Fourier series Part - A 1.Define Dirichet s conditions Ans: A function defined in c x c + can be expanded as an infinite trigonometric series of the form a + a n cos nx n 1 + b n sin nx, provided i) f x is singe vaued and periodic in c, c + ii) f x is continuous or piecewise continuous with finite number of finite discontinuous in c, c + iii) f x has no or finite number of maxima or minima in c, c +.. Write the Fourier series for the function f(x) in the interva c x c +. Where f x a + a n cos nx + b n sin nx n1 a 1 c+ c f x dx a n 1 b n 1 c+ c c+ c f x cos nx dx f x sin nx dx 3. Write the Fourier series for the function f(x) in the interva x.
f x a + a n cos nx + b n sin nx n1 Where a 1 a n 1 b n 1 f x dx f x cos nx f x sin nx dx dx 4. Write the Fourier series for the function f(x) in the interva x. Where f x a + a n cos nx + b n sin nx n1 a 1 a n 1 b n 1 f x dx f x cos nx f x sin nx dx dx 5. Write the Fourier series for the function f(x) in the interva x.
f x a + a n cos nx + b n sin nx n 1 Where a 1 f x dx a n 1 f x cos nxdx b n 1 f x sin nxdx 6.Write the Fourier series for the function f(x) in the interva x. Where f x a + a n cos nx + b n sin nx n 1 a 1 f x dx a n 1 f x cos nxdx b n 1 f x sin nxdx 7. Write the Fourier series for the even function f(x) in the interva x. (or) Expain haf range cosine series in the interva x. a f x dx a n f x cos nxdx
b n, Then the Fourier series is reduced to f x a + a n cos nx n1 8. Write the Fourier series for the odd function f(x) in the interva x. (or) Expain haf range sine series in the interva x. a, a n, b n Then the Fourier series is reduced to f x b n sin nx n1 f x sin nxdx, 9. Write the Fourier series for the even function f(x) in the interva x. (or) Expain haf range Fourier cosine series in the interva x a a n f x dx f x cos nx dx b n, Then the Fourier series is reduced to f x a + a n cos nx n1 1. Write the Fourier series for the odd function f(x) in the interva x. (or) Expain haf range Fourier sine series in the interva x
a, a n, b n Then the Fourier series is reduced to f x sin nx dx f x n1 b n sin nx 11. Define periodic function and give exampes. Ans: A function f x is said to be periodic, if and ony if f x + T f x where T is caed period for the function f x. Eg: sin x and cos x are periodic functions with period. 1. Write any two advantages of Fourier series. Ans: i) Discontinuous function can be represented by Fourier series. Athough derivatives of the discontinuous functions do not exist. ii) The Fourier series is usefu in expanding the periodic functions. Since outside the cosed interva, there exists a periodic extension of the function iii) Fourier series of a discontinuous function is not uniformy convergent at a points. iv) Expansion of an osciation function by Fourier series gives a modes of osciation which is extremey usefu in physics. 13. Define Parseva s identity for the Fourier series in the interva, 1 f x dx a + a n + b n n1 14. Define Parseva s identity for the Fourier series in the interva, 1 f x dx a + a n + b n n1
15. Define Parseva s identity for the Fourier series in the interva, 1 f x dx a + a n + b n n1 16. Define Parseva s identity for the Fourier series in the interva, 1 f x dx a + a n + b n n1 17. Define Parseva s identity for the haf range Fourier cosine series in the interva, f x dx a + a n n1 18. Define Parseva s identity for the haf range Fourier sine series in the interva, f x dx b n n1 19. Define Parseva s identity for the haf range Fourier cosine series in the interva, f x dx a + a n n1. Define Parseva s identity for the haf range Fourier sine series in the interva, f x dx b n n1
1. Define Root-Mean Square vaue of a function f x RMS vaue of a function f x b a f x dx b a. Find the vaue of a n in the cosine series expansion of f x k in the interva, 1. a n f x cos nx dx 1 k 5 1 1 kcos nx 1 dx cos nx 1 dx k 5 sin nx 1 1 n 1 k 5 1 n sin 1n 1 sin a n 3. Find the root mean square vaue of the function f x x in the interva, RMS vaue of a function f x b a f x dx b a x dx
