S. R. Zinka [email protected] May … · Helmholtz Wave EquationTE, TM, and TEM Modes – RectRectangular WaveguideTE, TM, and TEM Modes – CylCylindrical Waveguide

Helmholtz Wave Equation TE, TM, and TEM Modes – Rect Rectangular Waveguide TE, TM, and TEM Modes – Cyl Cylindrical Waveguide

Waveguides

S. R. [email protected]

Department of Electrical & Electronics EngineeringBITS Pilani, Hyderbad Campus

May 7, 2015

Vector Calculus EE208, School of Electronics Engineering, VIT

mailto:[email protected]


Outline

1 Helmholtz Wave Equation

2 TE, TM, and TEM Modes – Rect

3 Rectangular Waveguide

4 TE, TM, and TEM Modes – Cyl

5 Cylindrical Waveguide



Outline








Helmholtz Wave EquationIn a source-less dielectric medium,

∇ · ~D = 0

∇ ·~B = 0

∇× ~H = jω~D = jωε~E (1)

∇×~E = −jω~B = −jωµ~H (2)

Taking curl of (2) gives

∇×(∇×~E

)= ∇×

(−jωµ~H

)⇒ ∇

(∇ ·~E

)−∇2~E = −jωµ

(∇× ~H

)⇒ ∇

(∇ ·~E

)−∇2~E = −jωµ

(jωε~E

)⇒ ∇2~E = ∇

(∇ ·~E

)−ω2µε~E

⇒ ∇2~E =~0−ω2µε~E (3)

Similarly, it can be proved that

∇2~H = −ω2µε~H. (4)



Solution of Helmholtz Equation (in Cartesian System)Vector Helmholtz equation can be decomposed as shown below:

∇2Ex + ω2µεEx = 0

∇2~E + ω2µε~E = 0 ∇2Ey + ω2µεEy = 0

∇2Ey + ω2µεEy = 0

Since all the differential equations are similar, let’s solve just one equation using variable-separablemethod. If Exs can be decomposed into

Ex = A (x)B (y)C (z)

then substituting the above equation into Helmholtz equation gives

∇2Ex + ω2µεEx = 0

⇒ ∂2Ex

∂x2 +∂2Ex

∂y2 +∂2Ex

∂z2 + ω2µεEx = 0

⇒ B (y)C (z)∂2A∂x2 + A (x)C (z)

∂2B∂y2 + A (x)B (y)

∂2C∂z2 + ω2µεA (x)B (y)C (z) = 0

⇒ 1A (x)

∂2A∂x2 +

1B (y)

∂2B∂y2 +

1C (z)

∂2C∂z2 −γ2 = 0



Solution of Helmholtz Equation ... Contd

⇒ 1A (x)

∂2A∂x2 +

1B (y)

∂2B∂y2 +

1C (z)

∂2C∂z2 −γ2 = 0

⇒ 1A (x)

∂2A∂x2 +

1B (y)

∂2B∂y2 +

1C (z)

∂2C∂z2 −γ2

x−γ2y−γ2

z = 0 (5)

The above equation can be decomposed into 3 separate equations:

1A (x)

∂2A∂x2 − γ2

x = 0

1B (y)

∂2B∂y2 − γ2

y = 0

1C (z)

∂2C∂z2 − γ2

z = 0

It is sufficient to solve only one of the above equations and it’s solution is given as

⇒ ∂2A∂x2 − γ2

xA (x) = 0

⇒ A (x) = L1eγxx + L2e−γxx = L−eγxx + L+e−γxx (6)



Solution of Helmholtz Equation ... Contd

So, finally Ex is given as

Ex =(L−eγxx + L+e−γxx) (M−eγyy + M+e−γyy) (N−eγzz + N+e−γzz) (7)

with the condition thatγ2

x + γ2y + γ2

z = γ2. (8)



Outline








Maxwell’s Equations in Free Space

∇×~E = −jωµ~H

∂Ez

∂y+ jβzEy = −jωµHx

−jβzEx −∂Ez

∂x= −jωµHy

∂Ey

∂x− ∂Ex

∂y= −jωµHz

∇× ~H = jωε~E

∂Hz

∂y+ jβzHy = jωεEx

−jβzHx −∂Hz

∂x= jωεEy

∂Hy

∂x− ∂Hx

∂y= jωµεEz

Hx =j

β20 − β2

z

(ωε

∂Ez

∂y− βz

∂Hz

∂x

)Hy =

−jβ2

0 − β2z

(ωε

∂Ez

∂x+ βz

∂Hz

∂y

)Ex =

−jβ2

0 − β2z

(βz

∂Ez

∂x+ ωµ

∂Hz

∂y

)Ey =

jβ2

0 − β2z

(−βz

∂Ez

∂y+ ωµ

∂Hz

∂x

)

