St. Joseph’s College of Engineering & Page No. 1 ISO 9001:2008 St. Joseph’s Institute of Technology
TRANSFORMS & PDE (MA6351)
Question Bank with Answers
UNIT I PARTIAL DIFFERENTIAL EQUATIONS PART-A
1. Obtain partial differential equation by eliminating arbitrary constant ‘a’ and ‘b’ from (x–a) 2 +(y–b) 2= z A.U May/June2013
Solution: Given (x–a)2 +(y–b)2=z -----(1) Diff Partially w.r.t x and y 2(x–a) = p => x-a = p/2 2(y–b)=q => y–b= q/2 Substituting in (1), we get 2 2 4p q z+ =
2. Form the partial differential equation by eliminati ng the arbitrary constants ‘a’ & ‘b’ from z= (x2+a)(y2+b) A.U April/May 2011
Solution: Given z= (x2+a)(y2+b) -----(1) Diff partially w.r.t x and y
ax
y2
q,by
x2
p
),ax(y2qyz
),by(x2pxz
22
22
+=+=
+==∂∂
+==∂∂
Substituting in (1) we get pq= 4xyz 3. Find the PDE of all planes having equal intercepts on the x and y axis. Solution:
The equation of plane is 1 .............(1)x y z
a a b+ + =
Diff. w.r.t. x and y partially respectively
10 .............(2)
p
a b+ =
10 .............(3)
q
a b+ =
Solving (2) & (3) we get, p=q
4. Obtain the partial differential equation by eliminating arbitrary constants ‘a’ and ‘ b’ from
( ) ( )2 2 2 1x a y b z− + − + = .
Solution:
( ) ( )2 2 2 1x a y b z− + − + = - (1)
Differentiating (1) partially w.r. t x and y we get
( )2 2 0x a zp− + = x a zp⇒ − = − - (2)
( )2 2 0y b zq− + = y b zq⇒ − = − - (3)
Substituting (2) & (3) in (1) we get
2 2 2 2 2 1z p z q z+ + = i.e., ( )2 2 2 1 1z p q+ + =
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5. Eliminate the arbitrary function f from y
z fx
=
and form the PDE. (Nov/Dec 2012)
Solution:
2' .............(1)z y y
p fx x x
∂ − = = ∂
1' .............(2)
z yq f
y x x
∂ = = ∂
From (1) & (2) 0p y
px qyq x
−= ⇒ + =
6. Form the partial differential equation by eliminati ng the arbitrary constants ‘a’ & ‘b’ from z= ax+by
Solution: Given z= ax+by -----(1) Diff partially w.r.t x and y
,bqyz
,apxz
==∂∂
==∂∂
Substituting in (1) we get z = px+qy. 7. Form the partial differential equation by eliminati ng f from 2 1
2 logz x f xy
= + +
.
Solution:
Let 2 12 logz x f x
y
= + +
- (1)
Differentiate (1) partially w.r. t x and y
' 1 12 2 log
zp x f x
x y x
∂ = = + + ∂ - (2)
'2
1 12 log
zq f x
y y y
∂ −= = + ∂
- (3)
Eliminating 'f from (2) & (3)
( )22 1p xy
q x
− −∴ = ⇒ 2 22px x qy− = − ⇒ 2 22px qy x+ =
8. From the partial differential equation by eliminati ng g from ( )2 2 2, 0g x y z x y z+ + + + = .
Solution:
We know that if g(u, v) = 0 then u = f (v)
( )2 2 2x y z f x y z∴ + + = + + - ---- (1)
Differentiating (1) partially w.r. t x and y
We get
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( )( )'2 2 1x zp f x y z p+ = + + + ----- (2)
( )( )'2 2 1y zq f x y z q+ = + + + …..(3)
Divide (2) & (3) 1
1
x zp p
y zq q
+ +=
+ + ⇒ x qx zp zpq y py zq zpq+ + + = + + +
( ) ( )z y p x z q y x− + − = −
9. Find the complete integral of p+q = pq Solution:
Given p+q = pq -----(1) This is of the type f(p,q)=0 Let z= ax+by+c be the solution for (1) Partially Diff w.r.t ‘x ‘ and ‘y’ we get p= a and q = b
=> a+b = ab => b= a/(a–1)
Therefore z= ax+ (a/a–1)y+c is the complete solution.
10. Find the singular integral of pq2+qy+px =z
Solution:
This of Clairaut’s type and the complete solution is ab2+by+ax =z
Diff w.r.t ‘a ‘ and ‘b’ we get
0b2
a2x
bz
,0a2
b2x
az
=+=∂∂
=+=∂∂
� b
ay&
a
bx −=−= ⇒ 1
b
a
a
bxy ==
xy= 1 is singular integral. 11. Find the complete integral of p+q=0 Solution:
Given p +q = 0 -----(1) This is of the type f(p,q)=0 Let z= ax+by+c be the solution for (1) Partially Diff w.r.t ‘x ‘ and ‘y’ we get p=a and q=b
=> a+b = 0 => b= –a Therefore z= ax– ay +c be the complete solution.
12. Solve ( )1p q qz+ = .
Solution:
( )1p q qz+ = - (1)
This equation is of the form ( ), , 0f z p q =
( )z g x ay= + be the solution Let ( )x ay u z g u+ = =
dz adzp q
du du= =
(1) reduces to 1
1 1 11
dz dz dz dz dz dz dza a z a az a az z du
du du du du du du a za
+ = ⇒ + = ⇒ = − ⇒ = − ⇒ = −
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St. Joseph’s College of Engineering & Page No. 4 ISO 9001:2008 St. Joseph’s Institute of Technology
Integrating 1
log z u ba
− = +
i.e., 1
log z x ay ba
− = + +
is the complete solution.
13. Solve yp= 2yx+log q Solution:
The equation can be reduced to the type F1(x,p)= F2(y,q) � y(p–2x) = log q � p–2x = (log q)/ y =k
� p= 2x+k ; q = kye
� z = ∫ ∫+ qdypdx
� z= ∫ ∫++ dyedx)kx2( ky
z = (x2+kx)+ k
eky
+c
14. Solve '( 1)( 1) 0D D D z− − + = (Nov/Dec 2012) Solution:
'( 1)( 1) 0D D D z− − + = Here c1=1, c2=-1 , m1=0, m1=1
General solution 1 2( ) ( )x xz e f y e f y x−= + +
15. Solve 1p q+ = .
Solution:
This is of the form ( ), 0f p q = .
The solution is given by z ax by c= + + where
( )2111 ababba −=⇒−=⇒=+
∴ The complete solution is ( )2
1z ax a y c= + − +
16. Solve ( )( ) 21 2 x yD D D D z e −′ ′− − − − = .
Solution:
This is of the form ( ) ( ) ( )1 1 2 2 .... 0n nD m D C D m D C D m D C Z′ ′ ′− − − − − − =
Hence 1 1 2 11 1 1 1m c m c= = = = 2 2c =
Hence the C.F. is ( ) ( )21 2
x xz e y x e y xφ φ= + + +
P.I. ( )( )
2
1 2
x ye
D D D D
−
=′ ′− − − − ( )( )
221
2 1 1 2 1 2 2
x yx ye
e−
−= =+ − + −
Hence, the complete solution is ( ) ( )2 21 2
1
2x x x yZ e y x e y x eφ φ −= + + + +
17. Solve ( )3 2 33 2 0D DD D Z′ ′− + = .
Solution:
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Substituting D = m, & D1 = 1
The Auxiliary equation is m3 - 3m + 2 = 0
m = 1, 1, -2
Complimentary function is ( ) ( ) ( )1 2 3 2y x x y x y xφ φ φ+ + + + −
i.e., ( ) ( ) ( )1 2 3 2z y x x y x y xφ φ φ= + + + + −
18. Find the particular integral of yx22 ez)'D5'DD2D( −=++ Solution:
P.I = yx22
e'D5'DD2D
1 −
++
Replace D by 1 and 'D by 1 ,we get
= yx2
yxyx e2
xe
'D2D2
xe
121
1 −−− =+
=+−
19. Find the general solution of 0
y
z8
yx
z4
yx
z2
x
z3
3
2
3
2
3
3
3
=∂
∂+
∂∂
∂−
∂∂
∂−
∂
∂
Solution:
Auxiliary Equations is m3– 2m
2 –4m+8=0 � m= 2,2,–2
General solution z= ( ) ( ) ( )x2yfx2yxfx2yf 321 −++++ 20. Find the Particular integral of yx222 ez)1'D2D2'DD2'DD( −=+++++ Solution:
P.I = yx222
e)1'DD(
1 −
++
Replace D by 2 and D’ by –1 ,we get
P.I = yx2yx22
e4
1e
)112(
1 −− =+−
PART - B 1. a) Find the general solution of )y-z(x=)qz-y(x+)pz-x(y 222222
[A.U Apr/ May 2008 ]
Solution:
The equation is of the form Pp+Qq=R
where P= )z-x(y 22 ,Q= )z-y(x 22 ,R= )y-z(x 22
The auxiliary equation is )()()( 222222 yxz
dz
zxy
dy
yzx
dx
−=
−=
−
Using the multipliers x, y, z , we get each ratio is equal to
0)()()( 222222222=
−+−+−
++
yxzzxyyzx
zdzydyxdx
i.e. zdzydyxdx ++ =0
Integrating we get 222 zyx ++ 1c=
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St. Joseph’s College of Engineering & Page No. 6 ISO 9001:2008 St. Joseph’s Institute of Technology
using the another set of multipliers zyx
1,
1,
1
we get each of the above ratio is equal to 0)()()(
111
222222=
−+−+−
++
yxzxyz
dzz
ydyy
dxx
i.e. dzz
ydyy
dxx
111++ =0
Integrating we get 2c log =(xyz) log
The solution of the linear p.d.e is ϕ( u, v)=0
i.e. ϕ( 222 zyx ++ , xyz) =0
b) Solve yx3222 eyxz)'D6'DDD( ++=−+ Solution:
The auxiliary equation is 062 =−+mm 3,2 −=m
C.F= )3()2( 21 xyxy −++ φφ
P.I= )('6'
1 3222
yxeyxDDDD
++−+
= yxeDDDD
yx
D
D
D
DD
+
−++
−+
322
2
2
22 '6'
1
'6
'1
1
yxeyxD
D
D
D
D+
−
−++
−+= 32
1
2
2
2 )1(6)1(39
1'6
'1
1
652
1 354
yxexyx
+
+
−=
The general Solution is )3()2( 21 xyxyy −++= φφ +652
1 354
yxexyx
+
+
−=
2. a) Solve ( ) ( )2 2 2x yz p y zx q z xy− + − = −
This is a Lagrange’s linear equation of the form Pp Qq R+ =
The subsidiary equations are 2 2 2
dx dy dz
x yz y zx z xy= =
− − − - (1)
( ) ( ) 2 2 2 22 2
dx dy dy dz dx dz
y zx z xy x yz z xyx yz y zx
− − −= =
− − + − − +− − −
i.e., ( )
( ) ( )( )
( )( )
( )2 2 2 22 2 1
d x y d y z d x z
y z x y z x z y zx y z x y
− − −= =
− + − − + −− + −
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St. Joseph’s College of Engineering & Page No. 7 ISO 9001:2008 St. Joseph’s Institute of Technology
i.e., ( )
( )( )( )
( )( )( )
( ) ( )d x y d y z d x z
x y x y z y z x y z x z x y z
− − −= =
− + + − + + − + + - (2)
i.e., ( )( )
( ) ( )d x y d y z d x z
x y y z x z
− − −= =
− − − - (3)
Taking the first two ratios, and integrating ( ) ( )log log logx y y z a− = − +
x ya
y z
−∴ =
− - (4)
Similarly taking the last two ratios of (3) we get,
y zb
x z
−∴ =
− - (5)
But x y
y z
−−
and y z
x z
−−
are not independent solutions for 1x y
y z
−+
− gives
x z
y z
−−
which is the
reciprocal of the second solution.
Therefore solution given by (4) and (5) are not independent. Hence we have to search for
another independent solution.
