P.DE AND INTEGRAL TRASFORMS - Fmcetfmcet.in/AUTO/MA6351_uw.pdf · ma6351-transforms and partial...
Transcript of P.DE AND INTEGRAL TRASFORMS - Fmcetfmcet.in/AUTO/MA6351_uw.pdf · ma6351-transforms and partial...
MA6351-TRANSFORMS AND PARTIAL
DIFFERENTIAL EQUATIONS
SUBJECT NOTES
Department of Mathematics
FATIMA MICHAEL
COLLEGE OF ENGINEERING &
TECHNOLOGY
MADURAI – 625 020, Tamilnadu, India
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Basic Formulae
DIFFERENTIATION &INTEGRATION FORMULAE
0 Function
( )y f x
Differentiation
dy
dx
1 nx 1nnx
2 log x 1
x
3 sin x cos x
4 cos x sin x
5 axe a xe
6 C (constant) 0
7 tan x 2sec x
8 sec x sec tanx x
9 cot x 2cos ec x
10 cos ecx cos cotecx x
11 x 1
2 x
12 1sin x 2
1
1 x
13 1cos x 2
1
1 x
14 1tan x 2
1
1 x
15 1sec x 2
1
1x x
16 1cosec x 2
1
1x x
17 1cot x 2
1
1 x
18 xa logxa a
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19. If y uv , then dy du dv
v udx dx dx
20. If u
yv
, then2
du dvv u
dy dx dx
dx v
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1
2
1
2 2
2 2
1.1
2. , &
cos3. sin cos & sin
sin4. cos sin & cos
5. tan log sec log cos
6. sec tan
17. tan
18. log
2
nn
ax axx x ax ax
xx dx
n
e ee dx e e dx e dx
a a
axxdx x axdx
a
axxdx x axdx
a
xdx x x
xdx x
dx xdx
x a a a
dx x adx
x a a x
1
2 2
1
2 2
1
2 2
22 2 2 2 1
22 2 2 2 1
22 2 2 2 1
2 2
2 2
9. sin
10. sinh
11. cosh
12. sin2 2
13. sinh2 2
14. cosh2 2
15. log
216. log
a
dx xdx
aa x
dx xdx
aa x
dx xdx
ax a
x a xa x dx a x
a
x a xa x dx a x
a
x a xx a dx x a
a
dxx
x
xdxx a
x a
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3
2
3
2
17. log log
18.3
19.3
120. 2
xdx x x x
a xa x dx
a xa x dx
dx xx
21. 2 2
cos cos sinax
ax ee bxdx a bx b bx
a b
22.2 2
sin sin cosax
ax ee bxdx a bx b bx
a b
23.1 2 3........udv uv u v u v u v´ ´´ ´´´
24.0
( ) 2 ( )
a a
a
f x dx f x dx when f(x) is even
25. ( ) 0
a
a
f x dx when f(x) is odd
26. 2 2
0
cosax ae bxdx
a b
27. 2 2
0
sinax be bxdx
a b
TRIGNOMETRY FORMULA
1. sin2 2sin cosA A A
2 2
2
2
2.cos 2 cos sin
1 2sin
2cos 1
A A A
A
A
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3. 2 1 cos 2cos
2
xx & 2 1 cos 2
sin2
xx
4. sin( ) sin cos cos sin
sin( ) sin cos cos sin
cos( ) cos cos sin sin
cos( ) cos cos sin sin
A B A B A B
A B A B A B
A B A B A B
A B A B A B
15.sin cos sin( ) sin( )
2
1cos sin sin( ) sin( )
2
1cos cos cos( ) cos( )
2
1sin sin cos( ) cos( )
2
A B A B A B
A B A B A B
A B A B A B
A B A B A B
3
3
16. sin 3sin sin 3
4
1cos 3cos cos3
4
A A A
A A A
2 2
2 2
7.sin 2sin cos2 2
cos cos sin2 2
1 2sin 1 cos 2sin2 2
A AA
A AA
A AA
LOGRATHEMIC FORMULA
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log
log log log
log log log
log log
log 1 0
log 0
log 1
n
a
a
a
x
mn m n
mm n
n
m n m
a
e x
UNIT - 1
PARTIAL DIFFRENTIAL EQUATIONS
PARTIAL DIFFERENTIAL EQUTIONS
Notations
zp
x
zq
y
2
2
zr
x
2 zs
x y
2
2
zt
y
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
dx dy dz
P Q R
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dx dy dz x my nz
P Q R 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)
F( xm
p, ynq, z)=0 (1)
F( zkp, z
kq)=0
If 1& 1m n then
X = x1-m
& Y = y1-n
xm
p = P(1-m) & yn q = Q(1-n)
Using the above in (1)we get
F(P,Q) = 0 (or) F(P,Q,z) = 0
If 1k then Z = zk+1
1
k Qz q
k
Using the above in (1) We get
F(P,Q) = 0
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If m=1 & n=1 then
X= logx & Y= logy
xp = P & yq = Q
Using the above in (1) we get
F(P,Q) (or) F(P, Q, z) = 0
If k =-1 then Z = log z
p
Pz
& q
Qz
Using the above in (1)
we get
F(P,Q) = 0
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STANDARD TYPES
TYPE –1
TYPE –3(a)
TYPE –3(b)
TYPE –3(c)
TYPE –4
General form
F(p,q) = 0 (1)
General form
F(x,p,q) = 0 (1)
General form
F(y,p,q) = 0
(1)
General form
F(z,p,q) = 0
(1)
General form
F(x,y,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
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
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
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
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 (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
Singular Integral
Diff (2) partially
w.r.t cWe get,0
=1 (absurd
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
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)
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
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
from (3) and (4)
we get General
Integral
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HOMOGENEOUS LINEAR EQUATION
General form
33 2 2( ) ( , )aD bD D cDD 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 1 2 2 3 3( ) ( ) ( )y m x y m x y m x
Case (2)
If the roots are same then
C.F. = 2
1 2 3( ) ( ) ( )y mx x y mx x y mx
Case (3)
If the two roots are same and one is distinct, then
C.F = 1 2 3 3( ) ( ) ( )y mx x y mx y m x
Function PI =
1
1( , )
( , )F x y
F D D
F(x,y) = eax+by
Put D = a & D1
= b
F(x,y)= sin(ax+by)(or)
Cos (ax+by)
Put
22 2 2( ), ( )& ( )D a DD ab D b
F(x,y) = xr
ys
PI=
1
( , ) r sF D D x y
Expand and operating D & D
on xr
ys
F(x,y) = eax+by
f(x,y) Put D = D+a & D = D +b
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Particular Integral
F(x,y)=ex+y
cosh(x+y) F(x ,y)=2 21
2
x ye e
F(x,y)=ex+y
sinh(x+y) F(x, y) =2 21
2
x ye e
F(x,y)=sin x cos y
1( , ) sin( ) sin( )
2F x y x y x y
F(x,y)= cos x sin y
1( , ) sin( ) sin( )
2F x y x y x y
F(x,y)= cos x cos y
1( , ) s( ) s( )
2F x y co x y co x y
F(x,y)= sin x sin y
1( , ) cos( ) cos( )
2F x y x y x y
Note:
D represents differentiation with respect to ‘x ‘
D represents differentiation with respect to ‘y ‘
1
D
represents integration with respect to ‘x ‘
1
D
represents integration with respect to ‘y ‘
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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 z
p qx 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 &z z
p qx y
p = ( ).f xy y q = ( ).f xy x
p/q = y/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 &z z
p qx y
p = naxn-1
; q = nbyn-1
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1 1
1 1
;n n
n n
n n
p qa b
nx ny
p qz x y
nx ny
nz px qy
4. Eliminate the arbitrary function f from xy
z fz
and form the PDE
Answer:
xy
z fz
Diff. w .r. t. and y here &z z
p qx y
2
2
.