The above equations indicate the fact that if

longitudinal components Ez&Hz are known, one

can get all the transversal components



TE Modes – Ez = 0

Hx =j

β20 − β2

z

(ωε

∂Ez

∂y− βz

∂Hz

∂x

)=

jβ2

0 − β2z

(−βz

∂Hz

∂x

)Hy =

−jβ2

0 − β2z

(ωε

∂Ez

∂x+ βz

∂Hz

∂y

)=

−jβ2

0 − β2z

(βz

∂Hz

∂y

)Ex =

−jβ2

0 − β2z

(βz

∂Ez

∂x+ ωµ

∂Hz

∂y

)=

−jβ2

0 − β2z

(ωµ

∂Hz

∂y

)Ey =

jβ2

0 − β2z

(−βz

∂Ez

∂y+ ωµ

∂Hz

∂x

)=

jβ2

0 − β2z

(ωµ

∂Hz

∂x

)

Characteristic impedance of TE waved is given as

ZTE0 =

Ex

Hy= −

Ey

Hx=

ωµ

βz



TM Modes – Hz = 0

Hx =j

β20 − β2

z

(ωε

∂Ez

∂y− βz

∂Hz

∂x

)=

jβ2

0 − β2z

(ωε

∂Ez

∂y

)Hy =

−jβ2

0 − β2z

(ωε

∂Ez

∂x+ βz

∂Hz

∂y

)=

−jβ2

0 − β2z

(ωε

∂Ez

∂x

)Ex =

−jβ2

0 − β2z

(βz

∂Ez

∂x+ ωµ

∂Hz

∂y

)=

−jβ2

0 − β2z

(βz

∂Ez

∂x

)Ey =

jβ2

0 − β2z

(−βz

∂Ez

∂y+ ωµ

∂Hz

∂x

)=

jβ2

0 − β2z

(−βz

∂Ez

∂y

)

Characteristic impedance of TM waved is given as

ZTM0 =

Ex

Hy= −

Ey

Hx=

βz

ωε



TEM Modes – Ez = 0 & Hz = 0

Hx =j

β20 − β2

z

(ωε

∂Ez

∂y− βz

∂Hz

∂x

)Hy =

−jβ2

0 − β2z

(ωε

∂Ez

∂x+ βz

∂Hz

∂y

)Ex =

−jβ2

0 − β2z

(βz

∂Ez

∂x+ ωµ

∂Hz

∂y

)Ey =

jβ2

0 − β2z

(−βz

∂Ez

∂y+ ωµ

∂Hz

∂x

)

If longitudinal components Ez&Hz are zeros, all the transversal components become zero ... then

the wave does not exist at all !!!

So, something must be wrong ... isn’t it ?!



TEM Modes – Ez = 0 & Hz = 0

Hx =j

β20 − β2

z

(ωε

∂Ez

∂y− βz

∂Hz

∂x

)Hy =

−jβ2

0 − β2z

(ωε

∂Ez

∂x+ βz

∂Hz

∂y

)Ex =

−jβ2

0 − β2z

(βz

∂Ez

∂x+ ωµ

∂Hz

∂y

)Ey =

jβ2

0 − β2z

(−βz

∂Ez

∂y+ ωµ

∂Hz

∂x

)

Transverse components can exist only when βz = β0. In other words, βx = 0 and βy = 0.



TEM Modes – ZTEM0

Characteristic impedance of TE and TM waves are given as

ZTE0 =

Ex

Hy= −

Ey

Hx=

ωµ

βz

ZTM0 =

Ex

Hy= −

Ey

Hx=

βz

ωε

Then what is the Characteristic impedance of TEM wave?

⇓

Since TEM mode is TE as well as TM, we can actually use any one of the above equations ... also,for TEM waves βz = β0,

ZTE0 =

ωµ

βz=

ωµ

β0=

ωµ

ω√

µε=

√µ

ε

ZTM0 =

βz

ωε=

β0

ωε=

ω√

µε

ωε=

√µ

ε

Have you noticed that the Characteristic impedance of TEM wave is independent of waveguide

dimensions ?!