Using multipliers x, y, z in equation (1) each ratio is 3 3 3 3
xdx ydy zdz
x y z xyz
+ +=
+ + −
Using multipliers 1, 1, 1 each ratio is 2 2 2
dx dy dz
x y z xy yz zx
+ +=
+ + − − −
3 3 3 2 2 23
xdx ydy zdz dx dy dz
x y z xyz x y z xy yz zx
+ + + +=
+ + − + + − − −
( )( )( )
( )2 2 2
2 2 22 2 2
12
d x y z d x y z
x y z xy yz zxx y z x y z xy yz zx
+ + + +=
+ + − − −+ + + + − − −
Hence ( ) ( ) ( )2 2 21
2d x y z x y z d x y z+ + = + + + +
Integrating ( ) ( )22 2 21
2 2
x y zx y z k
+ ++ + = +
( ) ( )22 2 2
2 2 2 2 2 2
2
2 2 2 2
x y z x y z k
x y z x y z xy yz zx k
∴ + + = + + +
+ + = + + + + + +
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St. Joseph’s College of Engineering & Page No. 8 ISO 9001:2008 St. Joseph’s Institute of Technology
i.e., xy yz zx b+ + =
∴ the general solution is , 0x y
xy yz zxy z
φ −
+ + = −
b) Solve 22222 )( yxqpz +=+ [A.U Nov/ Dec 2007]
Solution:
The given equation can be written as )1.......()()( 2222 yxzqzp +=+
This is of the type of f (x, zmp)=f(y, zmq)
2
Q =zq &
2
P=zp 2zq =Q ,2zp=P and Z= z Put 2 ⇒
Equation (1) can be written as )y+4(x =Q+P 2222
� 2222 4y-Q4x-P =
The equation is of the form f(x,p) =f (y,q)
� 24a4y-Q4x-P 2222 == ;
22 44 a4y-Q anda4x-P 2222 ==
� 222 axP += and
222 ayQ −=
We know that dZ= Pdx + Qdy
+
+= dxaxdZ 222 dyay
− 222
Integrating we get Z = 22 axx + + +
−
a
xa 12
sinh2
−− −
a
yayy 122 cosh
222 axxz += + +
−
a
xa 12
sinh2
−− −
a
yayy 122 cosh
3. a) Find the singular solution of the equation 22qp+qypx=z + Solution:
The equation is of the form z= px+qy+f(p,q)
For complete integral replace ‘p’ by ‘a’ and ‘q’ by b ,we get 22ba+byax=z +
� Complete integral = 22ba+byax=z + -----(1)
Diff w.r.t ‘a’ and ‘b’ we get
)3......(02
)2......(02
2
2
=+=∂
∂
=+=∂∂
bayb
z
abxa
z
Solving we get )4......(2 2abx −= & )5......(2 2bay −=
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The equation (4) can be written as )6......(2 2b
xa
−=
Substituting (6) in (5) we get )7......(2
3/12
−=
y
xb and )8......(
2
3/12
−=
x
ya
Substituting (7) & (8) in (1) we get 0y27x16z 223 =+ which is singular integral.
b) Solve yx22 e)yx2sin(5z)'D2'DD5D2( +−++=+− Solution:
The auxiliary equation is 0252 2 =+− mm
2
1,2=m
C.F=
+++ xyxy2
1)2( 21 φφ
P.I= ( )yxeyxDDDD
+−+++−
)2sin(5'2'52
122
yxeDDDD
yxDDDD
+−
+−++
+−=
2222 '2'52
1)2sin(5
'2'52
1
yxeyx +−
+−−++
−+−=
)1(2)1(5)1(2
1)2sin(
2108
15
yxeyxDD
x +−++−
=9
1)2sin(
'545
yxeyxDD
DD
DD
DDx +−++++
−+
=9
1)2sin(
'54
'54
'54
)'54(5
yxeyxDD
DDx +−++−
+=
9
1)2sin(
'2516
)'54(5
22
[ ] yxeyxxyxx +−++−+=9
1)2cos(5)2cos(8
39
5
The General Solution is
+
+++= xyxyy2
1)2( 21 φφ [ ] yxeyxxyxx +−++−+
9
1)2cos(5)2cos(8
39
5
4. a) Solve 2 2( 2 ) ( )z yz y p xy zx q xy zx− − + + = − .
Solution:
This is a Lagrange’s linear equation of the form Pp Qq R+ =
The subsidiary equations are 2 22
dx dy dz
z yz y xy zx xy zx= =
− − + −
Taking x, y, z as multipliers, we have each fraction 0
xdx ydy zdz+ +=
0xdx ydy zdz∴ + + =
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St. Joseph’s College of Engineering & Page No. 10 ISO 9001:2008 St. Joseph’s Institute of Technology
Integrating
2 2 2
2 2 2
x y zc+ + =
i.e., 2 2 2
1x y z C+ + = - (1)
Again, taking the last two members, we have ( ) ( )
dy dz
x y z x y z=
+ −
i.e., dy dz
y z y z=
+ −
( ) ( )
( )( )
0
0
y z dy y z dz
ydy zdy ydz zdz
ydy zdy ydz zdz
ydy d yz zdz z
− = +
− = +
− + − =
− − =
Integrating we get 2 2
22y yz z C− − = - (2)
From (1) & (2) the general solution is
( )2 2 2 2 2, 2 0x y z y yz zφ + + − − =
b) Solve 7xyz)'D3D3'DD( 22 +=+−− ( Nov/ Dec 2009 )
Solution:
The given equation is 0z)'D3D3'DD( 22 =+−−
This can be factorized as 0= z3)- (-1)D' - D'-0)(D -(D
It is of the form 0)c'DmD)(c'DmD( 2211 =−−−− where m1 =1 c1 = 0 ,m2 = -1 c2 = 3.
C.F= )()( 23
1 xyfexyf x −++
P.I = 3)- D' )(DD' -(D
7xy
+
+7xy
3
'DD13
1
)'DD(
1+
+−−
−=
)7xy(...3
'DD
3
'DD1
)'DD(3
1)7xy(
3
'DD1
)'DD(3
1 21
+
+
+
++
+−−
=+
+−
−−
=−
)7('9
2
3
'1
)'(3
1+
+
++
−−
= xyDDDD
DD
+++++++
−−
= )7xy('DD9
2)7xy('D
3
1)7xy(D
3
1)7xy(
)'DD(3
1
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St. Joseph’s College of Engineering & Page No. 11 ISO 9001:2008 St. Joseph’s Institute of Technology
++++
−−
=9
2)yx(
3
1)7xy(
)'DD(3
1=
++++
−
−=
9
2)yx(
3
1)7xy(
D
'D1D3
1
++++
−
−=
−
9
2)yx(
3
1)7xy(
D
'D1
D3
1 1
++++
+
++
−=
9
2)(
3
1)7(....
''1
3
12
yxxyD
D
D
D
D
+++++++++
−=
9
2)yx(
3
1)7xy(
D
'D
9
2)yx(
3
1)7xy(
D3
1
++++++−
=3
11
9
2)(
3
1)7(
3
1x
Dyxxy
D
++++++−
= ∫ dxxyxxyD 3
1
9
2)(
3
1)7(
3
1
++++++
−=
329
2)(
3
1)7(
3
1 2 xxyxxy
D =
∫
++++++
−dx
xxyxxy
32(
9
2)(
3
1)7(
3
1 2
=
+++
+++
−
669
2
23
1)7
23
1 2322 xxxxy
xx
yx
++++
−=
6
x
3
xy
3
x
9
x65
2
yx
3
1 322
++++−−++=
6339
65
23
1)()(
322
23
1xxyxxyx
xyfexyfz x
5. a) Find the singular solution of the equation 22 qpqp+qypx=z +++ Solution:
The equation is of the form z = px +qy + f(p,q)
For complete integral replace ‘p’ by ‘a’ and ‘q’ by b ,we get 22 baba+byax=z +++
� Complete integral = 22 baba+byax=z +++ -----(1)
Diff w.r.t ‘a’ and ‘b’ we get )3......(02)2......(02 =++=∂∂
=++=∂∂
bayb
zbax
a
z
Solving we get )4......(2
)( bxa
+−= substituting in (3) we get )5......(
3
)2( yxb
−=
Substituting (5) in (4) we get )6......(3
)2( xya
−=
Substituting (5) & (6) in (1) we get 0= y-x-xy-3z 22 which is singular integral.
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St. Joseph’s College of Engineering & Page No. 12 ISO 9001:2008 St. Joseph’s Institute of Technology
b) Solve 4ez)'D2D2'D'DD2D( yx322 +=−−++ + Solution:
4ez)'D2D2'D'DD2D( yx322 +=−−++ +
( ) 4)'(2)'( 32 +=+−+ + yxezDDDD ( ) 4)2')('( 3 +=−++ + yxezDDDD C.F= )()( 2
21 xyexy x −+− φφ
P.I= ( )yxyx eeDDDDDD
00322
4'22''2
1 ++ +−−++
yxyx eDDDDDD
eDDDDDD
0022
322
4'22''2
1
'22''2
1 ++
−−+++
−−++=
yxyx eDD
xe 003 4
2'2226169
1 ++
−++
−−++=
yxyx ex
e 003 428
1 ++
−+=
xe yx 28
1 3 −= +
UNIT II FOURIER SERIES PART-A
1. State Dirichlet’s conditions for the existence of Fourier series of f(x) in the interval . (Nov./Dec.2011,Reg 2008)
A function f(x) can be expanded as a Fourier series in the interval if the following conditions are
satisfied. (i) f(x) is periodic, single valued and finite in
(ii) f(x) has finite number of finite discontinuities and no infinite discontinuity in
(iii) f(x) has a finite number of maxima and minima in
2. Does possess a Fourier expansion?
does not possess a Fourier expansion because the function has the infinite
discontinuity at the point .
3. If
2 n 22
2 2 2 2n 1
π ( 1) 1 1 1 πx 4 cosnx ,deduce that ... [ , ]
3 n 1 2 3 6in π π
∞
=
−= + + + + = −∑
2 n 2 2n
22 2
n 1 n 1
2
2 2 2
P ut x π then
π ( 1 ) π ( 1 )π 4 cosn 4
3 n 3 n
1 1 1 π...
1 2 3 6
π∞ ∞
= =
=
− −= + = +
∴ + + + =
∑ ∑
4. The cosine series for f(x) = x sin x in 0 < x < π is given as
x sin x = 22
1 ( 1)1 cos 2 cos
2 1
n
n
x nxn
∞
=
−− −
−∑ Deduce that .
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St. Joseph’s College of Engineering & Page No. 13 ISO 9001:2008 St. Joseph’s Institute of Technology
Given x sin x = 2
2
1 ( 1)1 cos 2 cos
2 1
n
n
x nxn
∞
=
−− −
−∑
Put x = , a point of continuity => 22
1 ( 1)sin 1 cos 2 cos
2 2 2 2 1 2
n
n
n
n
π π π π∞
=
− = − − − ∑
i.e. 22
( 1)1 2 cos
2 1 2
n n
n
π π∞ − = − − ∑
i.e.
5. Find the value of the Fourier series for f(x) = at x=1. (Nov./Dec.2011,Reg 2010).
x=1 is a point of discontinuity, value of Fourier series of f(x) is .
2)1()1( ++− ff
f(1-) =
0 0lim (1 ) lim (1 ) 1h h
f h h→ →
− = − = .
f(1+) = .∴f(x) at x = 1 is =1.5
6. Find the constant term of the Fourier series for the function f(x) = x2 , – ππππ < x < ππππ
Since f(x) is an even function in
– ππππ < x < ππππ
2 20
0 0
2 2 2( )
3a f x dx x dx
π
ππ
= = =∫ ∫�
� ∴ Constant term =
32
20 π=
a
7. To which value, the half range sine series corresponding to f(x) = x2 expressed in the interval (0, 5) converges at x =5 ?
At the end points half range sine series converges to zero. ∴ The half range sine series corresponding to f(x) = x2 converges to zero at x = 5.
8. What you meant by Harmonic analysis ? The process of finding the harmonics in the Fourier series expansion of a function numerically is known as
harmonic analysis. 9. Find the Fourier constant bn for x sin x in xπ π− < < , when expressed as a Fourier series. Solution:
( ) ( ) ( ) ( )sin sin sinf x x x f x x x x x f x= ⇒ − = − − ⇒ =
Here ( )f x is an even function 0nb∴ =
10. Find the root mean square value of the function f(x) = x in .
RMS value = RMS value = [ ]
2 21( )
b
ay f x dx
b a= =
− ∫
11. Obtain the constant term of the Fourier cosine series of y = f(x) in (0,6) using the following table
x: 0 1 2 3 4 5 y: 4 8 15 7 6 2
ao = 2 x Mean value of y =2(Σy)/6= 14 ∴ Constant term = 72
a0 =
12. Find the value of an in the cosine series expansion of f(x) = K in the interval (0,10)
10
0 0
2 1( ) cos cos 0
5 10n
n x n xa f x dx K dx
π π= = =∫ ∫
�
� �
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13. 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?
Since x = 2 is an end point then f(x) converges to 4
2
26
2
)2()2(=
+=
+− ff
14. State Parseval’s theorem in the interval (c,c+2l).
If the Fourier series corresponding to f(x) converges uniformly to f(x) in (c,c+2l)
then [ ] +=∫+
2
adx)x(f
1 20
2c
c
2�
�( )∑
∞
=
+1n
2n
2n ba
15. Write down the complex form of Fourier Series in (c,c+2l)
∑∞
∞−=
π
=n
xni
nec)x(f � where dxe)x(f2
1c
2c
c
xni
n ∫+ π−
=�
�
�
16. If f(x) = e x in xπ π− < < , find a n .
Solution:
1cosx
na e nxdxπ
ππ −
= ∫ ( )2
1cos sin
1
xenx n nx
n
π
ππ
−
= +
+ ( ) ( )2 2
11 1
1 1n ne e
n n
π π
π
− = − − −
+ +
( )( ) { }2
1
1
n
na e en
π π
π−−
= −+
.
