EEE-FORMULA-SHEET (1).pdf - Institute of Engineering Studies(IES,Bangalore MatheMatics Mathematics Formula Sheet Matrix:\u2022 If |A| = 0 \u2192 Singular

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Unformatted text preview: Institute of Engineering Studies (IES,Bangalore) MatheMatics Mathematics Formula Sheet Matrix :• If |A| = 0 → Singular matrix ; |A| ≠ 0 Non singular matrix • Scalar Matrix is a Diagonal matrix with all diagonal elements are equal • Unitary Matrix is a scalar matrix with Diagonal element as ‘1’ (AQ = (A∗ )T = A−1 ) • If the product of 2 matrices are zero matrix then at least one of the matrix has det zero • Orthogonal Matrix if AAT = AT .A = I ⇒ AT = A−1 • A = AT → Symmetric A = - AT → Skew symmetric Properties :- (if A & B are symmetrical ) • A + B symmetric • KA is symmetric • AB + BA symmetric • AB is symmetric iff AB = BA • For any ‘A’ → A + AT symmetric ; A - AT skew symmetric. • Diagonal elements of skew symmetric matrix are zero • If A skew symmetric A2n → symmetric matrix ; A2n−1 → skew symmetric • If ‘A’ is null matrix then Rank of A = 0. Consistency of Equations :• r(A, B) ≠ r(A) is consistent • r(A, B) = r(A) consistent & if r(A) = no. of unknowns then unique solution r(A) < no. of unknowns then ∞ solutions . Hermition , Skew Hermition , Unitary & Orthogonal Matrices :• • • • • • AT = A∗ → then Hermition AT = −A∗ → then Hermition Diagonal elements of Skew Hermition Matrix must be purely imaginary or zero Diagonal elements of Hermition matrix always real . A real Hermition matrix is a symmetric matrix. |KA| = K n |A| Eigen Values & Vectors :• Char. Equation |A – λI| = 0. Roots of characteristic equation are called eigen values . Each eigen value corresponds to non zero solution X such that (A – λI)X = 0 . X is called Eigen vector . • Sum of Eigen values is sum of Diagonal elements (trace) • Product of Eigen values equal to Determinent of Matrix . • Eigen values of AT & A are same |A| • λ is Eigen value of A then 1/ λ → A−1 & is Eigen value of adj A. • λ1 , λ2 …… λn are Eigen values of A then λ Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 1 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet KA → K λ1 , K λ2 ……..K λn m Am → λ1m , λm 2 ………….. λn . • • • • • A + KI → λ1 + k , λ2 + k , …….. λn + k (A − KI)2 → (λ1 − k)2 , ……… (λn − k)2 Eigen values of orthogonal matrix have absolute value of ‘1’ . Eigen values of symmetric matrix also purely real . Eigen values of skew symmetric matrix are purely imaginary or zero . λ1 , λ2 , …… λn distinct eigen values of A then corresponding eigen vectors X1 , X 2, .. … X n for linearly independent set . 2 adj (adj A) = |A|n−2 ; | adj (adj A) | = |A|(n−1) Complex Algebra :• Cauchy Rieman equations = ; =− 1 = r 1 =− r � Neccessary & Sufficient Conditions for f(z) to be analytic • • • ∫c f(z)/(Z − a)n+1 dz = f(z) = f(z0 ) + f ′ (z0 ) (z−z0 ) 1! 