.
.
xy z xpp f y
z z
xy z yqq f x
z z
p y z xp
q x z yq
0
pxz pqxy qyz pqxy
px qy
5. Find the complete integral of p + q =pq
Answer:
Put p = a, q = b
p + q =pq a+b=ab
b – ab = -a
1 1
a ab
a a
The complete integral is z= ax+
1
a
ay +c
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6. Find the solution of 1p q
Answer:
z = ax+by+c ----(1) is the required solution
given 1p q -----(2)
put p=a, q = b in (2)
2
2
1 1 (1 )
(1 )
a b b a b a
z a x a y c
7. Find the General solution of p tanx + q tany = tanz.
Answer:
1 2
1 2
tan tan tan
cot cot cot
cot cot cot cot
logsin logsin log logsin logsin log
sin sin
sin sin
sin sin, 0
sin sin
dx dy dz
x y z
x dx y dy z dz
take x dx y dy y dy zdz
x y c y z c
x yc c
y z
x y
y z
8. Eliminate the arbitrary function f from 2 2z f x y and form the PDE.
Answer:
2 2z f x y
2 2 2 22 ; ( 2 )
20
2
p f x y x q f x y y
p xpy qx
q y
9. Find the equation of the plane whose centre lie on the z-axis
Answer:
General form of the sphere equation is
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22 2 2x y z c r (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 y
p q
That is py -qx =0 which is a required PDE.
10. Eliminate the arbitrary constants 2 2z ax by a b and form the PDE.
Answer:
2 2z ax by a b
2 2
;p a q b
z px qy p q
11. Find the singular integral of z px qy pq
Answer:
The complete solution is z ax by ab
0 ; 0
;
( ) ( ) ( . )
0
z zx b y a
a b
b x a y
z y x x y y x
xy xy xy
xy
xy z
12. Find the general solution of px+qy=z
Answer:
The auxiliary equation is
dx dy dz
x y z
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From
dx dy
x y Integrating we get log x = log y + log c
on simplifying 1
xc
y.
2
dy dz yc
y z z
Therefore , 0
x y
y z is general solution.
13. Find the general solution of px2
+qy2
=z2
Answer:
The auxiliary equation is 2 2 2
dx dy dz
x y z
From 2 2
dx dy
x y Integrating we get 1
1 1c
y x
Also 2 2
dy dz
y z Integrating we get 2
1 1c
z y
Therefore 1 1 1 1
, 0y x z y
is general solution.
14. Solve 2 22 3 0D DD D z
Answer:
Auxiliary equation is
2 2 3 0m m
3 1 0m m
1, 3m m
The solution is 1 2 3z f y x f y x
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15. Solve 2 24 3 x yD DD D z e
Answer:
Auxiliary equation is 2 4 3 0m m
3 1 0m m
1, 3m m
The CF is 1 2 3CF f y x f y x
2 2
1
4 3
x yPI eD DD D
Put 1, 1D D Denominator =0.
2 4
x yxPI e
D D
2
x yxe
Z=CF + PI
1 2 3z f y x f y x
2
x yxe
16. Solve. 2 23 4 x yD DD D z e
Answer:
Auxiliary equation is 2 3 4 0m m
4 1 0m m
C.F is = f1
(y + 4x) + f2
(y - x)
2 2
1 1 1
3 4 1 3 4 6
x y x y x yPI e e eD DD D
17. Find P.I 2 2 24 4 x yD DD D z e
Answer:
2
2 2
1
4 4
x yPI eD DD D
Put 2, 1D D
2
2
1
2
x yPI eD D
2
2
1
2 2
x ye
2
16
x ye
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18. Find P.I 2 2 26D DD D z x y
Answer:
2
22
2
1
61
PI x yD D
DD D
2
2
11
Dx y
D D
32
2
1
3
xx y
D
4 5
12 60
x y x
19. Find P.I
2 2
2sin
z zx y
x x y
Answer:
2
1PI Sin x y
D DD Put
2 1, (1)( 1) 1D DD
1
1 1Sin x y
2
Sin x y
20. Solve 3 33 2 0D DD D Z
Answer:
Auxiliary equation is 3 3 2 0m m
21 2 0m m m
1 2 1 0m m m
1,1 2m m
The Solution is 1 2 3 2CF f y x x f y x f y x
FOR PRACTICE:
1. Eliminating arbitrary constants
2 2 2
2 2 21
x y z
a b c
2. Solve
2
2sin
zy
x
3. Find the complete the solution of p. d .e 2 2 4 0p q pq
4. Form p.d.e eliminating arbitrary function from 2 ,
2
xz xy
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5. Find the singular soln of 2 2 1z px qy p q
1. (i) Solve 2 2 2x y z p y z x q z x y
(ii) Solve 2 2 2 2 2 2x z y p y x z q z y x
2. (i) Solve z z
mz ny nx lz ly mxx y
(ii) Solve 3 4 4 2 2 3z y p x z q y x
3. (i) Solve 2 2 2 2 2x y z p xyq xz
(ii) Solve 2 2 2 2 2 0y z x p xyq zx
4. (i) Solve y z p z x q x y
(ii) Solve y z p z x q x y
5. Solve 2 2 3 23 2 sin(3 2 )x yD DD D e x y
6. Solve 2 2
2cos cos 2
z zx y
x x y
7. Solve 2 26 cosD DD D z y x
8. Solve 2 2 630 x yD DD D z xy e
9. Solve 2 26 5 sinhxD DD D z e y xy
10. Solve 2 2 24 4 x yD DD D z e
11. Solve 3 2 2 3 2 cos( )x yD D D DD D z e x y
12. Solve (i) 2 21z px qy p q
(ii) 2 2z px qy p q
13. Solve 2 2 21z p q
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14. Solve 2 2 2 2( ) 1z p x q
15. Solve (i) 2 2 2 2( )z p q x y
(ii) 2 2 2 2 2( )z p q x y
UNIT - 2
FOURIER SERIES
0
1
( ) cos sin2
n n
n
af x a nx b nx
(0,2 ) ( - , )
Even (or) Half range
Fourier co sine series
Odd (or) Half range
Fourier sine series
Neither even nor odd
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2
0
0
1( )a f x dx
0
0
2( )a f x dx
0 0a 0
1( )a f x dx
2
0
1( )cosna f x nxdx
0
2( )cosna f x nxdx
0na 1( )cosna f x nxdx
2
0
1( )snb f x innxdx
bn=0
0
2( )snb f x innxdx
1( )snb f x innxdx
0
1
( ) cos sin2
n n
n
a n x n xf x a b
(0,2 ) ( - , )
Even (or) Half range