17. Determine bn in the Fourier series expansion of
2)(
xxf
−=π
in 0< x < 2ππππ.
∫∫−
==ππ
ππ 2
0
2
0
sin2
1sin)(
1dxnx
xdx
xnxfbn
�
��
nnn
nx
n
nxx
12
2
1sin)1(
cos)(
2
12
02
==
−−−
−−=π
ππ
π
π
18. Find the constant term in the expression of x2cos as a Fourier series in the interval ),( ππ− . Solution:
12
2sin
2
1
2
2cos11cos
1 20 =
+=+
==−
−− ∫∫π
π
π
π
π
π πππx
xdxx
xdxa ∴∴∴∴Constant term = 2
1
20 =
a
19. Give the expression for the Fourier series coefficient bn for the function f(x) defined in 2 2x− ≤ ≤ .
Solution:
( )2
2
1sin
2 2n
n xb f x dx
π
−
= ∫
20. Determine the value of 0& aan in the Fourier series expansion of 3)( xxf = in ππ <<− x
Solution: 3)( xxf = ⇒⇒⇒⇒ )()()( 33 xfxxxf ==−=− ⇒⇒⇒⇒ )(xf is an odd function ∴ 00 == aan
PART – B 1.
a) If ( )
2
2x
f xπ −
=
in 0 2x π< < . Hence show that
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(i) 2
2 2 2
1 1 1.............
1 2 3 6
π+ + + = . (ii)
2
2 2 2 2
1 1 1 1...
1 2 3 4 12
π− + − + = .
Solution:
We know that
( ) ( )0
1
cos sin2 n n
af x a nx b nx
∞
= + +∑
22
0
0
12
xa dx
π ππ
− = ∫
( )2
2
0
1
4x dx
π
ππ
= −∫
( )( )
31
4 3
xππ
−=
−
3 3 21
4 3 3 6
π π ππ −
= + = −
22
0
1cosn
xa nxdx
π ππ α
− = ∫
( )2
2
0
1cos
4x nxdx
π
ππ
= −∫
( ) ( )2
2
2 3
0
1 sin cos sin2 2
4
nx nx nxx x
n n n
π
π ππ − − = − − − − +
2 2
1 cos2 20 2 0 0
4
n
n n
π ππ
π = + + − +
2 2
1 4 1
4 n n
ππ = =
( )2
2
0
1sin
4nb x nxdxπ
ππ
= −∫
( ) ( ) ( )2
2
2 3
0
1 cos sin cos2 2
4
nx nx nxx x
n n n
π
π ππ − − = − − − − +
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St. Joseph’s College of Engineering & Page No. 16 ISO 9001:2008 St. Joseph’s Institute of Technology
2 2
3 3
1 2 20
4 n n n n
π ππ
= − + + − =
( )2
21
1cos .
12f x n x
n
π ∞
= +∑
( )2
2 2
1 1cos cos .... 1
12 1 2x x
π = + + + −
0Put x=
( ) ( )2
2 2
1 10 ... 2
12 1 2f
π = + + + −
0x = is a pt of discontinuity.
( )2 2 21
02 4 4 4
fπ π π
∴ = + =
( )2 2
2 2
1 12 ...
4 12 1 2
π π ∴ => = + + +
2 2
2 2
1 1......
4 12 1 2
π π− = + + ∞
2
2 2
1 1.....
1 2 6
π∴ + + ∞ =
Put x π= in (1)
( ) ( ) ( )2
21
13
12
n
f xn
π ∞ −= + −∑
Here π is a pt of continuity.
( ) 0.f π∴ =
( )2
2 2 2
1 1 13 0 .....
12 1 2 3
π=> = − + − + ∞
2
2 2 2
1 1 1.....
12 1 2 3
π− = − + − + ∞
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2
2 2 2
1 1 1.....
12 1 2 3
π− = − + + ∞
b) Find the Fourier series for in π .
Solution:
Given
Since is an even function
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St. Joseph’s College of Engineering & Page No. 18 ISO 9001:2008 St. Joseph’s Institute of Technology
Substitute in equation (1) we get ∑∞
= π−
+−−+
π=
2n2
n
nxcos)1n(
]1)1[(22)x(f
2. a) Obtain the Fourier series to represent the function f(x) = | x | is xπ π− < < and deduce that 2
2 2 2
1 1 1.............
1 3 5 8
π+ + + = .
Solution:
Given ( )f x x=
( ) ( )f x f x− =
∴ The given function is an even function.
Hence 0nb =
( ) 0
1
cos2 n
af x a nx
∞
∴ = +∑
0
2na x dx
π
π= ∫
2
0
22x
π
ππ
= =
0
2cosna x nxdx
π
π= ∫
( ){ }2 2 2 20
2 sin cos 2 cos 1 21 1
nnx nx nx
n n n n n
π ππ π π = + = − = − −
0na∴ = if n is even
2
4na
n π= − if n is odd
( ) 21,3
4cos
2f x nx
n
ππ
∞ −∴ = +∑
2
4 cos3cos ...
2 3
xx
ππ− = − + +
0Put x=
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St. Joseph’s College of Engineering & Page No. 19 ISO 9001:2008 St. Joseph’s Institute of Technology
( ) 2 2
4 1 10 1 ...
2 3 5f
ππ = − + + +
Here 0 is a pt of continuity
( )0 0f∴ =
2 2
4 1 10 1 ...
2 3 5
ππ = − + + +
2 2
4 1 11 ...
2 3 5
ππ = + + +
2
2 2
1 11 ... .
8 3 5
π ∴ = + + +
b) Find the first two harmonic of the Fourier series of f (x) given by
x 0 1 2 3 4 5 f (x) 9 18 24 28 26 20
Solution:
Here the length of the in level is 2 6, 3l l= =
( ) 01 1 2 2
2 2cos sin cos sin
2 3 3 3 3
a x x x xf x a b a b
π π π π = + + + +
( )0
2 1252 41.66
6 6
ya = = =∑
x
3
xπ
2
3
xπ
y cos
3
xy
π sin
3
xy
π
2cos
3
xy
π
2sin
3
xy
π
0 0 0 9 9 0 9 0
1
3
π
2
3
π
18 9 15.7 -9 15.6
2 2
3
π
4
3
π
24 -12 20.9 -12 -20.8
3 π 2π 28 -28 0 28 0
4 4
3
π
8
3
π
26 -13 -22.6 -13 22.6
5 5
3
π
10
3
π
20 10 17.6 -10 -17.4
125 -25 -3.4 -7 0
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St. Joseph’s College of Engineering & Page No. 20 ISO 9001:2008 St. Joseph’s Institute of Technology
1
1
2cos 8.33
6 3
2sin 1.15
6 3
xa y
xb y
π
π
= = −
= = −
∑
∑
2
2 2cos 2.33
6 3
xa y
π = = − ∑
2
2 2sin 0
6 3
xb y
π= =∑
( ) 41.66 28.33cos 2.33cos 1.15sin .
2 3 3 3
x x xf x
π π π∴ = − − −
3. a) Find the Fourier series for the function
Solution:
Given
Therefore is an odd function. Therefore
Substitute in (1) we get
b) Find the first two harmonic of the Fourier series of f (x). Given by
x 0
3π
23π
π 4
3π
53π
2π
f (x) 1 1.4 1.9 1.7 1.5 1.2 1.0
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St. Joseph’s College of Engineering & Page No. 21 ISO 9001:2008 St. Joseph’s Institute of Technology
Solution:
∵ The last value of y is a repetition of the first; only the first six values will be used
The values of cos , cos2 , sin , sin 2y x y x y x y x as tabulated
x ( )f x cosx sinx cos2x sin 2x
0 1.0 1 0 1 0
3
π
1.4 0.5 0.866 -0.5 0.866
2
3
π
1.9 -0.5 0.866 -0.5 0.866
π 1.7 -1 0 1 0
4
3
π
1.5 -0.5 -0.866 -0.5 -0.866
5
3
π
1.2 0.5 -0.866 -0.5 -0.866
0 2 2.96
ya = =∑
, 1
cos2 0.37
6
y xa = = −∑
, 2
cos22 0.1
6
y xa = = −∑
1
sin2 0.17
6
y xb = =∑
, 2
sin 22 0.06
6
y xb = = −∑
4. a) Obtain the Fourier series for of period and defined as follows
Hence deduce that
Solution:
Given
The Fourier series is
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Deduction (i)
Put
Deduction (ii)
Put
b) Expand as a cosine series in and deduce the value of
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St. Joseph’s College of Engineering & Page No. 23 ISO 9001:2008 St. Joseph’s Institute of Technology
Solution: Given The cosine series is
2 2
0
4
if n is even
lif n is odd
n π
= −
Substituting in equation (1) we get
Deduction (i)
By Parseval’s identity
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St. Joseph’s College of Engineering & Page No. 24 ISO 9001:2008 St. Joseph’s Institute of Technology
Deduction (ii)
9616
15S
9616
11S
44 ππ=
⇒=
−
5. a) Find the complex form of the Fourier series of in
Solution:
The complex form of the Fourier series in is given by
Where
Hence (1) becomes
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St. Joseph’s College of Engineering & Page No. 25 ISO 9001:2008 St. Joseph’s Institute of Technology
b) Find the half range Fourier sine series for ( ) ( )f x x xπ= − in the interval (0, )π and deduce that
∞−+− ......5
1
3
1
1
1333
Solution:
The sine series is ∑∞
=
=1
sin)(n
n nxbxf
dxnxxfbn ∫=π
π 0
sin)(2
dxnxxx∫ −=π
ππ 0
2 sin)(2
π
πππ
032
2 cos)2(
sin)2(
cos)(
2
−+
−−−
−−=n
nx
n
nxx
n
nxxx
[ ]
=−−=
oddisnifn
evenisnif
nn
33 8
0)1(1
4
ππ
∑∞
=
=,....3,2,1
3sin
8)(
n
nxn
xfπ
Deduction:
Put 2
π=x is point of continuity
−+−=
...5
1
3
1
1
18
2 333ππ
f
−+−= ...5
1
3
1
1
18
4 333
2
ππ
32...
5
1
3
1
1
1 2
333
π=−+−
UNIT - III - APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS PART – A 1. State any two assumptions made in the derivation of the one dimensional wave equation.
i. The string is perfectly elastic and does not offer any resistance to bending.
ii. Deflection y and slope dy/dx at every point of the string are small, so that their higher powers may be neglected.
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St. Joseph’s College of Engineering & Page No. 26 ISO 9001:2008 St. Joseph’s Institute of Technology
2. Write down the partial differential equation governing one dimensional wave equation?
2
22
2
2
x
yC
t
y
∂∂
=∂∂
where y(x,t) is the displacement of the string .
3. In the wave equation
2
22
2
2
x
yC
t
y
∂∂
=∂∂
, what does C2 stands for?
C2 = T/m, where T is the tension and m is the mass of the string. 4. Write all possible solutions of the transverse vibration of the string in one dimension.
(i) y(x ,t) = (Aepx + Be – px) (Cepat + De – pat) (ii) y(x,t) = (A cos px + Bsin px) (C cos pat + sin pat). (iii) y(x,t) = (Ax +B) (Ct + D)
5. Write down the appropriate solution of the vibration of string equation. How is it chosen?
The appropriate solution of the vibration of string equation is y(x,t) = (A cos px+ B sin px) ( C cos pat+ D sin pat) . When we deal with the vibration of an elastic string y(x,t) representing the displacement of the string at any point x, it must be periodic in ‘t’. Hence the above solution which consists of periodic functions in‘t’ is the proper solution of the problems on vibration of strings.
6. A tightly stretched string with fixed end points x=0 and x=l is initially at rest in its equilibrium position. If it is set vibrating giving each point a velocity x(l-x), write down the boundary and initial conditions of the above problem.
lxforxlxt
xyandxy
tfortlyandty
≤≤−=∂
∂=
≥==
0)()0,(
0)0,(
00),(0),0(
λ
7. State Fourier law of heat conduction.
The rate at which heat flows through an area is jointly proportional to the area and to the temperature gradient normal to the area.
8. Write down the partial differential equation that r epresents one dimensional heat equation.
,2
22
x
u
t
u
∂
∂=
∂∂
α where 2α = c
k
ρ
9. What are the various possible solutions of one dimensional diffusion equation?
(i) u(x ,t) = tppxpx CeBeAe22
)( α−+
(ii) u(x ,t )= tpCepxBpxA22
)sincos( α−+
(iii) u(x ,t) = Ax + B 10. Why can’t u(x ,t) = tppxpx CeBeAe
22
)( α−+ be the correct solution in solving one dimension heat
equation?
As t→ ∞, u → ∞. It is not possible. 11. In steady state condition derive the solution of one dimension heat flow equation.
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St. Joseph’s College of Engineering & Page No. 27 ISO 9001:2008 St. Joseph’s Institute of Technology
The one dimension heat flow equation is 2
22
x
u
t
u
∂∂
=∂∂
α .
In steady state condition, 0=∂∂
t
u. ∴ 02
2
=∂∂x
u ⇒ A
x
u=
∂∂
⇒ BAxu += .
