MA6351-TRANSFORMS AND PARTIAL
DIFFERENTIAL EQUATIONS
SUBJECT NOTES
Department of Mathematics
FATIMA MICHAEL
COLLEGE OF ENGINEERING &
TECHNOLOGY
MADURAI – 625 020, Tamilnadu, India
Basic Formulae
DIFFERENTIATION &INTEGRATION FORMULAE
0
Function
y
Differentiation
f ( x)
dy
dx
1
xn
nx n
2
log x
1
x
3
sin x
cos x
4
cos x
5
e ax
a ex
6
C (constant)
0
7
tan x
sec2 x
8
sec x
sec x tan x
9
cot x
cos ec 2 x
10
cos ecx
cos ecx cot x
11
x
1
sin x
1
2 x
12
sin 1 x
1
1 x2
13
cos 1 x
1
1 x2
14
15
tan 1 x
sec 1 x
1
1 x2
1
x x2 1
16
cos ec 1 x
1
x x2 1
cot 1 x
1
1 x2
18
ax
a x log a
Page
2
17
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dy
dx
v
du
dv
u
dx
dx
20. If y
u
dy
, then
v
dx
3
uv , then
du
dv
u
dx
dx
v2
Page
19. If y
v
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2. e x dx
ex ,
3. sin xdx
e ax
a
e ax dx
& e
cos x & sin axdx
4. cos xdx sin x &
cos axdx
5. tan xdx
log sec x
log cos x
6. sec 2 xdx
tan x
7.
dx
dx
x a2
1
x
tan 1
a
a
8.
dx
dx
x a2
1
x a
log
2a
x a
9.
10.
11.
2
2
dx
a2
x2
dx sin
dx
a2
x2
dx
x2 a2
1
ax
dx
cos ax
a
sin ax
a
x
a
dx sinh
1
dx
cosh
1
x
a
x
a
12.
a2
x 2 dx
x 2
a
2
x2
a2
x
sin 1
2
a
13.
a2
x 2 dx
x 2
a
2
x2
a2
x
sinh 1
2
a
14.
x 2 a 2 dx
x 2
x a2
2
a2
x
cosh 1
2
a
15.
dx
x
16.
2 xdx
x2 a2
4
log x
log x 2
e ax
a
a2
Page
1. x dx
xn 1
n 1
n
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17. log xdx
x log x x
a x
3
2
18. a x dx
a x
3
2
19. a x dx
3
3
1
dx 2 x
x
20.
eax cos bxdx
21.
eax
a cos bx b sin bx
a 2 b2
eax
a sin bx b cos bx
a 2 b2
ax
22. e sin bxdx
uv u´v1 u´´v2 u´´´v3........
23. udv
a
a
24.
f ( x)dx
a
2 f ( x)dx when f(x) is even
0
a
25.
f ( x)dx 0 when f(x) is odd
a
26.
e
ax
cos bxdx
0
27.
e
ax
sin bxdx
0
a
a
2
b2
b
a
2
b2
TRIGNOMETRY FORMULA
1. sin 2 A
2sin A cos A
2.cos 2 A cos 2 A sin 2 A
1 2sin 2 A
Page
5
2 cos 2 A 1
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3. cos 2 x
1 cos 2 x
1 cos 2 x
& sin 2 x
2
2
4. sin( A B) sin A cos B cos A sin B
sin( A B ) sin A cos B cos A sin B
cos( A B ) cos A cos B sin A sin B
cos( A B ) cos A cos B sin A sin B
5.sin A cos B
1
sin( A B ) sin( A B)
2
cos A sin B
1
sin( A B ) sin( A B)
2
cos A cos B
1
cos( A B ) cos( A B )
2
sin A sin B
1
cos( A B) cos( A B)
2
1
3sin A sin 3 A
4
1
cos3 A
3cos A cos 3 A
4
6. sin 3 A
A
A
cos
2
2
A
A
cos A cos 2
sin 2
2
2
A
A
1 2sin 2
1 cos A 2sin 2
2
2
7.sin A 2sin
Page
6
LOGRATHEMIC FORMULA
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log mn
log
log m log n
m
n
log m log n
log m n
n log m
log a 1 0
log a 0
log a a 1
elog x
x
UNIT - 1
PARTIAL DIFFRENTIAL EQUATIONS
PARTIAL DIFFERENTIAL EQUTIONS
Notations
z
x
z
y
p
2
q
x
z
2
2
r
z
x y
2
s
z
y2
t
Formation PDE by Eliminating arbitrary functions
Suppose we are given f(u,v) = 0
Then it can be written as u = g(v) or v = g(u)
LAGRANGE’S LINEAR EQUATION
(Method of Multipliers)
General form
Pp + Qq = R
Subsidiary Equation
dz
R
7
dy
Q
Page
dx
P
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dx
P
dy
Q
dz
R
x my nz
P mQ nR
Where ( , m ,n) are the Lagrangian Multipliers
Choose , m, n such that P + mQ + nR = 0
Then dx + m dy + n dz = 0
On Integration we get a solution u = a
Similarly, We can find another solution v = a for another multiplier
The solution is (u, v) = 0
TYPE –2 (Clairut’s form)
General form
Z = px + qy + f(p,q)
(1)
Complete integral
Put p = a & q = b in (1), We get (2) Which is the Complete integral
Singular Integral
Diff (2) Partially w.r.t a We get (3)
Diff (2) Partially w.r.t b We get (4)
Using (3) & (4) Find a & b and sub in (2) we get Singular Integral
REDUCIBLE FORM
F(xm p ,ynq) = 0 (1) (or)
(1)
If m 1& n 1 then
X = x1-m & Y = y1-n
xm p = P(1-m) & yn q = Q(1-n)
If k
1 then Z = zk+1
Q
zk q
k 1
Using the above in (1) We get
F(P,Q) = 0 (or) F(P,Q,z) = 0
F(P,Q) = 0
8
Using the above in (1)we get
Page
F( xmp, ynq, z)=0
F( zkp, zkq)=0
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Using the above in (1) we get
F(P,Q) (or) F(P, Q, z) = 0
p
z
P &
q
z
Q
Using the above in (1)
we get
F(P,Q) = 0
9
xp = P & yq = Q
If k =-1 then Z = log z
Page
If m=1 & n=1 then
X= logx & Y= logy
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STANDARD TYPES
TYPE –3(a)
TYPE –3(b)
TYPE –3(c)
General form
F(z,p,q) = 0
(1)
Complete
Integral
General form
F(x,y,p,q) = 0
(1)
Complete
Integral
(1) Can be written
as,
f(x,p) =f(y,q) = a
Then, find p and q
sub in
dz = p dx + qD y
Integrating,
We get (2) which
is the
Complete integral
Singular Integral
Diff (2) partially
w.r.t cWe get,0
=1 (absurd
General form
F(p,q) = 0 (1)
General form
F(x,p,q) = 0 (1)
Complete
Integral
Put p = a and q =
b in (1)
Find b in terms
of a
Then sub b in
z = ax + by + c
we get (2)
which is the
Complete
Integral
Complete
Integral
General form
F(y,p,q) = 0
(1)
Complete
Integral
Put q = a in (1)
Then, find p and
sub in
dz = p dx + q dy
Integrating ,
We get (2)
which is the
Complete
integral
Put p = a in (1)
Then, find q and
sub in dz = p dx
+ q dy
Integrating ,
We get (2)
which is the
Complete
integral
Put q = ap in (1)
Then, find p and
sub in
dz = p dx + q dy
Integrating,
We get (2) which
is the
Complete
integral
Singular
Integral
Diff (2) partially
w.r.t cWe get,0
=1 (absurdThere
is no Singular
Integral
Singular
Integral
Diff (2) partially
w.r.t cWe get,0
=1 (absurdThere
is no Singular
Integral
Singular
Integral
Diff (2) partially
w.r.t cWe get,0
=1 (absurdThere
is no Singular
Integral
Singular
Integral
Diff (2) partially
w.r.t cWe get,0
=1 (absurdThere
is no Singular
Integral
General
Integral
Put c = (a) in
(2)We get
(3)Diff (3)
partially w.r.t
aWe get
(4)Eliminating a
from (3) and (4)
we get General
Integral
General
Integral
Put c = (a) in
(2)We get
(3)Diff (3)
partially w.r.t
aWe get
(4)Eliminating a
rom (3) and (4)
we get General
Integral
General
Integral
Put c = (a) in
(2)We get (3)
General
Integral
Put c = (a) in
(2)We get
(3)Diff (3)
partially w.r.t
aWe get
(4)Eliminating a
from (3) and (4)
we get General
Integral
Diff (3) partially
w.r.t aWe get (4)
There is no
Singular Integral
General Integral
Put c = (a) in
(2)We get (3)
Diff (3) partially
w.r.t aWe get
(4)Eliminating a
from (3) and (4)
we get General
Integral
10
Eliminating a
from (3) and (4)
we get General
Integral
TYPE –4
Page
TYPE –1
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HOMOGENEOUS LINEAR EQUATION
