III Semester b

7/28/2019 III Semester b

1/22

www.iannauniversity.comR.Anirudhan 1

III SEMESTER B.E/B.TECH(COMMON TO ALL BRANCHES)

MA 2211 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS

QUESTION BANK

UNIT-I FOURIER SERIES

PART A

1. Define Dirichlets condition (or) State the condition for f(x) to have

Fourier Series expansion .

SOL : (i) f(x) is periodic, single valued and finite.(ii) f(x) has a finite number of discontinuities in any one period

(iii) f(x) has a finite number of maxima and minima.

(iv) f(x) and f(x) are piecewise continuous.

2. State whether y = tan x can be expanded as a Fourier Series. If so how?

If not why?

SOL : tan x cannot be expanded as a Fourier series. Since tan x notsatisfied Dirichlets Conditions. (tan x has infinite number of infinite

discontinuities)

3. Find the sum of the Fourier Series for f(x) = x 0


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= - xx3

= - ( x+x3)

= - f(x)f(x) is an odd function.

Hence a0 = 0 an = 0

5. If f(x) = x2 + x is expressed as a Fourier series in the interval ( -2, 2) towhich value this series converges at x = 2.

SOL : x = 2 is a point of discontinuity in the extremumf(x) = [f(-2) + f(2)] / 2

= {[(-2)2 + (-2)] + [22 + 2]} / 2

= {[4-2] + [4 +2] } / 2

= 8 /2= 4

f(x) = 4

at x = 2

6. State Parsevals Identity for full range expansion of f(x) as Fourier Series

in (0,2l)

SOL : Let f(x) be a periodic function with period 2l defined in theinterval ( 0, 2l ) then

)(2

1

4)]([

2

1 21

2

2

0

2

0

2

nnn

l

baa

dxxfl

7. If the Fourier Series corresponding to f(x) =x in the interval (0, 2 ) is

)sincos(2 10 nxbnxa

ann n

without finding the values ofa0, an , bn

find the value of )(

2

2

1

2

2

0

nnn ba

a

SOL : By Parsevals Theorem

dxxfbaa

nnn

2

0

22

1

22

0 )}([1

)(2


3/22


= dxx2

0

21

=

2

0

3

3

1 x

= 23

8

8. Define Root Mean Square value (or) RMS value

SOL : The root mean square value of f(x) over the interval (a,b) isdedfined as

R.M.S. =ab

dxxf

b

a

2)]([

9. Find the constant term in the Fourier series corresponding to f (x) = cos2x

expressed in the interval (- ,).

SOL: Given f(x) = cos2x =2

2cos1 x

W.K.T f(x) =2

0a + nxbnxan

n

n

n sincos11

To find a0 = xdx

2cos1

= dxx

0

2

2cos12

=

02

2sin1

xx

=

1[( + 0) (0+0)]

= 1.

10. If f(x) = x2+x is expressed as a Fourier series in the interval (-2,2) to

which value this series converges at x = 2.SOL: x = 2 is a point of discontinuity in the extremum.


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[f(x)]x = 2 =2

)2()2( ff

=2

]22[)]2()2[( 22

= 2

]24[]24[

=2

8= 4.

11. Expand f(x) = x in 0 < x < 1 as a Fourier series.

SOL: Let f(x) =

l

xnb

l

xna

a

n

n

n

n

sincos

2 11

0

a0 = l

dxxfl

2

0

)(1

an = dxl

xn

xfl

l

2

0cos)(

1

bn = dxl

xnxf

l

l

sin)(

12

0

Here 2l = 1, l =2

1.

a0 =

1

0

2

1

1xdx = 2

1

0

2

2

x= 2

0

2

1= 1

an = dxxn

x

1

0

2

1cos

2

1

1

= 2 1

0

2cos xdxnx

= 2 1

0

224

2cos1

2

2sin

n

xn

n

xnx

= 2

2222 4

10410

nn

= 0

bn = dxxn

x

1

0

2

1sin

2

1

1


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bn = 21

0

224

2sin

2

2cos

n

xn

n

xnx

bn = -n

1

f(x) =2

1+ 0 +

1

1

n nsin

2

1

xn

f(x) = xnnn

2sin11

2

1

1

.