2πi n! [ f n (a) ] if f(z) is analytic in region ‘C’ & Z =a is single point + f ′′ (z0) (z−z0 )2 2! (z−z )n + …… + f n (z0 ) n!0 + ………. Taylor Series ⇓ fn (z0 ) n if z0 = 0 then it is called Mclauren Series f(z) = ∑∞ 0 a n (z − z0 ) ; when a n = n ! If f(z) analytic in closed curve ‘C’ except @ finite no. of poles then ∫c f(z)dz = 2πi (sum of Residues @ singular points within ‘C’ ) Res f(a) = lim( − () → = Φ(a) / φ′ (a) Calculus :- 1 = lim (−1)! → −1 −1 ((Z − a)n f(z) ) Rolle’s theorem :If f(x) is (a) Continuous in [a, b] Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 2 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) (b) Differentiable in (a, b) Mathematics Formula Sheet (c) f(a) = f(b) then there exists at least one value C ϵ (a, b) such that f ′ (c) = 0 . Langrange’s Mean Value Theorem :If f(x) is continuous in [a, b] and differentiable in (a, b) then there exists atleast one value ‘C’ in (a, b) f(b)−f(a) such that f ′ (c) = b−a Cauchy’s Mean value theorem :- If f(x) & g(x) are two function such that (a) f(x) & g(x) continuous in [a, b] (b) f(x) & g(x) differentiable in (a, b) (c) g ′ (x) ≠ 0 ∀ x in (a, b) Then there exist atleast one value C in (a, b) such that f ′ (c) / g ′ (c) = f(b)−f(a) g(b)−g(a) • b Properties of Definite integrals :- • • c a a ∫−a f(x). dx = 2 ∫0 f(x)dx = 0 • • • • b a < c < b ∫a f(x). dx = ∫a f(x). dx + ∫c f(x). dx a a ∫0 f(x)dx = ∫0 f(a − x)dx a a ∫0 f(x). dx = 2 ∫0 f(x)dx = 0 na a ∫0 f(x). dx = n ∫0 f(x)dx b f(x) is even f(x) is odd if f(x) = f(2a- x) if f(x) = - f(2a – x) if f(x) = f(x + a) b ∫a f(x). dx = ∫a f(a + b − x). dx Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 3 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) a a a • ∫0 x f(x). dx = ∫0 f(x). dx if f(a - x) = f(x) • π/2 ∫0 2 π/2 sinn x = ∫0 = • π/2 ∫0 (n−1)(n−3)(n−5)………2 n (n−2)(n−4)……….3 cosn x = sinm x . cosn x . dx = (n−1)(n−3)……1 n (n−2)(n−4)……….2 Mathematics Formula Sheet if ‘n’ odd 2 . � � if ‘n’ even {(m−1)(m−3)….(m−5)……(2 or 1)} {(n−1)(n−3)…….(2 or 1)}.K (m+n) (m+n−2)(m+n−4)………2 or 1 Where K = π / 2 when both m & n are even otherwise k = 1 Maxima & Minima :- A function f(x) has maximum @ x = a if f ′ (a) = 0 and f ′′ (a) < 0 A function f(x) has minimum @ x = a if f ′ (a) = 0 and f ′′ (a) > 0 Constrained Maximum or Minimum :- To find maximum or minimum of u = f(x, y, z) where x, y, z are connected by Φ (x, y, z) = 0 Working Rule :- (i) Write F(x, y, z) = f(x, y, z) + λ ϕ(x, y, z) (ii) Obtain Fx = 0, Fy = 0 , Fz = 0 (ii) Solve above equations along with ϕ = 0 to get stationary point . Laplace Transform :• • • • L � ()� = s n f(s) - s n−1 f(0) - s n−2 f ′ (0) …… f n−1 (0) L { t n f(t) } = (−1)n f(t) t t ∞ ⇔ ∫s f(s) ds dn dsn f(s) ∫0 f(u) du ⇔ f(s) / s . Inverse Transforms :- Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 4 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) s 1 • t sin at 2 2 2 = (s +a ) 2a s2 • (s2 +a2 )2 1 (s2 +a2 )2 • s s 2 − a2 • a s 2 − a2 • Mathematics Formula Sheet = = 1 2a [ sin at + at cos at] 1 2a3 [ sin at - at cos at] = Cos hat = Sin hat Laplace Transform of periodic function : L { f(t) } = T ∫0 e−st f(t)dt