General form
(aD3 bD2 D
3
cDD 2 dD ) z
f ( x, y)
(1)
To Find Complementary Function
Auxiliary Equation
Put D = m & D = 1 in (1)
Solving we get the roots m1 , m2 , m3
Case (1)
If the roots are distinct then
C.F. =
1
( y m1 x)
2
( y m2 x)
3
( y m3 x)
Case (2)
If the roots are same then
C.F. =
1
( y mx) x 2 ( y mx ) x 2 3 ( y mx )
Case (3)
If the two roots are same and one is distinct, then
( y mx) x 2 ( y mx)
3
( y m 3 x)
PI =
Function
F(x,y) = eax+by
1
F ( x, y )
F ( D, D1 )
Put D = a & D1 = b
F(x,y)= sin(ax+by)(or)
Cos (ax+by)
F(x,y) = x y
r
Put D2
(a2 ), DD
(ab) & D
PI= F ( D, D )
1
2
(b2 )
xr y s
s
Expand and operating D & D on xr ys
F(x,y) = eax+by f(x,y)
Put D = D+a & D = D +b
11
1
Page
C.F =
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Particular Integral
F(x,y)=ex+y cosh(x+y)
F(x ,y)=
1 2x
e
2
F(x,y)=ex+y sinh(x+y)
F(x, y) =
1 2x 2 y
e
e
2
e2 y
F(x,y)=sin x cos y
F ( x, y)
1
sin( x y ) sin( x y )
2
F(x,y)= cos x sin y
F ( x, y )
1
sin( x y ) sin( x y )
2
F(x,y)= cos x cos y
F ( x, y)
1
co s( x y) co s( x y)
2
F(x,y)= sin x sin y
F ( x, y )
1
cos( x y) cos( x
2
y)
Note:
D represents differentiation with respect to ‘x ‘
D represents differentiation with respect to ‘y ‘
Page
12
1
D represents integration with respect to ‘x ‘
1
D represents integration with respect to ‘y ‘
Department of Mathematics – FMCET – MADURAI
PARTIAL DIFFRENTIAL EQUATIONS
1. Eliminate the arbitrary constants a & b from
z = (x2 + a)(y2 + b)
Answer:
z = (x2 + a)(y2 + b)
Diff partially w.r.to x & y here
z
&q
x
p
z
y
p = 2x(y2 + b) ; q = (x2 + a) 2y
(y2 + b) = p/2x
;
(x2 + a) = q/2y
z = (p/2x)(q/2y)
4xyz = pq
2. Form the PDE by eliminating the arbitrary function from
z = f(xy)
Answer:
z = f(xy)
Diff partially w.r.to x & y here
p = f ( xy ). y
p/q = y/x
p
z
&q
x
z
y
q = f ( xy).x
px – qy = 0
3. Form the PDE by eliminating the constants a and b from z = axn + byn
Answer:
z = axn + byn
Diff. w .r. t. x and y here
p
z
&q
x
z
y
Page
13
p = naxn-1 ; q = nbyn-1
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p
nx n
a
z
;b
1
p n
x
nx n 1
nz px qy
q
ny n 1
q
yn
n 1
ny
4. Eliminate the arbitrary function f from z
f
xy
z
and form the PDE
Answer:
z
f
xy
z
Diff. w .r. t. and y here
z
&q
x
p
p
f
xy
z xp
.y
z
z2
q
f
xy
z yq
.x
z
z2
p
q
y z
.
x z
pxz
px
xp
yq
pqxy
qy
z
y
qyz
pqxy
0
5. Find the complete integral of p + q =pq
Answer:
Put p = a, q = b
b
a
1 a
a
a 1
The complete integral is z= ax+
a
a 1
y +c
14
b – ab = -a
a+b=ab
Page
p + q =pq
Department of Mathematics – FMCET – MADURAI
6. Find the solution of
Answer:
z = ax+by+c
given
put
p
q 1
----(1)
p
is the required solution
1 -----(2)
q
p=a, q = b in (2)
a
b
1
z
ax
(1
b 1
a )2 y
a
b
a )2
(1
c
7. Find the General solution of p tanx + q tany = tanz.
Answer:
dx
tan x
dy
tan y
cot x dx
take
cot y dy
cot x dx
log sin x
c1
dz
tan z
cot z dz
cot y dy
log sin y
cot y dy
log c1
log sin y
sin x
sin y
c2
sin x sin y
,
sin y sin z
cot zdz
log sin z
log c2
sin y
sin z
0
8. Eliminate the arbitrary function f from z
f x2
y 2 and form the PDE.
Answer:
z
f x2 y 2
x2
p
f
p
q
2x
2y
y 2 2x ; q
f
x2
y 2 ( 2 y)
py qx 0
Page
Answer:
General form of the sphere equation is
15
9. Find the equation of the plane whose centre lie on the z-axis
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x2
y2
2
z c
r2
(1)
Where ‘r’ is a constant. From (1)
2x+2(z-c) p=0
(2)
2y +2(z-c) q = 0
(3)
From (2) and (3)
x
p
That is
y
q
py -qx =0
which is a required PDE.
10. Eliminate the arbitrary constants z
ax by
a 2 b2 and form the PDE.
Answer:
a 2 b2
z ax by
p
a; q
z
b
px
p2
qy
11. Find the singular integral of
q2
z
px
qy
pq
Answer:
The complete solution is
z
a
0
x
b
z
x
;
( y) x
xy
xy
xy
b
z
z
ax
;
a
z
b
y
( x) y
xy
by
0
( y.
ab
y
a
x)
xy
0
12. Find the general solution of px+qy=z
dx
x
dy
y
dz
z
Page
The auxiliary equation is
16
Answer:
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dx
x
From
dy
y
Integrating we get
x
y
on simplifying
c1 .
dy dz
y z
x y
,
y z
Therefore
log x = log y + log c
0
y
c2
z
is general solution.
13. Find the general solution of px2+qy2=z2
Answer:
The auxiliary equation is
From
Also
dx
x2
dy
y2
dy
y2
dz
z2
Therefore
14. Solve D2
1
y
dx
x2
dy
y2
dz
z2
1
y
Integrating we get
Integrating we get
1 1
,
x z
2DD
1
y
3D 2 z
1
z
1
y
1
x
c1
c2
is general solution.
0
0
Answer:
m 2 2m 3 0
m 3 m 1
The solution is z
f1 y x
1, m 3
f2 y 3x
17
m
0
Page
Auxiliary equation is
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15. Solve D2
3D 2 z
4DD
Answer:
Auxiliary equation is
ex
y
m 2 4m 3 0
m 3 m 1
0
m 1, m 3
The CF is
PI
D
1
4 DD
2
x
PI
2D 4D
xe x y
2
CF
2
3D
ex
ex
f1 y x
Put D
y
1, D
f2 y 3x
1 Denominator =0.
y
Z=CF + PI
z
f1 y x
16. Solve. D2
f2 y 3x
4D 2 z
3DD
xe x
2
Answer:
Auxiliary equation is
ex
y
y
m2 3m 4 0
m 4 m 1
0
C.F is = f1(y + 4x) + f2(y - x)
PI
D
2
1
3DD
17. Find P.I D2
4D
2
ex
y
1
ex
1 3 4
4D 2 z
4DD
e2 x
y
1 x
e
6
y
y
Answer:
PI
1
4 DD
2, D
4D
1
1
D 2D
2
e
2
e2 x
2x y
y
1
2 2
e2 x
2
y
e2 x y
16
18
D
Put D
2
Page
PI
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18. Find P.I D2
6D 2 z
DD
x2 y
Answer:
PI
1
D
D2 1
D
6D
D2
1
D
1
2
D
D
2
z
x
2
x2 y
x2 y
x3
3
1
x2 y
2
D
19. Find P.I
2
x4 y
12
x5
60
2
z
x y
sin x
y
Answer:
1
Put D 2
Sin x y
D DD
Sin x y
1
Sin x y
2
1 1
PI
1,
2
20. Solve D3
3DD
Answer:
Auxiliary equation is
2D 3 Z
DD
(1)( 1) 1
0
m3 3m 2 0
m 1 m2 m 2
0
m 1 m 2 m 1
m 1,1 m
The Solution is
0
2
CF
f1 y x
x2
a2
y2
b2
x f2 y x
f3 y 2x
FOR PRACTICE:
1. Eliminating arbitrary constants
z
x2
sin y
3. Find the complete the solution of p. d .e p 2
q 2 4 pq
4. Form p.d.e eliminating arbitrary function from
z2
0
xy ,
x
2
19
2. Solve
1
Page
2
z2
c2
Department of Mathematics – FMCET – MADURAI
5. Find the singular soln of z
1. (i) Solve x2 y
z p y2 z x q
(ii) Solve x z 2
y 2 p y x2
z
x
2. (i) Solve mz ny
z y2
z
y
ly mx
y2
z 2 p 2 xyq
(ii) Solve y 2
z2
x2 p 2 xyq 2 zx 0
2 xz
4. (i) Solve y
z p
z x q
x y
(ii) Solve y
z p
z x q
x y
5. Solve D2
3DD
2D 2
x
2
2y
sin(3x 2 y)
2
z
x y
cos x cos 2 y
7. Solve D2
DD
6D 2 z
y cos x
8. Solve D2
DD
30D 2 z
xy e6 x
9. Solve D2
6DD
5D 2 z
e x sinh y xy
10. Solve D2
4DD
4D 2 z
11. Solve D3
D2 D
DD 2 D 3 z
12. Solve (i) z
px qy
(ii) z
px qy
13. Solve z 2 1 p 2
q2
y
e2 x
y
e2 x
y
cos( x y)
1 p2 q2
p 2q 2
20
z
e3 x
Page
2
x2
4x 2z q 2 y 3x
3. (i) Solve x2
