MA6351-TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS SUBJECT NOTES
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MA6351-TRANSFORMS AND PARTIAL
DIFFERENTIAL EQUATIONS
SUBJECT NOTES
Basic Formulae
DIFFERENTIATION &INTEGRATION FORMULAE
0 FunctionDifferentiation
y f (x )dydx
1 xn nxn1
2 log x 1x
3 sin x cos x
4 cos xsinx
5 eax
a ex
6 C (constant) 0
7 tanx sec2 x
8 sec x sec x tan x
9 cot x cos ec 2 x
10 cos ecx cos ecx cot x
11 1x2 x
12 sin 1 x 1
1
x2
13 cos 1 x 1
1x2
14 tan 1 x 1
1 x2
15 sec 1 x 1
x x2 1
16 cos ec 1 x 1
x x2 1
17 cot 1 x 1
1 x2
18 a x a x log a
Page2
v
du
u
dv
dy du
dv u
dy
19. If y uv , then v u 20. If y
, then
dxdx
dx dx
dx v
dx v2
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1. x n dxxn
1
n 1
2. e x dx e x , e ax dxe ax
& e ax dxeax
a a
3. sin xdx cos x & sin axdxcos ax
a
4. cos xdx sin x & cos axdxsin ax
a
5.tan xdx
log secx log cos x
6. sec 2 xdxtanx
7. dxdx 1 tan 1 x
x 2 a 2 aa
8. dxdx 1 log x a
x 2 a 22a x a
9. dx dx sin 1 x
aa 2 x2
10.
dx
dx sinh 1x
a 2 x2 a
11.
dx
dx cosh 1x
ax 2 a2
x a 2 x12. a 2 x 2 dx a 2 x2 sin 1
2 2 a
x a 2 x13. a 2 x 2 dx a 2 x2 sinh
1
2 2 a
x a 2 x14. x 2 a 2 dx x 2 a2 cosh
1
2 2 a
15.
dx log xx
16.2xdx
log x 2 a2
x 2 a2
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17. log xdx x log x x
18. a x 2 dx a x 3
3
19. a x2dx
a x 3
3
120. dx 2 xx
21. e ax cos bxdxeax a cos
bx b sin bx
a 2b2
22. e ax sin bxdx eax
a sin bx b cos bxa 2 b2
23. udv uv u´v1 u´´v2 u´´´v3........
a a
24. f ( x ) dx 2 f ( x ) dxwhen f(x) is even
a 0
a
25. f ( x ) dx 0 when f(x) is odda
26. e ax cos bxdxa
a 2 b20
27. e ax sin bxdxb
a 2 b20
TRIGNOMETRY FORMULA
1. sin 2 A 2sin A cos A
2.cos 2 A cos 2 A sin2 A
1 2 sin2 A
2 cos 2 A 1
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3. cos2 x1 cos 2x &
sin2 x1 cos 2x
2 2
4. sin( A B )
sin( A B )
cos( A B )
cos( A B )
5.sin A cos B
cos A sin B
cos A cos B
sin A sin B
sin A cos B cos A sin B
sin A cos B cos A sin B
cos A cos B sin A sin B
cos A cos B sin A sin B
12 sin( A B ) sin( A B)
12 sin( A B ) sin( A B)
12 cos( AB ) cos( A B)
12 cos( AB ) cos( A B)
6. sin 3 A1
3sin A sin 3A4
cos 3 A1
3 cos A cos 3A4
7.sin A 2 sin A cos A2 2
cos A cos 2A
sin2
2
1 2 sin 2A2
A
2
1 cos A 2 sin2 A
2
LOGRATHEMIC FORMULAPa
ge6
log mn log m log n
log m log m log nn
log m n n log m
log a 1 0
log a 0
log a a 1
e log x x
UNIT - 1
PARTIAL DIFFRENTIAL EQUATIONS
PARTIAL DIFFERENTIAL EQUTIONSNotation
s
z p z q
2
z r 2 z s
2
z tx y x2 x y y2
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 formPp + Qq = R
Subsidiary Equation
dx dy dz
P Q R
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dx dy dz x my nzP Q R P mQnR
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 integralPut p = a & q = b in (1), We get (2) Which is the Complete integral
Singular IntegralDiff (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( zkp, zkq)=0
F( xmp, ynq, z)=0 (1)
If m 1& n 1 then If k 1 then Z = zk+1
X = x1-m & Y = y1-n
z k qQ
xm p = P(1-m) & yn q = Q(1-n) k 1
Using the above in (1)we get Using the above in (1) We get
F(P,Q) = 0 (or) F(P,Q,z) = 0 F(P,Q) = 0
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If m=1 & n=1 then If k =-1 then Z = log zX= logx & Y= logy
p qxp = P & yq = Q P & Q
z z
Using the above in (1) we getUsing the above in (1)
F(P,Q) (or) F(P, Q, z) = 0 we getF(P,Q) = 0
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STANDARD TYPES
TYPE –1 TYPE –3(a) TYPE –3(b) TYPE –3(c) TYPE –4
General form General formGeneral form General form General form
F(y,p,q) = 0 F(z,p,q) = 0 F(x,y,p,q) = 0F(p,q) = 0(1) F(x,p,q) = 0 (1)
(1) (1) (1)Complete Complete Complete Complete CompleteIntegral Integral Integral Integral IntegralPut p = a and q =b in (1) Put q = a in (1) Put p = a in (1) Put q = ap in (1) (1) Can be writtenFind b in terms Then, find p and Then, find q and Then, find p and as,of a sub in sub in dz = p dx sub in f(x,p) =f(y,q) = aThen sub b in dz = p dx + q dy + q dy dz = p dx + q dy Then, find p and qz = ax + by + c Integrating , Integrating , Integrating, sub inwe get (2) We get (2) We get (2) We get (2) which dz = p dx + qD ywhich is the which is the which is the is the Integrating,Complete Complete Complete Complete We get (2) whichIntegral integral integral integral is the
Complete integral
Singular Singular Singular Singular Singular Integral
Diff (2) partiallyIntegral Integral Integral Integralw.r.t cWe get,0Diff (2) partially Diff (2) partially Diff (2) partially Diff (2) partially
w.r.t cWe get,0 w.r.t cWe get,0 w.r.t cWe get,0 w.r.t cWe get,0 =1 (absurd=1 (absurdThere =1 (absurdThere =1 (absurdThere =1 (absurdThere
There is nois no Singular is no Singular is no Singular is no SingularIntegral Integral Integral Integral Singular Integral
General General General GeneralGeneral IntegralIntegral Integral Integral Integral
Put c = (a) in Put c = (a) in Put c = (a) in Put c = (a) in Put c = (a) in