12. Expand f(x) = 1 in a sine series in 0 < x < .

SOL : The sine series of f(x) in (0, ) is given by

f (x) = nxbn

n sin1

where bn =

0

sin2

nxdx

= -

0cos2

nxn

= 0 if n is even

= n

4

if n is odd

f(x) = nxnoddn

sin4

=

4

1 12

12sin

n n

xn.

PART-B

1. a) Expand f (x) =eax in aFourier series in (0, 2 ).b) Find the sine series for f (x) =x in 0


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5. Find the half range sine series for f(x) =xcosx in (0, ).

6. Find the sine series of f(x) =x in 0


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0

cos)(2

)()]([ sxdxxfsFxfF cc

The inversion formula is

0

cos)(2

)( sxdssFxf c

5. State the Convolution theorem for Fourier Transform.

SOL : If F(s) and G(s) are the Fourier transform of f(x) and g(x)respectively. Then the Fourier transform of the convolution of f(x) and g(x)

is the product of their Fourier transform.)()()](*)([ SGSFxgxfF

6. Define Self Reciprocal

SOL : If a transformation of a function f(x) is equal to f(s) then the functionf(x) is called self reciprocal.7. Find the Fourier Sine transform of 1/x

SOL : We know that

0

sin)(2

)()]([ sxdxxfsFxfF sS

=

0

sin12

]1[ sxdx

xxFS

Let sx = x 0 0 s dx = d x

0

sin2

s

ds

0

sin2

d

=2

2

=

2

.

0

sin

d =

2

8. Find Fourier Cosine transform of e x.

SOL : We know that


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0

cos)(2

)]([ sxdxxfxfFc

0

cos2

][ sxdxeeF xxc

=

21

12

s

22

0

cosba

abxdxe ax

9.Find Fourier Sine transform of e -3x.

SOL: We know that

0

sin)(2

)]([ sxdxxfxfFS

0

33 sin2][ sxdxeeF xxS

=9

22 s

s.

22

0

sinba

bbxdxe ax

10. Find Fourier Cosine transform of xeax.

SOL : we know that

)]([)]([ xfFds

dxfF sC

][][ axsax

C eFds

dxeF

=

0

sin2

sxdxeds

d ax

=

222

as

s

ds

d

= 222

222

as

sa

11. Find Fourier Sine transform of xeax.


9/22


SOL : we know that

)]([)]([ xfFds

dxfF cs

][][ axcaxs eFdsdxeF

=-

0

cos2

sxdxeds

d ax

=-

222

as

a

ds

d

= 22222

as

as

12. Show that f(x) = 1, 0 < x < cannotbe represted by a Fourier integral.

SOL : dxxf

0

)( = 1

0

dx = 0x = and this value tends to as x .

i.e., dxxf

0

)( is not convergent.

Hence f(x) = 1 cannot be represented by a Fourier integral.

PARTB

1. Using Fourier integral of f(x) = x x for x < 10 x for x > 1

evaluate

SinCos

0

2. Find the Fourier transform of f(x) where1 - x for x < 1

f(x)

0 for x > 1


10/22


Hence deduce that 2

Sin t dt = / 2

t

3. Find the Fourier Sine transform of Sinx , 0 < x < f(x) = 0 < x 0. Deduce that

1 dx = ( / 4a

3) if a > 0

0 (x2

+ a2)

2

7. Find the Fourier Cosine transform of x e x 2 / 28. Derive the parsevals identity for Fourier transforms.

UNIT III BOUNDARY VALUE PROBLEMSUNITIII

APPLICATION OF PARTIAL DIFFERENTIAL EQUATIONS

2 MARKS QUESTIONS


11/22


1. Classify the partial differential equation xyy

u

x

u

yx

u

2

Solution: B = 1, A = 0, C = 0 0142 ACB Hyperbolic type.