6. Solve
q2 1
z2 x y
z2 q
nx lz
(ii) Solve 3z 4 y p
p2
px qy
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14. Solve z 2 ( p 2 x 2
q2 ) 1
15. Solve (i) z ( p 2
q2 )
(ii) z 2 ( p 2
q2 )
x2
x2
y2
y2
UNIT - 2
FOURIER SERIES
a0
2
(- ,
Even (or) Half range
Fourier co sine series
)
Odd (or) Half range
Fourier sine series
Neither even nor odd
21
(0,2 )
an cos nx bn sin nx
n 1
Page
f ( x)
Department of Mathematics – FMCET – MADURAI
a0
2
1
f ( x)dx
a0
2
0
an
1
bn
a0
0
an
0
a0
2
f ( x) cos nxdx
an
2
f ( x) cos nxdx
2
f ( x)s innxdx
bn=0
bn
an
2
f ( x)s innxdx
a0
2
an cos
n 1
bn
1
bn
1
f ( x)s innxdx
( - , )
Even (or) Half range
Fourier cosine series
1
f ( x) cos nxdx
n x
n x
bn sin
(0,2 )
an
1
0
f ( x)
1
f ( x)dx
0
0
a0
1
0
0
1
f ( x)dx
2
f ( x)dx
a0
0
2
f ( x) cos
0
2
f ( x)s in
0
n x
dx
n x
dx
2
f ( x)dx
0
an
2
n x
f ( x) cos
dx
0
bn=0
Odd (or) Half range
Fourier sine series
a0
an
0
Neither even nor odd
a0
0
an
bn
2
n x
f ( x)s in
dx
0
bn
1
1
1
f ( x)dx
f ( x) cos
n x
dx
f ( x)s in
n x
dx
Even and odd function:
Even function:
f(-x)=f(x)
Page
Odd function:
22
eg : cosx,x2 , , x , sin x , cos x are even functions
Department of Mathematics – FMCET – MADURAI
f(-x)=-f(x)
eg : sinx,x3 ,sinhx, tanx are odd functions
For deduction
In the interval (0,2 ) if x = 0 or x = 2 then
f(0) = f(2 ) =
f (0)
f (2 )
2
In the interval (0,2 ) if x = 0 or x = 2 then
f (2)
f (0)
f(0) = f(2 ) =
2
In the interval (- , ) if x = - or x = then
f(- ) = f( ) =
f(
) f( )
2
In the interval (- , ) if x = - or x = then
f(- ) = f( ) =
f ( ) f ()
2
HARMONIC ANALYSIS
f(x)=
y
a1
n
f(x)=
2
y cos x
n
, a2
2
y cos 2 x
n
b1
2
y sin x
n
, b2
2
y sin 2 x
n
a0
x
x
2 x
2 x
+ a1 cos
+b1 sin
+ a2 cos
+ b2 sin
………( form)
2
23
2
Page
a0
a0
+ a1 cosx +b1sinx + a2cos2x + b2sin2x ……… for form
2
Department of Mathematics – FMCET – MADURAI
2
y
y cos
n
a1
2
x
y cos
a2
n
2
2 x
n
y sin
b1
2
n
x
,
b2
y sin
2
2 x
n
1. Define R.M.S value.
If let f(x) be a function defined in the interval (a, b), then the R.M.S value of
f(x) is defined by
y
b
1
2
b aa
f ( x) dx
2. State Parseval’s Theorem.
Let f(x) be periodic function with period 2l defined in the interval (c, c+2l).
1
2l
c 2l
ao2
f ( x) dx
4
2
c
1
an2 bn2
2n1
Where ao , an & bn are Fourier constants
3. Define periodic function with example.
If a function f(x) satisfies the condition that f(x + T) = f(x), then we say f(x) is a periodic
function with the period T.
Example:i) Sinx, cosx are periodic function with period 2
ii) tanx is are periodic function with period
4. State Dirichlets condition.
(i) f(x) is single valued periodic and well defined except possibly at a
Finite number of points.
(ii) f (x) has at most a finite number of finite discontinuous and no infinite
Discontinuous.
5. State Euler’s formula.
24
(iii) f (x) has at most a finite number of maxima and minima.
Page
a0
Department of Mathematics – FMCET – MADURAI
Answer:
In (c, c 2l )
ao
2
f x
an cos nx bn sin nx
1
where ao
c 2l
f ( x )dx
c
1
c 2l
1
c 2l
an
f ( x ) cos nxdx
c
bn
f ( x ) sin nxdx
c
6. Write Fourier constant formula for f(x) in the interval (0, 2 )
Answer:
ao
1
2
f ( x)dx
0
an
1
2
f ( x) cos nxdx
0
bn
1
2
f ( x) sin nxdx
0
7. In the Fourier expansion of
1
f(x) =
1
2x
2x
,
x 0
in (-π , π ), find the value of bn
,0
x
Since f(-x)=f(x) then f(x) is an even function. Hence
bn = 0
8. If f(x) = x3 in –π < x < π, find the constant term of its Fourier series.
Answer:
Given f(x) = x3
f(-x) = (- x)3= - x3 = - f(x)
Hence f(x) is an odd function
Page
25
The required constant term of the Fourier series = ao = 0
Department of Mathematics – FMCET – MADURAI
9. What are the constant term a0 and the coefficient of cosnx in the Fourier
Expansion
f(x) = x – x3 in –π < x < π
Answer:
Given f(x) = x – x3
f(-x) = -x - (- x)3= - [x - x3] = - f(x)
Hence f(x) is an odd function
The required constant term of the Fourier series = a 0 = 0
10. Find the value of a0 for f(x) = 1+x+x2 in ( 0 ,2
)
Answer:
1
ao
2
f ( x)dx
0
1
2
1
2
(1 x x )dx
x
0
1
2
2
4
2
x2
2
3
8
3
2 2
x3
3
2
0
2
8
3
11. (i)Find bn in the expansion of x2 as a Fourier series in (
(ii)Find bn in the expansion of xsinx a Fourier series in (
, )
, )
Answer:
(i) Given f(x) = x2
f(-x) = x2 = f(x)
Hence f(x) is an even function
In the Fourier series bn = 0
(ii) Given f(x) = xsinx
f(-x) = (-x)sin(-x) = xsinx = f(x)
Hence f(x) is an even function
Page
26
In the Fourier series bn = 0
Department of Mathematics – FMCET – MADURAI
x
12. Obtain the sine series for f x
x
Given f x
0
0
x l/2
l x l/2
x l
x l/2
l x l/2
x l
Answer:
x
Given f x
0
x l/2
l x l/2
Fourier sine series is
x l
f x
bn sin
nx
l
l
2
nx
f ( x) sin dx
l 0
l
bn
2
l
2
l
l 2
0
l
nx
x sin dx
l
cos
lx
(l x) sin
l 2
nx
l
nx
dx
l
sin
(1)l 2
n
nx
l
l 2
cos
l (l x)
n2
nx
l
n
0
2
l
sin
l 2 ( 1)
nx
l
l
n2
l 2
l 2 cos n 2 l 2 sin n 2 l 2 cos nl 2 l 2 sin n 2
2 2
2 2
2n
n
2n
n
2 2l 2 sin n 2
2 2
l
n
Fourier series is
f x
4l sin n 2
2 2
n
4l
2
13. If f(x) is odd function in
sin n 2 n x
sin
n2
l
n 1
l , l . What are the value of a0 &an
Page
27
Answer:
If f(x) is an odd function, ao = 0, an = 0
Department of Mathematics – FMCET – MADURAI
14. In the Expansion f(x) = |x| as a Fourier series in (Answer:
Given f(x) = |x|
. ) find the value of a0
f(-x) = |-x| = |x| = f(x)
Hence f(x) is an even function
2
ao
xdx
0
2 x2
2
2
2
2
0
15. Find half range cosine series of f(x) = x, in 0
x
Answer:
xdx
0
an
2
1
x sin nxdx
2
0
cos nx
n
x
0
1
cos n
n
f x
Fourier series is
2
1
n
0 0
ao
2
n 0
sin nx
n2
(1)
n
1
n
0
n 1
an cos nx
n 0
1
n
n 1
cos nx
16. Find the RMS value of f(x) = x2, 0
Answer:
Given f(x) = x2
2
2
x 1
28
2 x2
2
Page
ao
2
Department of Mathematics – FMCET – MADURAI
R.M.S value
2l
1
1
2
f ( x) dx
l 0
y
1
x5
2
5
1
x2
120
2
dx
2
5
0
17. Find the half range sine series of
f ( x)
x in (0,
(1)
sin nx
n2
)
Answer:
2
bn
f ( x) sin nxdx
0
2
x sin nxdx
2
cos nx
n
x
0
( 1) n
n
2
2( 1) n
n
0
1
Half range Fourier sine series is
f x
n
2( 1) n 1
sin nx
n
0
18. Find the value of a0 in the cosine series of
f ( x)
x in (0, 5)
Answer:
5
ao
2
xdx
50
2 x2
5 2
5
0
2 52
5 2
5
19. Define odd and even function with example.
Answer:
(i) If f ( x)
f ( x) then the function is an even function.
eg : cosx ,x2 , x , sin x , cos x are even functions
(ii) If f ( x)
f ( x) then the function is an odd function.