(2)We get (2)We get (2)We get (3) (2)We get (2)We get (3)(3)Diff (3) (3)Diff (3)
Diff (3) partially(3)Diff (3)
Diff (3) partiallypartially w.r.t partially w.r.t partially w.r.t
w.r.t aWe get (4) w.r.t aWe getaWe get aWe get aWe get
(4)Eliminating a(4)Eliminating a (4)Eliminating a Eliminating a (4)Eliminating a
from (3) and (4)from (3) and (4) rom (3) and (4) from (3) and (4) from (3) and (4)
we get Generalwe get General we get General we get General we get General IntegralIntegral Integral Integral Integral
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HOMOGENEOUS LINEAR EQUATION
General form
3 2 23 (1)
cDD dD ) z f ( x, y)(aD bD D
To Find Complementary FunctionAuxiliary EquationPut 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 ( ymx)
Case (3)
If the two roots are same and one is distinct, then
C.F = 1 ( y mx ) x 2 ( y mx ) 3 ( y m 3 x)
PI =
1F ( x , y)Function 1
F ( D , D )
F(x,y) = eax+by Put D = a & D1 = b
F(x,y)= sin(ax+by)(or)Put 2 2
), DD(ab) & D2
2
)Cos (ax+by) D (a (b
F(x,y) = xr ysPI= F ( D, D ) 1 x r ys
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)=1
e 2 x e2 y2
F(x,y)=ex+y sinh(x+y) F(x, y) =1
e 2 x e2 y2
F ( x, y )1 sin(
x y ) sin( x y)F(x,y)=sin x cos y 2
F ( x, y )1 sin(
x y ) sin( x y)F(x,y)= cos x sin y 2
F ( x, y ) 1 co s( x y ) co s( x y)F(x,y)= cos x cos y 2
F ( x, y )1 cos(
x y ) cos( x y)F(x,y)= sin x sin y 2
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 pz
& qz
x yp = 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 pz
& qz
x 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 pz
& qz
x y
p = naxn-1 ; q = nbyn-1
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ap ;
bq
nx n 1 nyn 1
zp
x nq
yn
nx n 1 nyn 1
nz px qy
4. Eliminate the arbitrary function f from z f xy
and form the PDE
z
Answer:
z f xy
z
Diff. w .r. t. and y here pz
& qz
x y
p fxy
.yz xp
z z 2
q fxy
.xz yq
z z 2
p y . z xpqxz yq
pxz pqxy qyz pqxy
px qy 0
5.Find the complete integral of p + q =pq
Answer:Put p = a, q = b
p + q =pq a+b=ab
b – ab = -a ba a
1 a a 1
The complete integral is z= ax+ a a 1 y +c
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6. Find the solution of p q 1
Answer:z = ax+by+c ----(1) is the required solution
given p q
1 -----
(2)put p=a, q = b in (2)
)2a b 1 b 1 a b (1 a
)2 yz a x (1 a c7. Find the General solution of p tanx + q tany = tanz.
Answer:
dx dy dztan x tan y tan z
cot x dx cot y dy cot z dz
take cot x dx cot y dy cot y dy cot zdz
log sin x log sin y log c1 log sin y log sin z log c2
c1sin x
c2sin y
sin y sin z
sin x ,sin y 0sin y sin z
8. Eliminate the arbitrary function f
from z f x 2 y2 and form the PDE.
Answer:
z f x 2 y2
p f x 2 y 2 2 x ; q f x 2 y 2 ( 2 y)
p 2xpy qx 0
q2 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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x 2 y2 z c 2 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 y
pq
That ispy -qx =0 which is a required PDE.
10. Eliminate the arbitrary constants
z ax bya22
and form the PDE.
bAnswer:
z ax by a 2 b2
p a;q b
z px qy p 2 q2
11. Find the singular integral of z px qy pqAnswer:
z ax by abThe complete solution is
z0 x b ;
z0 y a
a bb x ; a y
z ( y )x ( x )y( y . x )
xy xy xy
xy
xy z 0
12. Find the general solution of px+qy=z
Answer:
dx dy dz
The auxiliary equation isx y z
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dx dy
From Integrating we getlog x = log y + log c
x y
on simplifyingx
c1 .y
dy dz yc2y z z
Thereforex , y 0
is general solution.
y z
13. Find the general solution of px2+qy2=z2
Answer:
dx
dy
dz
The auxiliary equation is
x 2 y 2 z2
From
dx
dy
Integrating weget
1 1
x 2 y2 y x
Alsody
dz
Integrating we get1 1
c2
y 2 z2 z y
Therefore1 1
,1 1
0is general solution.
y x z y
14. Solve D 2 2 DD 3D 2 z 0Answer:
m 2 2m 3 0Auxiliary equation ism 3 m 1 0
m 1, m 3
The solution is z f1 y x f 2 y 3x
c1
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15.Solve D 2 4 DD 3D 2 z ex y
Answer:
m 2Auxiliary equation is 4 m 3 0
m 3 m 1 0
m 1, m 3
The CF is CF f1 y x f 2 y 3x
PI1
ex y Put D 1, D 1Denominator =0.
D 2 4 DD 3D 2
PIx
ex y
2 D
4D
xex y
2Z=CF + PI
z f1 y x f 2 y 3xxex
y
2
16. Solve. D 2 3DD 4D 2 z ex y
Answer:
m 2 3m 4 0Auxiliary equation ism 4 m 1 0
C.F is = f1(y + 4x) + f2(y - x)
PI 1 ex y 1 e x y 1 ex y
D 2 3DD 4 D 2 1 3 4 6
17. Find P.I D 2 4 DD 4D 2 z e2 x y
Answer:
1e2 x yPI
D 2 4 DD 4D 2Put
D 2, D 1
PI
1
e2 x y
1
e
2 x y e2 x y
2 2 2 16D 2D 2
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18. Find P.I D 2 DD 6D 2 z x 2 yAnswer:
1x 2 yPI
D 2 1 D 6D 2
D D2
11
D x 2
yD 2 D
1x 2 y
x3 x 4 y x5
D2 3 12 60
19. Find P.I
2 z 2 z sin x yx 2 x y
Answer:
PI 1 Sin x yPut
D 21, DD(1)( 1) 1D 2 DD1
Sin x y
Sin x y
1 1 2
20. Solve D3
3DD 2 D 3 Z 0Answer:
m3 3m 2 0Auxiliary equation is
m 1 m 2 m 2 0
m 1 m 2 m 1 0
m 1,1 m 2
The Solution isCF f1 y x x f 2 y xf 3 y 2x
FOR PRACTICE:
1. Eliminating arbitrary constants x 2 y 2 z2
1a 2 b 2 c2
2. Solve
2
z sin yx2
3.