Fourier cosine series
Odd (or) Half range
Fourier sine series
Neither even nor odd
2
0
0
1( )a f x dx
0
0
2( )a f x dx
0 0a 0
1( )a f x dx
2
0
1( )cosn
n xa f x dx
0
2( )cosn
n xa f x dx
0na 1( )cosn
n xa f x dx
2
0
1( )sn
n xb f x in dx
bn=0
0
2( )sn
n xb f x in dx
1( )sn
n xb f x in dx
Even and odd function:
Even function:
f(-x)=f(x)
eg : cosx,x2
, , , sin , cosx x x are even functions
Odd function:
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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 ) = (0) (2 )
2
f f
In the interval (0,2 ) if x = 0 or x = 2 then
f(0) = f(2 ) = (0) (2 )
2
f f
In the interval (- , ) if x = - or x = then
f(- ) = f( ) = ( ) ( )
2
f f
In the interval (- , ) if x = - or x = then
f(- ) = f( ) = ( ) ( )
2
f f
HARMONIC ANALYSIS
f(x)= 0
2
a + a1 cosx +b1sinx + a2cos2x + b2sin2x ……… for form
0 2y
an
1
cos2
y xa
n, 2
cos 22
y xa
n 1
sin2
y xb
n, 2
sin 22
y xb
n
f(x)= 0
2
a + a1 cos
x
+b1
xsin
+ a2
2cos
x
+ b2
2 xsin
………( form)
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0 2y
an
1
cos
2
xy
an
2
2cos
2
xy
an
1
sin
2
xy
bn
,2
2sin
2
xy
bn
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
21( )
b
a
y f x dxb a
2. State Parseval’s Theorem.
Let f(x) be periodic function with period 2l defined in the interval (c, c+2l).
2 22 2 2
1
1 1( )
2 4 2
c l
on n
nc
af x dx a b
lWhere , &o n na a b 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.
(iii) f (x) has at most a finite number of maxima and minima.
5. State Euler’s formula.
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Answer:
( , 2 )
cos sin2
on n
In c c l
af x a nx b nx
2
2
2
1( )
1( )cos
1( )sin
c l
o
c
c l
n
c
c l
n
c
where a f x dx
a f x nxdx
b f x nxdx
6. Write Fourier constant formula for f(x) in the interval (0,2 )
Answer:
2
0
2
0
2
0
1( )
1( )cos
1( )sin
o
n
n
a f x dx
a f x nxdx
b f x nxdx
7. In the Fourier expansion of
f(x) =
21 , 0
21 ,0
xx
xx
in (-π , π ), find the value of nb
Since f(-x)=f(x) then f(x) is an even function. Hence nb = 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
The required constant term of the Fourier series = ao
= 0
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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 = 0a = 0
10. Find the value of a0
for f(x) = 1+x+x2
in ( 0 ,2 )
Answer:
2
0
1( )oa f x dx
22 2 32
0 0
2 3 2
1 1(1 )
2 3
1 4 8 82 2 2
2 3 3
x xx x dx x
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 nb = 0
(ii) Given f(x) = xsinx f(-x) = (-x)sin(-x)
= xsinx = f(x)
Hence f(x) is an even function
In the Fourier series nb = 0
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12. Obtain the sine series for
0 / 2
/ 2
x x lf x
l x l x l
Given
0 / 2
/ 2
x x lf x
l x l x l
Answer:
Given
0 / 2
/ 2
x x lf x
l x l x l
Fourier sine series is sinn
nxf x b
l
0
2
0 2
2
2 2
2 2
0 2
2( )sin
2sin ( )sin
cos sin cos sin2
(1) ( ) ( 1)
l
n
l l
l
l l
l
nxb f x dx
l l
nx nxx dx l x dx
l l l
nx nx nx nx
l l l llx l l l x ll n n n n
2 2 2 2
2 2 2 2
2
2 2 2 2
2 cos 2 sin 2 cos 2 sin 2
2 2
2 2 sin 2 4 sin 2
l n l n l nl l n
l n n n n
l n l n
l n n
Fourier series is 2 21
4 sin 2sin
n
l n n xf x
n l
13. If f(x) is odd function in ,l l . What are the value of a0
&an
Answer:
If f(x) is an odd function, ao
= 0, an
= 0
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14. In the Expansion f(x) = |x| as a Fourier series in (- . ) find the value of a0
Answer:
Given f(x) = |x|
f(-x) = |-x|
= |x| = f(x)
Hence f(x) is an even function
2 2
0 0
2 2 2
2 2o
xa xdx
15. Find half range cosine series of f(x) = x, in 0 x
Answer:
2 2
0 0
2 2 2
2 2o
xa xdx
2
0 0
1
2 1 cos sinsin (1)
1 11 cos0 0
n
n n
nx nxa x nxdx x
n n
n
n n n
Fourier series is
0
1
0
cos2
1cos
2
on
n
n
n
af x a nx
nxn
16. Find the RMS value of f(x) = x2
, 0 1x
Answer:
Given f(x) = x2
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R.M.S value
2 122 2
0 0
15
0
1 1( )
1 2
22
5 5
l
y f x dx x dxl
x
17. Find the half range sine series of ( )f x x in (0, )
Answer:
0
2( )sinnb f x nxdx
2
0 0
1
2 2 cos sinsin (1)
2 ( 1) 2( 1)n n
nx nxx nxdx x
n n
n n
Half range Fourier sine series is
1
0
2( 1)sin
n
n
f x nxn
18. Find the value of a0
in the cosine series of ( )f x x in (0, 5)
Answer:
55 2 2
0 0
2 2 2 55
5 5 2 5 2o
xa xdx
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 , , sin , cosx x x are even functions
(ii) If ( ) ( )f x f x then the function is an odd function.
eg : sinx,x3 ,sinhx, tanx are odd functions
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20. Write the first two harmonic.
Answer:
1 1 2 2cos sin cos 2 sin 22
o
The first twoharmonics are
af x a x b x a x b x
FOURIER SERIES
1. Expand
(0, )( )
2 ( ,2 )
xf x
x as Fourier series
and hence deduce that
2
2 2 2
1 1 1.........