Existence of TEM Modes

For TEM waves E and H fields exist in transverse plane only (since Ez = 0 & Hz = 0). Also, from themaxwell equation ∇ ·~B = 0, we know that magnetic fields are rotational fields. So, line integral of~H around any arbitrary closed loop

¸C~H · ~dl has to be non-zero. However, from Stokes’ theorem and

Ampere’s law, we know that

˛C~H · ~dl =

¨ (∇× ~H

)· ~ds =

¨~Je · ~ds.

Since¸

C~H · ~dl 6= 0,~Je has to be a non-zero value. However, we know that~Je is conduction current ...

So, we need at least one conductor inside the waveguide for TEM modes to exist.



Outline








Longitudinal Component – Ez

From (7), similar to Ex, Ez also can be written as

Ez =(L−eγxx + L+e−γxx) (M−eγyy + M+e−γyy) (N−eγzz + N+e−γzz) . (9)

If the wave is propagating along +z direction, N− = 0. Remaining constants can be derived usingboundary conditions. Since Ez is tangential to all the four surfaces, boundary conditions are

{Ez = 0, when x = 0 & x = aEz = 0, when y = 0 & y = b

. (10)




First let us see the boundary conditions imposed by x = 0 and x = a. Since Ez = 0, when x = 0,

L− + L+ = 0⇒ L− = −L+ . (11)

Similarly Ez = 0, when x = a, so using the above equation one gets

L−eγxa + L+e−γxa = 0⇒ L+(e−γxa − eγxa) = 0⇒ sinh γxa = 0.

Hence, from the boundary conditions, we have γxa = sinh−1 (0) = j sin−1 (0) = jmπ, where m =0,±1,±2, . . . So,

γx=jmπ

a= 0 + jβx. (12)

Similarly, from boundary conditions at y = 0 and y = b, we can derive that

M− = −M+ and γy=jnπ

b= 0 + jβy, where n = 0,±1,±2, . . . (13)




Substituting (11), (12), and (13) into (9) gives

Ez =(L−eγxx + L+e−γxx) (M−eγyy + M+e−γyy) (N−eγzz + N+e−γzz)

=(−L+eγxx + L+e−γxx) (−M+eγyy + M+e−γyy) (0 + N+e−γzz)

=(−L+

) (−M+

) (N+) (

eγxx − e−γxx) (eγyy − e−γyy) (e−γzz)=(2L+

) (2M+

) (N+)

sinh (γxx) sinh(γyy)

e−γzz

=(2L+

) (2M+

) (N+) (

j2)︸︷︷︸

A

sin (βxx) sin(

βyy)

e−γzz, ∵ sinh (jx) = j sin x

= A sin(mπ

ax)

sin( nπ

by)

e−γzz. (14)

We also know that β2x + β2

y + β2z = β2

0. So, γz = jβz where

βz =√

β20 − β2

x − β2y =

√β2

0 −(mπ

a

)2−( nπ

b

)2. (15)



Longitudinal Component – Hz

When electric field is zero, magnetic field is maximum. So, for magnetic fields, sin function simplybecomes cos function. So, we do not have to derive expression for Hz once again. So, Ez and Hz aresimply given as

Ez = Asin(mπ

ax)

sin( nπ

by)

e−γzz (16)

Hz = Bcos(mπ

ax)

cos( nπ

by)

e−γzz (17)



TE Modes in Rectangular Waveguide – Ez = 0

Now that we have derived expressions for longitudinal components, we can get transversal compo-nents by using the formulas for TE waves:

Hx =j

β20 − β2

z

(−βz

∂Hz

∂x

)=B

−jβz

β20 − β2

z

[−mπ

a× sin

(mπ

ax)

cos( nπ

by)

e−γzz]

Hy =−j

β20 − β2

z

(βz

∂Hz

∂y

)= B

−jβz

β20 − β2

z

[− nπ

b× cos

(mπ

ax)

sin( nπ

by)

e−γzz]

Ex =−j

β20 − β2

z

(ωµ

∂Hz

∂y

)= B

−jωµ

β20 − β2

z

[− nπ

b× cos

(mπ

ax)

sin( nπ

by)

e−γzz]

Ey =j

β20 − β2

z

(ωµ

∂Hz

∂x

)=B

jωµ

β20 − β2

z

[−mπ

a× sin

(mπ

ax)

cos( nπ

by)

e−γzz]

since Hz = B cos( mπ

a x)

cos( nπ

b y)

e−γzz.



