulated surfaces is 8 cm wide and so long compared to its width that it

may be considered infinite in length. If the temperature along one short edge y=0 is kept at

,80,8

sin100)0,(

x

xxu

while the other two long edges 0x and 8x as well as short

edges are kept at ,0 c write the boundary conditions.

Solution:

The Boundary conditions are

8sin100)0,(

0),(0),8(

0),0(

xxu

yxultyu

yu

y

34. A rectangular plate with insulated surfaces is 10 cm wide and so long compared to its width that it

may be considered infinite in length. The temperature along short edge

50,20 xx

)0,(xu

105),10(20 xx

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Other edges are kept at .0 c Write the boundary conditions.

Solution:


0),(

0),10(

0),0(

yxult

yu

yu

y

50,20 xx

)0,(xu

105),10(20 xx

35. An infinitely long rectangular plate with insulated surface is 10 cm wide. The two long edges are

one short edge are kept at c0 while the other short edge x=0 is kept at

50;20 xy

u =

105);10(20 yy

Solution:


0),(

0)10,(

0)0,(

yxult

xu

xu

x

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50,20 yy

),0( yu u

105),10(20 yy

36. A square plate is bounded by the lines .20,20,0,0 yxyx its faces are insulated. The

temperature along the upper horizontal edge is given by

200),20()20,( xxxxu While the other three edges are kept at c0 .Write the Boundary

conditions.

Solution:

Boundary conditions are

)20()20,(

0)0,(

0),20(

0),0(

xxxu

xu

yu

yu

37. A square plane of side 30 cm is bounded by .30,30,0 yxx the edges 0,30 yx are kept

c50 and 20,0 yx are kept at c0 , Formulate the problem.

Solution:


0)30,(

50)0,(

0),30(

0),0(

xu

xu

yu

yu

38. A rectangular plate with insulated surfaces is a cm wide and so long compared to its width that it

may be considered infinite in length without introducing an appreciable error. If the two long edges

0x and ax and the short edge at infinity are kept at temperature c0 , while the other shortedge 0y is kept at temperature T (constant) Write the Boundary conditions.

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Solution:


Txu

xu

yau

yu

)0,(

0),(

0),(

0),0(

39. An infinitely long plane uniform plate by two parallel edges and end right angles to them. Thebreadth is , and this end is maintained at a temperature 0u

and other edges are kept at c0 . Write the Boundary conditions:

Solution:


0)0,(

0),(

0),(

0),0(

uxu

xu

yu

yu

40. An infinitely long plate in the form of an area is enclosed between the lines andyy 0 for

positive values of x .The temperature is zero along the edges 0x is kept at temperature ky , Find

the steady state temperature distribution in the plate.

Solution:


kyyu

yu

xu

xu

),0(

0),(

0),(

0)0,(

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41. A square plate of length 20 cm has its faces insulated and its edges along

20&0,20,0 yyxx .If the temperature along the edge 20x is given by

100;10

yyT

u=

2010);20(20

yyT

While the other three edges are kept at c0 , write the Boundary conditions.

Solution:

0),0(

0)20,(

0)0,(

yu

xu

xu

100;10

yyT

),20( yu =

2010);20(20

yyT

42. A rectangular plate is bounded by the lines byyaxx &0,,0 .Its surfaces are insulated.

The temperatures along 0,0 yx are kept at c0 and others at c100 . Write the boundary

conditions.

Solution:


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cbxu

xu

cyau

yu

100),(

0)0,(

100),(

0),0(

43. Classify: yyxx uu

Solution:

04)1)(1(404

1,0,1

2

ACB

CBA


44. (a)Write the general solution of ),( txy of vibrating motion of a string of length ""l with fixed

end points and zero initial velocity.

Solution:

lctn

lxnbtxy

nn

cossin),(1

44.(b) Write the general solution of ),( txy of vibrating motion of length ""l and fixed end points and

zero initial shape.

Solution:

l

ctn

l

xnbtxy

n

n

sinsin),(

1

l

ctn

l

xnbtxy

n

n

sinsin),(

1

45. Classify the P.D.E 0 yyxx yfxf

Solution:

yCBxA ,0,

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044042

xyxyACB If

0&0

0&0

yx

yx

0 If 0x or 0y

0 If

0&0

0&0

yx

or

yx


If

0&0)(0&0)3(0)(0)2(

0&0)(0&0)1(

yxoryx

yorx

yxoryx

46. The ends A and B of a rod of 40 cm long are kept at c0 and the initial temperature is 23 x ,formulate the model.

Solution:


23)0,(

:

0),40(

0),0(

xxu

ditionsInitialcon

tu

tu

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47. A rod of length 20 cm whose one end is kept at c30 and the other end at ,70 c until steady

state prevails, find the steady state temperature.