1 3 5 8
2. Find the Fourier series for f(x) = x2
in (- . ) and also prove that
(i)
2
2 2 2
1 1 1.........
1 2 3 6 (ii)
2
2 2 2
1 1 1.........
1 2 3 12
3. (i) Expand f(x) = | cosx | as Fourier series in (- . ).
(ii) Find cosine series for f(x) = x in (0, ) use Parsevals identity to
Show that
4
4 4 4
1 1 1.........
1 2 3 90
4. (i) Expand f(x) = xsinx as a Fourier series in (0,2 )
(ii) Expand f(x) = |x| as a Fourier series in (- . ) and deduce to
2
2 2 2
1 1 1.........
1 3 5 8
5. If
0 , ( ,0)( )
sin , (0, )f x
x Find the Fourier series and hence deduce that
1 1 1 2
.........1.3 3.5 5.7 4
6. (i) Find the Fourier series up to second harmonic
X 0 1 2 3 4 5
Y 9 18 24 28 26 20
(ii)Find the Fourier series up to third harmonic
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X 0 π/3 2π/3 π 4π/3 5π/3 2π
F(x) 10 14 19 17 15 12 10
7. (i) Find the Fourier expansion of 2( ) ( )f x x in (0,2 ) and
Hence deduce that
2
2 2 2
1 1 1.........
1 2 3 6
(ii). Find a Fourier series to represent
2( ) 2f x x x with period 3
in the range (0,3)
(ii) Find the Fourier series of xf x e in ( , ) .
(ii) Find the Fourier series for
1 (0, )
2 ( ,2 )
inf x
in
and hence show that
2
2 2 2
1 1 1.........
1 3 5 8
8. (i) Find the the half range sine series for f x x x in the interval (0, ) and deduce
that 3 3 3
1 1 1....
1 3 5
(ii) Obtain the half range cosine series for 2
1f x x in (0,1)
and also deduce that
2
2 2 2
1 1 1.........
1 2 3 6
9. (i) Find the Fourier series for f(x) = x2
in (- . ) and also prove that
4
4 4
1 11 .........
1 2 90 (use P.I)
(ii) Find the Fourier series for f(x) = x in (- . ) and also prove that
4
4 4
1 11 .........
1 3 96 (use P.I)
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10(i)Obtain the sine series for
,02
,2
lcx x
f xl
c l x x l
(ii). Find the Fourier series for the function
,02
2 ,2
lkx x
f xl
k l x x l
11.(i).Find the Fourier series for the function 21 ( , )f x x x in and also
deduce that
2
2 2 2
1 1 1.........
1 2 3 6
(ii) Find the Fourier expansion of
f(x) =
21 , 0
21 ,0
xx
xx
in (-π , π ), and also deduce that 2
2 2 2
1 1 1.........
1 3 5 8
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UNIT - 3
APPLICATIONS OF P.D.E
S.
N
O
ONE DIMENSIONAL WAVE EQUATION
VELOCITY MODEL INITIAL POSITION MODEL
1 STEP-1
One Dimensional wave equation
is
2 22
2 21
y ya
t x
STEP-1
One Dimensional wave equation
is
2 22
2 21
y ya
t x
2 STEP-2
Boundary conditions
1. y(0,t) = 0 for 0t
2. y( , t) = 0 for 0t
3. y(x,0) = 0 for 0 < x <
4. 0t
y
t= f(x) for 0 < x <
STEP-2
Boundary conditions
1. y(0,t) = 0 for 0t
2. y( , t) = 0 for 0t
3. 0t
y
t = 0 for 0 < x <
4. y(x,0) = f(x) for 0 < x <
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3 STEP-3
The possible solutions are
y(x,t) = (A ex + 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 ex + 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-4
The suitable solution for the given
boundary condition is
y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)
(2)
5 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-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)
6
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
= n
Then (3) becomes,
( , ) sin( ) cos( ) sin( )n x n at n at
y x t B C D
(4)
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
= n
Then (3) becomes,
( , ) sin( ) cos( ) sin( )n x n at n at
y x t B C D
(4)
7 STEP-7
Using Boundary condition 3
y(x,0) = 0
Then (4) becomes,
STEP-7
Using Boundary condition 3
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( , ) sin( ) cos0 sin 0n x
y x t B C D
=0
sin( ) 0n x
B C
C = 0
Then (4) becomes,
( , ) sin( ) sin( )n x n at
y x t B D
The most general solution is
1
( , ) sin( )sin( )n
n
n x n aty x t B
(5)
0t
y
t= 0Then (4) becomes,
Differentiating (5) partially w.r.to ‘t’ and put t =0
0
sin( ) sin( ) cos( )t
y n x n at n at n aB C D
t
sin( ) 0n x n a
B D
D = 0
Then (4) becomes,
( , ) sin( ) cos( )n x n at
y x t B C
The most general solution is
1
( , ) sin( ) cos( )n
n
n x n aty x t B
(5)
8 STEP-8
Differentiating (5) partially w.r.to ‘t’
1
sin( ) cos( )n
n
y n x n at n aB
t
Using Boundary condition (4),
0t
y
t= f(x)
1
( ) sin( )n
n
n x n af x B
This is the Half Range Fourier Sine Series.
0
2( )sin( )n
n a n xB f x
0
2( )sin( )n
n xB f x dx
n a
STEP-8
Using Boundary condition (4),
y(x,0) = f(x)
1
( ,0) sin( )cos(0)n
n
n xy x B
1
( ) sin( )n
n
n xf x B
This is the Half Range Fourier Sine Series.