12. An insulated rod of length 60 cm has its ends A and B maintained at oooo20C20C20C20C and oooo80C80C80C80C respectively. Find the steady state temperature in the rod. (Nov. 2012)
Steady state heat equation is 2
20
d u
dx= . The solution is ( )u x Ax B= + ……..(1)
Given conditions are ( ) ( )0 20, 60 80u u= = .
Substituting in (1) we get 20u x= + , which is the steady state temperature in the rod.
13. Write down the partial differential equation that r epresents variable heat flow in two dimensions? Deduce the equations of steady state heat flow in two dimensions.
The heat flow equation in two dimensional Cartesian co-ordinate system is
∂∂
+∂∂
=∂∂
2
2
2
22
y
u
x
u
t
uα When
steady state conditions prevail in the plate the temperature at any point of the plate does not depend on
‘t’, but depends on x and y only (i.e.) 0=∂∂
t
u. Thus steady state temperature distribution in a two
dimensional plate is 02
2
2
2
=∂∂
+∂∂
y
u
x
u (Laplace equation).
14. Write any two solutions of the Laplace equation obtained by the method of separation of variables involving exponential terms in x and y.
u(x ,y) = (Aepx + Be – px) (C cos py + sin py) & u(x ,y)= (A cos px + Bsin px) (C e py + e - py)
15. Given the boundary conditions on a square or rectangular plate, how will you identify the proper solution?
If the non-zero temperature is prescribed either on x=0 or on x=a (vertical edges), the proper solution will be u(x ,y) = (Aepx + Be – px) (C cos py + sin py) If the non-zero temperature is prescribed either on y=0 or y=b (horizontal edges), the proper solution will be u(x ,y)= (A cos px + Bsin px) (C e py + e - py)
16. An infinitely long plate is bounded by two parallel edges and an end at right angles to them. The breath of the edge y = 0 is ππππ and it is maintained at constant temperature u0 at all points and the other edges are kept at zero temperatures. Formulate the boundary value problem to determine the steady state temperature.
2 0u∇ = subject to π
π≤≤==∞
≥==
xforuxuandxu
yforyuyu
0)0,(0),(
00),(,0),0(
0
17. A plate is bounded by the lines x=0, y=0, x=l, and y=l. Its faces are insulated. The edge coinciding with x-axis is kept at oooo100C100C100C100C . The edge coinciding with y-axis is kept at oooo50C50C50C50C . The other two
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St. Joseph’s College of Engineering & Page No. 28 ISO 9001:2008 St. Joseph’s Institute of Technology
edges are kept atoooo0C0C0C0C . Write the boundary conditions that are needed for solving two dimensional heat flow equation. (Nov. 2012)
Boundary conditions are
(i) ( ),0 100 ; 0u x C x l= < <�. (ii) ( )0, 50 ;0u y C y l= < <�
.
(ii) ( ), 0 ;0u x l C x l= < <�. (iv) ( ), 0 ; 0u l y C y l= < <� .
18. The boundary value problem governing the steady state temperature distribution in a flat thin
plate is given by 02
2
2
2
=∂
∂+
∂
∂
y
u
x
u, 0)0,( =xu , 0),0( =yu , 0),( =yau and
=a
xaxu
π3sin4),( . Find
Cn when the most general solution is ∑∞
=
=1
sinhsin),(n
n a
yn
a
xncyxu
ππ.
−=
==∑∞
= 4
3sinsin3
4sin4sinhsin),( 3
1
a
x
a
x
a
xn
a
xncaxu
nn
πππ
ππ
a
x
a
x
a
xc
a
xc
a
xc
πππ
ππ
ππ
π 3sinsin3........3sinh
3sin2sinh
2sinsinhsin 321 −=+++
1 3
1 3
2 4 5 6
sinh 3, sinh 3 1,
3 1
sinh sinh 3.......... 0
c c
c c
c c c c
π π
π π
∴ = =
⇒ = ⇒ =
= = = = =
19. Define thermally insulated ends.
The end at which the temperature gradient is zero is called thermally insulated ends. (i.e) there is no heat flow throws the ends.
20. When the heat flow is called two dimensional?
When the heat flow is along curves instead of along straight lines, all the curves lying in parallel planes, then the flow is called two dimensional.
PART B 1. A uniform string is stretched and fastened to two points ''l apart. Motion is started by displacing
the string into the form of the curve )( xlkxy −= and then releasing it from this position at
time 0=t . Find the displacement of the point of the string at a distance x from one end at timet .
Solution:
One dimensional wave equation is 2
22
2
2
x
ya
t
y
∂
∂=
∂
∂
Boundary conditions are (i) 0),0( =ty (ii) 0),( =tly (iii) 0)0,( =∂∂
xt
y (iv) )()()0,( xlkxxfxy −==
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St. Joseph’s College of Engineering & Page No. 29 ISO 9001:2008 St. Joseph’s Institute of Technology
The correct solution is
)sincos)(sincos(),( patDpatCpxBpxAtxy ++= (1)
Put x = 0 and apply the boundary condition (i) in (1)
=),0( ty 0)sincos)(( =+ patDpatBA ⇒ .
Equation (1) becomes )sincos(sin),( patDpatCpxBtxy += (2)
Put x = l and apply the boundary condition (ii) in (2)
0)sincos(sin),( =+= patDpatCplBtly ⇒ 0sin =pl πnpl =⇒ ⇒
Equation (2) becomes
+=l
atnD
l
atnC
l
xnBtxy
πππsincossin),( (3)
Differentiate Partially (3) w. r. to t
=∂∂
),( txt
y
+−l
atn
l
anD
l
atn
l
anC
l
xnB
πππππcossinsin (4)
Put t = 0 and apply the boundary condition (iii) in (4)
0sin)0,( ==∂∂
l
anD
l
xnBx
t
y ππ ⇒
Equation (3) becomes
l
atn
l
xnB
l
atnC
l
xnBtxy
n
ππ
ππ
cossin
cossin),(
=
= where BCBn =
By the superposition principle, the most general solution is
∑∞
=
=1
cossin),(n
n l
atn
l
xnBtxy
ππ (5)
Put t = 0 and apply the boundary condition (iv) in (5)
l
np
π=
0=A
0=D
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St. Joseph’s College of Engineering & Page No. 30 ISO 9001:2008 St. Joseph’s Institute of Technology
)(1.sin)0,(1
xfl
xnBxy
nn ==∑
∞
=
π (6)
By half range Fourier sine series ∑∞
=
=1
sin)(n
n l
xnbxf
π (7)
By comparing (6) and (7),we get
( ) ( ) ( )
[ ]
[ ]
n
n
n
l
l
l
n
B
oddisnifn
kl
evenisnifn
kl
n
l
l
k
n
ln
n
l
l
k
l
nl
xn
l
nl
xn
xl
l
nl
xn
xlxl
k
dxl
xnxlkx
l
dxl
xnxf
lb
=
=
−−=
−−=
+−+−+=
−+
−
−−
−−=
−=
=
∫
∫
33
2
33
2
33
3
33
3
33
3
03
33
2
222
0
0
8
0
)1(14
)1(122
1..200cos.2002
cos2
sin2
cos2
sin)(2
sin)(2
π
π
π
ππ
π
π
π
π
π
π
π
π
π
∴The equation (5) becomes ∑∞
=
=5,3,1
33
2
cossin8
),(n l
atn
l
xn
n
kltxy
πππ
2. If a string of length l is initially at rest in its equilibrium position and each of its points is given a velocity
v such that
<<−
<<=
lxl
xlc
lxcx
v
2;)(
20;
, find the displacement at any timet .
Solution:
One dimensional wave equation 2
22
2
2
x
ya
t
y
∂
∂=
∂
∂
The boundary conditions are
nn Bb =
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(i) 0),0( =ty (ii) 0),( =tly (iii) 0)0,( =xy
(iv)
<<−
<<=
∂∂
lxl
xlc
lxcx
xt
y
2);(
20;
)0,(
The correct solution is
)sincos)(sincos(),( patDpatCpxBpxAtxy ++= (1)
Put x = 0 and apply the boundary condition (i) in (1)
=),0( ty 0)sincos)(( =+ patDpatBA ⇒ .
Equation (1) becomes )sincos(sin),( patDpatCpxBtxy += (2)
Put x = l and apply the boundary condition (ii) in (2)
0)sincos(sin),( =+= patDpatCplBtly ⇒ 0sin =pl πnpl =⇒ ⇒
Equation (2) becomes
+=l
atnD
l
atnC
l
xnBtxy
πππsincossin),( (3)
Put t = 0 and apply the boundary condition (iii) in (3)
=)0,(xy 0sin =Cl
xnB
π ⇒ C = 0
Equation (3) becomes
l
atn
l
xnB
l
atnD
l
xnBtxy
n
ππ
ππ
sinsin
sinsin),(
=
= where BDBn =
By the superposition principle, the most general solution is
∑∞
=
=1
sinsin),(n
n l
atn
l
xnBtxy
ππ (4)
Differentiate Partially (4) w.r.to t
l
np
π=
0=A
C = 0
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St. Joseph’s College of Engineering & Page No. 32 ISO 9001:2008 St. Joseph’s Institute of Technology
=∂∂
),( txt
y
l
an
l
atn
l
xnB
nn
πππ∑∞
=1
cossin (5)
Apply the boundary condition (iv) in (5)
=∂∂
)0,(xt
y)(
2);(
20;
sin1
xf
lxl
xlc
lxcx
l
an
l
xnB
nn =
<<−
<<=∑
∞
=
ππ (6)
By half range Fourier sine series ∑∞
=
=1
sin)(n
n l
xnbxf
π (7)
By comparing (6) and (7), we get nnnn ban
lB
l
anBb
ππ
=⇒=
( )
−
−−
−−+
−
−
−
=
−+=
=
∫∫
∫
l
l
l
n
l
l
l
l
l
l
nl
xn
l
nl
xn
xl
l
nl
xn
l
nl
xn
xl
c
dxl
xnxldx
l
xnx
l
c
dxl
xnxf
lb
2
2
2
2
2
22
02
22
0
0
sin)1(
cos)(
sin1
cos2
sin)(sin2
sin)(2
π
π
π
π
π
π
π
π
ππ
π
2sin
4
2sin
2cos
22sin
2cos
2
2
22
22
22
22
22
ππ
ππ
ππ
ππ
ππ
n
n
clb
n
n
ln
n
ln
n
ln
n
l
l
c
n =
+++−=
2sin
4
2sin
433
2
22
ππ
πππ
n
an
cln
n
cl
an
lBn ==∴
∴The equation (4) becomes
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∑∞
=
=1
33
2
sinsin2
sin4
),(n l
atn
l
xnn
an
cltxy
ππππ
3. The ends A and B of a rod l cm long have their temperatures kept at C�30 and C�80 , until steady state conditions prevail. The temperature of the end B is suddenly reduced to C�60 and that of A is increased to C�40 . Find the steady state temperature distribution in the rod after time t.
Solution:
One dimensional heat equation 2
22
x
ua
t
u
∂
∂=
∂∂
(1)
In the steady state condition the solution is baxxu +=)( (2)
The boundary conditions are 80)()(30)0()( == luiiui
Apply (i) in (2), 30)0( == bu
Then the equation (2) becomes 30)( += axxu (3)
Apply (ii) in (3) 8030)( =+= allu l
a50
=⇒
Then the equation (3) becomes 3050
)( +=l
xxu
Now consider the unsteady state condition.