x 3 3 3 3 3 4. State the convergence theorem on Fourier series. Ans: i) The Fourier series of f x converges to f x at a points where f x is continuous. Thus, if f x is continuous at x x, the sum of the Fourier series when x x is f x. ii) At a point x x where f x has a finite discontinuity, the sum of the Fourier series 1 y 1 + y where y 1 and y are the two vaues of f x at the point of finite discontinuity. 5. The function f x 1 Expain why? x x 3 cannot be expanded as Fourier series. Ans: Given f x 1 x x 3 At x, f x becomes infinity. So it does not satisfy one of the Direchet s condition. Hence it cannot be expanded as a Fourier series. 6. Can you expand f x 1 x 1+x as a Fourier series in any interva. Let f x 1 x 1 + x This function is we defined in any finite interva in the range, it has no discontinuous in the interva
Differentiate f x 1 + x x 1 x x 1 + x f x 4x 1 + x f x is maximum or minimum when f (x) (i.e.,) when 4x i. e., x. So it has ony one extreme vaue (i.e.,) a finite number of maxima and minima in the interva, Since it satisfies a the Dirichet s conditions, it can be expanded in a Fourier series i aspecified interva in the range, 7. Find a, f(x) xsinx in ( < x < ). Since is f(x) xsinx is an even function a f x dx x sinx dx x( cos x) 1( sin x) ( cos ) 1( sin ) 9. Find a, f(x) x in(, ). a f x dx ( x )dx x3 x 3
3. Find b n, f(x) x 3 in (, ). Since f(x) x 3 is an odd function 3 3 b n x3 sin nxdx b n cos nx sin nx cos nx x3 3x n n + 6x n 3 1 b n cos n sin n cos n 3 3 n n + 6 n 3 1 n n 1 1 3 + 6 n n 3 31. Find a n in f x e x in (, ). a n 1 a n 1 e x f x cos nxdx e x cos nxdx sin nx n 4 sin n n 4 1 1 + n cos nx + nsinnx 1 e e cos n + nsinn + cos( n) + nsin( n) 1 + n 1 + n 1 e 1 + n 1 n + e 1 + n 1 n (1 + n ) e e n 1 (1 + n sin x ) 1 n 3. Find b n, f(x) ( x) in(, ).
b n 1 f x sin nxdx b n 1 ( x) sin nxdx 1 ( x) cos nx n sin nx ( 1) n 1 ( ) cos n n sin n 1 n ( ) cos n + 1 1 n 1 n + 1 n
1. State the Fourier integra theorem. Fourier Transform Part-A Ans: If is a given function f(x) defined in, and satisfies Dirichet s conditions, then f x 1 f t cos λ t x dx dλ. State the convoution theorem of the Fourier transform. Ans: if F(s) and G(s) are the functions of f(x) and g(x) respectivey then the Fourier transform of the convoution of f(x) and g(x) is the product of their Fourier transform F (f g x] F S. G(s) 1 f g x e isx dx 1 f x e isx dx 1 g x e isx dx 3. Write the Fourier transform pairs. Ans: F f x and F 1 F(s) are Fourier transform pairs. 4. Find the Fourier sine transform of f(x) e ax (a > ). Ans: The Fourier sine transform of f(x) is F s f x f x sin sx dx e ax sin sx dx s s + a Since b e ax sin bx dx b + a 5. What is the Fourier transform of f(x a) if the Fourier transform of f(x) is F(s). F f x a e ias F(s) 6. State the Fourier transforms of the derivatives of a function. F dn f(x) dx n is n F s.
7. Find the Fourier sine transform of e x. The Fourier sine transform of f(x) is F s f x f x sin sx dx e x sin sx dx s s + 1 Since e ax b sin bx dx b + a 8. Prove that F c f ax 1 a F c s a, a >. F c f ax f ax cos sx dx Put ax y when x, y. dx dy a When x, y 1 a f y cos sy a f y cos s a dy a y dy 1 a F c s a 9. Find Fourier sine transform of e ax. The Fourier sine transform of f(x) is The Fourier sine transform of f(x) is F s f x f x sin sx dx
e ax sin sx dx s s + a Since b eax sin bx dx b + a 1. if F s F f x, then prove that F xf x i d[f s ] ds F s 1 f x e isx dx d[f s ] ds d ds 1 f x e isx dx 1 f x s (eisx )dx 1 f x ix(e isx )dx if xf x F xf x i d[f s ] ds 11. Find Fourier sine transform of 1 x F s f x f x sin sx dx 1 sin sx dx x Since 1 x sin ax dx, a >. 1. Find the Fourier cosine transform of e x.