Find the complete the solution of p. d .e p 2 q 2
4 pq 0
4.Form p.d.e eliminating arbitrary function from z 2 xy,x2
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5. Find the singular soln of z px qy p 2 q
2 1
1.
(i) Solve x 2 y z p y 2 z x q z 2 x y
(ii) Solve x z 2 y 2 p y x 2 z 2 q z y 2 x2
2.
(i) Solve
mz ny z nx lz z ly mx
x y(ii) Solve
3z 4 y p
4 x 2 z q 2 y 3x
3.
(i) Solve x 2 y 2 z 2 p 2 xyq 2xz
(ii) Solve y 2 z 2 x 2 p 2 xyq 2 zx 0
4.
(i) Solve y z p z x q x y
(ii) Solve y z p z x q x y
5.
Solve D 2 3DD 2D 2 e3 x 2 y sin(3 x 2 y)
6.
Solve
2 z 2 z cos x cos 2 y
x 2 x y7.
Solve D 2 DD 6 D 2 z y cos x
8. Solve D 2 DD 30D 2 z xy e6 x y
9. Solve D 2 6 DD 5 D 2 z e x sinh y xy
10. Solve D 2
4DD 4D 2 z e2 x y
11. Solve D3 D 2 D DD 2 D 3 z e 2 x y cos( x y)
12. Solve (i)z px qy 1 p 2 q2
(ii) z px qy p 2 q2
13. Solve z 2 1 p 2 q2
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14.Solve z 2 ( p 2 x 2 q2 ) 1
15.Solve (i) z ( p2 q 2 ) x 2 y2
(ii) z 2 ( p 2 q 2 ) x 2 y2
UNIT - 2
FOURIER SERIES
f ( x )a0 a n cos nx bn sin nx2 n 1
(0,2 ) ( - , )
Even (or) Half range Odd (or) Half range
Neither even nor odd
Fourier co sine series Fourier sine series
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a01 2
f ( x ) dx a02
f ( x ) dxa0 0
a01 f ( x )
dx0 0
an
1 2
f ( x ) cos nxdxa
n
2f ( x ) cos nxdx
an 0 an
1f ( x ) cos nxdx0 0
bn1 2
f ( x ) s innxdx
bn=0
bn2
f ( x ) s innxdx bn1
f ( x ) s innxdx0 0
f ( x )a
0an cos
n xbn sin
n x2 n 1
(0,2 ) ( - , )
Even (or) Half range Odd (or) Half range
Neither even norodd
Fourier cosine series Fourier sine series
1 2 2 a0 0 1
a0 f ( x ) dx a0 f ( x ) dx a0
f ( x ) dx
0 0
1 2 n x 2 n x an 0 1 n xa f ( x ) cos dx a f ( x ) cos dx a f ( x ) cos dx
n0
n0
n
b 12
f (x) s in n x dx bn=0 b 2 f (x) s in n x dx b 1 f (x) s in n x dx
n
0n
0n
Even and odd function:
Even function:
f(-x)=f(x)
eg : cosx,x2 ,
, x , sin x , cos 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 )=
f (0) f (2 )
2
In the interval (0,2 ) if x = 0 or x = 2 then
f(0) = f(2 )=
f (0) f (2 )
2In 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)=
a0 + a1 cosx +b1sinx + a2cos2x + b2sin2x ……… for
form
2
a0 2y
a1 2y cos x
, a2 2y cos 2x
b1 2y sin x
, b2 2y sin 2x
n n n n n
f(x)= a0
+ a1 cos x
+b1 sin x
+ a2 cos 2 x
+ b2 sin 2 x
………( form)
2
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a0 2y
y cosx
2 x y sinx
y sin2 x
y cosn a1 2 b1 2, b2a 2n n2 n 2 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
1b
f ( x ) 2 dxf(x) is defined by yb a a
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 2 a2 1 2 2o
f ( x ) dx an bn
4 2 n 1c
Where ao , an & bn are Fourierconstants
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 2ii) 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:
In (c ,c 2l )
f xao an cos nx bn sin nx2
1 c 2 l
where ao f (x )dxc
1 c 2 l
an f ( x ) cosnxdxc
1 c 2 l
bn f (x ) sinnxdxc
6. Write Fourier constant formula for f(x) in the interval (0, 2 )Answer:
1 2
ao f (x )dx0
1 2
an f ( x ) cosnxdx0
1 2
bn f ( x ) sinnxdx0
7. In the Fourier expansion of
1
2 x
,
x 0
f(x) = 2 x
in (-π , π ), find the value of bn
1 , 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
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 = a0
= 0
10. Find the value of a0 for f(x) = 1+x+x2 in ( 0 ,2 )
Answer:
ao
1 2
f (x )dx0
1 2 1 x 2 x32
(1 x x 2 )dx x
02 3
0
12
4 2 8 3 2 28 2
2 3 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) = xsinxf(-x) = (-x)sin(-x) = xsinx
= f(x)
Hence f(x) is an even function
In the Fourier series bn = 0
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12. Obtain the sine series for f x
x 0 x l / 2
l x l / 2 x l
Given f x x 0 x l / 2
x l / 2 x ll
Answer:
x 0 x l / 2Given f xx l / 2 x ll
Fourier
sine series is
f x b sin
nx
n l2 l nx
b f ( x ) sin dxn
l l0
2 l 2
nx l
nx
x sindx (l x ) sin dxl l l0 l 2
nx nxl 2
nx nxl
cos sin cos sin2lx
l (1)l
2 ll (l x)
ll
2
( 1)l
l n n2 n n2
0 l 2
2 l 2 cos n 2 l 2 sin n 2 l 2 cos nl 2 l 2 sin n 2
l 2 n 2 n 2 2n 2 n2
2 2l 2 sin n 2 4l sin n 2
l 2 n2 2 n2
Fourier series is f x
4l sin n 2 sin n x
2 n 1 n 2 l
13. If f(x) is odd function inl , l. What are the value of a0 &an
Answer:
If f(x) is an odd function, ao = 0, an = 0
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7
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
a 2 xdx 2 x2 2 2
o2 20 0
15. Find
half range cosine series of f(x) = x, in
0 xAnswer:
2 2 x2 22
ao xdx2 20 0
a 2 x sin nxdx 1 x cos nx (1) sin nxn n n2
0 0
1 cos n 0 0 1 n 1 n 1
n n n
f xao an cos nx2 n 0
Fourier series is 1 n 1
cos nx2 n 0 n
16. Find the RMS value of f(x) = x2, 0 x 1Answer:
Given f(x) = x2
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R.M.S value
1 2 lf ( x ) 2 dx
1 1 x 2 2
dxyl 0 1 2 0
x5
1
2 25
05
17. Find the half range sine seriesof f ( x ) x
in (0, )
Answer:
bn
2 f ( x ) sinnxdx0
2 x sin nxdx 2 x cos nx (1) sin nx
0 n n20
2 ( 1) n 2( 1)n 1
n n
2( 1)n 1
Half range Fourier sine series is
fx sin nx
n 0 n18. Find the value of a0 in the cosine series of f ( x ) x in (0, 5)Answer:
2 5 2 x25
2 52a xdx 5
o5
0 5 2 5 20
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.