In unsteady state the correct solution is
tpaCepxBpxAtxu22
)sincos(),( −+= (4)
The boundary conditions are
3050
)()0,()(60),()(40),0()( +====l
xxuxuvtluivtuiii
Since we have all non zero boundary conditions, we write the temperature distribution function as
),()(),( txuxutxu ts += (5)
⇒ )(),(),( xutxutxu st −=
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To find )(xus
The solution is baxxus +=)( (6)
The boundary conditions are 40)(,60)0( == luu ss
∴ ,40)0( == bus
la
al
ballus
20
6040
60)(
=⇒
=+⇒
=+=
∴ the equation (6) becomes 4020
)( +=l
xxus (7)
To find ),( txut
Given the boundary conditions are
1030
4020
3050
)()0,()0,()(
06060)(),(),()(
04040)0(),0(),()(
−=
+−+=−=
=−=−=
=−=−=
l
x
l
x
l
xxuxuxuviii
lutlutluvii
ututxuvi
st
st
st
In unsteady state, the correct solution is tpaCepxBpxAtxu22
)sincos(),( −+= (8)
Apply the boundary condition (vi) in (8)
0),0(22
== − tpat ACetu
here 0=A
∴the equation (8) becomes tpat CepxBtxu
22
sin),( −= (9)
Apply the boundary condition (vii) in (9)
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St. Joseph’s College of Engineering & Page No. 35 ISO 9001:2008 St. Joseph’s Institute of Technology
0sin),(22
== − tpat pleBCtlu
l
np
pl
π=⇒
= 0sin
∴the equation (9) becomes t
l
na
t Cexl
xnBtxu
2
222
sin),(ππ −
=
t
l
na
n
tl
na
t
el
xnB
el
xnBCtxu
2
222
2
222
sin
sin),(
π
π
π
π
−
−
=
= where nBBC =
By the super position principle, the most general solution is ∑∞
=
−
=1
2
222
sin),(n
tl
na
nt el
xnBtxu
ππ (10)
Apply the boundary condition (viii) in (10)
∑∞
=
=−==1
)(1030
1.sin)0,(n
nt xfl
x
l
xnBxu
π
Half range Fourier sine series of )(xf is ∑∞
=
=1
sin)(n
n l
xnbxf
π (11)
From the equations (10) & (11) we get nn Bb =
−−
=
−
−
−
−=
−=
=
∫
∫
1.10
cos202
sin30
cos10
302
sin10302
sin)(2
02
22
0
0
ππ
π
π
π
π
π
π
π
n
ln
n
l
l
l
nl
xn
ll
nl
xn
l
x
l
dxl
xn
l
x
l
dxl
xnxf
lb
l
l
l
n
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St. Joseph’s College of Engineering & Page No. 36 ISO 9001:2008 St. Joseph’s Institute of Technology
[ ]
n
n
Bn
=
−+−
= )1(2120
π
∴the equation (10) becomes
[ ]∑∞
=
−
−+−
=1
2
222
sin)1(2120
),(n
tl
nan
t el
xn
ntxu
πππ
Then the required temperature distribution function is
[ ]∑∞
=
−
−+−+=1
2
222
sin)1(2120
4020
),(n
tl
nan e
l
xn
nl
xtxu
πππ
4. A rectangular plate is bounded by the lines byaxyx ==== ,,0,0 . Its surfaces are insulated. The
temperature along 0=x and 0=y are kept at C�0 and the others at C�100 . Find the steady state temperature at any point of the plate.
Solution:
We know that two dimensional heat equation 02
2
2
2
=∂
∂+
∂
∂
y
u
x
u
Let l be the length of the square plate
Given the boundary conditions are
( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 100
( ) ( , ) 100
i u y ii u x iii u a y
iv u x b
= = =
=
We write the temperature function as
),(),(),( 21 yxuyxuyxu += (A)
To find ),(1 yxu
Consider the boundary conditions
1 1 1
1
( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 0
( ) ( , ) 100
v u y vi u x vii u x b
viii u a y
= = =
=
The suitable solution which satisfying the given boundary conditions is
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St. Joseph’s College of Engineering & Page No. 37 ISO 9001:2008 St. Joseph’s Institute of Technology
)sincos)((),(1 pyDpyCBeAeyxu pxpx ++= − (1)
Apply the boundary condition (v) in (1)
0)sincos)((),0(1 =++= pyDpyCBAyu
here 0=+ BA
AB −=
then the equation (2) becomes
)sincos)((
)sincos)((),(1
pyDpyCeeA
pyDpyCAeAeyxupxpx
pxpx
+−=
+−=−
−
(2)
Apply the boundary condition (vi) in (2)
0)()0,(1 =−= − CeeAxu pxpx
here 0=C
then the equation (3) becomes
pyDeeAyxu pxpx sin)(),(1−−= (3)
Apply the boundary condition (vii) in (3)
0sin)(),(1 =−= − pbDeeAbxu pxpx
here 0sin =pb πnpb=⇒b
np
π=⇒
then the equation (4) becomes
b
yn
b
xnB
b
yn
b
xnAD
b
ynDeeAyxu
n
b
xn
b
xn
ππ
ππ
πππ
sinsinh
sinsinh
sin)(),(1
=
=
+=−
where ADBn 2=
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The most general solution is
b
yn
b
xnByxu
nn
ππsinsinh),(
11 ∑
∞
=
= (4)
Apply the boundary condition (viii) in (4)
)(100sinsinh),(1
1 yfb
yn
b
anByau
nn ===∑
∞
=
ππ (5)
To find nB , expand )(xf in half range sine series
We know that half range Fourier sine series of )(xf is
∑∞
=
=1
sin)(n
n b
ynbyf
π (6)
From the equations (5) & (6) we get
b
anb
Bb
anBb n
nnn ππ
sinhsinh =⇒=
( )
[ ]
=
−+=
+−=
−=
=
=
∫
∫
oddisnifn
evenisnifn
nn
b
b
b
nb
yn
b
dyb
yn
b
dyb
ynyf
bb
n
b
b
b
n
π
π
ππ
π
π
π
π
400
0
)1(1200
1cos200
cos200
sin1002
sin)(2
0
0
0
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St. Joseph’s College of Engineering & Page No. 39 ISO 9001:2008 St. Joseph’s Institute of Technology
∴ππ nn
Bn sinh
400= if n is odd
then the equation (6) becomes
b
yn
b
xn
b
ann
yxun
πππ
πsinsinh
sinh
400),(
5,3,11 ∑
∞
=
=
To find ),(2 yxu
Consider the boundary conditions
2 2 2
2
( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 0
( ) ( , ) 100
ix u y x u x xi u a y
xii u x b
= = =
=
and the suitable solution in this case is
))(sincos(),(2pypy HeGepxFpxEyxu −++= (7)
Apply the boundary condition )(ix in (7)
0)(),0(2 =+= − pypy HeFeEyu
here 0=E
then the equation (8) becomes
)(sin),(2pypy GeGepxFyxu −+= (8)
Apply the boundary condition )(x in (8)
0)(sin)0,(2 =+= HGpxFxu
here GHHG −=⇒=+ 0
then the equation (9) becomes
)(sin
)(sin),(2
pypy
pypy
eepxFG
GeGepxFyxu−
−
−=
−= (9)
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Apply the boundary condition )(xi in (9)
0)(sin),(2 =−= − pypy eepaFGyau
here 0sin =pa πnpa =⇒a
np
π=⇒
then the equation (10) becomes
a
xn
a
ynC
a
xn
a
ynFG
eea
xnFGyxu
n
a
yn
a
yn
ππ
ππ
π ππ
sinsinh
sinsinh2
)(sin),(2
=
=
−=
where FGBn 2=
The most general solution is
a
xn
a
ynCyxu
nn
ππsinsinh),(
12 ∑
∞
=
= (11)
Apply the boundary condition )(xii in (11)
)(100sinsinh),(1
2 xfa
xn
a
bnCbxu
nn ===∑
∞
=
ππ
To find nC , expand )(xf in half range sine series
We know that half range Fourier sine series of )(xf is
∑∞
=
=1
sin)(n
n a
xndxf
π (12)
From the equations (11) & (12) we get
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St. Joseph’s College of Engineering & Page No. 41 ISO 9001:2008 St. Joseph’s Institute of Technology
a
bnd
Ca
bnCd n
nnn ππ
sinhsinh =⇒=
a
a
a
n
a
na
xn
a
dxa
xn
a
dxa
xnxf
ad
0
0
0
cos200
sin1002
sin)(2
−=
=
=
∫
∫
π
π
π
π
( )
[ ]
=
−+=
+−=
oddisnifn
evenisnif
n
nn
a
a
n
π
π
ππ
400
0
)1(1200
1cos200
∴
a
bnn
Cn ππ sinh
400= if n is odd
then the equation (10) becomes
a
xn
a
yn
nnyxu
n
ππππ
sinsinhsinh
400),(
5,3,12 ∑
∞
=
=
Then the equation (A) becomes
a
xn
a
yn
nnb
yn
b
xn
nn
yxuyxuyxu
nn
ππππ
ππππ
sinsinhsinh
400sinsinh
sinh
400
),(),(),(
5,3,15,3,1
21
∑∑∞
=
∞
=
+=
+=
5. A rectangular plate with insulated surface is 10cm wide and so long compared to its width that it
may be considered infinite in length without introducing appreciable error. The temperature at
short edge 0=y is given by 20 , 0 5
20(10 ), 5 10
x xu
x x
≤ ≤=
− ≤ ≤ and all the other three edges are kept
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St. Joseph’s College of Engineering & Page No. 42 ISO 9001:2008 St. Joseph’s Institute of Technology
at C�0 . Find the steady state temperature at any point in the plate.
Solution:
We know that two dimensional heat equation 02
2
2
2
=∂
∂+
∂
∂
y
u
x
u
Given the boundary conditions are
≤≤−
≤≤=
=∞==
105),10(20
50,20)0,()(
0),()(0),10()(0),0()(
xx
xxxuiv
xuiiiyuiiyui
The suitable solution which satisfying the given boundary conditions is
))(sincos(),( pypy DeCepxBpxAyxu −++= (1)
Apply the boundary condition (i) in (1)
( ) 0),0( =+= − pypy DeCeAyu
here 0=A
Then the equation (1) becomes
( )pypy DeCepxByxu −+= sin),( (2)
Apply the boundary condition (ii) in (2)
( ) 010sin),10( =+= − pypy DeCepByu
here 010sin =p πnp =⇒ 1010
πnp =⇒
Then the equation (2) becomes
+=
−
1010sin),(ynyn
DeCea
xnByxu
πππ (3)
Apply the boundary condition (iii) in (3)
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( ) 010
sin),( =+=∞ ∞−∞ DeCexn
Bxuπ
here 0=C
Then the equation (3) becomes
10
10
sin
sin),(
yn
n
yn
ea
xnB
ea
xnByxu
π
π
π
π
−
−
=
=
The most general solution is
10
1 10sin),(
yn
nn e
xnByxu
ππ −∞
=∑= (4)
Apply the boundary condition (iv) in (4)
∑∞
=
==1
1.10
sin)0,(n
n
xnBxu
π
≤≤−
≤≤
105),10(20
50,20
xx
xx
To find nB , expand )(xf in Half range Fourier sine series.
We know that Half range Fourier Cosine series
∑∞
=
=1 10
sin)(n
n
xnbxf
π (5)
From (4) and (5) we get nn bB =
−+=
=
∫∫
∫10
5
5
0
10
0
10sin)10(20
10sin20
5
1
10sin)(
10
2
dxxn
xdxxn
x
dxxn
yfbn
ππ
π
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−
−−
−−+
−
−
−=
10
5
22
5
0
22
100
10sin
)1(
10
10cos
)10(
100
10sin
)1(
10
10cos
5
20
π
π
π
π
π
π
π
π
n
xn
n
xn
xn
xn
n
xn
x
2sin
800
2sin
100
2cos
50
2sin
100
2cos
504
22
2222
ππ
ππ
ππ
ππ
ππ
n
n
n
n
n
n
n
n
n
nbn
=
+++−
=
Then the equation (4) becomes
∑∞
=
−
=1
1022 10
sin2
sin800
),(n
yn xne
n
nyxu
πππ
π
UNIT IV – FOURIER TRANSFORMS PART - A
1. Write the Fourier transform pair. [Nov/Dec 2011, 2010] The Fourier transform pair is defined as
( )[ ] ( ) ( )sFdxxfisxexfF =∫∞
∞−=
π2
1 : ( ) ( ) ( )11
2isxF F f x e F s ds f x
π
∞− − = = ∫ −∞
2. State Fourier integral theorem. [Nov/Dec 2011] If ( )xf is piecewise continuous, continuously differentiable and absolutely integrable in ( )∞∞− , then
( ) ( ) ( ) dtdstxisetfxf −∫∞
∞−∫∞
∞−=
π21
3. Find the Fourier transform of
>
<=
2xfor;0
2xfor;1f(x)
The Fourier transform of the function )(xf is
[ ] ∫∞
∞−= dxisxexfxfF )(
2
1)(
π.
[ ]2 2 221 1 1
( ) 1.2 2 22 2
isx i s i se e eisxF f x e dxis is isπ π π
− ∴ = = = −∫ − −
2 21 1 1 2 sin 2
(2 sin 2 ) ( )2 2
i s i se e si s F s
is is sππ π
− − = = ∴ =
4. Find )(xf from the integral equation sf(x)cos x e
0s
∞ −=∫
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St. Joseph’s College of Engineering & Page No. 45 ISO 9001:2008 St. Joseph’s Institute of Technology
( ) cos0
sgiventhat f x sxdx e∞ −=∫ ,
20
2 2cos
0
2 2 2 2 1( ) ( ) cos cos
10
sc
s se sx dx e
f x F s sx ds e sx dsx
π π
π π π π
∞−
∞ − −=∫
∞ = = =∫ + ∫
5. Define Self-reciprocal Fourier transform and give an example. [ Nov / Dec 2013]
If ( )f s is the Fourier transform of( )xf , then ( )xf is said to be self-reciprocal under Fourier
transform.2 2
2 2x s
F e e − − =
.
6. Find the Fourier transform of |x| ae− , if a 0 [ Nov / Dec 2012 ]
[ ] 1 1( ) (cos sin )
2 2
a x a xisxF f x e e dx e sx i sx dxπ π
∞ ∞− −∴ = = +∫ ∫
−∞ −∞
1 1 2cos sin 2 cos
2 22 2 ( )0
a x a x aaxe sxdx i e sxdx e sxdxs aππ π
∞ ∞ ∞ − − − = + = =∫ ∫ ∫ + −∞ −∞
7. State Convolution theorem in Fourier Transform. [Nov / Dec 2012 ] The Fourier transform of the convolution of ( )xf and ( )xg is the product of their Fourier transforms
i.e. ( ) ( ){ } ( ){ } ( ){ }xgFxfFxgxfF =∗ 8. State Parseval’s identity on Fourier Transform. [Nov/Dec 2011 ]
)()]([,2)(2)( sFxfFwheredxxfdssF =∫∞
∞−∫∞
∞−=
9. Find ( ){ }xxfF from ( )( ) 2 2
2 aF f x
π s a = +
.