0
2( )sin( )n
n xB f x
9 STEP-9
The required solution is
1
( , ) sin( )sin( )n
n
n x n aty x t B
Where
0
2( )sin( )n
n xB f x dx
n a
STEP-9
The required solution is
1
( , ) sin( )sin( )n
n
n x n aty x t B
Where
0
2( )sin( )n
n xB f x dx
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ONE DIMENSIONAL HEAT EQUATION
TWO DIMENSIONAL HEAT FLOW EQUATION
1 The one dimensional heat equation is
22
2
u u
t x
The two Dimensional equation is
2 2
2 20
u u
x y
2 Boundary conditions
1.u(0,t) = 0 for 0t
2.u( ,t) = 0 for 0t
3.u(x,t) = f(x) for 0<x<
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<
3 The possible solutions are 2 2
2 2
( , ) ( )
( , ) ( cos sin )
( , ) ( )
x x t
t
u x t Ae Be Ce
u x t A x B x Ce
u x t Ax B C
The possible solutions are
( , ) ( )( cos sin )
( , ) ( cos sin )( )
( , ) ( )( )
x x
y y
u x y Ae Be C y D y
u x y A x B x Ce De
u x y Ax B Cy D
4, The most suitable solution is 2 2
( , ) ( cos sin ) tu x t A x B x Ce (2)
The most suitable solution is
( , ) ( cos sin )( )y yu x y A x B x Ce De (2)
5 Using boundary condition 1
u(0,t) = 0 2 2
(0, ) ( cos0 sin0) tu t A B Ce =0
2 2
( ) tA Ce =0
A = 0
Then (2) becomes 2 2
( , ) ( sin ) tu x t B x Ce (3)
Using boundary condition 1
u(0,y) = 0
(0, ) ( cos0 sin 0)( )y yu y A B Ce De
(0, ) ( )( )y yu y A Ce Be
A = 0
Then (2) becomes
( , ) ( sin )( )y yu x y B x Ce De (3)
6 Using boundary condition 2
u(l,t) = 0 2 2
( , ) ( sin ) tu t B Ce =0
2 2
( sin ) 0tB Ce
n
Then (3) becomes 2 2 2
2
( , ) sin( )
n tn x
u x t B Ce
The most general solution is 2 2 2
2
1
( , ) sin( )
n t
n
n
n xu x t B e
(4)
Using boundary condition 2
u(l,t) = 0
( , ) ( sin )( )y yu y B Ce De
0 ( sin )( )y yB Ce De
n
Then (3) becomes
( , ) ( sin )( )n y n y
n xu x y B Ce De
(4)
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7 Using Boundary condition 3
u(x,0) = f(x)
1
( ,0) sin( )n
n
n xu x B
1
( ) sin( )n
n
n xf x B
This the Half range Fourier sine series
0
2( )sin( )n
n xB f x dx
Using Boundary condition 3
u(x, ) =0
( , ) ( sin )( )n x
u x B Ce De
0 ( sin )( 0)n x
B C D
C=0
then (3) becomes
( , ) ( sin )( )n y
n xu x y B De
The most general solution is
1
( , ) sin( )n y
n
n
n xu x y B e
(5
8 The required solution is 2 2 2
2
1
( , ) sin( )
n t
n
n
n xu x t B e
Where
0
2( )sin( )n
n xB f x dx
Using Boundary condition 4 y(x,0) = f(x)
0
1
( ,0) sin( )n
n
n xu x B e
1
( ) sin( )n
n
n xf x B
This the Half range Fourier sine series
0
2( )sin( )n
n xB f x dx
The required solution is
1
( , ) sin( )n y
n
n
n xu x y B e
Where
0
2( )sin( )n
n xB f x dx
QUESTION WITH ANSWER
1. Classify the Partial Differential Equation i)
2 2
2 2
u u
x y
Answer:
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2 2
2 2
u u
x y here A=1,B=0,&C=-1
B2
- 4AC=0-4(1)(-1)=4>0
The Partial Differential Equation is hyperbolic
2. Classify the Partial Differential Equation
2u u uxy
x y y x
Answer:
2u u uxy
x y y x 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 22 2
2 2
u u u u
x y y x
Answer:
2 22 2
2 2
u u u u
x y y x 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 2 2
2 24 4 6 8 0
u u u u u
x x y y x y
Answer:
2 2 2
2 24 4 6 8 0
u u u u u
x x y y x y
here A= 4,B =4, & C = 1
B2
-4AC =16 -4(4)(1) = 0
The Partial Differential Equation is Parabolic
5. Classify the following second order Partial Differential equation
i) 2 22 2 3 0xx xy yy xy u xyu x u u u
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ii) 2 2 2 7 0xx yy x yy u u u u
Answer:
i) Parabolic ii) Hyperbolic (If y = 0)
iii)Elliptic (If y may be +ve or –ve)
6. In the wave equation
2 22
2 2
y yc
t x what does c
2
stands for?
Answer:
2 22
2 2
y yc
t x
here 2 T
am
T-Tension and m- Mass
7. In one dimensional heat equation ut
= α2
uxx
what does α2
stands for?
Answer:-
22
2
u u
t x
2
=k
c is called diffusivity of the substance
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
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
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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
(iii)
0
0t
y
t
(iv) y(x , 0 ) =
0
(2 ) 2
bx x
bx x
10. What are the possible solutions of one dimensional Wave equation?
The possible solutions are
Answer:
y(x,t) = (A ex
+ B e- x
) (C eat
+ 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
2 2
2 2
( , ) ( )
( , ) ( cos sin )
( , ) ( )
x x t
t
u x t Ae Be Ce
u x t A x B x Ce
u x t Ax B C
12. State Fourier law of heat conduction
Answer:
uQ kA
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)
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Q=Quantity of heat flowing
k – Thermal conductivity
A=area of cross section ;u
x=Temperature gradient
13. What are the possible solutions of two dimensional head flow equation?
Answer:
The possible solutions are
( , ) ( )( cos sin )
( , ) ( cos sin )( )
( , ) ( )( )
x x
y y
u x y Ae Be C y D y
u x y A x B x Ce De
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 ( , )b a x
u x t al
80 2020
60
x
( , ) 20u x t x
16. Write the D’Alembert’s solution of the one dimensional wave equation?
Answer:
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1 1( )
2 2
x at
x at
y x at x at v da
here x f x g x
v x ax f ag
17. What are the boundary conditions of one dimensional Wave equation?
Answer:
Boundary conditions
1. y(0,t) = 0 for 0t
2. y( , t) = 0 for 0t
3. y(x,0) = 0 for 0 < x <
4.
0t
y
t= f(x) for 0 < x <
18. What are the boundary conditions of one dimensional heat equation?
Answer:
Boundary conditions
1.u(0,t) = 0 for 0t
2.u( ,t) = 0 for 0t
3.u(x,t) = f(x) for 0<x<
19. What are the boundary conditions of one dimensional heat equation?
Answer:
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<
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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 ( , )b a x
u x t al
Here a=40°C & b=60°C
60 40 240 40
30 3t
xu x
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) =3
0 sinx
yl
. 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
0 < x < 1
(2 ) 0 < x < 1
cx
lV
cl x
l
Find the
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.
7. A rod 30 cm long has its end A and B kept at 20o
C and 80o
C, respectively until steady state
condition prevails. The temperature at each end is then reduced to 0o
C and kept so. Find
the resulting temperature u(x, t) taking x = 0.
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8. A bar 10 cm long , with insulated sides has its end A & B kept at 20o
C and 40o
C respectively
until the steady state condition prevails. The temperature at A is suddenly raised to 50o
C
and B is lowered to 10o
C. 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) = 100sin8
x 0 < x < 1. While two edges x = 0 and x = 8 as well as the other short
edges are kept at 0o
C. Find the steady state temperature.
10. 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
20 0 x 5
20(10 ) 5 x 10
xu
x and all other three edges are kept at 0
o
C. Find the steady
state temperature at any point of the plate.