eg : sinx,x3 ,sinhx, tanx are odd functions
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20. Write the first two harmonic.
Answer:
The first two harmonics are
f xao a1 cos x b1 sin x a2 cos 2 x b2 sin 2x2
FOURIER SERIES
1. Expand f ( x)x (0, )
as Fourier series2 x ( , 2 )
1 1 12
and hence deduce that .........12 32 5 2 8
2. Find the Fourier series for f(x) = x2 in (- . ) and also prove that
1 1 1 2 1 1 1 2
(i) ......... (ii) .........2
22 2
62
22 2
121 3 1 33. (i) Expand f(x) = | cosx | as Fourier series in (- . ).
(ii) Find cosine series for f(x) = x in (0, ) use Parsevals identity to
1 1 14
Show that .........14 2 4 34 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
1 1 12
.........
12 32 5 2 85. If
f ( x) 0 , ( , 0)Find the Fourier series and hence deduce that
sin x , (0, )
1 1 1.........
2
1.3 3.5 5.7 46. (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 f ( x ) (x)2 in (0, 2 ) and
1 1 1 2
Hence deduce that .........2
22 2
61 3
(ii). Find a Fourier series to
represent in the range (0, 3)
f ( x ) 2x x2 with period 3
(ii) Find the Fourier series of f x ex in ( , ) .
(ii) Find the Fourier seriesfor f x
1 in (0, )
2 in ( , 2 )
1 1 1 2
and hence show that .........2 2
52
81 3
8. (i) Find the the half range sine series for f x x xin the interval (0, ) and deduce
that 1 1 1 ....3 3 3
1 3 5
(ii) Obtain the half range cosine series for f x x 1 2 in (0,1)
1 1 1 2
and also deduce that .........2
22 2
61 39. (i) Find the Fourier series for f(x) = x2 in (- . ) and also prove that
1 14
1 ......... (use P.I)
2 4 9014
(ii) Find the Fourier series for f(x) = x in (- . ) and also prove that
1 14
1 ......... (use P.I)34 9614
31
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cx , 0 xl210(i)Obtain the sine series for f x
lc l x , x l2
kx , 0 xl
2(ii). Find the Fourier series for the functionf x
lk 2l x , x l2
11.(i).Find the Fourier series for the function f x 1 x x 2 in ( , ) and also
1 1 1 2
.........deduce that 2
22 2
1 3 6
(ii)Find the Fourier expansion of
12x ,
x 0 21 1 1
f(x) =in (-π , π ), and also deduce that .........2
x2 2 2
1 , 0 x 1 3 5 8
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UNIT - 3
APPLICATIONS OF P.D.E
S. ONE DIMENSIONAL WAVE EQUATIONN
O
VELOCITY MODEL INITIAL POSITION MODEL
1 STEP-1 STEP-1
One Dimensional wave equation One Dimensional wave equation
is2 y a2
2
y 1 is
2
y a2 2 y 1t 2 x2 t 2 x2
2 STEP-2 STEP-2
Boundary conditions Boundary conditions
1. y(0,t) = 0 for t 0 1. y(0,t) = 0 for t 02. y( , t) = 0 for t 0 2. y( , t) = 0 for t 0
3. y(x,0) = 0 for 0 < x <3.