Page
29
eg : sinx,x3 ,sinhx, tanx are odd functions
Department of Mathematics – FMCET – MADURAI
20. Write the first two harmonic.
Answer:
The first two harmonics are
ao
2
f x
a1 cos x b1 sin x a2 cos 2 x b2 sin 2 x
FOURIER SERIES
x
1. Expand f ( x)
(0, )
2
x
as Fourier series
( ,2 )
1
12
and hence deduce that
1
32
2. Find the Fourier series for f(x) = x2 in ((i)
1
12
1
22
2
1
.........
32
(ii)
6
1
12
2
1
.........
52
1
22
8
.
1
32
) and also prove that
2
.........
3. (i) Expand f(x) = | cosx | as Fourier series in ((ii) Find cosine series for f(x) = x in (0,
1
14
Show that
1
24
1
.........
34
12
. ).
) use Parsevals identity to
4
90
4. (i) Expand f(x) = xsinx as a Fourier series in (0, 2 )
(ii)
Expand
1
12
5. If f ( x)
f(x)
=
1
32
1
.........
52
0
,
sin x ,
(
|x|
as
a
Fourier
series
in
(-
.
)
and
deduce
to
2
8
,0)
(0, )
Find the Fourier series and hence deduce that
1
1
1
.........
1.3 3.5 5.7
2
4
X
0
1
2
3
4
5
Y
9
18
24
28
26
20
30
6. (i) Find the Fourier series up to second harmonic
Page
(ii)Find the Fourier series up to third harmonic
Department of Mathematics – FMCET – MADURAI
7.
X
0
π/3
2π/3
π
4π/3
5π/3
2π
F(x)
10
14
19
17
15
12
10
(i) Find the Fourier expansion of f ( x)
Hence deduce that
1
12
1
22
2
1
.........
32
(ii). Find a Fourier series to represent
x) 2 in (0, 2 ) and
(
6
2 x x 2 with period 3
f ( x)
in the range (0,3)
(ii) Find the Fourier series of f x
(ii) Find the Fourier series for
and hence show that
1
32
, ).
1
in
(0, )
2 in ( , 2 )
2
1
.........
52
8
(i) Find the the half range sine series for f x
that
1
13
1
33
1
....
53
(ii) Obtain the half range cosine series for f x
1
and also deduce that 2
1
1
22
1
.........
32
9. (i) Find the Fourier series for f(x) = x2 in (-
1
1
14
1
.........
24
1
.........
34
2
in (0,1)
2
6
. ) and also prove that
90
(use P.I)
. ) and also prove that
4
96
(use P.I)
31
1
14
x 1
4
(ii) Find the Fourier series for f(x) = x in (-
1
x in the interval (0, ) and deduce
x
Page
8.
1
12
f x
ex in (
Department of Mathematics – FMCET – MADURAI
cx
,0
10(i)Obtain the sine series for f x
c l x
l
2
x
,
l
2
x l
kx
,0
(ii). Find the Fourier series for the function f x
k 2l x
11.(i).Find the Fourier series for the function f x
deduce that
1
12
1
22
1
.........
32
,
x
l
2
l
2
x l
1 x x2 in (
, ) and also
2
6
(ii) Find the Fourier expansion of
1
2x
,
,0
x 0
x
1
in (-π , π ), and also deduce that 2
1
1
32
1
.........
52
2
8
32
f(x) =
2x
Page
1
Department of Mathematics – FMCET – MADURAI
UNIT - 3
APPLICATIONS OF P.D.E
S.
ONE DIMENSIONAL WAVE EQUATION
N
O
VELOCITY MODEL
2
is
2
t
y
2
2
a2
x
2
y
1
2
is
STEP-2
Boundary conditions
1. y(0,t)
= 0
for t 0
2. y( , t) = 0
for t 0
3. y(x,0)
= 0
for 0 < x <
4.
y
t
= f(x)
t 0
STEP-1
One Dimensional wave equation
for
0<x<
t
y
2
2
a2
y
x2
1
STEP-2
Boundary conditions
1. y(0,t)
=0
for
2. y( , t) = 0
for
3.
y
t
4. y(x,0)
=0
t 0
t 0
for 0 < x <
t 0
= f(x) for
0<x<
33
STEP-1
One Dimensional wave equation
Page
1
INITIAL POSITION MODEL
Department of Mathematics – FMCET – MADURAI
3
STEP-3
The possible solutions are
y(x,t) = (A e x + B e- x) (C e at + D e- at)
y(x,t) = (A cos x + B sin x )( C cos at + D sin
at)
y(x,t) = (Ax + B) ( Ct + D)
STEP-3
The possible solutions are
y(x,t) = (A e x + B e- x) (C e at + D e- at)
y(x,t) = (A cos x + B sin x )( C cos at + D
sin at)
y(x,t) = (Ax + B) ( Ct + D)
4
STEP-4
The suitable solution for the given
boundary condition is
y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)
(2)
STEP-5
Using Boundary condition 1
y(0,t) = 0
Then (2) becomes,
y(0,t) = (A cos 0 +B sin 0 ) ( C cos at + Dsin at) =0
(A) ( C cos at + D sin at)=0
A=0
Using A = 0 in (2)
y(x,t) = ( B sin x) ( C cos at + D sin at) (3)
STEP-4
The suitable solution for the given
boundary condition is
y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)
(2)
STEP-5
Using Boundary condition 1
y(0,t) = 0
Then (2) becomes,
y(0,t) = (A cos 0 +B sin 0 ) ( C cos at + D sin at)
=0
(A) ( C cos at + D sin at)=0
A=0
Using A = 0 in (2)
y(x,t) = ( B sin x) ( C cos at + D sin at) (3)
STEP-6
Using Boundary condition 2
y( ,t) = 0
Then (3) becomes,
y( ,t) = (B sin ) ( C cos at + D sin at)=0
(B sin ) ( C cos at + D sin at)=0
STEP-6
Using Boundary condition 2
y( ,t) = 0
Then (3) becomes,
y( ,t) = (B sin ) ( C cos at + D sin at)=0
(B sin ) ( C cos at + D sin at)=0
5
6
=
n
Then (3) becomes,
y ( x, t )
B sin(
n x
n at
n at
) C cos(
) D sin(
)
=
n
Then (3) becomes,
y ( x, t )
B sin(
n x
n at
n at
) C cos(
) D sin(
)
(4)
STEP-7
Using Boundary condition 3
y(x,0) = 0
Then (4) becomes,
STEP-7
Using Boundary condition 3
Page
7
34
(4)
Department of Mathematics – FMCET – MADURAI
y( x, t )
B sin(
B sin(
n x
) C cos 0 D sin 0 =0
n x
) C
B sin(
= 0Then (4) becomes,
t 0
Differentiating (5) partially w.r.to ‘t’ and put t =0
0
y
t
n x
n at
n at n a
) C sin(
) D cos(
)
t 0
n x
n a
B sin(
) D
0
C=0
Then (4) becomes,
y ( x, t )
y
t
n x
n at
) D sin(
)
B sin(
D=0
Then (4) becomes,
The most general solution is
y ( x, t )
Bn sin(
n 1
n x
n at
) sin(
)
(5)
y ( x, t )
B sin(
n x
n at
) C cos(
)
The most general solution is
y ( x, t )
Bn sin(
n 1
STEP-8
Differentiating (5) partially w.r.to ‘t’
y
t
Bn sin(
n 1
n x
n at n a
) cos(
)
STEP-8
Using Boundary condition (4),
y(x,0) = f(x)
y ( x, 0)
Using Boundary condition (4),
n 1
f ( x)
= f(x)
n 1
Bn sin(
n 1
9
n x
)
2
n x
f ( x)sin(
)
0
Bn
n a
2
n x
) cos(0)
This is the Half Range Fourier Sine Series.
n x n a
)
This is the Half Range Fourier Sine Series.