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Unit - 4
FOURIER TRANSFORMS
FORMULAE
1. Fourier Transform of f(x) is isx
-
1[ ( )] f(x)e
2F f x dx
2. The inversion formula -isx
-
1( ) ( )e
2f x F s ds
3. Fourier cosine Transform Fc
[f(x)] = Fc
(s) =
0
2( )cosf x sxdx
4. Inversion formula f(x) =
0
2( )coscF s sxds
5. Fourier sine Transform (FST) Fs
[f(x)] = Fs
(s) =
0
2( )sinf x sxdx
6. Inversion formula f(x) =
0
2( )sinsF s sxds
7. Parseval’s Identity
2
2( ) ( )f x dx F s ds
8. Gamma function 1
0
1, 1 &
2
n xn x e dx n n n
9. 2 2
0
cosax ae bxdx
a b
10 2 2
0
sinax be bxdx
a b
11. 0
sin
2
axdx
x
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12. 2 2
0
&2
x xe dx e dx
13. cos & sin2 2
iax iax iax iaxe e e eax ax
ORKING RULE TO FIND THE FOURIER TRANSFORM
Step1: Write the FT formula.
Step2: Substitute given f(x) with their limits.
Step3: Expand isxe 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 , in the formula
Step3: Expand isxe 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
2( ) ( )f x dx 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
Step2: Simplify, we get ( )C
F S
WORKING RULE TO FIND INVERSE FCT
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Step1: Write the inverse FCT formula & Sub ( )C
F 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 ( )SF S
WORKING RULE TO FIND INVERSE FCT
Step1: Write the inverse FST formula & Sub ( )sF S with limit in the formula
Step2: Simplify, we get f(x)
WORKING RULE FOR f(x) = axe
Step:1 First we follow the above FCT & FST working rule and then we get this
result
Fc
(e-ax
) = 2 2
2 a
a s F
s
(e-ax
) = 2 2
2 s
a s
By Inversion formula, By Inversion formula,
2 2
0
cos
2
axsxds e
a s a
2 2
0
sin2
axssxds e
a s
TYPE-I : If problems of the form i)2 2
x
x a ii)
2 2
1
x a , then use Inversion formula
TYPE-II: If problems of the form i)
2
22 2
0
xdx
x a ii)
22 2
0
dx
x a, then use Parseval’s Identity
TYPE-III
2 2 2 2
0
dx
x a x b , then use
0 0
( ) ( ) ( ) ( )C Cf x g x dx F f x F g x dx
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UNIT - 4
FOURIER TRANSFORM
1. State Fourier Integral Theorem.
Answer:
If ( )f x is piece wise continuously differentiable and absolutely on , then,
( )1( )
2
i x t sf x f t e dt ds.
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2. StateandproveModulation
theorem.
1cos
2F f x ax F s a F s a
Proof:
( ) ( )
1cos cos
2
1
22
1 1 1 1
2 22 2
1 1
2 2
1cos
2
isx
iax iaxisx
i s a x i s a x
F f x ax f x ax e dx
e ef x e dx
f x e dx f x e dx
F s a F s a
F f x ax F s a F s a
3. State Parseval’s Identity.
Answer:
If F s is a Fourier transform of f x , then
2 2
F s ds f x dx
4. State Convolution theorem.
Answer:
The Fourier transform of Convolution of f x and g x is the product of their Fourier
transforms.
F f g F s G s
5. State and prove Change of scale of property.
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Answer:
If ,F s F f x then
1s
aF f ax Fa
1
2
1; where
2
sa
isx
i t
F f ax f ax e dx
dtf t e t ax
a
1 sF f ax Faa
6. Prove that if F[f(x)] = F(s) then ( ) ( ) ( )
nn n
n
dF x f x i F s
ds
Answer:
1
2
isxF s f x e dx
Diff w.r.t s ‘n’ times
1
2
1( ) ( )
2
nn isx
n
n n isx
dF s f x ix e dx
ds
f x i x e dx
1 1( )
( ) 2
1( ) ( )
2
nn isx
n n
nn n isx
n
nnn
n
dF s x f x e dx
i ds
di F s x f x e dx
ds
dF x f x i F s
ds
7. Solve for f(x) from the integral equation
0
( )cos sf x sxdx e
Answer:
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0
( )cos sf x sxdx e
0
2cos
2
c
s
c
F f x f x sx dx
F f x e
0
0
2
0
2( ) cos
2 2cos
2 2 1cos
1
c
s
s
f x F f x sx ds
e sx ds
e sx dsx
2 2
0
cos
1,
ax ae bx dx
a b
a b x
8. Find the complex Fourier Transform of
1( )
0 0
x af x
x a
Answer:
1
2
isxF f x f x e dx
;x a a x a
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11
2
1(cos sin )
2
a
isx
a
a
a
F f x e dx
sx i sx dx
00
2 2 sin(cos )
2 2
2 sin
aasx
sx dxs
as
s
[Use even and odd property second term become zero]
9. Find the complex Fourier Transform of ( )0 0
x x af x
x a
Answer:
1
2
1
2
1(cos sin )
2
isx
a
isx
a
a
a
F f x f x e dx
xe dx
x sx i sx dx
;x a a x a
2
0 0
2
2 2 cos sin( ( sin ) (1)
2 2
2 cos sin
aai sx sx
x i sx dx xs s
as as asi
s
[Use even and odd property first term become zero]
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10. Write Fourier Transform pair.
Answer:
If ( )f x is defined in , , then its Fourier transform is defined as
1
2
isxF s f x e dx
If F s is an Fourier transform of f x , then at every point of Continuity of f x , we
have
1
2
isxf x F s e ds .
11. Find the Fourier cosine Transform of f(x) = e-x
Answer:
0
0
2
2cos
2cos
2 1
1
c
x x
c
x
c
F f x f x sx dx
F e e sx dx
F es
2 2
0
cosax ae bx dx
a b
12. Find the Fourier Transform of
,( )
0,
imxe a x bf x
otherwise
Answer:
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1
2
1 1
2 2
isx
b bi m s ximx isx
a a
F f x f x e dx
e e dx e dx
1 1 1
2 2
bi m s x
i m s b i m s a
a
ee e
i m s i m s
13. Find the Fourier sine Transform of 1
x.
Answer:
0
0
2sin
2 sin 2
2
1
2
s
s
F f x f x sx dx
sxdx
x
Fx
14. Find the Fourier sine transform of xe
Answer:
0
0
2
2sin
2sin
2
1
s
x x
s
x
s
F f x f x sx dx
F e e sx dx
sF e
s
2 2
0
sinax be bx dx
a b
15. Find the Fourier cosine transform of 2 2x xe e
Answer:
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0
2coscF f x f x sx dx
2 2
0
2
0 0
22 2 cos
2cos 2 cos
x x x x
c
x x
F e e e e sx dx
e sx dx e sx dx
2 2 2 2
2 2 1 2 1 12 2
4 1 4 1s s s s
16. Find the Fourier sine transform of
1 , 0 1( )
0 1
xf x
x
Answer:
0
1
0 1
11
00
2sin
2sin sin
2 2 cos1sin 0
2 1 cos
s
s
F f x f x sx dx
F f x f x sx dx f x sx dx
sxsx dx
s
s
s
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17. Obtain the Fourier sine transform of
, 1
( ) 2 , 1 2
0, 2
x o x
f x x x
x.
