SYLLABUS COPY FOR SECOND YEAR B. PHARM SEMESTER III ANATOMY, PHYSIOLOGY AND
III Semester b
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Transcript of III Semester b
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III SEMESTER B.E/B.TECH(COMMON TO ALL BRANCHES)
MA 2211 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS
QUESTION BANK
UNIT-I FOURIER SERIES
PART A
1. Define Dirichlets condition (or) State the condition for f(x) to have
Fourier Series expansion .
SOL : (i) f(x) is periodic, single valued and finite.(ii) f(x) has a finite number of discontinuities in any one period
(iii) f(x) has a finite number of maxima and minima.
(iv) f(x) and f(x) are piecewise continuous.
2. State whether y = tan x can be expanded as a Fourier Series. If so how?
If not why?
SOL : tan x cannot be expanded as a Fourier series. Since tan x notsatisfied Dirichlets Conditions. (tan x has infinite number of infinite
discontinuities)
3. Find the sum of the Fourier Series for f(x) = x 0
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= - xx3
= - ( x+x3)
= - f(x)f(x) is an odd function.
Hence a0 = 0 an = 0
5. If f(x) = x2 + x is expressed as a Fourier series in the interval ( -2, 2) towhich value this series converges at x = 2.
SOL : x = 2 is a point of discontinuity in the extremumf(x) = [f(-2) + f(2)] / 2
= {[(-2)2 + (-2)] + [22 + 2]} / 2
= {[4-2] + [4 +2] } / 2
= 8 /2= 4
f(x) = 4
at x = 2
6. State Parsevals Identity for full range expansion of f(x) as Fourier Series
in (0,2l)
SOL : Let f(x) be a periodic function with period 2l defined in theinterval ( 0, 2l ) then
)(2
1
4)]([
2
1 21
2
2
0
2
0
2
nnn
l
baa
dxxfl
7. If the Fourier Series corresponding to f(x) =x in the interval (0, 2 ) is
)sincos(2 10 nxbnxa
ann n
without finding the values ofa0, an , bn
find the value of )(
2
2
1
2
2
0
nnn ba
a
SOL : By Parsevals Theorem
dxxfbaa
nnn
2
0
22
1
22
0 )}([1
)(2
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= dxx2
0
21
=
2
0
3
3
1 x
= 23
8
8. Define Root Mean Square value (or) RMS value
SOL : The root mean square value of f(x) over the interval (a,b) isdedfined as
R.M.S. =ab
dxxf
b
a
2)]([
9. Find the constant term in the Fourier series corresponding to f (x) = cos2x
expressed in the interval (- ,).
SOL: Given f(x) = cos2x =2
2cos1 x
W.K.T f(x) =2
0a + nxbnxan
n
n
n sincos11
To find a0 = xdx
2cos1
= dxx
0
2
2cos12
=
02
2sin1
xx
=
1[( + 0) (0+0)]
= 1.
10. If f(x) = x2+x is expressed as a Fourier series in the interval (-2,2) to
which value this series converges at x = 2.SOL: x = 2 is a point of discontinuity in the extremum.
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[f(x)]x = 2 =2
)2()2( ff
=2
]22[)]2()2[( 22
= 2
]24[]24[
=2
8= 4.
11. Expand f(x) = x in 0 < x < 1 as a Fourier series.
SOL: Let f(x) =
l
xnb
l
xna
a
n
n
n
n
sincos
2 11
0
a0 = l
dxxfl
2
0
)(1
an = dxl
xn
xfl
l
2
0cos)(
1
bn = dxl
xnxf
l
l
sin)(
12
0
Here 2l = 1, l =2
1.
a0 =
1
0
2
1
1xdx = 2
1
0
2
2
x= 2
0
2
1= 1
an = dxxn
x
1
0
2
1cos
2
1
1
= 2 1
0
2cos xdxnx
= 2 1
0
224
2cos1
2
2sin
n
xn
n
xnx
= 2
2222 4
10410
nn
= 0
bn = dxxn
x
1
0
2
1sin
2
1
1
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bn = 21
0
224
2sin
2
2cos
n
xn
n
xnx
bn = -n
1
f(x) =2
1+ 0 +
1
1
n nsin
2
1
xn
f(x) = xnnn
2sin11
2
1
1
.