Bn
Bn sin(
t 0
f ( x)
Bn
Bn sin(
2
n x
f ( x)sin(
)
0
n a0
f ( x)sin(
n x
)dx
STEP-9
The required solution is
y ( x, t )
Bn sin(
n 1
Where Bn
STEP-9
The required solution is
n x
n at
) sin(
)
2
n a0
f ( x)sin(
n x
)dx
y ( x, t )
Bn sin(
n 1
n x
n at
) sin(
)
Where Bn
2
n x
f ( x)sin(
)dx
0
35
y
t
(5)
Page
8
n x
n at
) cos(
)
Department of Mathematics – FMCET – MADURAI
ONE DIMENSIONAL HEAT
EQUATION
The one dimensional heat equation is
u
t
2
Boundary conditions
1.u(0,t) = 0
for t 0
2.u( ,t) = 0
for t 0
3.u(x,t) = f(x) for 0<x<
3
The possible solutions are
x
x
Be
)Ce
t
u ( x, y ) ( Ae
2 2
2 2
t
2 2
t
=0
2 2
t
u ( x, y )
u(, t ) ( B sin )Ce
u (0, y )
2 2
t
u (, y )
( A cos 0 B sin 0)(Ce
( A)(Ce
y
y
Be
( B sin x)(Ce
( B sin )(Ce
0 ( B sin )(Ce
)
y
De
y
)
(2)
y
De
y
)
)
y
De
y
De
y
)
(3)
y
y
De
y
)
)
n
2 2 2
n
Then (3) becomes
t
2
u ( x, y )
The most general solution is
n x
Bn sin(
)e
1
n
2
( B sin
n y
n x
)(Ce
De
n y
)
(4)
t
(4)
36
2 2 2
n
y
=0
0
n x
B sin(
)Ce
u ( x, t )
De
Using boundary condition 2
u(l,t) = 0
Then (3) becomes
u( x, t )
( A cos x B sin x)(Ce
u ( x, y )
(3)
Using boundary condition 2
u(l,t) = 0
t
y
A=0
Then (2) becomes
u( x, t ) (B sin x)Ce
(B sin )Ce
n
(2)
u (0, y )
=0
2 2
)(C cos y D sin y )
Using boundary condition 1
u(0,y) = 0
A=0
Then (2) becomes
6
x
The most suitable solution is
u(0, t ) ( A cos0 B sin 0)Ce
t
Be
u ( x, y ) ( Ax B)(Cy D)
Using boundary condition 1
u(0,t) = 0
2 2
x
u ( x, y ) ( A cos x B sin x)(Ce
t
The most suitable solution is
( A)Ce
0
The possible solutions are
2 2
u( x, t ) ( A cos x B sin x)Ce
5
u
y2
Boundary conditions
1.u(0,y) = 0
for 0<y<
2.u( ,y) = 0
for 0<y<
3.u(x, ) = 0
for 0<x<
4.u(x,0) = f(x) for 0<x<
u ( x, t ) ( A cos x B sin x)Ce
u ( x, t ) ( Ax B)C
4,
2
u
x2
u
x2
2
2
u ( x, t ) ( Ae
The two Dimensional equation is
2
Page
1
TWO DIMENSIONAL HEAT FLOW
EQUATION
Department of Mathematics – FMCET – MADURAI
7
Using Boundary condition 3
u(x,0) = f(x)
u ( x, 0)
Bn sin(
n 1
f ( x)
Bn sin(
n 1
Using Boundary condition 3
u(x, ) =0
n x
)
u ( x, ) ( B sin
n x
)
0 ( B sin
n x
)(C
De )
D 0)
C=0
then (3) becomes
This the Half range Fourier sine series
Bn
n x
)(Ce
2
n x
f ( x)sin(
)dx
0
u ( x, y )
( B sin
n x
)( De
n y
)
The most general solution is
u ( x, y )
n
8
The required solution is
u ( x, t )
n
n
(5
y(x,0) = f(x)
t
2
u ( x, 0)
Bn sin(
n 1
Where Bn
n y
Using Boundary condition 4
2 2 2
n x
Bn sin(
)e
1
n x
Bn sin(
)e
1
2
n x
f ( x)sin(
)dx
0
f ( x)
Bn sin(
n 1
n x 0
)e
n x
)
This the Half range Fourier sine series
2
n x
f ( x)sin(
)dx
0
Bn
The required solution is
u ( x, y )
Bn sin(
n 1
n x
)e
n y
Where Bn
2
n x
f ( x)sin(
)dx
0
QUESTION WITH ANSWER
2
2
u
y2
37
Answer:
u
x2
Page
1. Classify the Partial Differential Equation i)
Department of Mathematics – FMCET – MADURAI
2
2
u
x2
u
here A=1,B=0,&C=-1
y2
B2 - 4AC=0-4(1)(-1)=4>0
The Partial Differential Equation is hyperbolic
2
2. Classify the Partial Differential Equation
u
x y
u
y
u
x
xy
Answer:
2
u
x y
u
y
u
x
xy here A=0,B=1,&C=0
B2-4AC=1-4(0)(0)=1>0
The Partial Differential Equation is hyperbolic
3. Classify the following second order Partial Differential equation
2
2
u
x2
u
y2
u
y
2
u
x
2
Answer:
2
2
u
x2
u
y2
u
y
2
u
x
2
here A=1,B=0,&C=1
B2-4AC=0-4(1)(1)=-4<0
The Partial Differential Equation is Elliptic
4. Classify the following second order Partial Differential equation
2
4
2
u
u
4
2
x
x y
2
u
6
y2
u
x
u
y
8
0
Answer:
2
4
2
u
u
4
2
x
x y
2
u
6
y2
u
x
8
u
y
0
here A= 4,B =4, & C = 1
B2-4AC =16 -4(4)(1) = 0
Page
5. Classify the following second order Partial Differential equation
i) y2uxx 2xyuxy x2uyy 2ux 3u 0
38
The Partial Differential Equation is Parabolic
Department of Mathematics – FMCET – MADURAI
ii) y 2uxx
uyy ux 2 uy 2 7 0
Answer:
i) Parabolic ii) Hyperbolic (If y = 0)
iii)Elliptic (If y may be +ve or –ve)
2
6. In the wave equation
t
y
2
2
c2
y
x2
what does c2 stands for?
Answer:
2
t
y
2
2
c2
y
x2
here a 2
T
m
T-Tension and m- Mass
7. In one dimensional heat equation ut = α2 uxx what does α2 stands for?
Answer:-
u
t
2
=
2
2
u
x2
k
is called diffusivity of the substance
c
Where k – Thermal conductivity
- Density
c – Specific heat
8. State any two laws which are assumed to derive one dimensional heat equation
Answer:
i) Heat flows from higher to lower temp
Page
39
ii) The rate at which heat flows across any area is proportional to the
area
and to the temperature gradient
normal to the curve. This constant of
proportionality is known as the conductivity of the material. It is known as
Fourier law of heat conduction
Department of Mathematics – FMCET – MADURAI
9. A tightly stretched string of length 2 is fastened at both ends. The midpoint of the
string is displaced to a distance ‘b’ and released from rest in this position. Write the
initial conditions.
Answer:
(i) y(0 , t) = 0
(ii) y(2 ,t) = 0
y
t
(iii)
0
t
0
(iv) y(x , 0 ) =
10.
b
x
b
(2
0
x)
x
x
2
What are the possible solutions of one dimensional Wave equation?
The possible solutions are
Answer:
y(x,t) = (A e
x
+ B e- x) (C e
at
+ D e- at)
y(x,t) = (A cos x + B sin x )( C cos at + D sin at)
y(x,t) = (Ax + B) ( Ct + D)
11. What are the possible solutions of one dimensional head flow equation?
Answer:
The possible solutions are
u ( x, t ) ( Ae
x
Be
x
)Ce
2 2
t
u ( x, t ) ( A cos x B sin x)Ce
u ( x, t ) ( Ax B)C
2 2
t
12. State Fourier law of heat conduction
Answer:
u
x
(the rate at which heat flows across an area A at a distance from one end of a bar is
proportional to temperature gradient)
40
kA
Page
Q
Department of Mathematics – FMCET – MADURAI
Q=Quantity of heat flowing
k – Thermal conductivity
A=area of cross section
;
u
=Temperature gradient
x
13. What are the possible solutions of two dimensional head flow equation?
Answer:
The possible solutions are
u ( x, y ) ( Ae
x
Be
x
)(C cos y D sin y )
u ( x, y ) ( A cos x B sin x)(Ce
y
De
y
)
u ( x, y ) ( Ax B)(Cy D)
14. The steady state temperature distribution is considered in a square plate with sides x
= 0 , y = 0 , x = a and y = a. The edge y = 0 is kept at a constant temperature T and the
three edges are insulated. The same state is continued subsequently. Express the
problem mathematically.
Answer:
U(0,y) = 0 , U(a,y) = 0 ,U(x,a) = 0, U(x,0) = T
15. An insulated rod of length 60cm has its ends A and B maintained 20°C and
80°C respectively. Find the steady state solution of the rod
Answer:
Here a=20°C & b=80°C
In Steady state condition The Temperature u ( x, t )
b a x
a
l
80 20 x
20
60
u( x, t )
x 20
16. Write the D’Alembert’s solution of the one dimensional wave equation?
Page
41
Answer:
Department of Mathematics – FMCET – MADURAI
x at
1
2
y
here
x at
x at
x
f x
v x
ax f
1
v( )d
2a x at
g x
ag
17. What are the boundary conditions of one dimensional Wave equation?
Answer:
Boundary conditions
1. y(0,t)
2. y( , t)
3. y(x,0)
4.
y
t
= 0
= 0
= 0
for
for
for
= f(x)
for
t 0
t 0
0<x<
0<x<
t 0
18. What are the boundary conditions of one dimensional heat equation?
Answer:
Boundary conditions
1.u(0,t) = 0
for t
0
2.u( ,t) = 0
for t
0
3.u(x,t) = f(x) for 0<x<
19. What are the boundary conditions of one dimensional heat equation?
Answer:
Boundary conditions
for 0<y<
3.u(x, ) = 0
for 0<x<
4.u(x,0) = f(x)
for 0<x<
42
2.u( ,y) = 0
for 0<y<
Page
1.u(0,y) = 0
Department of Mathematics – FMCET – MADURAI
20.T he ends A and B has 30cm long have their temperatures 30c and 80c until steady
state prevails. If the temperature A is raised to40c and Reduced to 60C, find the
transient state temperature
Answer:
Here a=30°C & b=80°C
In Steady state condition The Temperature u ( x, t )
b a x
a
l
Here a=40°C & b=60°C
ut
60 40 x
40
30
2
x 40
3
PART-B QUESTION BANK
APPLICATIONS OF PDE
1. 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 3x (l-x). Find the
displacement.