Answer:
0
1 2
0 1
1 2
2 2
0 1
2sin
2sin 2 sin
2 cos sin cos sin2
sF f x f x sx dx
x sx dx x sx dx
sx sx sx sxx x
s s s s
2 2 2
2
2 cos sin sin 2 cos sin
2 2sin sin 2
s
s s s s sF f x
s s s s s
s s
s
18. Define self reciprocal and give example.
If the transform of f x is equal to f s , then the function f x is called self
reciprocal.
2
2x
e is self reciprocal under Fourier transform.
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19. Find the Fourier cosine Transform of
0( )
0
x xf x
x
Answer:
0 0
2 2 2
0
2
2 2cos cos
2 sin cos 2 cos 1sin
2 sin cos 1
cF f x f x sx dx x sx dx
sx sx sx s
s s s s s
s s s
s
20. Find the Fourier sine transform of 2 2
x
x a.
Answer:
L et
axf x e
2 2
2ax
s
sF e
s a
Using Inverse formula for Fourier sine transforms
2 2
0
2 2
0
2 2sin
( ) sin , 02
ax
ax
se sx ds
s a
sie sx ds e a
s a
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Change x and s, we get 2 2
0
sin2
asxsx dx e
x a
2 2 2 2
0
2sin
2
2 2
s
as as
x xF sx dx
x a x a
e e
FOURIER TRANSFORM
PART-B
1. (i)Find the Fourier Transform of
21 1( )
0 1
x if xf x
if x and hence
deduce that (i)3
0
cos sin 3cos
2 16
x x x xdx
x (ii)
2
3
0
sin cos
15
x x xdx
x
(ii). Find the Fourier Transform of
2 2
( )0 0
a x x af x
x a . hence
deduce that 3
0
sin cos
4
x x xdx
x
2. Find the Fourier Transform of
1( )
0
if x af x
if x a and hence evaluate
0
sin)
xi dx
x ii)
2
0
sin xdx
x
4. Find Fourier Transform of
1 1( )
0 1
x if xf x
if x and hence evaluate
i)
2
0
sin xdx
x ii)
4
0
sin xdx
x
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5. Evaluate i)
2
22 2
0
xdx
x a ii) 2
2 20
dx
x a
6 i). Evaluate (a) 2 2 2 2
0
dx
x a x b(b)
2
2 2 2 2
0
x dx
x a x b
ii). Evaluate (a) 2 2
0 1 4
dx
x x(b)
2
2 2
0 4 9
t dt
t t
7. (i)Find the Fourier sine transform of
sin ;( )
0 ;
x when o xf x
whenx
(ii) Find the Fourier cosine transform of
cos ;( )
0 ;
x when o x af x
whenx a
8. (i) Show that Fourier transform
2
2
x
e is
2
2
s
e
(ii)Obtain Fourier cosine Transform of
2 2a xe and hence find Fourier sine Transform x
2 2a xe
9. (i) Solve for f(x) from the integral equation
0
( )cosf x xdx e
(ii) Solve for f(x) from the integral equation
0
1 ,0 1
( )sin 2 ,1 2
0 , 2
t
f x tx dx t
t
10. (i) Find Fourier sine Transform of xe , x>0 and hence deduce that
2
0
sin
1
x xdx
x
(ii) Find Fourier cosine and sine Transform of4xe , x>0 and hence deduce
that8 8
2 2
0 0
cos2 sin 2( ) ( )
16 8 16 8
x x xi dx e ii dx e
x x
11.(i)Find &ax ax
S cF xe F xe
(ii) Find &ax ax
S c
e eF F
x x
(iii) Find the Fourier cosine transform of ( ) cosaxf x e ax
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Z - TRANSFORMS
Definition of Z Transform
Let {f(n)} be a sequence defined for n = 0, 1,2 … and f(n) = 0 for n< 0 then its
Z – Transform is defined as
( ) ( ) n
n
Z f n F z f n z (Two sided z transform)
0
( ) ( ) n
n
Z f n F z f n z (One sided z transform)
Unit sample and Unit step sequence
The unit sample sequence is defined as follows
1 0( )
0 0
for nn
for n
The unit step sequence is defined as follows
1 0( )
0 0
for nu n
for n
Properties
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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 aT
at
z zeZ e f t F z
(ii) If Z {f(n)} = F(z),
then n z
Z a f n Fa
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)] = 2z [ F(z) – f(0)-f(1)
1z ]
(iii)Z[f(n +k)] = kz [ F(z) – f(0)-f(1)
1z - f(2)2z ………- f(k-1)
( 1)kz ]
(iv) Z[f(n -k)] = kz F(z)
4. Initial Value Theorem
If Z[f(n)] = F(z) then f(0) = lim ( )z
F z
5. Final Value Theorem
If Z[f(n)] = F(z) then 1
lim ( ) lim( 1) ( )n z
f n z F z
PARTIAL FRACTION METHODS
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Model:I
1 A B
z a z b z a z b
Model:II
2 2
1
( )
A B C
z a z b z bz a z b
Model:III
22
1 A Bz C
z a z bz a z b
Convolution of Two Sequences
Convolution of Two Sequences {f(n)} and {g(n)} is defined as
0
{ ( )* ( )} ( ) ( )n
K
f n g n f K g n K
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 1z both terms
Step: 3 Apply 1z formula
Step: 4 Simplifying we get answer
Note:
12 1
1 .......1
nn a
a a aa
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1
2 11 .......
1 ( )
n
n aa a a
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)] = 2z [ F(z) – y(0)- y(1)
1z ]
iv) Z[y(n +3)] = 3z [ F(z) – y(0)- y(1)
1z + y(2)2z ]
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.
f(n)
Z[f(n)]
1. 1
1
z
z
2. an
z
z a
3. n
2( 1)
z
z
4. n2
2
3( 1)
z z
z
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6. 1
n log
( 1)
z
z
7. 1
1n log
( 1)
zz
z
8. 1
1n
1log
( 1)
z
z z
9. ean
( )a
z
z e
10. 1
!n
1
ze
11. Cos n
2
( cos )
2 cos 1
z z
z z
12. sin n
2
sin
2 cos 1
z
z z
13.
cos2
n
2
2 1
z
z
sin2
n
2 1
z
z
14. nna
2( )
az
z a
f(t) Z(f(t)
1 t
2( 1)
Tz
z
2. t2
2
3
( 1)
( 1)
T z z
z
3 eat
( )aT
z
z e
4. Sin t
2
sin
2 cos 1
z T
z z T
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5. cos t
2
( cos )
2 cos 1
z z T
z z T
TWO MARKS QUESTIONS WITH ANSWER
1. Define Z transform
Answer:
Let {f(n)} be a sequence defined for n = 0, 1,2 … and f(n) = 0 for n< 0 then
its Z – Transform is defined as
( ) ( ) n
n
Z f n F z f n z (Two sided z transform)
0
( ) ( ) n
n
Z f n F z f n z (One sided z transform)
Find the Z Transform of 1
Answer:
0
n
n
Z f n f n z
0
1 (1) n
n
Z z
1 21 ....z z
111 z
1 11 1
11
z z
z z z
11
zZ
z
2. Find the Z Transform of n
Answer:
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0
0
1 2 3
0
2 22
1 1
2
0 2 3 ...