The Fourier sine transform of f(x) is F c f x f x cos sx dx e x cos sx dx 1 s + 1 a Since e ax cos bx dx b + a 13. If F s s is the Fourier sine transform of f(x), show that F s f x cos ax 1 F s s + a + F s s a F s f x cosax f x cos ax sin sx dx. 1 f x sin a + s x sin a s x dx. 1 f x sin a + s x dx. 1 f x sin a s x dx 1 F s s + a + F s s a 14. If F(s) is the Fourier transform of f(x), write the formua for the Fourier transform of f(x)cos ax in terms of F. F f x cos ax 1 f x cos ax e isx dx 1 f x eiax + e iax e isx dx 1 f x ei(s+a)x dx + 1 f x ei(s a)x dx
F f x cos ax 1 F s + a + F(s a) 15. Find the Fourier transform of f x 1 in x < a in x > a We know that the Fourier transform fo f(x) is given by F f(x) 1 f x e isx dx a 1 e isx dx a e isx a e ias e ias 1 is a 1 is sin as s 16. State any two properties of Fourier transform i) Linearity property: If f(x) and g(x)are any two functions then F af(x) + bg(x) af f(x) + bf g(x), where a and b are constants. ii)shifting property: If F f(x) F s, then F f(x a) e ias F(s) 17. Define the infinite Fourier cosine transform and its inverse The infinite Fourier cosine transform is given by F c f x F c s f x cos sx dx Then the inverse Fourier cosine transform is given by F c 1 F c s f(x) F c s cos sx ds 18. State Parseva s identity for Fourier transform. If F s is the Fourier transform of f(x), then f x dx F s ds
19. Define the Fourier sine transform of f(x) in (, ). Aso give the inversion formua The finite Fourier sine transform of a function f x in, is given by Then the inversion formua is given by f s n f x sin nx dx, f x n1 f s n sin nx.. Define the Fourier cosine transform of f(x) in (, ). Aso give the inversion formua The finite Fourier cosine transform of a function f x in, is given by Then the inversion formua is given by f c n f x c nx dx, f x 1 f c() + n 1 f c n cos nx. 1. Define the infinite Fourier sine transform and its inverse The infinite Fourier sine transform is given by F s f x F s s f x sin sx dx Then the inverse Fourier cosine transform is given by F s 1 F s s f(x) F s s sin sx ds. Prove that if f(x) is an even function of x, its Fourier transform F(s) wi aso be an even function of s. By definition F s 1 f x e isx dx 1
Changing s into s in both sides of (1), F s 1 f x e isx dx In the right hand side integra in (), put x u. then dx du; when x, u and when x, u. So () becomes F s 1 f u e isu. ( du) 1 f x e isx dx 1 f u e isu du [Changing the dummy variabe u into x] 1 f x e isx dx F S by 1. Here F S is an even function of s. 3. Prove that if f(x) is an odd function of x, its Fourier transform F(s) wi aso be an odd function of s. By definition F s Changing s into s in both sides of (1), 1 f x e isx dx 1 F s 1 f x e isx dx In the right hand side integra in (), put x u. then dx du; when x, u and when x, u.
So () becomes F s 1 f u e isu. ( du) 1 f x e isx dx 1 f u e isu du [Changing the dummy variabe u into x] 1 f x e isx dx F S by 1. Here F S is an odd function of s. 4. Find the Fourier transform of the function defined by f x, 1,, x < a a < x < b x > b F f(x) 1 f x e isx dx 1 a b f x e isx dx + f x e isx dx + f x e isx dx a b b 1 eisx dx a 1 e isx is b a 1 e isb e isa 5. Show that f(x) 1, < x < cannot be represented by a Fourier integra. f x dx dx x is
i. e., f x dx is not convergent Hence f(x) 1 cannot be represented by a Fourier integra. 6. Find the Fourier cosine transform of f x 1, < x < a, x > a We know that F c f x F c s f x cos sx dx a 1 cos sx dx a sin sx s sin sa s sin sa s 7. Find the Fourier transform of f x 1 in x 1 in x > 1 We know that the Fourier transform fo f(x) is given by F f(x) 1 f x e isx dx 1 1 e isx dx 1 e isx 1 e is e is 1 is 1 1 is
sin s s 8. Find the Fourier transform of e a x, a >. We know that the Fourier transform fo f(x) is given by F f(x) 1 f x e isx dx 1 f x e isx dx + 1 1 e (is+a)x dx + 1 f x e isx dx e (is a)x dx 1 e (is+a)x + is + a e (a is )x is a 1 1 is + a 1 is a 1 is a is a is a a s + a 9. State the Fourier transforms of derivates of a function. F df(x) dx 1 f x e isx dx 1 e isx d f(x) 1 eisx f x f x ise isx dx
is f x e isx dx (assuming f(x) as x ±) (assuming f (x) as x ±) (assuming f (x) as x ±) In genera, isf s. F d f(x) dx 1 1 e isx d f (x) f x e isx dx 1 eisx f x f x ise isx dx is is is eisx f x is f x e isx dx e isx d f(x) f x ise isx dx f x e isx dx is F s. F dn f(x) dx n is n F s. 3. Find the finite Fourier cosine transform of f x e ax in (, ). f c n f x cos nx dx, n being a positive integer