2. A string is stretched and fastened to two points and apart. Motion is started by displacing
the string into the form y = K(lx-x2) from which it is released at time
t = 0. Find the
displacement at any point of the string.
3. A taut string of length 2l is fastened at both ends. The midpoint of string is taken to a
height b and then released from rest in that position. Find the displacement of the string.
4. A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in its
position given by y(x, 0) = y0 sin 3
x
l
. If it is released from rest find the displacement.
5. A string is stretched between two fixed points at a distance 2l apart and points of the
string are given initial velocities where V
cx
l
c
(2l x)
l
0< x < 1
Find the
0< x < 1
displacement.
6. Derive all possible solution of one dimensional wave equation. Derive all possible solution
of one dimensional heat equation. Derive all possible solution of two dimensional heat
equations.
Page
43
7. A rod 30 cm long has its end A and B kept at 20oC and 80oC, respectively until steady state
condition prevails. The temperature at each end is then reduced to 0oC and kept so. Find
the resulting temperature u(x, t) taking x = 0.
Department of Mathematics – FMCET – MADURAI
8. A bar 10 cm long , with insulated sides has its end A & B kept at 20oC and 40oC respectively
until the steady state condition prevails. The temperature at A is suddenly raised to 50oC
and B is lowered to 10oC. Find the subsequent temperature function u(x , t).
9. A rectangular plate with insulated surface is 8 cm wide so long compared to its width that
it may be considered as an infinite plate. If the temperature along short edge y = 0 is u (
x ,0) = 100sin
x
8
0 < x < .1While two edges x = 0 and x = 8 as well as the other short
edges are kept at 0oC. Find the steady state temperature.
A rectangular plate with insulated surface is10 cm wide so long compared to its width that
it may be considered as an infinite plate. If the temperature along short edge y = 0 is given
by u
20 x
0
x
5
20(10 x)
5
x
10
and all other three edges are kept at 0o C. Find the steady
44
state temperature at any point of the plate.
Page
10.
Department of Mathematics – FMCET – MADURAI
Unit - 4
FOURIER TRANSFORMS
FORMULAE
1
2
1. Fourier Transform of f(x) is F[ f ( x)]
1
2
2. The inversion formula f ( x)
3. Fourier cosine Transform
f(x)eisx dx
-
F (s)e-isx ds
-
Fc [f(x)] = Fc(s) =
2
f ( x) cos sxdx
0
4. Inversion formula
f(x) =
2
Fc ( s) cos sxds
0
5. Fourier sine Transform (FST)
6. Inversion formula
f(x) =
F [f(x)] = F (s) =
s
2
s
2
f ( x)sin sxdx
0
Fs ( s)sin sxds
0
2
7. Parseval’s Identity
f ( x) dx
8. Gamma function
x n 1e x dx
n
0
ax
e
ax
a
cos bxdx
a
0
10
sin ax
dx
x
0
2
b2
b
sin bxdx
a
0
11.
1
2
2
2
b2
45
e
, n 1 n n &
Page
9.
2
F ( s) ds
Department of Mathematics – FMCET – MADURAI
12.
e
x2
dx
2
0
13.
cos ax
&
eiax
e
2
e
x2
dx
iax
& sin ax
eiax e
2
iax
ORKING RULE TO FIND THE FOURIER TRANSFORM
Step1: Write the FT formula.
Step2: Substitute given f(x) with their limits.
Step3: Expand
eisx
as cos sx + isin sx and use Even & odd property
Step4: Integrate by using Bernoulli’s formula then we get F(s)
WORKING RULE TO FIND THE INVERSE FOURIER TRANSFORM
Step1: Write the Inverse FT formula
Step2: Sub f(x) & F(s) with limit
Step3: Expand
e
isx
,
in the formula
as cos sx -isin sx and equate real part
Step4: Simplify we get result
WORKING RULE FOR PARSEVAL’S IDENTITY
If F(s) is the Fourier transform of f(x) then
2
f ( x) dx
2
F ( s) ds is known as Parseval’s identity.
Step1: Sub f(x) & F(s) With their limits in the above formula
Step2: Simplify we get result
WORKING RULE TO FIND FCT
Step1: Write the FCT formula & Sub f(x) with its limit in the formula
Page
WORKING RULE TO FIND INVERSE FCT
46
Step2: Simplify, we get FC ( S )
Department of Mathematics – FMCET – MADURAI
Step1: Write the inverse FCT formula & Sub FC ( S ) with its limit in the formula
Step2: Simplify, we get f(x)
WORKING RULE TO FIND FST
Step1: Write the FST formula & Sub f(x) with its limit in the formula
Step2: Simplify, we get FS ( S )
WORKING RULE TO FIND INVERSE FCT
Step1: Write the inverse FST formula & Sub Fs ( S ) with limit in the formula
Step2: Simplify, we get f(x)
WORKING RULE FOR f(x) = e
ax
Step:1 First we follow the above FCT & FST working rule and then we get this
result
Fc(e-ax) =
2
a
a
2
s
By Inversion formula,
cos sx
ds
a2 s2
0
2a
e
2
Fs(e-ax) =
2
s
a
2
s2
By Inversion formula,
s
ax
0
TYPE-I : If problems of the form i)
a
x
x
2
a
2
s
ii)
2
x2
TYPE-II: If problems of the form i)
0
x
2
a
2 2
2
sin sxds
1
x
2
a2
2
e
ax
, then use Inversion formula
dx
dx ii)
0
x
2
a2
2
, then use Parseval’s Identity
TYPE-III
x
a
x2 b2
, then use
f ( x) g ( x)dx
0
FC f ( x) FC g ( x) dx
0
47
0
2
Page
dx
2
Department of Mathematics – FMCET – MADURAI
UNIT - 4
FOURIER TRANSFORM
1. State Fourier Integral Theorem.
Answer:
f ( x) is piece wise continuously differentiable and absolutely on
1
2
then,
f t ei ( x t ) s dt ds .
48
f ( x)
,
Page
If
Department of Mathematics – FMCET – MADURAI
2. StateandproveModulation
1
F s a
2
theorem. F f x cos ax
F s a
1
2
f x cos ax eisx dx
1
2
eiax e
f x
2
iax
1 1
2 2
f x ei ( s
dx
F f x cos ax
1
F s a
2
F f x cos ax
1
F s a
2
Proof:
eisx dx
a) x
1 1
2 2
f x ei ( s
a) x
dx
1
F s a
2
F s a
3. State Parseval’s Identity.
Answer:
If F s is a Fourier transform of f x , then
F s
2
ds
f x
2
dx
4. State Convolution theorem.
Answer:
The Fourier transform of Convolution of f x
and g x is the product of their Fourier
F sGs
5. State and prove Change of scale of property.
Page
F f g
49
transforms.
Department of Mathematics – FMCET – MADURAI
Answer:
If
F s
1
F
a
F f x , then F f ax
F f ax
1
2
f ax eisx dx
1
2
f t e
i
s
a
t
dt
;
a
s
a
where t ax
1 s
F
a
a
F f ax
6. Prove that if F[f(x)] = F(s) then
n
F x f ( x)
dn
( i) n F (s)
ds
n
Answer:
1
2
F s
s ‘n’ times
1
2
n
f x ix eisx dx
1
2
f x (i)n ( x)n eisx dx
1 dn
F s
(i )n ds n
1
2
( x)n f x eisx dx
dn
( i) n F s
ds
1
2
( x) n f x eisx dx
n
n
F x f x
dn
i
F s
ds n
n
f ( x)cos sxdx e
7. Solve for f(x) from the integral equation
0
Answer:
s
50
dn
F s
ds n
Page
Diff w.r.t
f x eisx dx
Department of Mathematics – FMCET – MADURAI
f ( x)cos sxdx e
s
0
2
Fc f x
f x cos sx dx
0
2
Fc f x
2
f ( x)
e
s
Fc f x cos sx ds
0
2
2
e
e s cos sx ds
ax
cos bx dx
0
a
a
2
b2
0
a 1, b
2
e s cos sx ds
2
1
x
0
x
2
1
8. Find the complex Fourier Transform of f ( x)
1
x
a
0
x
a
0
Answer:
f x eisx dx
x
a;
a
x a
51
1
2
Page
F f x
Department of Mathematics – FMCET – MADURAI
F f x
1
2
a
1
2
a
1 eisx dx
a
(cos sx i sin sx)dx
a
a
2
2
2
2
(cos sx)dx
0
a
sin sx
s
0
2 sin as
s
[Use even and odd property second term become zero]
9. Find the complex Fourier Transform of f ( x )
x
x
a
0
x
a
0
Answer:
1
2
i
2
1
2
a
1
2
a
x eisx dx
x
a
a
x a
x (cos sx i sin sx)dx
a
a
( x(i sin sx)dx
0
a;
2i
x
2
cos sx
s
(1)
sin sx
s2
a
0
as cos as sin as
s2
[Use even and odd property first term become zero]
52
2
2
f x eisx dx
Page
F f x
Department of Mathematics – FMCET – MADURAI
10. Write Fourier Transform pair.
Answer:
If f ( x) is defined in
,
, then its Fourier transform is defined as
1
2
F s
f x eisx dx
If F s is an Fourier transform of f x , then at every point of Continuity of f x , we
have
1
2
f x
F s e
isx
ds
.