Numerical Methods :- 1−e−sT Bisection Method :(1) Take two values of x1 & x2 such that f(x1 ) is +ve & f(x2 ) is –ve then x3 = +ve then root lies between x3 & x2 otherwise it lies between x1 & x3 . x1 +x2 2 find f(x3 ) if f(x3 ) Regular falsi method :- x −x 0 Same as bisection except x2 = x0 - f(x 1)−f(x 1 Newton Raphson Method :xn+1 = xn – Pi cards Method :x Taylor Series method :dy dx = f(x, y) Euler’s method :y1 = y0 + h f(x0 , y0 ) (1) h 2 f(x0 ) f(xn ) f′ (xn ) yn+1 = y0 + ∫x f(x, yn ) 0 0) ← dy dx = f(x, y) y = y0 + (x- x0 ) (y ′ )0 + ← dy dx y1 = y0 + [f(x0 , y0 ) + f(x0 + h, y1 ) (x− x0 )2 2! (y)′′ 0 + …………. (x− x0 )n n! (y)n0 = f(x, y Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 5 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) h (2) (1) y1 = y0 + [f(x0 , y0 ) + f(x0+h, y1 ) ] Mathematics Formula Sheet 2 : : Calculate till two consecutive value of ‘y’ agree y2 = y1 + h f(x0 + h, y1 ) (1) h 2 y2 = y0 + [f(x0 + h, y1 ) + f(x0 + 2h, y2 ) ……………… Runge’s Method :k1 = h f(x0 , y0 ) h 2 k 2 = h f( x0 + , y0 + k1 2 ) k ′ = h f(x0 +h , y0 + k1 ) 1 6 finally compute K = (K1 + 4K 2 + K 3 ) k 3 = h ( f (x0 +h , y0 + k ′ )) Runge Kutta Method :k1 = h f(x0 , y0 ) h 2 k 2 = h f( x0 + , y0 + h 2 k 3 = h f(x0 + , y0 + k1 2 k2 2 ) ) 1 6 finally compute K = (K1 + 2K 2 + 2K 3 + K 4 ) k 3 = h f (x0 +h , y0 + k 3 ) ∴ approximation vale y1 = y0 + K . Trapezoidal Rule :x +nh ∫x 0 0 f(x). dx = h 2 [ ( y0 + yn ) + 2 (y1 + y2 + ……. yn−1 )] f(x) takes values y0 , y1 ….. @ x0 , x1 , x2 …….. Simpson’s one third rule :- Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 6 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) x +nh f(x). dx ∫x 0 0 h 3 = Mathematics Formula Sheet [ ( y0 + yn ) + 4 (y1 + y3 + ……. yn−1 ) + 2 (y2 + y4 + ⋯ … . + yn−2 )] Simpson three eighth rule :x0 +nh ∫x 0 3h 8 f(x). dx = [ ( y0 + yn ) + 3 (y1 + y2 + y4 + y5 + ……. yn−1 )+ 2 (y3 + y6 + ⋯ … . + yn−3 ) ] Differential Equations :Variable & Seperable :- f(y) dy = ϕ(x) dx General form is Sol: ∫ f(y) dy = ∫ ϕ(x) dx + C . Homo generous equations :dy dx General form = f(x,y) ϕ(x,y) dy dx ⇒ Sol : Put y = Vx =V+x Reducible to Homogeneous :dy dx General form (i) a a′ ≠ b b′ Sol : Put ⇒ (ii) dy dx a a′ = = = Sol : Let x=X+h a a′ = dy dx b b′ = = & solve y=Y+k Choose h, k such that dy dx becomes homogenous then solve by Y = VX 1 m ax+by+c m(ax+by)+c Put ax + by = t ⇒ dv dx ax+by+c a′ x+b′ y+c′ ax+by+(ah+bk+c) a′ x+b′ y+(a′ h+b′ k+c′ ) b b′ f(x, y) & ϕ(x, y) Homogenous of same degree dy dx =� − �/b Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 7 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) Then by variable & seperable solve the equation . Mathematics Formula Sheet Libnetz Linear equation :General form dy dx +py = Q where P & Q are functions of “x” I.F = e∫ p.dx Sol : y(I.F) = ∫ Q. (I. F) dx + C . Exact Differential Equations :- M → f (x, y) General form M dx + N dy = 0 If ∂M ∂y = ∂N ∂x N → f(x, y) then Sol : ∫ M. dx + ∫(terms of N containing x ) dy = C ( y constant ) Rules for finding Particular Integral :1 f(D) eax = = x eax 1 f′ (a) = x2 1 f(b2 ) 1 f(D) 1 f(a) eax if f (a) = 0 1 f′′ (a) sin (ax + b) = eax 1 f(−a2 ) = x x m = [f(D)]y x m 1 f(D) eax f(x) = eax Vector Calculus :- sin (ax + b) 1 f′ (−a2 ) = x2 if f ′ (a) = 0 sin (ax + b) 1 f′′ (−a2 ) 1 f(D+a) sin (ax + b) f(- a2 ) ≠ 0 f(- a2 ) = 0 Same applicable for cos (ax + b) f(x) Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 8 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) Green’s Theorem :∂Ψ Mathematics Formula Sheet ∂ϕ ∫C (ϕ dx + φ dy) = ∫ ∫ � ∂x − ∂y � dx dy This theorem converts a line integral around a closed curve into Double integral which is special case of Stokes theorem . Series expansion :Taylor Series :f(x) = f(a) + f(x) = f(0) + f′ (a) 1! f′ (0) 1! (x-a) + x + (1 + x)n = 1+ nx + x2 f′′ (0) 2! n(n−1) 2 ex = 1 + x + 2 ! + …….. Sin x = x - Cos x = 1 - x3 3! x2 2! + x5 5! f′′ (a) 2! (x − a)2 + …………+ x 2 + …………+ fn (0) n! x 2 + …… | nx| < 1 fn (a) n! (x − a)n x n + ……. (mc lower series ) - …….. x4 + 4 ! - …….. Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 9 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet Electromagnetic Fields Vector Calculus:→ A. (B × C) = C. (A × B) = B. (C × A) → A×(B×C) = B(A.C) – C(A.B) → Bac – Cab rule (A.B) → Scalar component of A along B is AB = A Cos θAB = A . aB = � B = A Cos θAB . aB = → Vector component of A along B is A |B| (A.B) B |B|2 Laplacian of scalars :• ∮ A. ds = ∫(∇.) → Divergence theorem • L∮ A.dI = ∫(∇× ) → Stokes theorem • ∇2 A = ∇ (∇ .A) - ∇ × ∇ × A • ∇ .A = 0 → solenoidal / Divergence loss ; ∇ .A > 0 → source ; ∇ .A < 0 ⇒ sink • ∇ × A = 0 → irrotational / conservative/potential. • ∇2 A = 0 → Harmonic . Electrostatics :• Force on charge ‘Q’ located @ r F = • • • • • Q 4πε0 ′ E @ point ‘r’ due to charge located @ E due to ∞ line charge @ distance ‘ ρ ‘ E = ρ Qk (r−rk ) Q Q � 12 ; F12 = 4πε1 R23 . R |r−rk |3 0 1 (r−rk ) ∑N � = Q 4πε0 K=1 |r−rk 3 k ρL . aρ (depends on distance) 2πε ρ ∑N k=1 0 E due to surface charge ρs is E = 2εs an . an → unit normal to surface (independent of distance) 0 For parallel plate capacitor @ point ‘P’ b/w 2 plates of 2 opposite charges is ρ ρ E = s an - � s � (− ) 2ε0 2ε0 ‘E’ due to volume charge E = → Electric flux density D = ε0 E Flux Ψ = s ∫ D .ds Q 4πε0 R2 ar . D → independent of medium Gauss Law :→ Total flux coming out of any closed surface is equal to total charge enclosed by surface . Ψ = Q enclosed ⇒ ∫ D . ds = Q enclosed = ∫ ρv . dv ρv = ∇. D → Electric potential VAB = B • VAB = - ∫A Q 4πε0 r2 w Q B = - ∫A E. dI (independent of path) ar . dr ar = VB - VA (for point charge ) Potential @ any point (distance = r), where Q is located same where , whose position is vector @ r ′ Q V= |r−r′ | → V(r) = 4πε0 Q 4πε0 r + C . [ if ‘C’ taken as ref potential ] → ∇ × E = 0, E = - ∇V Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 10 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) 1 1 → For monopole E ∝ 2 ; Dipole E ∝ 3 . • • • • r 1 Electro-Magnetic Theory Formula Sheet r 1 V∝ ; V ∝ 2 r r Electric lines of force/ flux /direction of E always normal to equipotential lines . 