2. In the wave equation for?stantcdoesWhat, 22

22

2

2

x

yc

t

y

lengthunitpermass

Tension

m

Tc

2

3. What are the possible solutions of the one dimensional wave equation?1. ))((),( 4321 cxccxctxy

2. )sincos)(sincos(),( 8765 pctcpctcpxcpxctxy

3. ))((),( 1211109pctpctpxpx ecececectxy

4. In a heat equation2

22

x

u

t

u

What does 2 stand for?

cp

k2 where k: Thermal conductivity

c: Specific heat and

p density 2 is called the diffusivity of the material.

5. State any two laws which are assumed to derive one dimensional heat equation.i. The amount of heat required to produce a given temperature change ina body is proportional to the mass of the body and to the temperature

change.

ii. The rate at which heat flows across any area is proportional to the areaand the temperature gradient normal to the area.

6. State Fourier law of heat conduction.The rate at which heat flows across an area A at a distance x from one end

of a bar is given by Q = KA

x

u, Where K : Thermal conductivity,

x

u: Temperature gradient at x.

7. Write down the three possible solutions of ODHE and state the suitable solution.1. 321 )(),( ccxctxu


12/22


2. ippxpx ececectxu22

654 )(),(

3. ipecpxcpxctxu22

987 )sincos(),(

Suitable solution: ipepxBpxAtxu22

)sincos(),(

8. An insulated rod of length 60 cm has its ends A and B maintained at 20 C and 80 Crespectively. Find the steady state solution of the rod.

Steady state solution xl

abaxu

)(

A = 20 C , B = 80 C l=60

Therefore u(x) = 20 + x for 0 < x < l

9. Write down the three possible solution of Laplace equation 0 yyxx uu 1. )sincos)((),( 4321 pycpycecectxu

pxpx

2. ))(sincos(),( 8765pypy ececpxcpxctxu

3. ))((),( 1211109 cyccxctxu

10.Write the boundary condition for the problem: A rectangular plate is bounded by thelines x = 0 ,y=0,x=a and y=b. Its surfaces are insulated. The temperature along x=0and y=0 are kept at 0 C and the other at 100 C.

The boundary conditions are

i) byou y0for0),(

ii) axu x0for0)0,(

iii) byau y0for100),(

iv) abxu x0for100),(

PART-B

1. A rectangular plate with insulated surfaces is 10 cm wide and so longcompared to its width that it may be considered infinite in length with out

introducing an appreciable error. If the temperature along one short edge y =0 is T(x,0) = 4(10xx2 ) for 0


13/22


3. A string is tightly stretched and its ends are fastened at two points x = o andx = l. The mid point of the string is displaced transversely through a smalldistance b and the string is released from rest in that position. Find an

expression for the transverse displacement of the string at any time during

the subsequent motion.

4. A rod of length l has its ends A and B are kept at 0c and 100c until steadystate condition prevail. If the temperature at B is suddenly reduced to 0c andkept so while that of A is maintained, find the temperature u(x,t) at adistance x from A and at time t.

5. A bar 10 cm long with insulated sides, has its ends A and B kept at 20 c and30c respectively until steady state condition prevail. T he temperature at Ais then suddenly raised to 50 c and at the same instant that at B is lowered to

10 c. Find the subsequent temperature at any point of the bar at any time.

UnitIV

PARTIAL DIFFERENTIAL EQUATIONS

PARTA

1. Form the p.d.e by eliminating the arbitrary constants a and b frombyaxz .

Soln: Given byaxz ..(1)

Diff. partially w.r.t. x we get ax

z

i.e., ap

Diff. partially w.r.t. y we get by

z

i.e., bq

Substituting in (1) we get qypxz .

2.

Form the p.d.e by eliminating the arbitrary constants a and b from))(( 2222 byaxz .

Soln: Given ))((2222 byaxz (1)

Diff. partially w.r.t. x we get ))(2( 22 byxx

zp

)(

2

22 byx

p ..(2)

Diff. partially w.r.t. y we get ))(2( 22 axyy

zq

)(

2

22 axy

q ..(3)


14/22


Substituting (2) and (3) in (1) we get the required p.d.e.

xy

pq

x

p

y

qz

422 1.e., pqxyz4 .