Solution:


)2(70),20(

)1(30),0(

tu

tu

The steady state solution is

)3()( baxxu


30

300.)0(

b

bau


2

4020

70302070)20()20(

a

a

abau

Substitute a and b in (3), we get

302)( xxu

48. A bar of length 50 cm has its ends kept at c20 and c100 until steady state prevails. Find thetemperature t any point

Solution:

)2(100),50(

)1(20),0(

ctu

ctu

The steady state solution is

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)3()( baxxu


20

20)0()0(

b

bau


6.1

6.15

8

8050

1002050

10050.)50(

a

a

a

a

bau

Substitute a and b in (3), we get

206.1)( xxu

49. Write the general solution of y(x,t) of vibrating motion of length l and fixed end points and zero

initial shape.

Solution:

l

ctn

l

xnbtxyn

sinsin),(

1

.

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50. Classify the P.D.E

.02)1(2

22 xyyxyxx uuyxyuux

Solution:

.04)1(444

1;2;

222222

22

xyxyxACB

yCxyBxA


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UNIT VZ transform

Two marks Q&A

1. Find the Z-transform of a nb )0,( ba .

Solution:

0

1}{n

nn zababZ

n

n z

ba

1 bz

az

z

b

a

1

2. Prove that )]0()([)]([ fzfzTtfZ .

Solution:

0

)()]([n

nzTnTfTtfZ

n

n

zTnf

])1([

0

)1(

0

])1([

n

n

zTnfz

k

n

zTkfz

)([

0

where k=n+1

)]0()([ fzfz .

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3. Find the Z-transform of n?

Solution:

00

][n

nn

n

z

nnznZ

...........321

32 zzz

2

2

11

1.....

321

1

zzzzz

22

2

)1()1(

z

z

zz

z.

4.state the initial and final value theorem of Z-transform.

Solution:

If )(][ zunZ ,then

(i)Initial value theorem: )(limlim0

zuuz

nn

(ii)Final value theorem: )()1(limlim1

zuzuz

nn

.

5. find

!n

aZ

n

in z-transform.

Solution:

!n

aZ

n

0

1

0 !

)(

! n

nn

n

n

n

azz

n

a

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.........)(

!11

211

azaz

z

a

az ee 1

6. FindiateZ using Z-transform.

Solution:

1*iatiat eZeZ

iattzez

Z )1( [by shifting property]

iatzezz

z

1[since z(1)=z/z-1]

1

iat

iat

ze

ze.

7. State and prove that initial value thorem in Z-transform.

Solution:

If )()( zFnfZ then )0(lim)0()(lim0ffzF

tz .

We know that

0 )()( n

n

znfnfZ

= f(0)+f(1) 1z +f(2) 2z +

......../)2(/)1()0([lim)(lim 2

zfzffzFzz

= f(0)

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)(lim)(lim0

nfzFnz

8. Find the Z-transform of (n+1)(n+2).

Solution:

]23[)]2)(1[( 2 nnZnnZ

= )1(2)(3)( 2 ZnZnZ

=)1(

2)1(

3)1(

)1(23

z

z

z

z

z

zz.

9. Find the Z-transform of n?

Solution:

00

)(n

nn

n

z

nnznZ

........

321

12zzz

111

11

1111

1

2

2

zorz

z

zzz

2

2

)1(

1

z

z

z

2)1(

z

z.

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10. Find the Z-transform of .cos n

Solution:

sincos

sincos

sincos])[(

iz

izx

iz

z

ez

zeZ

i

ni

22 sin)cos(

)sincos(

z

izz

Equating real part in both sides

22 sin)cos(

)cos()(cos

z

zznZ .

11. Prove that1

])1[(

z

zZ n .Also find the region of convergence.

Solution:

1

320

11......

1111)1(})1{(

zzzz

zZ n

n

nn

)1(11

1

z

z

z

Here the region of convergence is 11

z or 1z

1

])1[(

z

zZ n .

12. Define Z-transform.

Solution:

The z-transform of a sequence {x(n)}is defined byn

n

znxzX

0

)()( ,where z is a complex variable.

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13. What is the region of convergence?

Solution:

The region in which the seriesn

n

znx

0

)( is convergent is called the region of convergence.