e ax cos nx dx e ax a + nx acos nx + nx sin nx e a a + n a cos n a + n ( a) Partia Differentia Equations Part-A a a + n 1 + e a ( 1) n+1 1. Form a partia differentia equation by eiminating arbitrary constants a and b from z x + a + (y + b) Given z x + a + (y + b) (1) Substituting ()& (3) in (1), we get z p + q 4 4. Sove (D DD + D )z A.E is m m + 1 p z x + a () x q z y + b (3) y m 1 m 1, 1 y f 1 y + x + xf (y + x) 3. Form a partia differentia equation by eiminating arbitrary constants a and b from x a + (y b) z cot α Given x a + y b z cot α 1
Partiay differentiating w.r.t x and y we get x a zpcot α y b zqcot α 3 x a zpcot α 4 y b zqcot α 5 Substituting (4)& (5) in (1), we get z p cot 4 α + z q cot 4 α z cot α p + q tan α 4. Find the compete soution of the differentia equation p + q 4pq Given p + q 4pq (1) Let us assume that z ax + by + c () be the soution of (1). Partiay differentiating () w.r.t x and y we get Substituting (3) in (1) we get p z x a q z y b 3 a + b 4ab (4) From (4) we get a 4b± 16b 4a b Substituting (5) in () we get 4b ±b 4 a b ± b 4 a 5 5.Find the soution of px + qy z The S.E is Taking 1 st two members, we get z b ± b 4 a x + by + c dx x dy y dz z
dx x dy y Integrating we get 1 x 1 y + c 1 i.e., 1 y 1 x c 1 taking ast two members, we get dy y dz z 1 y 1 z + c i.e., 1 z 1 y c The compete soution is Φ 1 y 1 x, 1 z 1 y 6. Find the partia differentia equation of a panes having equa intercepts on the x and y axis. The equation of the pane is x a + y a + z b 1 (1) Partiay differentiating (1) w.r.t x and y we get 1 a + p b p b a () 1 a + q b q b a (3) From () and (3) we get p q
7. Form the partia differentia equation by eiminating the arbitrary function from Φ z xy, x z Given Φ z xy, x z 1 Let u z xy v x z Then the given equation is of the form Φ u, v () The eimination of Φ from the equation (), we get, u x u y v x v y i.e., zp y zq x z px z xq z i.e., zp y xq z px z zq x z i.e.,px q xy z zx 8. Find the singuar integra of the partia differentia equation z px + qy + p q The compete integra is z ax + by + a b (1) Now, z a z x + a x a b y b b y Substituting () in (1), we get z x + y + x 4 y 4 x 4 + y 4 i.e., y x 4z which is the singuar integra.
9. Find the compete soution of x p + y q z Given x p + y q z 1 Equation (1) can be written as xp + yq z Put og x X and og y Y xp P, were P z X yq Q, were Q z Y (3) P + Q z 4 This is of the form F z, P, Q Let z f X + ay be the soution of (4). Put u X + ay.then z f u P dz du, dz Q a du 5 Substituting (5) in (4), we get dz du 1 + a z dz du z 1 + a Separating the variabes we get dz z du 1 + a Integrating we get og z og z u 1 + a + b X + ay 1 + a + b Repacing X by og x and Y by og y, we get
Which gives the compete soution of (1). 1.Sove p + q m p + q m 1 og x + aog y og z + b 1 + a Let us assume that z ax + by + c be a soution of (1).... () Partiay differentiating () w.r.t x and y, we get Substituting (3) in (1) we get Hence z ax + by + c is the soution of (1). z z p a, q b (3) x y a + b m 4 11. Form a partia differentia equation by eiminating arbitrary constants a and b from z ax n + by n Given z ax n + by n 1 Substituting () and (3) in (1), we get p z x a. nxn 1 a p () nxn 1 q z y a. nyn 1 b q (3) nyn 1 z p nx n 1 xn + q yn nyn 1 z 1 n px + qy 1. Sove D 3 + 3D D + 3DD + D 3 z. A.E is m 3 + 3m + 3m + 1 m + 1 3
i.e., m 1, 1, 1. z f 1 y x + xf y x + x f 3 (y x) 13. Form a partia differentia equation by eiminating arbitrary constants a and b from z x + a y + b Given z x + a y + b 1 p z x x y + b q z y y x + a p x y + b () q y x + a (3) Substituting () and (3) in (1) we get z p q x y pq 4xyz 14. Sove D DD + D 1 z. The given equation can be written as D 1 D D + 1 z. We know that the compementary function corresponding to the factors Here α 1 1, α 1, m 1, m 1. D m 1 D α 1 D m D α z is z e α 1x f 1 y + m 1 x + e α x f (y + m x) z e x f 1 y + e x f (y + x) 15. Form the partia differentia equation by eiminating the arbitrary function from z f xy z z f xy z
p f xy z q f xy z zy xyp z (1) zx xyq z () From (1) we get f xy z pz zy xyp (3) Substituting (3) in (), we get q pz zx xyq zy xyp z 16. Form a partia differentia equation by eiminating arbitrary constants a and b from Given x a + y b + z 1 x a + y b + z 1 1 Partiay differentiating (1) w.r.t. x and y, we get i.e., x a zp () Substituting () and (3) in (1), we get x a + zp y b + zq y b zq (3) z p + z q + z 1 17. Find the compete integra of p + q pq Given p + q pq (1) p + q + 1 1 z Let us assume that z ax + by + c be a soution of (1).... () Partiay differentiating () w.r.t x and y, we get z z p a, q b (3) x y