By property [ ] ( )[ ]xfFds
dixxfF −=)( [ ] [ ]xaxa eF
ds
dixeF −− −=∴
( ) ( )2 22 2 2 2 2 2
2 2 .2 2 2d a a s asi i ids s a s a s aπ π π
− = − = = − == + + +
10. Find the Fourier sine transform of
a xe
x
− where 0>a .
[ ]
0
2( ) ( )sinSF f x f x sxdx
π
∞
= ∫ .
0
2sin
ax ax
S
e eF sxdx
x xπ
∞− − =
∫
Differentiate both sides with respect tos ,
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St. Joseph’s College of Engineering & Page No. 46 ISO 9001:2008 St. Joseph’s Institute of Technology
2 2sin sin
0 0
2 2 2cos cos
2 20 0
ax ax axd e d e eF sxdx sxdxsds x ds x x s
axe aaxx sxdx e sxdxx s a
π π
π π π
− − − ∞ ∞ ∂ = =∫ ∫
∂ −∞ ∞ −= = =∫ ∫
+
Now integrating both sides w. r. tos ,
=+
=
−−
∫ a
sds
as
a
x
eF
ax
S1
22tan
22
ππ
11. Find the Fourier sine transform of axe− , 0>a . Hence find
− xaxesF . [ May / June 2012 ]
[ ] ∫∞
=0
sin)(2
)( sxdxxfxfsFπ
.
22
2
0sin
2
as
ssxdxaxeaxe
sF
+π=∫
∞ −π
=
−
We know that by property ( )[ ] ( )[ ]xfFds
dxxfF CS −=
+
−=
+
−+−=
+−=
−∴
222
222222
)2.(22222
2
as
as
as
ssas
as
s
ds
daxxesFπππ
12. Write the Fourier sine transform pair and Fourier Cosine transform pair. The Fourier sine transform pair is defined as
( )[ ] ( )sFsxdxxfxfF ss =π
= ∫∞
0
sin)(2
: [ ] ( ) ( )21 sin0
F F s f x sx F s dss sπ
∞− = = ∫
The Fourier cosine transform pair is defined as
( )[ ] ( )sFsxdxxfxfF cc =π
= ∫∞
0
cos)(2
: [ ] ( ) ( )1
0
2cosc cF F s f x sx F s ds
π
∞− = = ∫
13. Prove that ( )( )c c
1 sF f ax F , 0
a aa = ≠
( )[ ] ∫∞
π=
0
cos)(2
sxdxaxfaxfFc , a
dtdxdtadxtaxPut === ,,
∞=∞=== txandtxwhen ,0,0 ( )0
2( ) cosc
st dtF f ax f t
a aπ
∞ = ∫
( )0
1 2( ) cosc
sF f ax f t t dt
a aπ
∞ = ∫ =
a
sF
a c1
.
14. State Parseval’s identity in Fourier sine and cosine Transform.
∫∫∞∞
=0
2
0
2 )()( dxxfdssFc
& ∫∫∞∞
=0
2
0
2 )()( dxxfdssFs
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15. Find the Fourier Sine transform of
x1
. [Nov/Dec 2011, Nov / Dec 2009]
0 0
1 2 sin 2 sin 2
2 2ssx t
F dx dtx x t
π ππ π π
∞ ∞ = = = = ∫ ∫
16. If )())(()())(( sFxfFandsFxfF sscc == , prove that [ ]1( ( )sin ) ( ) ( )
2c s sF f x ax F a s F a s= + + −
[Nov/Dec 2011, MA1201] Solution:
0 0
0 0
2 1 2( ( )sin ) ( ) sin cos ( )[sin( ) sin( ) ]
2
1 2 2( )[sin( ) ] ( )[sin( ) ]
2
cF f x ax f x ax sx dx f x s a x a s x dx
f x a s x dx f x a s x dx
π π
π π
∞ ∞
∞ ∞
= = + + −
= + + −
∫ ∫
∫ ∫
[ ]1( ( )sin ) ( ) ( )
2c s sF f x ax F a s F a s= + + − .
17. State and prove the change of scale property of Fourier Transform. [Apr/May 2011,May/June 2013] Statement:
If F [f(x)] = F(s) then 1
( ( )) , 0s
F f ax F aa a
= ≠
Proof:
( )[ ] ∫∞
∞−
= dxeaxfaxfF isx)(2
1
π ,
If a > 0 a
dtdxdtadxtaxPut === ,,
∞=⇒∞=−∞=⇒−∞= txandtxwhen ( ) 1( )
2
si t
a dtF f ax f t e
aπ
∞
−∞
⇒ = ∫ =
a
sF
a
1. –(1)
If a < 0 a
dtdxdtadxtaxPut === ,,
when x t and x t= −∞ ⇒ = ∞ = ∞⇒ = −∞
( ) 1 1( ) ( )
2 2
s si t i t
a adt dtF f ax f t e f t e
a aπ π
−∞ ∞
∞ −∞
−⇒ = = ∫ ∫ =
a
sF
a
1. ---(2)
From (1) & (2) we get 1
( ( )) , 0s
F f ax F aa a
= ≠
18. Find the Fourier cosine transform of
<<−
=otherwise,0
1x0;x1f(x)
2.
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[ ] ∫∞
π=
0
cos)(2
)( sxdxxfxfFC
[ ]
−
π=
+
−π
=
−−+
−−−−
π=
−π
=∴ ∫
332
1
032
2
1
0
2
cossin22
sin2cos22
sin)2(
cos)2(
sin)1(
2
cos)1(2
)(
s
sss
s
s
s
s
s
sx
s
sxx
s
sxx
sxdxxxfFc
19. Find the Fourier Sine Transform of 0,)( >= − xexf x [Apr/May2011, MA1201] We know that,
[ ] 2( ) ( )sin
0F f x f x sxdxs π
∞= ∫ .
( ) 2
0
2 2 2sin sin cos
2 110
xe sx xF e e sxdx sx s sxs ssπ π π
∞−∞ − − = = − − =∫ + +
20 If F(s) is the Fourier transform of f (x), then show that )()}({ sFeaxfF ias=−
[ Nov/Dec 2013, 2010 ,May/June 2012]
F {f(x-a)}= ∫∞
∞−
− dxeaxf isx)(2
1
π= ∫
∞
∞−
+ dtetf stai )()(2
1
π where x - a = t
= 1
( )2
ias itse f t e dtπ
∞
−∞∫ = F(s) = iase F{f(x)}
PART - B
1 Show that the Fourier transforms of
>>
<−=
0;0
;)(
22
ax
axxaxf is 3
2 sinsa sacossa2
s−−−−
ππππ . Hence
deduce that4
cossin
03
π=
−∫∞
dtt
ttt. Using Parseval’s Identity, show that
15
cossin
0
2
3
π=
−∫∞
dtt
ttt.
The Fourier transform of the function )(xf is [ ] ∫
∞
∞−
= dxexfxfF isx)(2
1)(
π.
Given 22)( xaxf −= in axa <<− and 0 otherwise.
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[ ]
[ ]
−
π=
+−−
+−
π=
−−+
−−−
−π
=
+−π
=
−+−π
=
+−π
=−π
=∴
∫
∫ ∫
∫∫
− −
−−
3
32
032
22
0
22
2222
2222
cossin22)(
000sin2cos2
02
sin)2(
cos)2(
sin)(
2
0cos)(22
1
sin)(cos)(2
1
)sin)(cos(2
1)(
2
1)(
s
sasasasF
s
sa
s
saa
s
sx
s
sxx
s
sxxa
sxdxxa
sxdxxaisxdxxa
dxsxisxxadxexaxfF
a
a
a
a
a
a
a
a
a
a
isx
By inverse Fourier transform ∫∞
∞−
−= dsesFxf isx)(2
1)(
π
−
−
=
−
−
−
=
−
−
=
−
=
∫
∫ ∫
∫
∫
∞
∞
∞−
∞
∞−
∞
∞−
∞
∞−
−
03
33
3
3
0coscossin
22
)(
sincossin
coscossin2
)sin(coscossin2
cossin22
2
1)(
sxdss
sasasaxf
sxdss
sasasaisxds
s
sasasa
dssxisxs
sasasa
dses
sasasaxf isx
π
π
π
ππ
Put 1&0 == ax , we get
4
cossin.
cossin41
03
03
π
π
=
−
−
=
∫
∫∞
∞
dss
sssei
dss
sss
Replacesbyt , we get 4
cossin
03
π=
−∫∞
dtt
ttt.
By Parseval’s Identity,
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)()]([,2)(2)( sFxfFwheredxxfdssF =∫∞
∞−∫∞
∞−=
a
a
a
a
a
a
xxaxads
s
sasasa
dxxxaadss
sasasa
dxxxaadss
sasasa
dxxadss
sasasa
0
5324
2
30
0
42242
30
42242
3
2222
3
532
cossin8
)2(2cossin8
2
)2(cossin8
)(cossin2
2
+−=
−
+−=
−
×
+−=
−
−=
−
∫
∫∫
∫∫
∫∫
∞
∞
−
∞
∞−
−
∞
∞−
π
π
π
π
.15
dtt
tcosttsin
15
a
a
dt
a
t
tcosttsin
0t,0sandt,swhen,a
dtds,
a
tstasPut
15
ads
s
sacossasasin
15
a3a10a15
85
aa
3
2a
8ds
s
sacossasasin
2
30
5
2
3
30
52
30
555555
2
30
π=
−
⇒π
=
−
==∞=∞===⇒=
π=
−
+−π=
+−
π=
−
∫∫
∫
∫
∞∞
∞
∞
2 a) Find the Fourier sine transform of
x
e xa−
where 0>a . [Nov/Dec 2011, 181301]
∫∞ −
−
=∴
=
0
sin2
)(
))((
dssxx
exf
x
exfF
ax
ax
S
π
Differentiating both sides w.r.to ‘s’
( )22
022
00
2sincos
2
cos2
)(cos2
sa
asxssxa
sa
e
ds
df
dssxedsxsxx
e
ds
df
ax
axax
+⇒
+−
+=
==
∞−
∞−
∞ −
∫∫
ππ
ππ
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Integrating w.r.to ‘s’
π
=∴
=⇒==
+
π
⇒+
π
⇒
+π=
−
−−∫
a
sxf
ccfxPut
ca
sc
a
s
aa
sa
dsaxf
1
1122
tan2
)(
0)0(,0
tan2
tan1
.22
)(
b) Show that 2
2x
e−
is self-reciprocal with respect to the Fourier cosine Transform.
The Fourier cosine transform of the function )(xf is [ ] ∫
∞
=0
cos)(2
)( sxdxxfxfFC π.
∫∫∫∞
−∞
∞−
−∞
−−===
∴
0
2
2
2
2
0
2
2
2
2
)(ofR.P2
1cos
2
12cos
2dxeesxdxesxdxeeF isx
xxxx
Cπππ
∫∫∞
∞−
+−∞
∞−
−== dxedxee
isxx
isxx
2
2
2
2
2
1ofR.P
2
1ofR.P
ππ
∫∫∞
∞−
−
−−∞
∞−
−−
== dxedxe
isisxisx
x2
2
2
222
2
2
1ofR.P
2
1ofR.P
ππ
∫∞
∞−
−−
−
π= dxeeofR.P
isxs
22
22
2
1
Put y2
isx=
−, then dydx 2= , ∞=⇒∞=−∞=⇒−∞= yxandyx
( )[ ] 2
2
2
2
22
22
2
2
1ofR.P
1ofR.P2
2
1ofR.P
s
C
s
ys
ys
exfFe
dyeedyee
−−
∞
∞−
−−
∞
∞−
−−
=⇒=
== ∫∫
ππ
ππ
∴ 2
2x
e−
is self-reciprocal with respect to Fourier transform
3 a) Find the Fourier transform of
≥
<−=
1;0
1;1)(
xif
xifxxf . Hence Evaluate dt
t
t∫∞
04
4sin.
The Fourier transform of the function )(xf is [ ] ∫
∞
∞−
= dxexfxfF isx)(2
1)(
π.
Given xxf −= 1)( in 11 <<− x and 0 otherwise.