1 1 11 1
1
1
n
n
n
n
n
n
Z f n f n z
Z n nz
nz z z z
zz z
z z z z
z
z
3. Find the Z Transform of n2
.
Answer:
2 dZ n Z nn z Z n
dz, by the property,
2 2
2 4 3
1 2 1( )
( 1)1 1
z z zd z z zz z
dz zz z
4. State Initial & Final value theorem on Z Transform
Initial Value Theorem
If Z [f (n)] = F (z) then f (0) = lim ( )z
F z
Final Value Theorem
If Z [f (n)] = F (z) then 1
lim ( ) lim( 1) ( )n z
f n z F z
6. State convolution theorem of Z- Transform.
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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 nna
Answer:
0
0
1 2 3
0
2
2
0 2 3 ...
1
n
n
n n n
n
n
n
Z f n f n z
Z na na z
a a a an
z z z z
a a az
z z z a
8. Find Z – Transform of cos
2
n and
sin2
n
Answer:
We know that
0
n
n
Z f n f n z
2
coscos
2 cos 1
z zZ n
z z
2
22
cos2
cos2 1
2 cos 12
z zz
Z nz
z z
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Similarly 2
sinsin
2 cos 1
zZ n
z z
22
sin2sin
2 12 cos 1
2
zz
Z nz
z z
9. Find Z – Transform of 1
n
Answer:
0
n
n
Z f n f n z
0
1 2 3
1
1
1 1
1....
1 2 3
1 1log 1 log
log1
n
n
n
n
Z zn n
z z zz
n
z
z z
z
z
10. Find Z – Transform of 1
!n
Answer:
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1
0
0
1 2 3
0
1
1 1
! !
11 ....
! 1! 2! 3!
n
n
n
n
n
n
z z
Z f n f n z
Z zn n
z z zz
n
e e
11. Find Z – Transform of 1
1n
Answer:
0
0
( 1)
0
2 31
1 1
1 1
1
1
....2 3
1log 1
log1
n
n
n
n
n
n
Z f n f n z
Z zn n
z zn
z zz z
zz
zz
z
12. Find Z – Transform of an
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Answer:
0
0 0
1 2 3
1
1
1 ...
1
n
n
nnn
nn n
Z f n f n z
a aZ a
z z
a a a
z z z
a
z
z a z
z z a
13. State and prove First shifting theorem
Statement: If Z f t F z , then ( )at aTZ e f t F ze
Proof:
0
( ) ( )at anT n
n
Z e f t e f nT z
As f(t) is a function defined for discrete values of t, where t = nT,
then the Z-transform is
0
( ) ( ) ( )n
n
Z f t f nT z F z ).
0
( ) ( ) ( )n
at aT aT
n
Z e f t f nT ze F ze
14. Define unit impulse function and unit step function.
The unit sample sequence is defined as follows:
1 0( )
0 0
for nn
for n
The unit step sequence is defined as follows:
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1 0( )
0 0
for nu n
for n
15. Find Z – Transform of
atZ e
Answer:
0 0
n nat anT n aT n aT
n n
n
aT
Z e e z e z z e
z zz a
z e z a
[Using First shifting theorem]
16. Find Z – Transform of
2tZ te
Answer:
2
2
2
2
2
22
1
1
T
T
t
z ze
z ze
T
T
TzZ te Z t
z
Tze
ze
[Using First shifting theorem]
17. Find Z – Transform of cos2tZ e t
Answer:
2
coscos 2 cos 2
2cos 1T
T
t
z ze
z ze
z zZ e t Z t
z z
2
cos
2cos 1
T T
T T
ze ze T
ze T ze
[Using First shifting theorem]
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18. Find Z – Transform of
2 t TZ e
Answer:
Let f (t) = e2t
, by second sifting theorem
2( )
2
2 2
( ) ( ) (0)
11
1 1
t T
T
T T
Z e Z f t T z F z f
zez z
ze ze
19. Find Z – Transform of sinZ t T
Answer:
Let f (t) = sint , by second sifting theorem
2
2 2
sin( ) ( ) ( ) (0)
sin sin0
2cos 1 2cos 1
Z t T Z f t T z F z f
z t z tz
z t z z t z
20. Find Z – transform of 1 2n n
Answer:
0
n
n
Z f n f n z
2
2 2
2
3 2
1 2 2 2
3 2 3 2 1
3 211 1
Z n n Z n n n
Z n n z n z n z
z z z z
zz z
Department of Mathematics – FMCET – MADURAI
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QUESTION BANK
Z-TRANSFORMS
1. (i)Find
21 8
(2 1)(4 1)
zZ
z z &
21 8
(2 1)(4 1)
zZ
z zby convolution theorem.
(ii) Find
21
( )( )
zZ
z a z b &
21
( 1)( 3)
zZ
z zby convolution theorem
2. (i) Find
21
2( )
zZ
z a &
21
2( )
zZ
z aby convolution theorem
3. (i ) State and prove Initial & Final value theorem.
(ii) State and prove Second shifting theorem
(i) Find the Z transform of 1
( 1)( 2)n n&
2 3
( 1)( 2)
n
n n
4. (i) Find
21
2( 4)
zZ
z by residues.
(ii) Find the inverse Z transform of
2
21 ( 1)
z z
z zby partial fractions.
5. (i) Find 1
2 2 2
zZ
z z &
21
2 7 10
zZ
z z
6. (i)Find the Z transform of 1
( )!
f nn
Hence find 1
( 1)!Z
n and
1
( 2)!Z
n.
(ii) Find 1
!Z
n and also find the value of sin( 1)n and cos( 1)n .
7. (i)Solve 2 6 1 9 2 0 0& 1 0ny n y n y n with y y
(ii) Solve 2 4 1 4 0y n y n y n y(0) = 1 ,y(1) =0
8. (i )Solve 3 1 4 2 0, 2 (0) 3& (1) 2y n y n y n n given y y
(ii) Solve 3 3 1 2 0, 0 4, 1 0& 2 8y n y n y n y y y ,
9. (i)Find cos & sinZ n Z n and also find cos & sinn nZ a n Z a n