12. Expand f(x) = 1 in a sine series in 0 < x < .
SOL : The sine series of f(x) in (0, ) is given by
f (x) = nxbn
n sin1
where bn =
0
sin2
nxdx
= -
0cos2
nxn
= 0 if n is even
= n
4
if n is odd
f(x) = nxnoddn
sin4
=
4
1 12
12sin
n n
xn.
PART-B
1. a) Expand f (x) =eax in aFourier series in (0, 2 ).b) Find the sine series for f (x) =x in 0
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5. Find the half range sine series for f(x) =xcosx in (0, ).
6. Find the sine series of f(x) =x in 0
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0
cos)(2
)()]([ sxdxxfsFxfF cc
The inversion formula is
0
cos)(2
)( sxdssFxf c
5. State the Convolution theorem for Fourier Transform.
SOL : If F(s) and G(s) are the Fourier transform of f(x) and g(x)respectively. Then the Fourier transform of the convolution of f(x) and g(x)
is the product of their Fourier transform.)()()](*)([ SGSFxgxfF
6. Define Self Reciprocal
SOL : If a transformation of a function f(x) is equal to f(s) then the functionf(x) is called self reciprocal.7. Find the Fourier Sine transform of 1/x
SOL : We know that
0
sin)(2
)()]([ sxdxxfsFxfF sS
=
0
sin12
]1[ sxdx
xxFS
Let sx = x 0 0 s dx = d x
0
sin2
s
ds
0
sin2
d
=2
2
=
2
.
0
sin
d =
2
8. Find Fourier Cosine transform of e x.
SOL : We know that
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0
cos)(2
)]([ sxdxxfxfFc
0
cos2
][ sxdxeeF xxc
=
21
12
s
22
0
cosba
abxdxe ax
9.Find Fourier Sine transform of e -3x.
SOL: We know that
0
sin)(2
)]([ sxdxxfxfFS
0
33 sin2][ sxdxeeF xxS
=9
22 s
s.
22
0
sinba
bbxdxe ax
10. Find Fourier Cosine transform of xeax.
SOL : we know that
)]([)]([ xfFds
dxfF sC
][][ axsax
C eFds
dxeF
=
0
sin2
sxdxeds
d ax
=
222
as
s
ds
d
= 222
222
as
sa
11. Find Fourier Sine transform of xeax.
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SOL : we know that
)]([)]([ xfFds
dxfF cs
][][ axcaxs eFdsdxeF
=-
0
cos2
sxdxeds
d ax
=-
222
as
a
ds
d
= 22222
as
as
12. Show that f(x) = 1, 0 < x < cannotbe represted by a Fourier integral.
SOL : dxxf
0
)( = 1
0
dx = 0x = and this value tends to as x .
i.e., dxxf
0
)( is not convergent.
Hence f(x) = 1 cannot be represented by a Fourier integral.
PARTB
1. Using Fourier integral of f(x) = x x for x < 10 x for x > 1
evaluate
SinCos
0
2. Find the Fourier transform of f(x) where1 - x for x < 1
f(x)
0 for x > 1
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Hence deduce that 2
Sin t dt = / 2
t
3. Find the Fourier Sine transform of Sinx , 0 < x < f(x) = 0 < x 0. Deduce that
1 dx = ( / 4a
3) if a > 0
0 (x2
+ a2)
2
7. Find the Fourier Cosine transform of x e x 2 / 28. Derive the parsevals identity for Fourier transforms.
UNIT III BOUNDARY VALUE PROBLEMSUNITIII
APPLICATION OF PARTIAL DIFFERENTIAL EQUATIONS
2 MARKS QUESTIONS
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1. Classify the partial differential equation xyy
u
x
u
yx
u
2
Solution: B = 1, A = 0, C = 0 0142 ACB Hyperbolic type.