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[ ]
−
=
−−
−=
−−−−=
+−=
−+−=
+−=−=∴
∫∫ ∫
∫∫
− −
−−
2
22
1
02
1
0
1
1
1
1
1
1
1
1
cos12)(
10
cos0
2cos)1(
sin)1(
2
0cos)1(22
1sin)1(cos)1(
2
1
)sin)(cos1(2
1)1(
2
1)(
s
ssF
ss
s
s
sx
s
sxx
sxdxxsxdxxisxdxx
dxsxisxxdxexxfF isx
π
ππ
ππ
ππ
By inverse Fourier transform ∫∞
∞−
−= dsesFxf isx)(2
1)(
π
−
−
=
−
−
−
=
−
−
=
−
=
∫
∫ ∫
∫∫
∞
∞
∞−
∞
∞−
∞
∞−
∞
∞−
−
02
22
22
0coscos1
21
)(
sincos1
coscos11
)sin(coscos11cos12
2
1)(
sxdss
sxf
sxdss
sisxds
s
s
dssxisxs
sdse
s
sxf isx
π
π
πππ
Put 0=x , we get
( )1.
22sin2
.
cos121
02
2
02
π
π
=
−
=
∫
∫
∞
∞
dss
sei
dss
s
Let ,2
ts= then dtds 2= ∞=⇒∞==⇒= tsandts 00
( ) 2
sin
22
2
sin2
02
2
02
2 ππ=
⇒=
∫∫∞∞
dtt
tdt
t
t
By Parseval’s identity ∫∫∞
∞−
∞
∞−
= dxxfdssF22
)()(
( )
( )
( )
2 12
21
2 12
20
2 12
20
2 1 cossds 1 x dx
s
2 1 cossds 2 1 x dx
s
2 1 cossds 2 1 x dx
s
∞
−∞ −
∞
−∞
∞
−∞
− ∴ = − π
− = − π
− = − π
∫ ∫
∫ ∫
∫ ∫
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( )2 1
22
0 0
2 1 coss2 ds 2 1 x 2x dx
s
∞ − = + − π ∫ ∫
12 3 2
20 0
2 1 coss x 2xds x
s 3 2
∞ − = + − π ∫
[ ]( )
∫∫∞∞
=
⇒
−+−
−+=
−
0
2
2
2
0
2
2 32sin2
200013
11
cos12 ππ
dss
sds
s
s
Let ,2
ts= then dtds 2= ∞=⇒∞==⇒= tsandts 00
( ) 3
sin
3
sin
32
2
sin22
04
4
0
2
2
2
0
2
2
2 πππ=⇒=
⇒=
∫∫∫∞∞∞
dtt
tdt
t
tdt
t
t
b) Evaluate ∫
∞
++02222 )()( bxax
dx using Fourier Cosine Transforms of , , 0a x b xe and e a b− − > .
The Fourier cosine transform of the function )(xf is [ ] ∫
∞=
0C sxdxcos)x(f
2)x(fF
π.
[ ]
22
0
2
cos2
as
a
sxdxeeF axaxC
+=
=∴ ∫∞
−−
π
π and
[ ]
22
0
2
cos2
bs
b
sxdxeeF bxbxC
+=
=∴ ∫∞
−−
π
π
Let ( ) axexf −= and ( ) bxexg −=
By Convolution theorem, ( )[ ] ( )[ ] ( ) ( )∫∫∞∞
=00
dxxgxfdsxgFxfF cc
( )( )( )∫∫
∫∫∞
+−∞
∞−−
∞
=++π
=+π+π
∴
002222
002222
2..
22
dxedsasas
abei
dxeedsas
b
as
a
xba
bxax
( )( )( )
( )
( )( )( ) ( )
( )
+−−
=++
+−=
++
+−∞+−∫∞
∞+−∫∞
ba
ee
asas
dsab2
ba
e
asas
dsab2
0baba
02222
0
xba
02222
π
π
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( )( ) ( )
( )( ) ( )baab2asas
ds
ba
1
asas
dsab2
02222
02222
+=
++
+−−
=++
∫∞
∫∞
π
π
Replacesbyx , we get
( )( ) ( )baabaxax
dx
+=
++∫∞
202222
π
4 a) Find the Fourier transforms of )( xf defined by
>
<=
ax,0
ax,1)x(f and hence find the value
of ∫∞
0
dtt
tsin also prove that∫
∞ π=
02
2
2dt
t
tsin.
The Fourier transform of the function )(xf is [ ] ∫
∞
∞−= dxe)x(f
2
1)x(fF isx
π.
Given 1)( =xf in axa <<− and 0 otherwise.
[ ]
s
sasFasi
is
is
ee
is
e
is
e
is
edxexfF
iasiasiasiasa
a
isxa
a
isx
sin2)(sin2
1
2
1
2
1
2
1
2
1.1
2
1)(
π=⇒
π=
−
π=
−
π=
π=
π=∴
−−
−−∫
By inverse Fourier transform ∫∞
∞−
−= dsesFxf isx)(2
1)(
π
−=
−=
−==∴
∫∞
∫∞
∞−∫∞
∞−
∫∞
∞−∫∞
∞−
−
0sxdscoss
sasin2
1sxdssin
s
sasinisxdscos
s
sasin1
ds)sxsinisx(coss
sasin1dse
s
sasin2
2
1)x(f
0
isx
ππ
πππ
∫∞
=0
sxdscoss
sasin2)x(f
π
Put 1&0 == ax we get
∫∫∞∞
=⇒=00 2
sinsin21
ππ
dss
sds
s
s, Replace sby t, we get ∫
∞
=0 2
sin πdt
t
t.
By Parseval’s identity ∫∫∞
∞−
∞
∞−
= dxxfdssF22
)()(
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St. Joseph’s College of Engineering & Page No. 55 ISO 9001:2008 St. Joseph’s Institute of Technology
[ ]
[ ]2
sin2
sin2
2)1(1
sin2
sin21
sin2
0
2
0
22
11
21
1
2
2
πππ
ππ
∫∫∫
∫∫∫∞∞∞
∞−
−
∞
∞−−
∞
∞−
=
⇒=
⇒−−=
=⇒=∴
dss
sds
s
sds
s
s
xdss
sdxds
s
s
Replace sby t, we get
∫∫∞∞
=⇒=
02
2
0
2
2
sin
2
sin ππdt
t
tdt
t
t.
b) Find Fourier Sine and Cosine transforms of axe− and hence find the Fourier sine transforms of
22 ax
x
+ and Fourier cosine transforms of
22
1
ax +.
The Fourier sine transform of the function )(xf is [ ] ∫
∞
=0
sin)(2
)( sxdxxfxfFS π.
[ ]22
0
2sin
2
as
ssxdxeeF axax
s+π
=π
= ∫∞
−−
The Fourier cosine transform of the function )(xf is [ ] ∫∞
π=
0
cos)(2
)( sxdxxfxfFc .
[ ]22
0
2cos
2
as
asxdxeeF axax
c+π
=π
=∴ ∫∞
−−
asas
ascs
asas
c
eaa
ee
ads
d
axF
ds
d
ax
xF
a
ee
asxdx
axaxF
−−
−
−−
∞
π=−
π−⇒
π
π−=
+−=
+
π⇒
π
π⇒
+π=
+∴ ∫
2)(
22
21
.22
2cos
121
2222
02222
5
a) Find the function f(x) if its sine transform is
s
e as−
[ Nov / Dec 2013 ]
Let
s
e as−
=(f(x))Fs
Then f(x) = )1(sin2
0
−−−−−∫∞ −
dssxs
e as
π
Differentiating both side with respect to ‘x’ we get
∫∫∞
−∞ −
==00
cos2
)()(cos2
dssxedsssxs
e
dx
df asas
ππ
+
=
+−
+==
∞−
220
22
2)sincos(
2
ax
asxxsxa
xa
e
dx
df as
ππ
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Integrating w.r.t ‘ x ’
f(x) = ca
x
a
x
aa
ax
dxa +
=
=+
−−∫ 1122
tan2
tan122
πππ -------(2)
put x= 0, f(0) = 0 from (1) from (2) , we get f(0) = c =0,
∴
= −
a
xxf 1tan
2)(
π
b) Express
>
<=
1||,0
1||,1)(
x
xxf as Fourier integral, hence evaluate ∫
∞
0
cossinds
s
sxs and find the values of
∫∞
0
sinds
s
s
We know that the Fourier integral formula for f(x)= ∫ ∫
∞ ∞
∞−
−0
)(cos)(1
dsdtxtstfπ
f(x)=
1
100
1
1
)(sin1)(cos
1
−
∞∞
−∫∫ ∫
−=−
s
xtsdsdtxts
ππ
∫∫∞∞
=
++−=⇒
00
cossin2)1(sin)1(sin1)( ds
s
sxsds
s
xsxsxf
ππ…….(1)
This is fourier integral representation of f(x).
From (1)
>
<==⇒
=⇒
∫
∫∞
∞
1||,0
1||,2)(
2
cossin
)(cossin2
0
0
x
xxfds
s
sxs
xfdss
sxs
ππ
π
Put x= 0, 2
sin
0
π=∫
∞
dss
s
UNIT V Z – TRANSFORMS AND DIFFERENCE EQUATIONS
PART - A 1. What is the Z- transform of discrete unit step function
Solution: Discrete Unit step function is
( )1, 0
0, 0
nu n
n
≥=
<
( ) 1 21
0
11. 1 ..
1n
n
Z u n z z zz
∞− − −
−=
= = + + + = −∑ =1
z
z−if 1z >
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St. Joseph’s College of Engineering & Page No. 57 ISO 9001:2008 St. Joseph’s Institute of Technology
2. Find
1
2nZ
.
Solution: 1 2 32 3
0
1 1 1 1 11
2 2 2 2 2n
n nZ z z z z
∞− − − − = = + + + +
∑ …
2 3
1 1 11 ...
2 4 8z z z= + + + +
1
2
z
z=
−
2, 2
2 1
zz
z= >
−
3. Find ( )1Z u n−
Solution:
( ) ( ) ( )0 1
2 3 21
1 1 1
1 1 1 1 1 1.... 1 ....
n n
n n
n
n
Z u n u n z u n z
zz z z z z z
∞ ∞− −
= =
∞−
=
− = − = −
= = + + + = + + +
∑ ∑
∑
1 1
1 1 1 11
z
z z z z
− −− = − =
1
1
z
z z = −
1
1z=
− if 1z >
4. Find the Z- transform of unit impulse function
Solution: Unit impulse function is
( ) 1, 0
0, 0
n n
n
δ = =
= ≠
( ) ( ) 0
0
1n
n
Z n n z zδ δ∞
− −
=
= = = ∑ .
5. If ( ) ( )Z f n U z= then find ( )nZ a f n
Solution: ( ) ( )( ) ( )
0 0
( ) / /nn n n
n n
Z a f n a f n z f n a z U z a∞ ∞
−
= =
= = = ∑ ∑ .
6. If ( ) ( )Z f n U z= , then show that ( ) ( )kZ f n k z U z+ =
Solution:
( ) ( ) ( ) ( )
0 0
n kn kZ f n k f n k z z f n k z∞ ∞
− +− ++ = + = + ∑ ∑
( ) ( )0
k r kz f r z z U z∞
+ − += =∑
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St. Joseph’s College of Engineering & Page No. 58 ISO 9001:2008 St. Joseph’s Institute of Technology
7. If ( ) ( )z f n U z= , then
( ) ( )1f nZ z U z dz
n−
= −
∫ .
Solution:
( ) ( ) 1
0 0
( )n n
n n
f n f nZ z f n z dz
n n
∞ ∞− − −
= =
= = −
∑ ∑ ∫ , since 1
nnz
z dzn
−− −= −∫
( ) ( )( )1 1
0
n n
n
f n z dz z f n z dz∞
− − − −
=
= − = −∑ ∑∫ ∫ ( )1z U z dz−= −∫ .
8. If ( ) ( ) ,Z f n U z= then find ( )Z n f n
Solution:
( ) ( ) ( ) 1n nZ nf n nf n z z nf n z− − −= = − − ∑ ∑
= ( ) ( ) ( ) ( )n nd d dz f n z z f n z z U z
dz dz dz− −− = − = −∑ ∑ .
9. Find 3nZ a + .
Solution:
( )3 3 3
0
n n n n
n
Z a a z a Z a∞
+ + −
=
= = ∑ = 3 za
z a−
10. Find
( )( )1
1 2
zZ
z z−
− −
Solution:
Let ( )( )
( )1 2
zU z
z z
=
− −
Then ( )
( )( )1
1 2 1 2
U z A B
z z z z z= = +
− − − −
2, 1
1, 1
z B
z A
= =
= = −
( ) 1 1
1 2
U z
z z z
−∴ = +
− −
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St. Joseph’s College of Engineering & Page No. 59 ISO 9001:2008 St. Joseph’s Institute of Technology
( ) ( ) 1 2 2 11 2
n nz zU z u n
z z
−∴ = + ⇒∴ = − + = −
− −
11. Find ( )nZ .
Solution:
{ } ....31
321
21
11
⋅+
⋅+⋅=−∑∞
==
zzznz
nnnZ
( )
221 2 1 1 1 11 3 1 .
2 21 1
z z
z z z z z zz z
− = + + ⋅ + ⋅⋅ ⋅⋅ = − = = − −
12. Find Z-Transform of 1
n.
Solution:
2 3
1
1 1 1 1 1 1 1....