TM Modes in Rectangular Waveguide

Similar to TE mode, TM mode’s transversal components are given as shown below:

Hx =j

β20 − β2

z

(ωε

∂Ez

∂y

)= A

jωε

β20 − β2

z

[ nπ

b× sin

(mπ

ax)

cos( nπ

by)

e−γzz]

Hy =−j

β20 − β2

z

(ωε

∂Ez

∂x

)=A

−jωε

β20 − β2

z

[mπ

a× cos

(mπ

ax)

sin( nπ

by)

e−γzz]

Ex =−j

β20 − β2

z

(βz

∂Ez

∂x

)=A

−jβz

β20 − β2

z

[mπ

a× cos

(mπ

ax)

sin( nπ

by)

e−γzz]

Ey =j

β20 − β2

z

(−βz

∂Ez

∂y

)= A

−jβz

β20 − β2

z

[ nπ

b× sin

(mπ

ax)

cos( nπ

by)

e−γzz]

since Ez = A sin( mπ

a x)

sin( nπ

b y)

e−γzz.



Modes in Rectangular Waveguide – IntuitiveExplanation

If you are fed up with the all the equations given in this PDF file, you may feel free to refer to much

simpler qualitative explanation given in another PDF file written on rectangular waveguides.

There, you can also find formulas for λg, ωcutoff, etc.



Parallel Plate Waveguide

All the equations derived for rectangular waveguides can also be used for parallel plate

waveguide. After all, parallel plate waveguide is nothing but a rectangular waveguide with a→ ∞.

One major difference though is, in parallel plate waveguide in addition to TE and TM modes, TEM

modes also exist ... do you know why?



Free Space Waveguide

Once again, the equations derived for rectangular waveguides can be used for free space too. It is

because, free space waveguide is nothing but a rectangular waveguide with a→ ∞ and b→ ∞ .

One major difference though is, similar to parallel plate waveguide, in free space, TEM modes also

exist ... do you know why?



Outline








Maxwell’s Equations in Free Space

∇×~E = −jωµ~H

1ρ

(∂Ez

∂φ− ρ

∂Eφ

∂z

)= −jωµHρ(

∂Eρ

∂z− ∂Ez

∂ρ

)= −jωµHφ

1ρ

(∂(ρEφ

)∂ρ

−∂Eρ

∂φ

)= −jωµHz

∇× ~H = jωε~E

1ρ

(∂Hz

∂φ− ρ

∂Hφ

∂z

)= jωεEρ(

∂Hρ

∂z− ∂Hz

∂ρ

)= jωεEφ

1ρ

(∂(ρHφ

)∂ρ

−∂Hρ

∂φ

)= jωεEz

Eρ =−j

β20 − β2

z

(βz

∂Ez

∂ρ+

ωµ

ρ

∂Hz

∂φ

)Eφ =

−jβ2

0 − β2z

(βz

ρ

∂Ez

∂φ−ωµ

∂Hz

∂ρ

)Hρ =

jβ2

0 − β2z

(ωε

ρ

∂Ez

∂φ− βz

∂Hz

∂ρ

)Hφ =

−jβ2

0 − β2z

(ωε

∂Ez

∂ρ+

βz

ρ

∂Hz

∂φ

)

Once again, the above equations indicate the fact

that if longitudinal components Ez&Hz are

known, one can get all the transversal

components



TE Modes – Ez = 0

Eρ =−j

β20 − β2

z

(βz

∂Ez

∂ρ+

ωµ

ρ

∂Hz

∂φ

)=

−jβ2

0 − β2z

(ωµ

ρ

∂Hz

∂φ

)Eφ =

−jβ2

0 − β2z

(βz

ρ

∂Ez

∂φ−ωµ

∂Hz

∂ρ

)=

−jβ2

0 − β2z

(−ωµ

∂Hz

∂ρ

)Hρ =

jβ2

0 − β2z

(ωε

ρ

∂Ez

∂φ− βz

∂Hz

∂ρ

)=

jβ2

0 − β2z

(−βz

∂Hz

∂ρ

)Hφ =

−jβ2

0 − β2z

(ωε

∂Ez

∂ρ+

βz

ρ

∂Hz

∂φ

)=

−jβ2

0 − β2z

(βz

ρ

∂Hz

∂φ

)