3. Form the p.d.e by eliminating the arbitrary funtion from )( 22 yxfz Soln: Given )(

22 yxfz

)2)((' 22 xyxfx

zp

, )2)((' 22 yyxf

y

zq

y

x

q

p

2

2 0 qxpy .

4. Form the p.d.e by eliminating the arbitrary funtion from0),( zyxyx

Soln. w.k.t if 0),( vu then )(vu

)( zyxyx .(1)Diff. partially w.r.t. x we get

)01)(('1x

zzyx

)('

1

1zyx

p

(2)

Diff. partially w.r.t. y we get

)10)(('10y

zzyx

)('

1

1zyx

q

(3)

From (2) and (3), we get 02 qp .

5. Solve 1 qp Soln. Given 1 qp

This is of the form 0),( qpF . Hence the complete integral is cbyaxz

Where 1 ba 2)1( ab . Therefore the complete solution is

cyaaxz 2)1( (1)

Diff. partially w.r.t. c we get 0=1. There is no singular integral. Taking )(afc

where f is arbitrary.

)()1( 2 afyaaxz (2)

Diff. P.w.r.t. a we get )('2

1

)1(20 afyaax

.(3)

Eliminating a between (2) and (3) we get the general solution.

6. Find the complete integral of 22 qpqypxz Soln. Given

22 qpqypxz (1)

This equation of the form ),( qpfqypxz [Clairauts type]


15/22


Therefore the complete integral is22 babyaxz .

7. Find the complete integral of pxq 2 .Soln. Given pxq 2

This equation of the form 0),,( qpxf

Let aq . Thenx

ap

2 .

But advdxx

adz

2

Integrating on both sides we get, bayxa

z log2

8. Solve yxqp Soln: Given yxqp

)( qyxp

This equation is of the form ),(),( qypxf

aqyxp )( say

xap and ayq But qdypdxdz dyaydxxadz )()( Integrating on both sides we get,

bayyx

axz 22

22

.(1)

Diff.(1) P.w.r.t. b we get )(ab .(2)

)(22

22

aayyx

axz (3)

Diff(3) P.w.r.t. a we get

)('0 ayx .(4)

Eliminating (3) and (4) we the general solution.

9. Solve 0]'''[ 3223 zDDDDDD Soln. The auxiliary equation is 0123 mmm

0)1)(1( 2 mm 1m , im General Solution is

)()()( 321 ixyixyxyz

10.Find the particular integral of )2cos(]'2'3[ 22 yxzDDDD Soln. Given )2cos(]'2'3[

22 yxzDDDD

)2cos('2'3

1..

22yx

DDDDIP


16/22


)2cos()4(2)2(31

1...... yx

)2cos(

861

1..... yx

)2cos(3

1..... yx

.

1. Solve p(1-q2)=q(1-z)2. Solve (D2-DD1-2D2) z=2x+3y+e3x+4y3. Solve z2=p2+q2+14. Find the general integral of x(y2+z)p+y(x2+z)q=z(x2-y2)5. Solve p=2qx6. Solve z2=xypq7. Find the PDE by eliminating f from

(a) f(xy+z2,x+y+z)=0(b) z=xy+f(x2+y2+z2)

(c)z= x2f(y)+y2g(x)

UNIT-V

ZTRANSFORMS

UNIT V

ZTRANSFORM AND DIFFERNCE EQUATIONS

PART A

1. Prove that Z[an] =az

z

if az .

Solution:


17/22


az

z

z

az

z

a

z

a

z

a

z

a

zaaZ

n

n

n

n

nn

1

1

2

0

0

1

.....1

][

2. Prove that Z(n)= 21zz

Solution:

2

2

2

2

32

0

0

1

11

11

1

.....1

31

211

.....321

0

][

z

z

z

z

z

zz

zzz

zzz

z

n

znnZ

nn

n

n

3. Find Z[(-1)n]Solution:

We know that Z[an] =azz

Z[(-1)n] =

1)1(

z

z

z

z

4. Find Z

)1(

1

nn


18/22


Solution:

1

log)1(

1log

1log

1

11

)1(

1

1

11

)1(

1

11

10

)()1(1

1)1(

1

z

zz

z

zz

z

z

nZnZ

nnZ

nnnn

Bgetwenput

Agetwenput

nBnA

n

B

n

A

nn

5. Find Z[ann]Solution:

We know that Z[anf(n)]=F[Z/a]

2

2

/

2

/

)(

)1(

][][

az

az

a

az

a

z

z

z

nZnaZ

azz

azz

n

6. Find Z[etsin2t]Solution:

12cos2

2sin

12cos2

2sin

]2[sin]2sin[

)]([)]([

22

2

Tzeez

Tze

Tzz

Tz

tZteZ

tfZtfeZ

TT

T

zez

zez

tzez

at

T

T

aT


19/22


7. Prove that Z[f(n+1)]=zF(z)-zf(0)Solution:

)0()(

)0()(

1)(

)1(

)1()]1([

0

1

0

)1(

0

zfzzF

fzmfz

nmwherezmfz

znfz

znfnfZ

m

m

m

m

n

n

n

n

8. Find )]([ knZ Solution:

k

n

n

zknZ

knfor

knforkn

zknknZ

1)]([

0

1)(

)()]([0

9. Find )]2(2[ nZ n Solution:

zz

z

nZnZ

zz

zzn

4

2

1

1

)]2([)]2(2[

2

2/

2

2/

10.If Z(f(n))=F(z), then f(0)= )(lim zFz

Solution:


20/22


)0()(lim

.....)2()1()0(lim))((lim

.....)2()1(

)0(

)()]([

2

2

0

fzF

zf

zffnfZ

z

f

z

ff

znfnfZ

z

zz

n

n

11.Define convolution of sequences.Solution:

The convolution of two sequences {x(n)} and {y(n)} is defined as

(i) {x(n) * y(n)} =

K

KngKf )()( if the sequences are non-causal.

(ii) {x(n) * y(n)} =

n

K KngKf0 )()( if the sequences are causal.

The convolution of two function f(t) and g(t) is defined as

(i) f(t) * g(t)= TKngKTfn

K

0

)()( , where T is the sampling period.

12.Find

)3()2( 2

31

zz

zZ by convolution theorem.

Solution:

n

m

mmn

n

m

mnm

nn

m

m

n

z

zZ

z

zZ

z

z

z

zZ

zz

zZ

0

0

1

2

21

2

31

2

31

32)1(3

32)1(

32)1(

)3()2(

3)2()3()2(

13.Find the Z transform of nanu(n)Solution:


21/22


21

1

211

111

11

1

1

1

1

)1(

)1(

)(1)1(

1

)(

az

az

azaz

aazz

azdz

dz

az

z

dz

dznuaZ n

14.Define Unit step sequence.Solution:

The unit step sequence u(n) is defined as

00

01)(

nfor

nfornu

15.Find Z[cos (t+T)]Solution:

1cos2

)1cos(

1cos2

cos

1cos2

cos2cos

1cos2)cos(

1cos2

)cos(

0cos][cos))(cos(

)0()]([

)0()()((

2

2

2

2

2323

2

2

2

Tzz

Tzz

Tzz

zTz

Tzz

zTzzTzz

zTzzTzz

zTzz

Tzzz

ztzZTtZ

zftfZz

zfzzFTtfZ

PARTB

1. Find Z -1 4Z22ZZ

35Z

2+8Z4


22/22


2. Find Z -1 Z by using residue theorem.( z1) ( z2)

3. Use residue theorem find Z-1 Z2Z

2+ 4

4. Solve by Z transform yn + 2 + yn = 3 , given that y0 = y1 = 05. Solve the difference equation x (n+2)6x(n+1) +8x(n) = 4n ,

x(0) = 0 , x(1) = 1 by Z transform.

6. Solve U(x+2)4U(x) = 9x2 , given that U(0) = 0 , U(1) = 0 by Ztransforms.

7. Solve yn+25yn+1 + 6yn = 6 n , y0 = 1 , y1 = 08. Solve by Z transforms yn+2 + yn = n 2n

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III Semester b