Substituting (3) in (1) we get a + b ab 3 b a a 1 (4) Substituting (4) in () we get z ax + a a 1 y + c which is the compete integra. 18. Sove D 3 3DD + D 3 z. A.E. is m 3 3m + m 1, 1 and. The soution is z f 1 y + x + xf y + x + f 3 (y x). 19. Find the genera soution of 4 z 1 z + 9 z x x y y. A.E. is 4m 1m + 9 m 1 ± 144 144 8 m 3, 3 The genera soution is z f 1 y + 3 x + xf y + 3 x. Form the partia differentia equation by eiminating the arbitrary function from z f y x z f y x (1) Partiay differentiating (1) w.r.t x and y, we get
p f y x q f y x y x () 1 x (3) From (1) we get f y x px y (4) Substituting (4) in (3), we get q px y px + qy is the required partia differentia equation. 1. Form the partia differentia equation by eiminating the arbitrary function from Given z x + y + f xy 1 1 x z x + y + f xy. Partiay differentiating (1) w.r.t x and y, we get () 3 gives p 1 q 1 y x p 1 + f xy y p 1 f xy y q 1 + f xy x q 1 f xy x 3 px x qy y or px qy x y is the required partia differentia equation.. Form the partia differentia equation by eiminating the arbitrary function from Given z f xy 1 z f xy. Partiay differentiating (1) w.r.t x and y, we get p f xy y q f xy x 3 () 3 gives p q y x px qy or px qy is the required partia differentia equation.
3. Find the partia differentia equation of a panes through the origin. The genera equation of the pane is ax + by + cz + d (1) If (1) passes through the origin, then d. So (1) becomes ax + by + cz () Partiay differentiating () w.r.t x and y we get Substituting (3) and (4) in (1), we get a + cp a cp 3 b + cq b cq (4) cpx cqy + cz px + qy z is the required partia differentia equation. 4. Write down the compete soution of z px + qy + c 1 + p + q. The given equation is Cairaut s type. So repacing p by a and q by b, the compete soution is z ax + by + c 1 + a + b 5. Find the differentia equations of a spheres of radius c having their centers in the xoy pane. Let the center of the sphere be (a, b, ), a point in the xoy pane. c is the given radius. The equation of the sphere is x a + y b + z c (1) Partiay differentiating (1) w.r.t x and y we get Substituting () and (3) in (1), we get x a + zp x a zp () y b + zq y b zq (3)
z p + z q + z c z p + q + 1 c which is the required partia differentia equation. 6. Eiminate f from xyz f(x + y + z) Given xyz f x + y + z 1. Partiay differentiating (1) w.r.t x and y we get y z + xp f x + y + z 1 + p () x z + yq f x + y + z 1 + q (3) () 3 gives 1+p 1+q y z+xp x z+yq 1 + p x z + yq 1 + q y z + xp p xy xz + q yz xy xz yz px y z + qy z x z(x y) is the required partia differentia equation. 7. Mention three types of partia differentia equation. i) A soution of a partia differentia equation which contains the maximum possibe number of arbitrary constants is caed compete integra. ii) A soution obtained by giving particuar vaues to the arbitrary constants in a compete integra is caed a particuar integra. iii) A soution of a partia differentia equation which contains the maximum possibe number of arbitrary functions is caed genera integra. 8. Define singuar integra. Let F x, y, z, p, q 1 be a partia differentia equation and its compete integra be Φ x, y, z, a, b Differentiating () partiay w.r.t. a and b in turn, we get Φ a (3) Φ b (4) The eiminant of a and b from (), (3) and (4), if it exits, is caed singuar integra.
9. Sove p + q 1 Given p + q 1 1 Let us assume that z ax + by + c be a soution of (1).... () Partiay differentiating () w.r.t x and y, we get Substituting (3) in (1), we get z z p a, q b (3) x y a + b 1 b 1 a (4) Substituting (4) in (), we get z ax + 1 a y + c is the compete integra of (1). 3. Find the particuar integra of D 3 3D D 4DD + 1D 3 z sin(x + y) Particuar integra sin (x+y ) D 3 3D D 4DD +1D 3 (1) Repacing D by 1 andd by in(1) sin(x + y) D + 3D + 16D 48D sin(x + y) 15D 45D sin(x + y) 15 D 3D D + 3D sin(x + y) 15 D 9D cos x + y + 6 cos(x + y) 15 D 9D 7cos x + y 15 1 9( 4) cos x + y 75
Z Transform Part-A 1. Find Z n. Z n z Z n nz n n n 1 z 1 + z + 3 z + 4 z 3 + 1 z 1 1 z z 1 [The region of convergence is 1 z. Form the difference equation from y n a + b3 n n z n 1 + z z + 3 z 3 + 1 z < 1 or z > 1] y n a + b3 n, y n +1 a + b3 n +1`, y n + a + b3 n+ y n + 4y n +1 + 3y n 3. Find the vaue of Z f(n) when f n na n z z 1 Z na n Z(n) z z a 4. Find Z an n! z z 1 z z a in Z-transform. Since Z a n f(t) F z a z a z a 1 az z a. Since Z n z z 1 Z an n! Z 1 n! z z a Since Z a n f(t) F z a 5.Find Z e iat using Z-transform. e 1 z z z a a z e Z e iat Z e iat. 1 6. Express Z f n + 1 in terms of F z. Z f n + 1 zf z zf. Z 1 z ze iat Since Z e iat. f(t) F(ze iat ) z z 1 z ze iat zeiat ze iat 1 7. State and prove initia vaue theorem in Z-transform.