y= 0 for 0 < x <
y4. = f(x) for 0 < x <
t t 0
t t 4. y(x,0) = f(x) for 0 < x <0
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3 STEP-3The 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-4The 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 1y(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-6
6Using Boundary condition 2
y( ,t) = 0Then (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,
y ( x , t ) B sin( n x ) C cos( n at ) D sin( n at )
(4)
7 STEP-7Using Boundary condition 3
y(x,0) = 0Then (4) becomes,
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-4
The suitable solution for the givenboundary condition is
y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)
(2)STEP-5
Using Boundary condition 1y(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) = 0Then (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,
y ( x , t ) B sin( n x ) C cos( n at ) D sin( n at )
(4)
STEP-7
Using Boundary condition 3
34
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y ( x, t ) B sin( n x ) C cos 0 D sin 0 =0 y = 0Then (4) becomes,t t 0
B sin(n x) C 0 Differentiating (5) partially w.r.to ‘t’ and put t =0y B sin(
nx ) C sin(
n at ) D cos(
n at
C = 0 t t 0Then (4) becomes,n x n aB sin( ) D 0
n x naty ( x , t ) B sin( ) D sin( ) D = 0
Then (4) becomes,The most general solution is
y ( x , t ) Bn sin( n x ) sin( nat ) (5) y ( x , t ) B sin( n x ) C cos( n
at )
n 1
The most general solution is
y ( x , t ) B sin( n x) cos( nat ) (5)
nn 1
8 STEP-8 STEP-8
Differentiating (5) partially w.r.to ‘t’ Using Boundary condition (4),
y B sin( n x ) cos( n at ) n a y(x,0) = f(x)t n 1 n x
n
y ( x , 0) Bn sin( ) cos(0)Using Boundary condition (4), n 1
y = f(x) f ( x ) B sin( n x )t
n
t 0 n 1
f ( x ) B sin( n x) n a This is the Half Range Fourier Sine Series.n 1 2 n x
nBn f ( x) sin( )
This is the Half Range Fourier Sine Series. 0
B n a 2 f ( x) sin(n x )n
0
2 n xB f ( x ) sin( )dxn
na
0
9 STEP-9 STEP-9
The required solution is The required solution is
y ( x , t ) B sin( n x ) sin( n at ) y ( x , t ) B sin( n x ) sin( n at )n n
n 1 n 1
Where B 2 f ( x ) sin( n x)dx Where B 2 f ( x ) sin( n x)dxn
n a 0n
0
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ONE DIMENSIONAL HEAT TWO DIMENSIONAL HEAT FLOWEQUATION EQUATION
1 The one dimensional heat equation is The two Dimensional equation is
u 2 2u
2
u 2u 0t x2 x 2 y2
2 Boundary conditions Boundary conditions1.u(0,t) = 0 for t 0 1.u(0,y) = 0 for 0<y<2.u( ,t) = 0 for t 0 2.u( ,y) = 0 for 0<y<3.u(x,t) = f(x) for 0<x< 3.u(x, ) = 0 for 0<x<
4.u(x,0) = f(x) for 0<x<
3 The possible solutions are The possible solutions are
u ( x , t ) ( Ae x Be x )Ce 2 2
t u ( x , y ) ( Ae x Be x )(C cos y D sin y)
u ( x , t ) ( A cos x B sin x )Ce 2 2t u ( x , y ) ( A cos x B sin x )(Ce y De y )
u ( x , t ) ( Ax B )C u ( x , y ) ( Ax B )(Cy D)
4, The most suitable solution is The most suitable solution isu ( x, t ) ( A cos x B sin x)Ce 2 2 (2) u ( x , y ) ( A cos x B sin x )(Ce y De y ) (2)
t
5 Using boundary condition 1 Using boundary condition 1u(0,t) = 0 u(0,y) = 0
u (0, t ) ( A cos 0 B sin 0)Ce 2 2t =0 u (0, y ) ( A cos 0 B sin 0)(Ce y De y )( A)Ce 2 2t =0 u (0, y ) ( A)(Ce y Be y )
A = 0 A = 0Then (2) becomesThen (2) becomes
x )(Ce y y )u ( x, t ) ( B sin
x)Ce 2
2
t(3)
u ( x , y ) ( B sin De (3)
6 Using boundary condition 2 Using boundary condition 2u(l,t) = 0 u(l,t) = 0
u ( , t ) ( B sin )Ce 2 2t =0u ( , y ) ( B sin )(Ce y De y )2 2
( B sin )Ce t 00 ( B sin )(Ce y De y )n
n
Then (3) becomesThen (3) becomes
n x2 n 2 2t
u ( x, t ) B sin( )Ce2
n x n y n y
u ( x , y ) ( B sin )(Ce De ) (4)
The most general solution is
n x2 n 2 2t
36Pa
ge
u ( x , t ) Bn sin( )e2
(4)n 1
7 Using Boundary condition 3 Using Boundary condition 3
u(x,0) = f(x) u(x, ) =0
u ( x , 0) B sin( n x ) u ( x , ) ( B sin n x )(Ce De )n
n 1
f ( x ) Bn sin( n x ) 0 ( B sinn x
)(C D 0)n 1
C=0This the Half range Fourier sine seriesthen (3) becomes2
f ( x ) sin(n x
)dx xBn n n y
0 u ( x , y ) ( B sin )( De )
The most general solution is
n xn y
u ( x , y ) B sin( )e (5n
n 1
8 The required solution is Using Boundary condition 4
n x2 n 2 2t y(x,0) = f(x)
u ( x, t ) Bn sin( )e 2u ( x , 0) B sin(
n x)e0
n 1n
2 n x n 1
Where Bn f ( x ) sin( )dx
f ( x ) Bn sin(n x
)0n 1
This the Half range Fourier sine series
2 n xB f ( x ) sin( )dx
n0
The required solution is
n xn y
u ( x , y ) B sin( )en
n 1
Where B 2 f ( x ) sin( n x )dxn
0
QUESTION WITH ANSWER
1. Classify the Partial Differential Equation i) 2 u 2u
x 2 y2
Answer:
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2
u
2
u here A=1,B=0,&C=-1
x 2 y2
B2 - 4AC=0-4(1)(-1)=4>0The Partial Differential Equation is hyperbolic
2. Classify the Partial Differential Equation
2u u uxy
x y y xAnswer:
2u u uxy here A=0,B=1,&C=0
x y y x
B2-4AC=1-4(0)(0)=1>0
The Partial Differential Equation is hyperbolic
3. Classify the following second order Partial Differential equation
2
u
2
u u 2 u 2
x 2 y 2 y x
Answer:
2
u
2
u u 2 u 2here A=1,B=0,&C=1
x 2 y 2 y xB2-4AC=0-4(1)(1)=-4<0
The Partial Differential Equation is Elliptic
4. Classify the following second order Partial Differential equation
4
2
u4
2 u
2
u6
u
8
u
0x 2 x y y 2 x yAnswer:
4
2 u
4
2 u 2u
6
u
8
u
0x 2 x y y 2 x yhere 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 equationi)
y 2uxx 2xyuxy x 2u yy 2ux 3u 0
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ii)y 2 uxx u yy ux 2 uy
2 7 0
Answer:
i) Parabolic ii) Hyperbolic (If y =
0) iii)Elliptic (If y may be +ve or –
ve)
2 y 2 y6. In the wave equation c2 what does c2 stands for?
t2 2
x
Answer:2 y
c22 y
t 2 x2
here a2T
T-Tension and m- Massm
7. In one dimensional heat equation ut = α2 uxx what does α2
stands for? Answer:-
u 22 u
t x2
2 = kc is called diffusivity of the substance
Where k – Thermal conductivity
- Densityc – 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)y
0t
t 0
b x 0 x(iv) y(x , 0 )
=b (2 x) x 2
10. 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) 2 2 t
u (x ,t ) (AxB C)
12. State Fourier law of heat conduction
Answer:
Q kA ux
(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=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
al
80 20 x
60
u ( x, t ) x 20
20
16.Write the D’Alembert’s solution of the one dimensional wave equation?