2. In the wave equation for?stantcdoesWhat, 22
22
2
2
x
yc
t
y
lengthunitpermass
Tension
m
Tc
2
3. What are the possible solutions of the one dimensional wave equation?1. ))((),( 4321 cxccxctxy
2. )sincos)(sincos(),( 8765 pctcpctcpxcpxctxy
3. ))((),( 1211109pctpctpxpx ecececectxy
4. In a heat equation2
22
x
u
t
u
What does 2 stand for?
cp
k2 where k: Thermal conductivity
c: Specific heat and
p density 2 is called the diffusivity of the material.
5. State any two laws which are assumed to derive one dimensional heat equation.i. The amount of heat required to produce a given temperature change ina body is proportional to the mass of the body and to the temperature
change.
ii. The rate at which heat flows across any area is proportional to the areaand the temperature gradient normal to the area.
6. State Fourier law of heat conduction.The rate at which heat flows across an area A at a distance x from one end
of a bar is given by Q = KA
x
u, Where K : Thermal conductivity,
x
u: Temperature gradient at x.
7. Write down the three possible solutions of ODHE and state the suitable solution.1. 321 )(),( ccxctxu
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2. ippxpx ececectxu22
654 )(),(
3. ipecpxcpxctxu22
987 )sincos(),(
Suitable solution: ipepxBpxAtxu22
)sincos(),(
8. An insulated rod of length 60 cm has its ends A and B maintained at 20 C and 80 Crespectively. Find the steady state solution of the rod.
Steady state solution xl
abaxu
)(
A = 20 C , B = 80 C l=60
Therefore u(x) = 20 + x for 0 < x < l
9. Write down the three possible solution of Laplace equation 0 yyxx uu 1. )sincos)((),( 4321 pycpycecectxu
pxpx
2. ))(sincos(),( 8765pypy ececpxcpxctxu
3. ))((),( 1211109 cyccxctxu
10.Write the boundary condition for the problem: A rectangular plate is bounded by thelines x = 0 ,y=0,x=a and y=b. Its surfaces are insulated. The temperature along x=0and y=0 are kept at 0 C and the other at 100 C.
The boundary conditions are
i) byou y0for0),(
ii) axu x0for0)0,(
iii) byau y0for100),(
iv) abxu x0for100),(
PART-B
1. A rectangular plate with insulated surfaces is 10 cm wide and so longcompared to its width that it may be considered infinite in length with out
introducing an appreciable error. If the temperature along one short edge y =0 is T(x,0) = 4(10xx2 ) for 0
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3. A string is tightly stretched and its ends are fastened at two points x = o andx = l. The mid point of the string is displaced transversely through a smalldistance b and the string is released from rest in that position. Find an
expression for the transverse displacement of the string at any time during
the subsequent motion.
4. A rod of length l has its ends A and B are kept at 0c and 100c until steadystate condition prevail. If the temperature at B is suddenly reduced to 0c andkept so while that of A is maintained, find the temperature u(x,t) at adistance x from A and at time t.
5. A bar 10 cm long with insulated sides, has its ends A and B kept at 20 c and30c respectively until steady state condition prevail. T he temperature at Ais then suddenly raised to 50 c and at the same instant that at B is lowered to
10 c. Find the subsequent temperature at any point of the bar at any time.
UnitIV
PARTIAL DIFFERENTIAL EQUATIONS
PARTA
1. Form the p.d.e by eliminating the arbitrary constants a and b frombyaxz .
Soln: Given byaxz ..(1)
Diff. partially w.r.t. x we get ax
z
i.e., ap
Diff. partially w.r.t. y we get by
z
i.e., bq
Substituting in (1) we get qypxz .
2.
Form the p.d.e by eliminating the arbitrary constants a and b from))(( 2222 byaxz .
Soln: Given ))((2222 byaxz (1)
Diff. partially w.r.t. x we get ))(2( 22 byxx
zp
)(
2
22 byx
p ..(2)
Diff. partially w.r.t. y we get ))(2( 22 axyy
zq
)(
2
22 axy
q ..(3)
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Substituting (2) and (3) in (1) we get the required p.d.e.
xy
pq
x
p
y
qz
422 1.e., pqxyz4 .