im If Z f n F z then z F z f im t f(t) We know that Z f n n f n z n f + f 1 z 1 + f z + F z f + f 1 z im im F z f Since z 8. Find the Z-transform of (n + 1)(n + ) + f z + 1 z z im z F z f im t f(t) Z (n + 1)(n + ) Z n + 3n + 9. Find the Z-transform of cos nx Z e inx ix n Z e Z n + 3Z n + Z z(z + 1) z 1 3 + 3 z z 1 + z z 1 z z e ix Since Z an z z a z z cos x + i sin x z z cos x + i sin x Z[cos nx + isin nx] Z[cos nx] + iz[sin nx] Equating rea and imaginary parts we get 1. Find the Z-transform of a n Z(cos nx) z z cos x i sin x z cos x + sin x z z cos x z zcos x + 1 isin x z zcos x + 1 z z cos x z zcos x + 1 Z a n a n z n n
1 + az 1 + a z + a 3 z 3 + 1 + az 1 + az 1 + az 1 3 + Since 1 + x + x + x 3 + (1 x) 1 1 z 1 az 1 z a 11. Let Z f n F z and k > then, Prove that Z f n k z k F z Z f n k f n k z n f n k z n n nk Since f 1, f, are not defined f z k + f 1 z k+1 + f z k+ + z k f + f 1 z 1 + f z + z k f n z n n z k F z 1. Let Z f n F z and k > then, Prove that Z f n + k z k F z f o f 1 z 1 f(k 1)z (k 1) Z f n + k Mutipy and divide by z k in RHS, we get n f n + k z n n f n + k z n z k z k z k f n + k z (n+k) n z k f k z k + f k + 1 z (k+1) + f k + z (k+) + 13. Find the Z-transform of 1 n z k f n z n n k 1 n f n k z n z k F z f o f 1 z 1 f(k 1)z (k 1)
14. Find the Z-transform of n dz n z dz Z 1 n n 1 n z n 1 z 1 + z z 3 + 1 1 1 + z z z + 1 Z n Z n. n z d dz Since Z nf t z df(z) dz z z 1 z z 1 z 1 z z 1 4 z 15. State convoution theorem for Z-transform z 1 z z 1 3 z z + 1 z 1 3 If f n and g(n) are two casua sequences, 16. If F z z im z e T, find We know that Z f(n) g(n) Z f(n). Z g(n) F z. G(z) t f(t) im f t im t im z 1 z 1 (z 1)F z z 1 17. Find the Z-transform of nf t if Z f t F(z) Differentiate both sides w.r.t. z 1. F z Z f t df(z) dz z z e T n f nt z n nf nt z n 1 n
z df(z) dz nf nt z n n Z nf t z df(z) dz Z nf t 18. If Z f t F(z) then prove that Z e at f t F(ze at ) Z e at f t 19. Find the Z-transform of e at cos bt n e ant f nt z n at n f nt ze n F ze at Since F z f nt z n Z e at cos bt Z[cos bt] z ze at Since Z e at f t F(ze at ) n z z cos bt z zcos bt + 1 z ze at. If F z 1z z 1 (z ) find f. f ze at ze at cos bt z e at ze at cos bt + 1 im F z im z z im 1z z z 1 1 z z 1z z 1 z 1. Find Z-transform of 1, n 1. n im z 1 1 z 1 z Z 1 n 1 n z n n1
1 z + 1 z + 1 3z 3 + og 1 1 z if 1 z < 1 og z z 1 if z > 1. Find Z-transform of 1 n! Z 1 n! n 1 n! z n n 1 n! 1 n z 1 + z 1 1! + z 1 + z 1 3 z 1 + e! 3! Z 1 n! e 1 z 3. Find the Z-transform of ab n Z ab n a n n b z a 1 + b z + b z ab n z n n + a 1 b 1 b if z < 1 z az if z > b z b 4. Find the inverse Z-transform by using convoution theorem z z a Z 1 z z a Z 1 z z a. z z a
5. Define Z-transform Z 1 z z a Z 1 n a n k. a k k z z a an a n n k n + 1 a n u(n) The two sided Z-transform F(z) for a sequence f(x) is defined as a n F z f x z n n The one sided Z-transform F(z) for a sequence f(x) is defined as 6. Find the vaue of Z a n f x 7. Find the vaue of Z 1. F z a F z Z a n f x n n f x z a n f x z n a n f x z n n Since F z Z 1 n 1. z n n 1 + 1 z + 1 z + 1 z 3 + 1 1 z 8. Find the Z-transform of δ(n k) 1 z z 1 if if z > 1. 1 z < 1 f x z n
Z δ(n k) 9. Find the Z-transform of u(n 1) n δ(n k)z n 1, if k is positive integer zk Z u(n 1) 1 z 1 1 z 3. Find the Z-transform of 3 n δ(n 1) n1 1. z n 1 z + 1 z + 1 z 3 + 1 1 z 1 if if z > 1 1 z < 1 Z 3 n δ(n 1) Z δ(n k) z z 3 1 3 z z z z 3