Answer:
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1
1 1 x at
y x at x at v ( )d2 2a x at
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 t 02. y( , t) = 0 for t 03. y(x,0) = 0 for 0 < x <
4.y
= f(x) for 0 < x <t
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
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 TheTemperature u ( x , t)
b a xa
lHere a=40°C & b=60°C
u 60 40 x 40 2 x 40t 30 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 itsequilibrium position. If it is set vibrating giving each point a velocity 3x (l-x). Find thedisplacement.
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) = y sin3 x. If it is released from rest find the displacement.
0 l5. A string is stretched between two fixed points at a distance 2l apart and points of the
cx0 < x < 1
V lstring are given initial velocities where Find the
c (2l x) 0 < x < 1l
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 20oC and 80oC, respectively until steady statecondition prevails. The temperature at each end is then reduced to 0oC and kept so. Find
the resulting temperature u(x, t) taking x = 0.43
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8. A bar 10 cm long , with insulated sides has its end A & B kept at 20 oC and 40oCrespectively until the steady state condition prevails. The temperature at A is suddenlyraised 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 thatit may be considered as an infinite plate. If the temperature along short edge y = 0 is u (
x ,0) = 100 sin
x
0 < x < .1While two edges x = 0 and x = 8 as well as the other short8edges are kept at 0oC. Find the steady state temperature.
10.A rectangular plate with insulated surface is10 cm wide so long compared to its widththat it may be considered as an infinite plate. If the temperature along short edge y = 0 isgiven
by u 20x 0 x5
and all other three edges are kept at 0o C. Find the steady20(10 x) 5 x 10
state temperature at any point of the plate.
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Unit - 4
FOURIER TRANSFORMS
FORMULAE
1. Fourier Transform of f(x) is F [ f ( x)]1
f(x)eisxdx2 -
2. The inversion formula f ( x)1
F ( s)e-isxds2 -
3. Fourier cosine Transform Fc
[f(x)] = Fc(s) =
2f ( x
) cossxdx0
4. Inversion formula f(x) =2
Fc ( s
) cossxds
0
5. Fourier sine Transform (FST) Fs [f(x)] = Fs(s)=
2f ( x
) sinsxdx
0
6. Inversion formula f(x) =2
Fs ( s
) sinsxds0
7. Parseval’s Identity f x dx F s ds
1
8. Gamma function n x n 1e x dx, n 1 n n
&0 2
9. e
ax cosbxdx
a
a 2 b20
10 e
ax sin bxdx
b
a 2 b20
11.
sinax dx
20 x
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12. e x 2 dx & e x
2 dx
20
13. cos axe iax e iax
& sin axe iax e iax
2 2
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 , in the formula
Step3: Expand e isx 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 F ( s ) 2 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 F ( S )C
WORKING RULE TO FIND INVERSE FCT
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Step1: Write the inverse FCT formula & Sub F ( S ) with its limit in the formulaC
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 workingrule
and then we get this
result
F (e-ax) = 2 a F (e-ax) = 2 s
c a 2 s2s a 2 s2
By Inversion formula, By Inversion formula,
cos sx ds e ax s sin sxds e ax
a 2 s20 a 2 s 2 2a 0 2
TYPE-I : If problems of the form i)
x
ii)
1 , then use Inversion formulax 2 a2 x 2 a2
TYPE-II: If problems of the form i)
x2
dx ii)dx
,then use Parseval’s Identity
0 x 2a2
2
0 x 2 a2
2
TYPE-III
dx, then use f ( x ) g ( x ) dx FC f ( x ) FC g ( x )dx
0 x 2 a 2 x 2 b20 0
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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,
f ( x )1
f t e i ( x t )s dt ds .2
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2. StateandproveModulation
theorem. F f x cos ax 1 F s a F s aProof:
2
F f x cos ax1
f x cos ax e isxdx2
1f x
e iax e iax e isx dx22
1 1 f x e i ( s a )x dx 1 1 f x e i ( s a )x dx2 22 2
1 F s a 1 F s a2 2
F f x cos ax 1 F s a F s a2
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
transforms.
F f g F s G s5. State and prove Change of scale of property.
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9
Answer:
1If F s F f x , then F f ax F s a
a
F f ax1
f ax e isxdx2
1f t ei
s a
t dt; where t
ax
a2
F f ax 1 F sa a
6. Prove that if F[f(x)] = F(s) then F x n f ( x ) ( i ) n d n
F ( s)dsn
Answer:
F s1
f x e isx dx2
Diff w.r.t s ‘n’ times
d n F s 1 f x ix n e isxdxdsn 2
1 fx (i ) n ( x ) n e isxdx
2
1 d nF s
1( x ) n f x e isx d x
(i ) n dsn 2
( i ) nd n
F s1
( x ) n f x e isx d x
dsn 2
Fx n
f x in d n
F sdsn
7. Solve for f(x) from the integral equation f ( x ) cos sxdx e s
0Answer:
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f ( x ) cos sxdx e s
0
Fc f x
2
f x cos sx dx0
F f x 2 e s
c
f ( x ) 2 F f x cos sx dsc
0
e ax cos bx dxa
2 2 e s cos sx ds a 2 b20
0
2 2 1a 1, b x
e s cos sx ds0 x2 1
1 x a8. Find the complex Fourier Transform of f (x)
Answer:
0 x a 0
F f x1
f x e isx dx2 x a; a x a
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1 a
F f x 1 e isx dx2 a
1 a
(cos sx i sin sx ) dx2 a
2a
2sin sx
a
(cos sx ) dx s
02 0 22 sin as
s
[Use even and odd property second term become zero]
x x a9. Find the complex Fourier Transform of f ( x)
x a 00Answer:
F f x1
f x e isx dx2
1 a
x e isx dx2 x a; a x aa
1 a
x (cos sx i sin sx ) dx
2 a
2 a 2icos sx
sin sx a
( x (i sin sx ) dx x (1)
s s22 20 0
i 2 as cos as sin as
s2
[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
F s1
f x e isx dx2
If F s is an Fourier transform of fx , then at every point of Continuity of f x , we
have f x1
F s e isx ds .2
11. Find the Fourier cosine Transform of f(x) = e-x Answer:
Fc f x2
f x cos sx dx0
F e x2 e x cos sx dx
c 0 e ax cos bx dx0
F e x 2 1c
s2 1
a
a 2
b2
12. Find the Fourier
Transform of Answer:
f ( x)e imx , a x b
0, otherwise
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F f x1
f x e isx dx2
1 b 1 b
e imx e isx dx e i ms x dx
2 a
2a
1 ei m s xb
1 1 i m s b i m s a
e ei m s i m s2a
2
13. Find the Fourier sine Transform of 1
x .
