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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
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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) = An−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]
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(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
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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
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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
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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
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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
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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
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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! (xa) +
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
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ElectroMagnetic 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
B2 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
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1
1
→ For monopole E ∝ 2 ; Dipole E ∝ 3 .
•
•
•
• r
1 ElectroMagnetic 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 dielectricdielectric 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
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Site: Google+: FB: Institute of Engineering Studies (IES,Bangalore) ElectroMagnetic 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
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•
• δave (z) = 1 E20 −2αz
e
2 η ElectroMagnetic 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 MediumI Dielectric , Medium2 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
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ElectroMagnetic 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 .
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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 Xdirection
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 − �
• ElectroMagnetic Theory ...
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 Winter '17