14. Find )]1([ nuZ

Solution:

))(()]1([ 1 nuZznuZ

)1(

1

)1(

1

zz

z

z 1zif

15. Find )]2(3[ nZ n .

Solution:

3

)]2([)]2(3[ zz

n nZnZ

3

2

3

2 1][

zz

zz z

z

2

9

z

16. Find

12

21

zZ

Solution:

1

1

11

2

1

2

1

2

1

1

12

2

n

nn

n

z

Zz

Z

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2

1

111

zz

Z

2

111

z

zzZ

1

12

1

2

1

n

nn

n

17.Find

1

11

zZ

Solution:

1

1

1

1 11

z

z

zZ

zZ

1

11

z

zzZ

.1

1

1

nnz

zZ

,....3,2,1,)1( 1 nn

18. Find ]cos[ naZ n

Solution:

])1([]cos[ nnn aZnaZ

a

zz

nZ ])1[(

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az

z

z

z

a

zz )1(

19.Find ][ bateZ

Solution:

][][ atbbat eZeeZ

at

b

ez

ze

20.Find the Z-transform of the convolution of x(n)= )()()( nubnandynua nn

Solution:

Z[x(n)*y(n]=Z[x(n)]Z[y(n)]=bz

zaz

z

))((

2

bzazz

21.State second shifting theorem in Z-transform.

Solution:

If Z[f(n)]=F(z),then

Z[f(n+k)]= ])1(

....)2()1(

)0()([12

k

k

z

kf

z

f

z

ffzFz

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22.Find ]44[ 21

zz

z

Z

Solution:

]44

[2

1

zz

zZ = . ]

)2([

2

1

z

zZ .

)2

1(

1

2

1 )2()2()2

1(]

)2(

)2([

nn nnz

zZ

1)2()2()2

1(

nn nn

23.Find ])2)(1(

[1

zz

zZ

Solution:

])2)(1(

[1

zz

zZ = 12]

12[1

n

z

z

z

zZ

24. Find the z-transform of )()3

1( nun

Solution:

Z[ )()3

1( nun ]=Z )](3[ nu

n

zzz

z3)

1(

13

3

z

z

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3

1 z

z

25. . Find the z-transform ofnn2

Solution:

Z[ nn2 ]=

2

)( zz

nZ

2

2

2 )2(

2]

)1([

z

z

z

zz

z

26.State the damping rule in Z-transform

Solution:

If Z[x(n)]=X(z),then

(i) andazXnxaZ n )()]([

)()]([)(a

zXnxaZii n

27.If )(][ zYyZ n ,then write down the values of of ][][ knkn yandZyZ

Solution:

)(][ zYzyZ k

kn

and

])1(

......)2()1(

)0()([][12

k

k

knz

ky

z

y

z

yyzYzyZ

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28. If Z[x(n)]=X(z),then what are the values of Z[x(n+1)] and Z[x(n+2)]

Solution:

Z[x(n+1)]=z{X(z)-x(0)] and

Z[x(n+2)]= ])1(

)0()([2

z

xxzXz

29.State convolution theorem for Z-transform

Solution:

If

n

kknykxnynxzYzXZ

thenzYandZnxzXZ

0

*1

11

)()()()()]()([

)]([)()]([

30.Find ])4)(3(

[1

zz

zZ

Solution:

])4)(3(

[1

zz

zZ = ]

3)4([1

z

z

z

zZ

= ])3(

[])4(

[ 11

z

zZ

z

zZ

nn 34

31.Solve 012 01 givenyyy nn ,

Solution:

Taking Z-transform on both sides

])2)(1(

[)(1

2)Y(z)-(z

zz

zzY

z

z

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12)( z

z

z

z

zY

)]([1 zYZyn = 12 n

32. Solve 202 01 givenyyy nn

Solution:

Taking Z-transform

(z-2)z(y(n))=2z

Z[y(n)]=22z

z

Y(n)=11

22.2]2

2[

nn

z

zz

33.Findnn 3*3 using Z-transform

Solution:

Z[ nn 3*3 ]=

2

2

)3()3().3(

z

zZZ nn

nn 3*3 = ])3(

)33([]

)3([

2

1

2

21

z

zzZ

z

zZ

= ])3(

3[]

3[

2

11

z

zZ

z

zZ

= )1(333 nn nnn

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34.Find the Z-transform of nn *2

Solution:

Z[ nn *2 ]=Z[2

2

2 )1)(2()1(.

2)()2(

zz

z

z

z

z

znZn ]

35. Find the Z-transform of u(n-2)

Solution:

Z[u(n-2)]= )1(

1

1.

1

)]([ 22

zzz

z

znuZz

36. Find the Z-transform of )1(2 nun

Solution:

Z[ )1(2 nun ]=Z[2

)]1([ zz

nu

]

Z[

1

1)]([

1)1(

z

nuZ

z

nu

Z[2

2

)12

(

1)]1(2[

zz

nun

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