Answer:
Fs f x2
f xsin sx dx
0
2 sin sx dx 2
0 x 2
Fs
1x 2
14. Find the Fourier sine transform of e x Answer:
F
s f x2
f x sin sx dx0
F e x 2 e x sin sx dxs
e ax sin bx dxb
0a 2 b2
02 sF e x
s
s2 1
15. Find the Fourier cosine transform of e 2 x 2e x
Answer:
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Fc f x2
f x cos sx dx0
Fe 2 x 2e x 2 e 2 x 2e x cos sx dxc
0
2e 2 x cos sx dx 2 e x cos sx dx
0 0
2 2 2 1 2 2 1 1
s 2 4 s2 1 s 2 4 s2 1
16. Find the Fourier sine transformof f ( x)
1 , 0 x 1
0 x 1Answer:
Fs f x2
f x sin sx dx0
1
Fs f x2
f x sin sx dx f xsin sx dx
0 1
12 2 cos sx1sin sx dx 0
s0
2 1 cos ss
1
0
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x , o x 117. Obtain the Fourier sine transform of f ( x ) 2 x , 1 x 2 .
0, x 2
Answer:
F
s f x
2
f x sin sx dx0
1 22x sin sx dx 2 x sin sx dx
0 1
sin sx 1 22 xcos sx 2 x cos sx sin sx
s20 s s2
1s
Fs f x
2 cos s sin s sin 2 scos
s sin s
s s 2 s 2 s s2
22 sin s
sin 2s
s2
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.
e x2 2 is self reciprocal under Fourier transform.
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x 0 x19. Find the Fourier cosine Transform of f ( x)
0 xAnswer:
Fc f x2
f x cos sx dx2
x cos sx dx0 0
2 x sin sx cos sx 2 sin s cos s 1s s2
0 s s 2 s2
2s sin s cos s 1
s2
x
20. Find the Fourier sine transform of .x 2a2
Answer:
L et f x eax
F e ax2 s
ss 2 a2
Using Inverse formula for Fourier sine transforms
e ax
2 2 s sin sx ds
0 s 2 a2
(ie)
s sin sx ds
e ax , a 00 s 2 a2 2
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xsin sx dx e asChange x and s, we get
x 2 a2 20
Fs
x 2 x sin sx dxx 2 a 2 0 x 2 a2
2e
as
e as2 2
FOURIER TRANSFORMPART-B
1. (i)Find the Fourier Transform of f ( x)1 x 2 if x 1
and hence0 if x
1
deduce that (i) x cos x sin x cos xdx 3 (ii) sin x x cos x 2 dx
16 x30 x3 2 0 15
(ii). Find the Fourier Transform off ( x)
a 2 x 2 x a . hence0 x a 0
deduce that sin x x cos x dxx3
0 4
2. Find the Fourier Transform off ( x)
1 if x a and hence evaluate
0 if x a
i )
sinx
dx
sinx 2
ii) dxx0 x0
4. Find Fourier Transform of f ( x)1 x if x 1 and hence
evaluate0 if x 1
sinx 2
sinx 4
i) dx ii) dx0 x x0
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5. Evaluate i) x2
dx ii) dx22
0 x 2 a2
0 x 2 a2
6 i). Evaluate (a)
dx(b)
x 2 dx
0 x 2 a 2 x 2 b2 0 x 2 a 2 x 2b2
ii). Evaluate (a)
dx
(b)
t 2 dt
0 x 2 1 x2 4 0 t 2 4 t 2 9
7. (i)Find the Fourier sine transform of f ( x)sin x; when o x
0 ; whenx
cos x; when o x a(ii) Find the Fourier cosine transform of f ( x)
0 ; whenx ax2 s2
8. (i) Show that Fourier transform e 2 is e 2
(ii)Obtain Fourier cosine Transform of e a 2 x2
9. (i) Solve for f(x) from the integral equation
and hence find Fourier sine Transform x a 2 x2
e
f ( x ) cos x dx e
0
1 , 0 t 1
(ii) Solve for f(x) from the integral equation f ( x ) sin tx dx2 ,1 t 20 0 , t 2
10. (i) Find Fourier sine Transform of e x , x>0 and hence deducethat x sin x
dx
0 1 x2
(ii) Find Fourier cosine and sine Transform of e 4 x , x>0 and hence deduce
that (i ) cos 2x dx e 8 (ii )
x sin2xdx e 8
x 2 16 8 80 0 x2 16
11.(i)Find F xe ax & F xe ax
S c
(ii) FindF
S
e ax
& Fc
e ax
x x
(iii) Find the Fourier cosine transform of f ( x ) e ax cos ax
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9
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 itsZ – Transform is defined as
Z f ( n ) F z f ( n ) z n (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
u ( n)1 for n 0
0 for n 0
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 Z e at f t F z aTz ze
(ii) If Z {f(n)} = F(z),
then Z an f n F za
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)] = z2 [ F(z) – f(0)-f(1) z 1 ]
(iii)Z[f(n +k)] = zk [ F(z) – f(0)-f(1) z 1 - f(2) z 2 ………- f(k-1) z ( k 1) ]
(iv)Z[f(n -k)] = z k F(z)
4. Initial Value Theorem
If Z[f(n)] = F(z) then f(0) = lim F ( z)z
5. Final Value Theorem
If Z[f(n)] = F(z) then lim f ( n) lim( z1) F ( z)
n z 1
PARTIAL FRACTION METHODS
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1
Model:I
1 A Bz a z b z a z b
Model:II
1 A B C
z a z b 2 z a z b ( z b)2
Model:III
1 A Bz C
z a z 2 b z a z 2 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 both terms
Step: 3 Apply z 1
formulaStep: 4 Simplifying we get answer
Note:
1 a a 2 ....... an 1 an 1
1 a
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1 a a 2 ....... a n1 a n 1
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)] = z2 [ F(z) – y(0)- y(1) z 1 ]
iv) Z[y(n +3)] = z3 [ 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
f(n)
No.Z[f(n)
]
1. 1 z
z 1
2. an zz a
3. n z
( z 1)2
4. n2 z 2 z( z 1)3
63
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6. 1
log
z
n ( z 1)
7. 1
z log
z
n 1 ( z 1)
8.1 1 log z
n 1 ( z 1)z
9. ean z
( z ea )
10. 1 1
e zn!