1 1 1 Energy Density WE = ∑N Q V = ∫ D. E dv = ∫ ε0 E 2 dv 2 k=1 k k 2 2 ∂ρv ∂t Continuity Equation ∇.J = ρv = ρ v 0 e −t/Tr . where Tr = Relaxation / regeneration time = ε/σ (less for good conductor ) Boundary Conditions :Et1 = Et2 • Tangential component of ‘E’ are continuous across dielectric-dielectric Boundary . • Tangential Components of ‘D’ are dis continues across Boundary . D • Et1 = Et2 ; D1t = ε1 / ε2 . • • • 2t Normal components are of ‘D’ are continues , where as ‘E’ are dis continues. ε tan θ ε ε D1n - D2n = ρs ; E1n = 2 E2n ; tan θ1 = ε1 = εr1 H1t = H2t B12 = B1n = B2n µ1 µ2 H1n = ε1 2 B2 t µ2 µ1 2 r2 H2n Maxwell’s Equations :→ faraday law Vemf = ∮ E. dI = - d dt ∫ B. ds ∂B → Transformer emf = ∮ E. dI = - ∫ ds ⇒ ∇ × E = ∂t s ∂B ∂t → Motional emf = ∇ × Em = ∇ × (μ × B). →∇×H=J+ ∂D ∂t Electromagnetic wave propagation :• ∇ × H = J + ̇ D = εE ̇ ∇ × E = - B = μH J = σE ∇. D = ρv ∇.B = 0 • Ey Hz =- Ez Hy = �µ/ε ; E.H = 0 For loss less medium ∇2 E - ρ2 E = 0 µϵ σ 2 α = ω � ��1 + � � − 1� 2 ωϵ ∇2 E = με̈ ∇2 H = με̈ E ⊥ H in UPW ρ = �jωµ(σ + jωϵ) = α + jβ. Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 11 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet σ 2 µϵ β = ω � ��1 + � � + 1� 2 ωϵ • E(z, t) = E0 e−αz cos(ωt – βz) ; H0 = E0 / η . • |η| = • • • • • • • • • jωµ η= � σ+jωϵ �µ/ε �1+� |η | < θη tan 2θη = σ/ωε. 1/4 σ 2 � � ωϵ | Np | = 20 log10 = 8.686 dB η= α + jβ α → attenuation constant → Neper /m . For loss less medium σ = 0; α = 0. β → phase shift/length ; μ = ω / β ; λ = 2π/β . Js =� � = σ / ωϵ = tan θ → loss tanjent θ = 2θη Jd ωϵE If tan θ is very small (σ < < ωϵ) → good (lossless) dielectric If tan θ is very large (σ >> ωϵ) → good conductor Complex permittivity ϵC = ϵ �1 − � = ε′ - j ε′′ . Tan θ = ε′′ ε′ = σ ωϵ . Plane wave in loss less dielectric :- ( σ ≈ 0) 1 • α = 0 ; β = ω√µϵ ; ω = ; λ = 2π/β ; η = �µr /εr ∠0. • √µϵ E & H are in phase in lossless dielectric Free space :- (σ = 0, μ = µ0 , ε = ε0 ) • α = 0 , β = ω �µ0 ε0 ; u = 1/ �µ0 ε0 , λ = 2π/β ; η = �µ0 /ε0 < 0 = 120π ∠0 Here also E & H in phase . Good Conductor :σ > > ωϵ σ/ωϵ → ∞ ⇒ σ = ∞ • • • • • • ε = ε0 ; μ = µ0 µr α = β = �πfµσ ; u = �2ω/µσ ; λ = 2π / β ; η = � Skin depth δ = 1/α 1 1+j η = √2 ejπ/4 = σδ σδ Skin resistance R s = Rs .l w l R dc = . σs R ac = 1 σδ =� Wµ σ ∠450 πfµ σ Poynting Vector :∂ 1 • ∫ (E × H) ds = - ∫ [ εE 2 + μH 2 ] dv – ∫ σ E 2 dv dt 2 S v Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 12 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) • • δave (z) = 1 E20 −2αz e 2 |η| Electro-Magnetic Theory Formula Sheet cos θη az Total time avge power crossing given area Pavge = ∫ Pave (s) ds S Direction of propagation :- ( ) ak × aE = aH aE × aH = ak → Both E & H are normal to direction of propagation → Means they form EM wave that has no E or H component along direction of propagation . Reflection of plane