11. Find the Fourier cosine Transform of f(x) = e-x
Answer:
2
Fc f x
f x cos sx dx
0
Fc e
x
2
e x cos sx dx
0
e
ax
cos bx dx
0
Fc e
x
2
a
a
2
b2
1
s2 1
12. Find the Fourier Transform of
f ( x)
eimx , a
0,
x b
otherwise
Page
53
Answer:
Department of Mathematics – FMCET – MADURAI
1
2
F f x
f x eisx dx
b
1
2
1
2
eimx eisx dx
a
ei m
1
1
ei m s b ei m s a
2 im s
a
Answer:
2
dx
a
13. Find the Fourier sine Transform of
Fs f x
s x
b
ei m s x
im s
1
2
b
1
.
x
f x sin sx dx
0
2
Fs
1
x
sin sx
dx
x
0
2
2
2
14. Find the Fourier sine transform of e
Answer:
2
Fs f x
x
f x sin sx dx
0
2
e x sin sx dx
e ax sin bx dx
0
Fs e
x
2
0
s
b
a
2
b2
s2 1
15. Find the Fourier cosine transform of e
Answer:
2x
2e
x
54
x
Page
Fs e
Department of Mathematics – FMCET – MADURAI
2
Fc f x
f x cos sx dx
0
Fc e
2x
2e
x
2
2x
e
2e
x
cos sx dx
0
2
e
2x
cos sx dx 2 e x cos sx dx
0
0
2
2
s
2
2
4
1
s 1
16. Find the Fourier sine transform of
2
2
2
1
s
f ( x)
2
1
2
4 s 1
1,
0
x 1
0
x 1
Answer:
Fs f x
2
f x sin sx dx
0
1
f x sin sx dx
0
2
1
1
1sin sx dx 0
0
f x sin sx dx
2
cos sx
s
1
0
55
2 1 cos s
s
Page
Fs f x
2
Department of Mathematics – FMCET – MADURAI
f ( x)
17. Obtain the Fourier sine transform of
x, o x 1
2 x, 1 x 2
.
0,
x 2
Answer:
Fs f x
2
f x sin sx dx
0
2
1
2
x sin sx dx
0
2
Fs f x
2
2
2 x sin sx dx
1
cos sx
x
s
cos s
s
sin sx
s2
sin s
s2
1
0
cos sx
2 x
s
sin 2s cos s
s2
s
sin sx
s2
2
1
sin s
s2
2sin s sin 2s
s2
18. Define self reciprocal and give example.
If the transform of f x
2
is self reciprocal under Fourier transform.
56
e
is called self
Page
reciprocal.
x2
is equal to f s , then the function f x
Department of Mathematics – FMCET – MADURAI
19. Find the Fourier cosine Transform of f ( x)
x
0
0
x
x
Answer:
2
Fc f x
2
f x cos sx dx
0
2
x cos sx dx
0
sin sx
s
x
2
cos sx
s2
s sin s
cos s
s
2
s
0
sin s
cos s
s2
1
s2
1
2
20. Find the Fourier sine transform of
Answer:
x
x
2
a2
.
f x e ax
L et
Fs e
2
ax
s
s2 a2
Using Inverse formula for Fourier sine transforms
2
0
s
(ie)
0
s
2
a
2
s
s2 a2
sin sx ds
sin sx ds
2
e ax , a 0
57
2
ax
Page
e
Department of Mathematics – FMCET – MADURAI
x
Change x and s, we get
2
0
Fs
x
x a
2
2
sin sx dx
x
x2 a2
sin sx dx
x2 a2
0
2
2
e
2
e as
as
e
2
as
FOURIER TRANSFORM
PART-B
1 x 2 if x
1. (i)Find the Fourier Transform of f ( x )
x cos x sin x
x
cos
dx
3
x
2
(ii). Find the Fourier Transform of
deduce that
0
2. Find the Fourier Transform of
sin x
i)
dx
x
0
x
0
x
a
a
0
2
dx
15
. hence
4
f ( x)
1 if x
a
0 if x
a
and hence evaluate
2
ii)
0
f ( x)
2
0
x2
0
sin x x cos x
x3
sin x
dx
x
4. Find Fourier Transform of
sin x
dx
x
a2
f ( x)
sin x x cos x
dx
x3
3
(ii)
16
and hence
1
1
x
0
if x
1
if x
1
and hence evaluate
4
ii)
0
sin x
dx
x
58
0
if x
Page
deduce that (i)
i)
0
1
Department of Mathematics – FMCET – MADURAI
5. Evaluate
x2
i)
0
x2 a
dx
ii)
dx
x2 a2
0
ii). Evaluate (a)
0
x2
x2
0
dx
6 i). Evaluate (a)
7.
2 2
(b)
x2 b2
dx
1 x2 4
2
a2
x2
0
(b)
0
x 2 dx
a 2 x2 b2
t 2 dt
4 t2 9
t2
sin x; when o
(i)Find the Fourier sine transform of f ( x)
0
8. (i) Show that Fourier transform
e
x
2
is e
(ii)Obtain Fourier cosine Transform of e
s
2
; whenx
cos x; when o
(ii) Find the Fourier cosine transform of f ( x)
2
x
0
x a
; whenx a
2
a2 x2
and hence find Fourier sine Transform x e
9. (i) Solve for f(x) from the integral equation
a2 x2
f ( x) cos x dx e
0
1 ,0 t 1
(ii) Solve for f(x) from the integral equation
f ( x) sin tx dx
0
2 ,1 t
0
,t
2
10. (i) Find Fourier sine Transform of e x , x>0 and hence deduce that
0
(ii) Find Fourier cosine and sine Transform of e
11.(i)Find FS xe
ax
e
ax
(ii) Find FS
x
8
e 8 (ii)
0
& Fc xe
ax
e
ax
& Fc
x sin 2 x
dx
x 2 16
8
e
, x>0 and hence deduce
8
x
(iii) Find the Fourier cosine transform of f ( x)
e
ax
cos ax
59
0
cos 2 x
dx
x 2 16
x sin x
dx
1 x2
Page
that (i)
4x
2
Department of Mathematics – FMCET – MADURAI
Z - TRANSFORMS
Definition of Z Transform
Let {f(n)} be a sequence defined for
Z – Transform is defined as
Z f (n)
F z
f (n) z
n
n = 0, 1,2 … and f(n) = 0 for n< 0 then its
(Two sided z transform)
n
Z f ( n)
F z
f (n) z
n
(One sided z transform)
n 0
Unit sample and Unit step sequence
The unit sample sequence is defined as follows
(n)
1 for n 0
0 for n 0
The unit step sequence is defined as follows
0 for n 0
60
Properties
1 for n 0
Page
u(n)
Department of Mathematics – FMCET – MADURAI
1. Z – Transform is linear
(i)
Z {a f(n) + b g(n)} = a Z{f(n)} + b Z{g(n)}
2. First Shifting Theorem
(i)
If Z {f(t)} = F(z),
then
(ii)
at
Z e
f t
F z
z zeaT
If Z {f(n)} = F(z),
then Z a n f n
z
a
F
3. Second Shifting Theorem
If Z[f(n)]= F(z) then
(i)Z[f(n +1)] = z[ F(z) – f(0)]
(ii)Z[f(n +2)] = z 2 [ F(z) – f(0)-f(1) z 1 ]
(iii)Z[f(n +k)] = z k [ F(z) – f(0)-f(1) z 1 - f(2) z 2 ………- f(k-1) z
(iv) Z[f(n -k)] = z
k
( k 1)
]
F(z)
4. Initial Value Theorem
If Z[f(n)] = F(z) then f(0) = lim F ( z )
z
5. Final Value Theorem
PARTIAL FRACTION METHODS
lim( z 1) F ( z )
z 1
Page
n
61
If Z[f(n)] = F(z) then lim f (n)
Department of Mathematics – FMCET – MADURAI
Model:I
1
A
B
z a z b
z a
z b
A
B
z a
z b
Model:II
1
z a z b
2
C
( z b) 2
Model:III
1
z a z
A
2
z a
b
Bz C
z2 b
Convolution of Two Sequences
Convolution of Two Sequences {f(n)} and {g(n)} is defined as
n
{ f (n) * g (n)}
f ( K ) g (n K )
K 0
Convolution Theorem
If Z[f(n)] = F(z) and Z[g(n)] = G(z) then Z{f(n)*g(n)} = F(z).G(z)
WORKING RULE TO FIND INVERSE Z-TRANSFORM USING CONVOLUTION THEOREM
Step: 1 Split given function as two terms
Step: 2 Take z
1
Step: 3 Apply z
both terms
1
formula
Step: 4 Simplifying we get answer
Note:
1
62
1 an
1 a
Page
1 a a 2 ....... a n
Department of Mathematics – FMCET – MADURAI
1
a
a
2
.......
a
n
1
n 1
a
1 ( a)
Solution of difference equations
Formula
i) Z[y(n)] = F(z)
ii) Z[y(n +1)] = z[ F(z) – y(0)]
iii) Z[y(n +2)] = z 2 [ F(z) – y(0)- y(1) z 1 ]
iv) Z[y(n +3)] = z 3 [ F(z) – y(0)- y(1) z 1 + y(2) z 2 ]
WORKING RULE TO SOLVE DIFFERENCE EQUATION:
Step: 1 Take z transform on both sides
Step: 2 Apply formula and values of y(0) and y(1).