11. Cos n z ( z cos )
z 2 2 z cos 1
12. sin n z sin
z 2 2 z cos 1
13.
cos
n z2
2 z2 1
sin n z
2 z2 1
14. nanaz
( 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 eaT )
4. Sin t z sin T
z 2 2 z cos T 1
64
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5. cos t z ( z cos T )
z 2 2 z cos T 1
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
Z f ( n ) F z f ( n ) z n (Two sided z transform)n
Z f ( n ) F z f ( n ) z n (One sided z transform)n 0
Find the Z Transform of 1
Answer: Z f n f n z nn 0
Z 1 (1)z n 1 z 1 z 2 ....
n 0
1 z 11
11
1z
Z 1z
z 1
z 11
zz z 1
2.Find the Z Transform ofn Answer:
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5
Z f n f n z nn 0
Z n nz n
n 0
nz n 0 z 1 2z 2 3z3 ...n 0
z 1 1 z 1 2 1 1 1 2 1 z 2
z z z z 1
z
z 1 2
3. Find the Z Transform of n2. Answer:
Z n 2 Zn n z
dZ n
, by theproperty,dz
d zz 1 2 z 2 z 1
z2
zz ( z)dz z 1 2 z 1 4 ( z 1)3
4. 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 TheoremIf Z [f (n)] = F (z) then
lim f ( n) lim( z 1) F ( z)n z 1
6. State convolution theorem of Z- Transform.
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6
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
nan Answer:
Z f n f n z nn 0
Z na n na n z nn 0
n a n 0 a 1 2a 2 3a 3 ...n 0 z z z z
a a2
az1
z z z a 2
8. Find Z – Transform of
cos
n
and sin
n
2 2Answer:
We know that Z f n f n z n
n 0
Z cos nz z cos
z 2 2 z cos1
z z cos z2
Z cos n2
2 z 2 2 z cos 1 z2 12
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Similarly Z sin nz sin
z 2 2 z cos1
z sin zZ sin n
22
z 2 2 z cos 1z2 1
2
9. Find Z – Transform of
1
nAnswer:
Z f n f n z nn 0
Z1 1 z
n
n n
0
n
1 z n
z 1 z 2 z 3
....n 1
n1 2 3
log 1 1 log z 1 1
z z
logz
z 1
10.Find Z – Transform of1n!
Answer:
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8
Z f n f n z n
n 0
Z1 1
z nn ! n 0 n!
1 z n 1 z 1 z 2 z 3 ....n 0 n! 1! 2! 3!
e z 1
1
e z
11. Find Z – Transform of n 1 1
Answer:
Z f n f n z nn 0
Z1 1
z n
n 1 n 0 n 1
z1
z ( n 1)
n 1n 0
z z 1 z 2 z 3 ....2 3
z log 1 1z
z logz
z 1
12. Find Z – Transform of an
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9
Answer:
Z f n f n z nn 0
a n an
Z an
n 0 z n n 0z
1a 1 a 2
z z
a1
1z
a 3
...z
z a1
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 nn 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 nF ( zeaT )
n 0
14. 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
The unit step sequence is defined as follows:
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0
u ( n)1
for n 0
0 for n 0
15.at
Find Z – Transform of Z eAnswer:
Z e at e anT z n e aTn
z n z eaT n
n 0 n 0
zz an z
z eaT z a[Using First shifting theorem]
16.
Find Z – Transform of
Z te 2t
Answer:
Z te 2t Z tTz
2Tz ze z 1 2
z ze2T
Tze2T
ze2T 1 2
[Using First shifting theorem]
17.
Find Z – Transform of
Z et cos 2t
Answer:
z z cosZ e t cos 2t Z cos 2t T
z ze z 2 2 cos z 1z ze T
ze T ze T cosT
ze 2T
2 cos T ze T 1
[Using First shifting theorem]
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18. Find Z – Transform of Z e2 t T
Answer:
Let f (t) = e2t , by second sifting theorem
Z e 2( t T )Z f (t T ) z F ( z ) f (0)
z
ze 2T
1 z
1
ze 2 T 1 ze 2T 1
19. Find Z – Transform of Z sin t TAnswer: Let f (t) = sint , by second sifting theorem
Z sin(t T ) Zf (t T ) z F ( z ) f (0)
z
z sin t
0
z 2 sin t
z 22 cos t z 1 z 2 2 cos t z 1
20. Find Z – transform of n 1 n 2
Answer:
Z f n f n z nn 0
Z n 1 n 2 Z n 2 2 n n 2
Z n 2 3n 2 z n 2 3 z n 2 z 1
z 2 z3
z2
zz 1 3 z 1 2 z 1
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QUESTION BANK
Z-TRANSFORMS
(i)Find Z 18z2 &
Z 1
8z2
1. by convolution theorem.
(2z
1)(4 z 1) (2 z 1)(4 z 1)
(ii) Find Z 1
z2 &
Z 1
z2
by convolution theorem
( z a )( z b) ( z 1)( z 3)
2. (i) Find Z 1z2
& Z 1 z2
by convolution theorem
( z a)2( z a)2
3. (i ) State and prove Initial & Final value theorem.
(ii) State and prove Second shifting theorem
(i) Find the Z transform of
1&
2 n 3( n 1)( n 2) ( n 1)( n 2)
4. (i) Find Z 1 z2 byresidues.
( z2 4)
(ii) Find the inverse Z transform of
z 2 zby partial fractions.z 1 ( z2 1)
5. (i) Find Z 1 z & Z
1 z2
z 2 2 z 2 z 2 7 z 10
6. (i)Find the Z transform of f ( n)
1
Hence find Z
1
and Z
1
.n!( n 1)! ( n 2)!
(ii) Find Z
1 and also find the value of sin(
n 1)and
cos( n 1) .
n!
7. (i)Solve y n 2 6 y n 1 9 y n 2n with y 0 0 & y 1 0
(ii) Solve y
n 2 4 y n 1 4 y n0 y(0) = 1 ,y(1)
=0
8.(i )Solve y n 3 y n 1 4 y n 2 0, n 2 given y (0) 3& y(1) 2
(ii) Solve y
n 3 3 y n 1 2 y n 0, y 0 4, y 1 0 & y 2 8,
9. (i)Find Z n
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