wave :(a) Normal incidence E η −η Reflection coefficient Γ = Er0 = η2 + η1 E i0 Txn coefficient Τ = Et0 = η i0 2η2 2 + η1 2 1 Medium-I Dielectric , Medium-2 Conductor : > :Γ>0 , there is a standing wave in medium & Txed wave in medium ‘2’. Max values of | E1 | occurs −nλ Zmax = - nπ/β1 = 2 1 ; n = 0, 1, 2…. Zmin = −(2n+1)π 2β1 = −(2n+1)λ1 4 < :- E1max occurs @ β1 Zmax = −(2n+1)π 2 β1 Zmin = nπ ⇒ Zmin = H1 min occurs when there is |t1 |max |E | |H | 1+|Γ| s−1 S = |E1 |max = |H1 |max = ;|Γ|= 1 min 1−|Γ| 1 min −nπ β1 = ⇒ Zmax = −nλ1 2 −(2n+1)π 2β1 = −(2n+1)λ1 4 s+1 Since |Γ| < 1 ⇒ 1 ≤ δ ≤ ∞ Transmission Lines :• Supports only TEM mode • LC = με ; G/C = σ /ε . • • • • d2 Vs dz2 - r 2 Vs = 0 ; d2 Is dz2 - r 2 Is = 0 Γ = �(R + jωL)(G + jωC) = α + jβ V(z, t) = V0+ e−αz cos (ωt- βz) + V0− eαz cos (ωt + βz) V− Z0 = − I−0 = 0 R+jωL γ = γ G+jωC =� Lossless Line : (R = 0 =G; σ = 0) R+jωL G+jωC Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 13 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet → γ = α + jβ = jω√LC ; α = 0, β = w √LC ; λ = 1/ f √LC , u = 1/ √LC Z0 = �L/C Distortion less :(R/L = G/C) G R R G → α = √RG ; β = ωL� = ωC� = ω √LC R G L C → Z0 = � = � ; λ = 1/f √LC ; u = 1 √LC i/p impedance : + Z tan hl Zin = Z0 � + Z 0 tan hl� for lossless line • • • • • • Zin = 0 L + Z0 tan βl Z0 � + Z tan βl � 0 L • γ = jβ ⇒ tan hjβl = j tan βl Z −Z VSWR = ΓL = ZL+Z0 L 0 CSWR = - ΓL Transmission coefficient S = 1 + Γ V I 1+| ΓL | Z SWR = Vmax = Imax = = ZL | min |Zin |max = |Zin |min = Vmax Imin Vmin Imax min 1−|ΓL = SZ0 = Z0 /S Shorted line :- ΓL = -1 , S = ∞ • = Vp ; uz0 = 1/C , u /z0 = 1/L ΓL = -1 , S = ∞ 0 = Z0 ZL (ZL > Z0 ) (ZL < Z0 ) Zin = Zsc = jZ0 tan βl Zin = Zsc = j Z0 tan βl. Zin may be inductive or capacitive based on length ‘0’ If l < λ / 4 → inductive (Zin +ve) λ < l < λ/2 → capacitive (Zin -ve) 4 Open circuited line :Zin = Zoc = -jZ0 cot βl Γl = 1 s = ∞ Zsc Zoc = Z02 λ 4 l < λ / 4 capacitive < l < λ/2 inductive Matched line : (ZL = Z0 ) Zin = Z0 Γ = 0 ; s =1 No reflection . Total wave Txed . So, max power transfer possible . Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/Practice Branches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email: [email protected] Pg.No. 14 of 143 Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) Behaviour of Transmission Line for Different lengths : =∞ � =0 l = λ /4 → l = λ /2 : Zin = Z0 ⇒ → impedance inverter @ l = λ /4 =0 � =∞ impedance reflector @ l = λ /2 Wave Guides :TM modes : (Hz = 0) Ez = E0 sin � �x sin � � y e−nz mπ 2 � a ∴ γ = �� h2 = k 2x + k 2y nπ 2 b + � � − ω2 µε m→ no. of half cycle variation in X-direction n→ no. of half cycle variation in Y- direction . 2 2 1 ��mπ� + �nπ� a b √µϵ nπ 2 � � → Evanscent mode b nπ 2 Cut off frequency ωC = • • k2 < mπ 2 � � a mπ 2 k2 > � a + • • • • fc = u′p 2 where k = ω √µϵ γ = 0; α = 0 = β ; γ=α; β=0 � + � � → Propegation mode γ = jβ α = 0 b mπ 2 � a β = �k 2 − � • Electro-Magnetic Theory ...
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