Appications of partia differentia equations Part-A 1. Cassify the partia differentia equation 3u xx + 4u xy + 3u y u x. Given 3u xx + 4u xy + 3u y u x. B 4AC 16 >, Hyperboic A 3, B 4, C. The ends A and B of a rod of ength 1 cm ong have their temperature kept at o C and 7 o C. Find the steady state temperature distribution on the rod. When the steady state condition exists the heat fow equation is i.e., u(x) c 1 x + c... (1) u x The boundary conditions are (a) u(), (b) u(1) 7 Appying (a) in (1), we get u c Substituting c in (1), we get u x c 1 x + () Appying (b) in (), we get u 1 c 1 1 + 7 c 1 5 Substituting c 1 5 in (), we get u x 5x + 3.Write the one dimensiona wave equation with initia and boundary conditions in which the initia position of the string is f(x) and the initia veocity imparted at each point x is g(x). the one dimensiona wave equation is The boundary conditions are y t y α x i y, t ii y, t, iii y x, f x iv y (x, ) t g(x)
4.In steady state conditions derive the soution of one dimensiona heat fow equation. when steady state conditions exist the heat fow equation is independent of time t. The heat fow equation becomes u t u x The soution of heat fow equation is u x c 1 x + c 5. What is the basic difference between the soutions of one dimensiona wave equation and one dimensiona heat equation. Soution of the one dimensiona wave equation is of periodic in nature. But soution of one dimensiona heat equation is not of periodic in nature. 6. What are the possibe soutions of one dimensiona wave equation? y x, t c 1 e px + c e px c 3 e pat + c 4 e pat (1) y x, t c 5 cos px + c 6 sin px c 7 cos pat + c 8 sin pat () y x, t c 9 x + c 1 c 11 t + c 1 (3) 7. Write down possibe soutions of the Lapace equation. u x, y c 1 e px + c e px c 3 cos py + c 4 sin py (1) u x, y c 5 cos px + c 6 sin px c 7 e py + c 8 e py () 8. State Fourier aw of conduction. Fourier aw of heat conduction: u x, y c 9 x + c 1 c 11 y + c 1 (3) The rate at which heat fows across an area A at a distance x from one end of a bar is given by
Q KA u x x where K is the therma conductivity and u x x means the temperature gradient at x. 9. In the wave equation y t α y x what does α stands for? α T Tension mass 1. State any two aws which are assumed to derive one dimensiona heat equation. (i) The sides of the bar are insuated so that the oss or gain of heat from the sides by conduction or radiation is negigibe. (ii) The same amount of heat is appied at a points of the face. 11. Cassify the partia differentia equation u xx + xu xy. Here A 1, B x, C B 4AC x i Paraboic if x. ii Hyperboic if x > and x <. 1. A rod 3 cm ong has its ends A and B kept at o C and 8 o C respectivey unti steady state conditions prevai. Find the steady state temperature in the rod. Let 3 cm When the steady state condition prevai the heat fow equation is i.e., u x ax + b (1) u x When the steady state condition exists the boundary conditions are Appying () in (1) we get u(), u() 8... ()
u() b... (3) and u() a + 8 Substituting (3) and (4) in (1) we get a 6 13. Cassify the partia differentia equation a) A y, B xy, C x Paraboic b) A y, B, C 1 Eiptic. 6 3 (4) u x x + a y u xx xyu xy + x u yy + u x 3u b y u xx + u yy + u x + u y + 7 B 4AC 4x y 4x y B 4AC 4y < 14. A rod 6 cm ong has its ends A and B kept at 3 o C and 4 o C respectivey unti steady state conditions prevai. Find the steady state temperature in the rod. Let 6 cm When the steady state condition prevai the heat fow equation is i.e., u x ax + b (1) u x When the steady state condition exists the boundary conditions are Appying () in (1) we get u() 3, u() 4... () u() b 3... (3)
and u() a + 3 4 Substituting (3) and (4) in (1) we get a 1 1 6 1 6 (4) 15. Cassify the partia differentia equation u x 1 x + 6 a) 4 u x + 4 u x y + u u 6 y x u 8 16u y b) u x + u u y x + u y a) A 4, B 4, C 1 B 4AC 16 16 Paraboic. b) A 1, B, C 1 B 4AC 4 < Eiptic. 16. Cassify the partia differentia equation a) u x u y b) u x y u x u y + xy a) Here A 1, B, C 1 B 4AC 4 > Hyperboic. b) Here A, B 1, C B 4AC 1 > Hyperboic.