3. Form the p.d.e by eliminating the arbitrary funtion from )( 22 yxfz Soln: Given )(
22 yxfz
)2)((' 22 xyxfx
zp
, )2)((' 22 yyxf
y
zq
y
x
q
p
2
2 0 qxpy .
4. Form the p.d.e by eliminating the arbitrary funtion from0),( zyxyx
Soln. w.k.t if 0),( vu then )(vu
)( zyxyx .(1)Diff. partially w.r.t. x we get
)01)(('1x
zzyx
)('
1
1zyx
p
(2)
Diff. partially w.r.t. y we get
)10)(('10y
zzyx
)('
1
1zyx
q
(3)
From (2) and (3), we get 02 qp .
5. Solve 1 qp Soln. Given 1 qp
This is of the form 0),( qpF . Hence the complete integral is cbyaxz
Where 1 ba 2)1( ab . Therefore the complete solution is
cyaaxz 2)1( (1)
Diff. partially w.r.t. c we get 0=1. There is no singular integral. Taking )(afc
where f is arbitrary.
)()1( 2 afyaaxz (2)
Diff. P.w.r.t. a we get )('2
1
)1(20 afyaax
.(3)
Eliminating a between (2) and (3) we get the general solution.
6. Find the complete integral of 22 qpqypxz Soln. Given
22 qpqypxz (1)
This equation of the form ),( qpfqypxz [Clairauts type]
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Therefore the complete integral is22 babyaxz .
7. Find the complete integral of pxq 2 .Soln. Given pxq 2
This equation of the form 0),,( qpxf
Let aq . Thenx
ap
2 .
But advdxx
adz
2
Integrating on both sides we get, bayxa
z log2
8. Solve yxqp Soln: Given yxqp
)( qyxp
This equation is of the form ),(),( qypxf
aqyxp )( say
xap and ayq But qdypdxdz dyaydxxadz )()( Integrating on both sides we get,
bayyx
axz 22
22
.(1)
Diff.(1) P.w.r.t. b we get )(ab .(2)
)(22
22
aayyx
axz (3)
Diff(3) P.w.r.t. a we get
)('0 ayx .(4)
Eliminating (3) and (4) we the general solution.
9. Solve 0]'''[ 3223 zDDDDDD Soln. The auxiliary equation is 0123 mmm
0)1)(1( 2 mm 1m , im General Solution is
)()()( 321 ixyixyxyz
10.Find the particular integral of )2cos(]'2'3[ 22 yxzDDDD Soln. Given )2cos(]'2'3[
22 yxzDDDD
)2cos('2'3
1..
22yx
DDDDIP
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)2cos()4(2)2(31
1...... yx
)2cos(
861
1..... yx
)2cos(3
1..... yx
.
1. Solve p(1-q2)=q(1-z)2. Solve (D2-DD1-2D2) z=2x+3y+e3x+4y3. Solve z2=p2+q2+14. Find the general integral of x(y2+z)p+y(x2+z)q=z(x2-y2)5. Solve p=2qx6. Solve z2=xypq7. Find the PDE by eliminating f from
(a) f(xy+z2,x+y+z)=0(b) z=xy+f(x2+y2+z2)
(c)z= x2f(y)+y2g(x)
UNIT-V
ZTRANSFORMS
UNIT V
ZTRANSFORM AND DIFFERNCE EQUATIONS
PART A
1. Prove that Z[an] =az
z
if az .