2 3n
n
Z zn n z z z
∞−
=
= = + + +
∑
1 1
1 log log1
z zlog
z z z
− = − − = − = −
13. Find Z-Transform of na .
Solution:
{ } ∑∑∞
=
∞
=
−
−=
==00 n
n
n
nnn
az
z
z
azaaZ
14. Find Z-Transform of
1
!Z
n
.
Solution:
11 1 1 1
1 ....2! ! 1! 2!1
n zZ z en n z zn
∞ −= = + + =∑ =
15. Solve 1 2nn ny y+ − = , given that ( )0 1y = .
Solution:
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St. Joseph’s College of Engineering & Page No. 60 ISO 9001:2008 St. Joseph’s Institute of Technology
{ } { }1 2nn nZ y y Z+ − =
( ) ( ) ( )02
zzY z y Y z
z− − =
− , ( ) { }nY z Z y=
( )( )12
zY z z z
z− = +
− ⇒ ( ) ( )
( )( )2
2 1
z z zY z
z z
− +=
− − ⇒ ( ) ( )
( )( )1
2 1
z zY z
z z
−=
− −
( )1 2
2n
n
zy Z
z−
= = −
16. Using Convolution theorem , evaluate
−−−−−−−−−−−−
)3()1(
21
zzz
Z
Solution:
We know that ( )
( )1 11
zZ z
z−
= −
and ( )
1 33
nzZ
z−
= −
Now ( ) ( )
21 1 .
1 3 1 3
z z zZ Z
z z z z− − = − − − −
0 0
1 3 1.3 3 3n n
n n m n m
m m
− −
= =
= ∗ = =∑ ∑
( )( )
1
1 1 1 1
0
11 3 1 3 / 31 1 3 3 13
3 3 .13 1 3 / 3 2 2
13
n
n n nm n nnn n
m
+
+ + + +
=
− − − − = = = = = − − −
∑
17. Find { }2 sin 2tZ e t−
Solution:
{ } { }( ) sin 22 sin 2 sin 2 2 2 2 cos2 1 2
z TtZ e t Z t Tz ze z z T Tz ze
− = =→ − + →
2 sin 2
2 4 22 cos 2 1
Tze TT Tz e ze T
=− +
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18. Find { }sinZ at
Solution:
{ } ( )sin sin0
nZ at anT zn
∞ −= ∑=
( )sinaT
Z nθ
θ=
= sin sin
2 22 cos 1 2 cos 1aT
z z aT
z z z z aTθ
θ
θ=
= = − + − +
19. State initial and final value theorem of Z - transform.
Solution: Initial value Theorem
( ) ( ) , 0If Z f n U z n= ≥ then ( ) ( ) ( )0n 0
Limit Limitf n f U z
z= =
→ →∞
Final value Theorem
( ) ( ) , 0If Z f n U z n= ≥ then ( ) ( ) ( )11
Limit Limitf n z U z
n z= −
→∞ →
20. Form the difference equation from .2 .3n nny A B= +
Solution:
Given .2 .3n nny A B= + , 1 2 .2 3 .3n n
ny A B+ = + , 2 4 .2 9 .3n nny A B+ = +
Elimination of A and B forms the difference equation
1
2
1 1
2 3 0
4 9
n
n
n
y
y
y+
+
=
2 15 6 0n n ny y y+ +− + = .
PART - B
1. a) Solve ( ) ( )2 15 6 5 , 0 0, 1 0nn n ny y y y y+ +− + = = = using Z-transform.
Solution:
2 15 6 5nn n ny y y+ +− + = .Taking Z-transform on both sides
( )( ) ( )( ) ( )2 10 0 5 0 65
zz Y z z z Y z Y z
z−− − × − − + =
−
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St. Joseph’s College of Engineering & Page No. 62 ISO 9001:2008 St. Joseph’s Institute of Technology
( )2 5 65
zz z Y z
z − + = −( )
( ) ( ) ( )( )( )2
1 1
5 2 3 5 2 35 5 6
Y z A B C
z z z z z z zz z z= = = + +
− − − − − −− − +
( )( ) ( )( ) ( )( )1 2 3 5 3 5 2A z z B z z C z z= − − + − − + − −
15, 1 6
6z A= = ⇒ =
3, 1 2 1/ 2z C C= = − ⇒ = −
2, 1 3 1/ 3z B B= = ⇒ =
( ) 1 1 1 1 1 1
6 5 3 2 2 3
Y z
z z z z∴ = + −
− − −
( ) 1 1 1
6 5 3 2 2 3
z z zY z
z z z= + −
− − −
1 1 15 2 3 .
6 3 2n n n
ny∴ = + −
b) State and Prove Second Shifting Property of Z-Transform.
Statement:
Let ( ) ( )nZ u U z= and 0k > then ( ) ( ) 1 2 ( 1)0 1 2 1...k k
n k kZ u Z U z u u z u z u z− − − −+ − = − − − − −
Proof
( ) ( )
0 0
n kn kn k n k n k
n n
Z u u z z u z∞ ∞
− +−+ + +
= =
= =∑ ∑
1
0 0
kk n n
n nn n
z u z u z∞ −
− −
= =
= −
∑ ∑
( ) 1 2 ( 1)0 1 2 1...k k
kz U z u u z u z u z− − − −− = − − − − −
2. a) Solve ( ) ( ) ( )3 3 1 2 0,y n y n y n+ − + + = given that ( ) ( ) ( )0 4, 1 0, 2 8y y y= = = Using Z-transform.
Solution:
( ) ( ) ( )3 3 1 2 0y n y n y n+ − + + = . Taking z-transform on both sides
[ ] ( ) ( )3 13 2 0n n nZ y Z y Z y+ +− + =
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St. Joseph’s College of Engineering & Page No. 63 ISO 9001:2008 St. Joseph’s Institute of Technology
( ) ( ) ( )3 1 20 1 2 03 2 0z y z y y z y z z y z y Y z− − − − − − − + =
( )( ) ( )( ) ( )3 24 0 8 3 4 2 0z Y z z z Y z Y z−− − − − − + =
( ) 3 3 33 2 8 4 12 4 4Y z z z z z z z z − + = + − = −
( )( ) ( )
2 2
23
4 4 4 4
3 2 1 2
Y z z z
z z z z z
− −= =
− + − +
( ) ( )23 3 2 1 2z z z z∴ − + = − +
( )( ) ( ) ( )
2
2 2
4 4
1 21 2 1
Y z z A B C
z z zz z z
−∴ = = + +
− +− + −
( )( ) ( ) ( )224 4 1 2 2 1z A z z B z C z∴ − = − + + + + −
1, 0 3 0z B B= = ⇒ =
2 12 9 12 / 9 4 / 3z c c= − = ⇒ = =
2, 4 4 4 / 3 8 / 3z A C A= + ⇒ = − =
( ) 8 1 4 10
3 1 3 2
Y z
z z z∴ = + + +
− +
( ) 8 1 4 1
3 1 3 2Y z
z z∴ = + +
− +
( )8 42 .
3 3n
ny = + −
b) If ( )
( )
2
4
2 3 12,
1
z zU z
z
+ +=
−find 2u and 3u .
Solution:
( )( )
2
0 4
lim lim 2 3 120
1
it it z zu U z
z z z
+ += = =
→∞ →∞ −
( )( )
2
1 0 4
lim lim 2 3 120 0
1
it it z zu z U z u
z z z
+ + = − = − = →∞ →∞ −
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( )2 12 0 1
lim itu z U z u u z
z− = − − →∞
( )
22
4
lim 2 3 120 0 2
1
it z zz
z z
+ += − − =
→ ∞ −
( )3 1 23 0 1 2
lim itu z U z u u z u z
z− − = − − − →∞
( )
23
4 2
lim 2 3 12 20 0
1
it z zz
z zz
+ += − − −
→ ∞ −
( )( )
4 2 4 3 2
342
2 3 12 2 4 6 4 1lim
1
z z z z z z zitz
z z z
+ + − − + − + =
→∞ −
( )3
4
lim11 8 2 11.
1
it zz z
z z = + − = →∞ −
3. a) Find
−−−
)14)(12(
8 21
zz
zZ by convolution theorem.
Solution:
2 21 1
1 12 4
8
(2 1)(4 1) ( )( )
z zZ Z
z z z z− −
= − − − −
1 1
1 12 4
.( ) ( )
z zZ Z by C T
z z− −
= ∗ − −
1 1
2 4
n n = ∗
0 0
1 1 1 1 1
2 4 2 2 4
n r r n r rn n
r r
− −
= =
= =
∑ ∑
2
0 0
11 1 1 1 1 1 14 1 ....
12 2 2 2 2 2 22
r
n n r n nn n
r r= =
= = = + + + +
∑ ∑
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St. Joseph’s College of Engineering & Page No. 65 ISO 9001:2008 St. Joseph’s Institute of Technology
11
11 1 1 12
2 112 2 2 2
12
n
n n n
+ − = = − −
1 1 1 1 1 12 2
2 2 2 2 2 4
n n n n n = − = −
b) Find ( )( )1 2Z n n n − −
Solution:
( )( )1 2Z n n n − − { }3 23 2Z n n n= − +
{ } { } { }3 23 2Z n Z n Z n= − +
( )( )
( ) ( )
3 2
4 3 2
3 14 2
1 1 1
z zz z z z
z z z
++ += − +
− − −
( )( ) ( )( ) ( )
23 2
4 4
4 3 1 1 2 1 6
1 1
z z z z z z z z z
z z
+ + − + − + −= =
− −
4. a) Find ( )cosnZ r nθ and hence deduce
2
nZ cos
π
Solution:
( ) ( )nin ii
zZ e Z e
z eθ θ
θ− −
− ∴ = = −
( ) ( )( )( )
i
ini i i
z z ezZ e
z e z e z e
θθ
θ θ θ−
− −
−∴ = =
− − −
( )( )
( )( )22
cos sin cos sin
2 cos 11i i
z z i z z iz
z zz z e eθ θ
θ θ θ θ
θ−
− + − − = =− +− + +
( ) ( )( ) ( )2 2
cos sincos sin
2 cos 1 2 cos 1
z z zz n i n i
z z z z
θ θθ θ
θ θ
−∴ + = −
− + − +
equating real and imaginary parts,
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( ) ( )2
coscos
2 cos 1
z zZ n
z z
θθ
θ−
=− +
{ } { } ( )
2 2
coscos cos
2 cosn
zz
r
z z rZ r n Z n
z rz r
θθ θ
θ→
− = = − +
{ }[ ]11cos2
)cos(
1cos2
)cos(cos
2cos
2
2
22
2
2
22 +=
+−
−=
+θ−
θ−=θ=
π
π
π
π=θ
π=θ
z
z
zz
zz
zz
zznZ
nZ
b) Find the inverse Z-Transform of
( )( )3
1
1
z z
z
+
− by residue method.
Solution:
F (z) = 3
2
)1( −
+
z
zz
∴ Zn-1F(z) = 3
1
)1( −
++
z
zz nn
Here z = 1 is a pole of order 3
[ ] ( )
−
+×−=
+
→= 3
13
2
2
11
)1(1
!2
1
z
zzz
dz
dLtRES
nn
zz
= 2
2
1 !2
1
dz
dLt
z→[ ]nn zz ++1
dz
dLtz !2
11→
= [ ]1)1( −++ nn nzzn
12
1→
=zLt [ ]21 )1()1( −− −++ nn znnznn
2
1= [ ])1()1( −++ nnnn
2
1= [ ]nnnn −++ 22( = 2n
5. a) Form the difference equation corresponding to the family of curves 2xxy ax b= + .
Solution:
2 (1)xxy ax b= + −
( ) 11 1 2x
xy a x b ++ = + +
( )12 2 2 (2)x x xy a b a b+∆ = + − = + −
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( ) ( )2 12 2 2 (3)x x xy a b a b b+∆ = + − + = −
2
(3),2x
yfrom b
∆=
2
(2) 22
x
x
ysub in y a
∆∆ = +
2a y y∴ = ∆ −∆
(1)sub in ( )2
2 2 .2
x xx x
yy y y
∆= ∆ −∆ +
( ) ( ) ( ) ( )22 1 11 1 2 0x x x x x xx y x y or x y y y x y y y+ + += − ∆ + ∆ − − + + − − =
( ) ( )2 11 3 2 2 0x x xie x y x y xy+ +− − − + =
b) Find
( ) ( )1
21 1
zZ
z z−
+ − using the method of partial fraction.
Solution:
Let ( )( )( )2
1 1
zF z
z z=
+ −
( )( )( ) ( ) ( ) ( )2 2
1
1 11 1 1
F z A B C
z z zz z z= = + +
+ −+ − −
( ) ( ) ( ) ( )( )( )
2
2
1 1 1 1
1 1
A z B z z C z
z z
− + − + + +=
+ −
( ) ( )( ) ( )21 1 1 1 1A z B z z C z= − + − + + +
Putting 1z= and 1z= − we get 1
4A= and
1
2c =
Equating the coefficient of on both side
0=A+B
1
4B A= − = −
( )( ) ( ) ( )2
1 1 14 4 2
1 1 1
F z
z z z z= − +
+ − −
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