17. State one dimensiona heat equation with the initia and boundary conditions. The one dimensiona heat equation is Initia condition means condition at t α u x u t Boundary condition means condition at x and x 18. Write the boundary conditions and initia conditions for soving the vibration of string equation, if the string is subjected to initia dispacement f(x) and initia veocity g(x). Initia dispacement f(x) i y wen x ii y wen x iii y wen t t iv y f(x) wen t Initia veocity g(x) i y wen x ii y wen x iii y g(x) wen t t iv y wen t 19. Write down the differentia equation for two dimensiona heat fow equation for the unsteady state. The two dimensiona heat fow equation for the unsteady state is given by α u x + u y u t. What is the steady state heat fow equation in two dimensions in Cartesian form? The two dimensiona heat fow equation for the steady state is given by
u x + u. Lapace equation y 1. Write down the poar form of two dimensiona heat fow equation in steady state. r u u + r r r + u. Lapace equation θ. Write a the soutions of Lapace equation in poar coordinates. u r, θ c 1 r p + c r p c 3 cos pθ + c 4 sin pθ (1) u r, θ c 5 cos pog r + c 6 pog r c 7 e pθ + c 8 e pθ () u r, θ c 9 og r + c 1 c 11 θ + c 1 (3) 3. Expain the initia and boundary vaue probems. The vaues of a required soution, on the boundary of some domain wi be given. These are caed boundary conditions. In other cases, when time t is one of the variabes, the vaues of the soution at t may be presented. These are caed initia conditions. The partia differentia equation together with these conditions constitutes a boundary vaue probem or an initia vaue probem, according to the nature of the condition. 4. What is meant by steady state condition in heat fow? Steady state condition in heat fow means that the temperature at any point in the body does not vary with time. i.e., it is independent of time t. 5. Distinguish between steady and unsteady states in heat conduction probems. In unsteady state, the temperature at any point of the body depends on the position of the point and aso the time t. In steady state, the temperature at any point depends ony on the position of the point and is independent of the time t. 6. Sove using separation of variabes method yu x + xu y. Given yu x + xu y (1)
Let u X x. Y y be the soution of (1). From () we get u x X Y 3 u y XY 4 Substituting (3) and (4) in (1), we get y. X Y + x. XY y. X Y x. XY X xx Y yy k X kxx or Y kyy dx dx kxx or dy dy kyy dx X kx dx or dy Y ky dy og X k x + k 1 or og Y k y + k Substituting (5) in () we get X c 1 e kx or Y c e ky 5 u x, y c 1 c e k(x y ) 7. By the method of separation of variabes sove x q + y 3 p Given x z z + y3 y x 1 Let z X x Y y () be the soution of (1). Then z x X Y 3 z y XY 4 Substituting (3) and (4) in (1), we get
y 3. X Y + x. XY y 3. X Y x. XY X x X Y y 3 Y k X kx X or Y ky 3 Y dx dx kx X or dy dy kyy3 dx X kx dx or dy Y ky3 dy og X k x3 3 + k 1 or og Y k y4 4 + k Substituting (5) in () we get X c 1 e kx 3 3 or Y c e ky 4 4 5 u x, y c 1 c e kx 3 3 e ky 4 4 8. State the assumptions made in the derivation of one dimensiona wave equation. (i) The mass of the string per unit ength is constant (ii) The string is perfecty eastic and does not offer any resistance to bending (iii) The tension caused by stretching the string before fixing it at the end points is so arge that the action of the gravitationa force on the string can be negected. (iv) The string performs a sma transverse motion in a vertica pane, that is every partice of the string moves stricty verticay so that the defection and the sope at every point of the string remain sma in absoute vaue. 9. Write down the boundary condition for the foowing boundary vaue probem If a string of ength initiay at rest in its equiibrium position and each of its point is given the veocity y(x, t)? y t v 3 x sin t The boundary conditions are < x <, determine the dispacement function i) y, t, t > ii) y, t, t >
iii) y x,, < x < iv) y(x, ) t v sin 3 x, < x < 3. If the ends of a string of ength are fixed and the midpoint of the string is drawn aside through a height h and the string is reeased from rest, write the initia conditions. Soutions: i y, t ii y, t iv y x, iii y(x, ) t x, < x < x, < x <