Solution:
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az
z
z
az
z
a
z
a
z
a
z
a
zaaZ
n
n
n
n
nn
1
1
2
0
0
1
.....1
][
2. Prove that Z(n)= 21zz
Solution:
2
2
2
2
32
0
0
1
11
11
1
.....1
31
211
.....321
0
][
z
z
z
z
z
zz
zzz
zzz
z
n
znnZ
nn
n
n
3. Find Z[(-1)n]Solution:
We know that Z[an] =azz
Z[(-1)n] =
1)1(
z
z
z
z
4. Find Z
)1(
1
nn
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Solution:
1
log)1(
1log
1log
1
11
)1(
1
1
11
)1(
1
11
10
)()1(1
1)1(
1
z
zz
z
zz
z
z
nZnZ
nnZ
nnnn
Bgetwenput
Agetwenput
nBnA
n
B
n
A
nn
5. Find Z[ann]Solution:
We know that Z[anf(n)]=F[Z/a]
2
2
/
2
/
)(
)1(
][][
az
az
a
az
a
z
z
z
nZnaZ
azz
azz
n
6. Find Z[etsin2t]Solution:
12cos2
2sin
12cos2
2sin
]2[sin]2sin[
)]([)]([
22
2
Tzeez
Tze
Tzz
Tz
tZteZ
tfZtfeZ
TT
T
zez
zez
tzez
at
T
T
aT
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7. Prove that Z[f(n+1)]=zF(z)-zf(0)Solution:
)0()(
)0()(
1)(
)1(
)1()]1([
0
1
0
)1(
0
zfzzF
fzmfz
nmwherezmfz
znfz
znfnfZ
m
m
m
m
n
n
n
n
8. Find )]([ knZ Solution:
k
n
n
zknZ
knfor
knforkn
zknknZ
1)]([
0
1)(
)()]([0
9. Find )]2(2[ nZ n Solution:
zz
z
nZnZ
zz
zzn
4
2
1
1
)]2([)]2(2[
2
2/
2
2/
10.If Z(f(n))=F(z), then f(0)= )(lim zFz
Solution:
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)0()(lim
.....)2()1()0(lim))((lim
.....)2()1(
)0(
)()]([
2
2
0
fzF
zf
zffnfZ
z
f
z
ff
znfnfZ
z
zz
n
n
11.Define convolution of sequences.Solution:
The convolution of two sequences {x(n)} and {y(n)} is defined as
(i) {x(n) * y(n)} =
K
KngKf )()( if the sequences are non-causal.
(ii) {x(n) * y(n)} =
n
K KngKf0 )()( if the sequences are causal.
The convolution of two function f(t) and g(t) is defined as
(i) f(t) * g(t)= TKngKTfn
K
0
)()( , where T is the sampling period.
12.Find
)3()2( 2
31
zz
zZ by convolution theorem.
Solution:
n
m
mmn
n
m
mnm
nn
m
m
n
z
zZ
z
zZ
z
z
z
zZ
zz
zZ
0
0
1
2
21
2
31
2
31
32)1(3
32)1(
32)1(
)3()2(
3)2()3()2(
13.Find the Z transform of nanu(n)Solution:
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21
1
211
111
11
1
1
1
1
)1(
)1(
)(1)1(
1
)(
az
az
azaz
aazz
azdz
dz
az
z
dz
dznuaZ n
14.Define Unit step sequence.Solution:
The unit step sequence u(n) is defined as
00
01)(
nfor
nfornu
15.Find Z[cos (t+T)]Solution:
1cos2
)1cos(
1cos2
cos
1cos2
cos2cos
1cos2)cos(
1cos2
)cos(
0cos][cos))(cos(
)0()]([
)0()()((
2
2
2
2
2323
2
2
2
Tzz
Tzz
Tzz
zTz
Tzz
zTzzTzz
zTzzTzz
zTzz
Tzzz
ztzZTtZ
zftfZz
zfzzFTtfZ
PARTB
1. Find Z -1 4Z22ZZ
35Z
2+8Z4
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2. Find Z -1 Z by using residue theorem.( z1) ( z2)
3. Use residue theorem find Z-1 Z2Z
2+ 4
4. Solve by Z transform yn + 2 + yn = 3 , given that y0 = y1 = 05. Solve the difference equation x (n+2)6x(n+1) +8x(n) = 4n ,
x(0) = 0 , x(1) = 1 by Z transform.
6. Solve U(x+2)4U(x) = 9x2 , given that U(0) = 0 , U(1) = 0 by Ztransforms.
7. Solve yn+25yn+1 + 6yn = 6 n , y0 = 1 , y1 = 08. Solve by Z transforms yn+2 + yn = n 2n