Step: 3 Simplify and we get F(Z)
Step:4 Find y(n) by using inverse method
Z - Transform Table
No.
1.
f(n)
1
Z[f(n)]
z
z 1
2.
an
z
z a
3.
n
z
n2
z2 z
( z 1)3
Page
4.
63
( z 1) 2
Department of Mathematics – FMCET – MADURAI
6.
1
n
7.
log
1
z
( z 1)
n 1
1
z
log
z
( z 1)
ean
z
1
9.
( z 1)
z log
n 1
8.
z
( z ea )
10.
1
1
n!
11.
ez
Cos n
z ( z cos )
z 2 z cos 1
2
sin n
z
cos
sin
14.
z sin
2 z cos
n
2
1
z2
z2 1
n
2
z
z
na n
2
1
az
( z a)2
f(t)
Z(f(t)
1
t
Tz
( z 1) 2
2.
t2
T 2 z ( z 1)
( z 1)3
3
eat
z
( z e aT )
4.
Sin t
z
2
z sin T
2 z cos T 1
64
13.
2
Page
12.
Department of Mathematics – FMCET – MADURAI
5.
cos t
z ( z cos T )
z 2 z cos T 1
2
TWO MARKS QUESTIONS WITH ANSWER
1.
Define Z transform
Answer:
Let {f(n)} be a sequence defined for
its Z – Transform is defined as
Z f (n)
F z
n
f (n) z
n = 0, 1,2 … and f(n) = 0 for n< 0 then
(Two sided z transform)
n
Z f ( n)
F z
f (n) z
(One sided z transform)
n
n 0
Find the Z Transform of 1
Z f n
Answer:
n
f nz
n 0
Z1
(1) z
n
1 z
1
z
2
....
n 0
1
1 z1
1
z
1
Z 1
1
z 1
z
1
z
z 1
z
z 1
2. Find the Z Transform of n
Page
65
Answer:
Department of Mathematics – FMCET – MADURAI
Z f n
f nz
n
n 0
Z n
nz
n
n 0
n
nz
0 z
1
2z
2
3z 3 ...
n 0
z
1
1 z
1
2
1
1
1
z
z
2
1 z
z z 1
2
z
z 1
3.
2
Find the Z Transform of n2.
Answer:
Z n2 Z nn
d
z
dz
4.
z
d
Z n
dz
z
z 1
2
( z)
, by the property,
z 1
2
z2 z 1
z 1
4
z2 z
( z 1)3
State Initial & Final value theorem on Z Transform
Initial Value Theorem
If Z [f (n)] = F (z) then f (0) = lim F ( z )
z
Final Value Theorem
n
lim( z 1) F ( z )
z 1
66
If Z [f (n)] = F (z) then lim f (n)
Page
6. State convolution theorem of Z- Transform.
Department of Mathematics – FMCET – MADURAI
Answer:
Z[f(n)] = F(z) and Z[g(n)] = G(z) then Z{f(n)*g(n)} = F(z) · G(z)
7. Find Z –Transform of
na n
Answer:
Z f n
n
f nz
n 0
Z na n
na n z
n
n 0
a
n
z
0
n
n
0
2
a
a
1
z
z
8. Find Z – Transform of
a
2
z
2
a
3
z
3
...
az
2
z a
sin
and
Z f n
n
2
f nz
n
n 0
z z cos
z 2 2 z cos
Z cos n
z z cos
Z cos n
2
z
2
2 z cos
1
z2
2
2
1
z2 1
67
We know that
n
2
1
Page
Answer:
cos
a
z
Department of Mathematics – FMCET – MADURAI
Z sin n
Similarly
z sin
Z sin n
2
z
9. Find Z – Transform of
2
z sin
2 z cos
z2
z
2
2 z cos
1
2
1
z2 1
1
n
Answer:
Z f n
f nz
n
n 0
1
n
1
z
n
n 0
1
z
n 1n
n
z1 z2 z3
....
1 2 3
1
log 1
z
z 1
log
z
z
z 1
10. Find Z – Transform of
Answer:
1
1
n!
68
log
n
Page
Z
Department of Mathematics – FMCET – MADURAI
Z f n
n
f nz
n 0
Z
n
1
n!
1
z
n
!
0
n
n
1
z
0 n!
n
z1 z2
1
1! 2!
e
z
1
e
11. Find Z – Transform of
Answer:
Z f n
z3
....
3!
1
z
1
n 1
n
f nz
n 0
n 1
1
n 0
n 1
1
z
n 0
n 1
1
z z
z
z
n
z
z2
2
log 1
z log
( n 1)
z3
....
3
1
z
z
z 1
12. Find Z – Transform of an
69
1
Page
Z
Department of Mathematics – FMCET – MADURAI
Answer:
Z
f n
n
f n z
n 0
Z a
n
n 0
an
zn
n 0
1
a
z
1
z
n
2
a
z
a
z
3
...
1
a
z
1
a
z
1
a
z
z
z
a
13. State and prove First shifting theorem
Statement: If Z
f t
F z
, then
Z e at f (t ) F zeaT
Proof:
Z e at f (t )
e
anT
f (nT ) z
n
n 0
As f(t) is a function defined for discrete values of t, where t = nT,
then the Z-transform is
Z f (t )
f (nT ) z
n
F ( z ) ).
n 0
Z e at f (t )
f (nT ) ze aT
n
F ( ze aT )
n 0
Define unit impulse function and unit step function.
The unit sample sequence is defined as follows:
(n)
1 for n
0
0 for n
0
70
The unit step sequence is defined as follows:
Page
14.
Department of Mathematics – FMCET – MADURAI
1 for n 0
u(n)
15.
0 for n 0
Z eat
Find Z – Transform of
Answer:
Z eat
eanT z
n
n
eaT z
n 0
n
z eaT
n
n 0
z
z eaT
z
z an
z a
[Using First shifting theorem]
16.
Find Z – Transform of
Z te
2t
Answer:
Z te
2t
Z t
Tz
z ze
2T
z 1
2
z ze 2 T
Tze 2T
ze 2T
1
2
[Using First shifting theorem]
17.
Find Z – Transform of
Z et cos 2t
Answer:
Z et cos 2t
Z cos 2t
z z cos
z ze
ze
ze
2T
T
T
z
ze
2
T
2 cos
z 1
z ze
T
cos T
2cos T ze
T
1
Page
71
[Using First shifting theorem]
Department of Mathematics – FMCET – MADURAI
18.
Find Z – Transform of
Z e2 t
T
Answer:
Let f (t) = e2t , by second sifting theorem
Z e2(t
T)
Z f (t T )
z F ( z ) f (0)
ze 2T
z
1
ze 2T 1
19.
Find Z – Transform of
Answer:
z
1
ze
2T
1
Z sin t T
Let f (t) = sint , by second sifting theorem
Z sin(t T ) Z f (t T ) z F ( z ) f (0)
z
20. Find Z – transform of
z2
z2
z 2 sin t
2cos t z 1
3z n
2z 1
z sin t
0
2cos t z 1
n 1 n 2
Answer:
Z f n
f nz
n
n 0
Z n2
Z n2
z2
2n n 2
z n2
3n 2
z
z 1
3
3
z
z 1
2
2
z
z 1
72
n 1 n 2
Page
Z
Department of Mathematics – FMCET – MADURAI
QUESTION BANK
Z-TRANSFORMS
1. (i)Find Z
8z 2
(2 z 1)(4 z 1)
1
(ii) Find Z
2. (i) Find Z
3.
1
1
& Z
z2
( z a)( z b)
z2
( z a) 2
&Z
& Z
1
1
1
8z 2
by convolution theorem.
(2 z 1)(4z 1)
z2
by convolution theorem
( z 1)( z 3)
z2
by convolution theorem
( z a)2
(i ) State and prove Initial & Final value theorem.
(ii) State and prove Second shifting theorem
(i) Find the Z transform of
z2
1
( z 2 4)
by residues.
(ii) Find the inverse Z transform of
5. (i) Find Z
z
2z 2
1
z2
&Z
(ii) Find Z
7.
1
n!
z2
(i)Solve y n 2
(i )Solve y n
6y n 1
4y n 1
3y n 1
(ii) Solve y n 3
9. (i)Find Z cos n
1
1
Hence find Z
n!
(n 1)!
and also find the value of sin(n 1)
(ii) Solve y n 2
8.
z2
7 z 10
1
6. (i)Find the Z transform of f (n)
z2 z
by partial fractions.
z 1 ( z 2 1)
3y n 1
& Z sin n
9y n
4y n
4y n 2
2y n
2n with y 0
and Z
1
(n 2)!
.
and cos(n 1) .
0& y 1
0
0 y(0) = 1 ,y(1) =0
0, n 2
0, y 0
given y(0) 3& y(1)
4, y 1
and also find Z a n cos n
0& y 2
& Z a n sin n
2
8,
73
(i) Find Z
Page
4.
1
2n 3
&
(n 1)(n 2)
(n 1)(n 2)
Department of Mathematics – FMCET – MADURAI
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