ZTE0 =

Eρ

Hφ= −

Eφ

Hρ=

ωµ

βz



TM Modes – Hz = 0

Eρ =−j

β20 − β2

z

(βz

∂Ez

∂ρ+

ωµ

ρ

∂Hz

∂φ

)=

−jβ2

0 − β2z

(βz

∂Ez

∂ρ

)Eφ =

−jβ2

0 − β2z

(βz

ρ

∂Ez

∂φ−ωµ

∂Hz

∂ρ

)=

−jβ2

0 − β2z

(βz

ρ

∂Ez

∂φ

)Hρ =

jβ2

0 − β2z

(ωε

ρ

∂Ez

∂φ− βz

∂Hz

∂ρ

)=

jβ2

0 − β2z

(ωε

ρ

∂Ez

∂φ

)Hφ =

−jβ2

0 − β2z

(ωε

∂Ez

∂ρ+

βz

ρ

∂Hz

∂φ

)=

−jβ2

0 − β2z

(ωε

∂Ez

∂ρ

)


ZTM0 =

Eρ

Hφ= −

Eφ

Hρ=

βz

ωε



TEM Modes – ZTEM0

Characteristic impedance of TE and TM waves are given as

ZTE0 =

Eρ

Hφ= −

Eφ

Hρ=

ωµ

βz

ZTM0 =

Eρ

Hφ= −

Eφ

Hρ=

βz

ωε

Then what is the Characteristic impedance of TEM wave?

⇓

Since TEM mode is TE as well as TM, we can actually use any one of the above equations ... also,for TEM waves βz = β0,

ZTE0 =

ωµ

βz=

ωµ

β0=

ωµ

ω√

µε=

√µ

ε

ZTM0 =

βz

ωε=

β0

ωε=

ω√

µε

ωε=

√µ

ε

Have you noticed that the Characteristic impedance of TEM wave is independent of waveguide

dimensions ?!



Outline








Helmholtz Wave Equation - Cylindrical CoordinateSystem

Since all the electromagnetic field components have to satisfy the Helmholtz wave equation, one canwrite the wave equation for Ez as

∇2Ez + ω2µεEz = 0

⇒[

1ρ

∂

∂ρ

(ρ

∂Ez

∂ρ

)+

1ρ2

∂2Ez

∂φ2 +∂2Ez

∂z2

]+ ω2µεEz = 0.

Solving the above differential equation gives

Ez = [AJn (βcρ) + BYn (βcρ)] [C sin (nφ) + D cos (nφ)][Ee−jβzz + Fejβzz

](18)



Longitudinal Components – Ez & Hz

In (18), the term Yn (βcρ) → ∞ at ρ = 0; So, this term is physically unacceptable for cylindricalwaveguides (however, this term is required for coaxial lines) ... also, if we assume that the wave ispropagating in +z direction, then F = 0 ... So,

Ez = AJn (βcρ) [C sin (nφ) + D cos (nφ)] e−jβzz. (19)

Similarly, one can write equation for Hz as

Hz = BJn (βcρ) [C′ sin (nφ) + D′ cos (nφ)] e−jβzz. (20)



Cylindrical Waveguide – TE modes

Eρ =−j

β20 − β2

z

(ωµ

ρ

∂Hz

∂φ

)= B

−jβ2

0 − β2z

[ωµ

ρ

[Jn (βcρ) [nC′ cos (nφ)− nD′ sin (nφ)] e−jβzz

]]Eφ =

−jβ2

0 − β2z

(−ωµ

∂Hz

∂ρ

)=B

−jβ2

0 − β2z

[−ωµ

[βcJ′n (βcρ) [C′ sin (nφ) + D′ cos (nφ)] e−jβzz

]]Hρ =

jβ2

0 − β2z

(−βz

∂Hz

∂ρ

)= B

jβ2

0 − β2z

[−βz

[βcJ′n (βcρ) [C′ sin (nφ) + D′ cos (nφ)] e−jβzz

]]Hφ =

−jβ2

0 − β2z

(βz

ρ

∂Hz

∂φ

)= B

−jβ2

0 − β2z

[βz

ρ

[Jn (βcρ) [nC′ cos (nφ)− nD′ sin (nφ)] e−jβzz

]]

where Hz = BJn (βcρ) [C′ sin (nφ) + D′ cos (nφ)] e−jβzz.

Cutoff propagation constant can be derived from the above equations by using boundary conditions.Since Eφ is tangential to the boundary of the cylindrical waveguide, boundary conditions are

{Eφ = 0, when ρ = a . (21)

So, boundary condition for TE modes is

J′n (βca) = 0. (22)



Cylindrical Waveguide – TM modes

Eρ =−j

β20 − β2

z

(βz

∂Ez

∂ρ

)= A

−jβ2

0 − β2z

(βz

[βcJ′n (βcρ) [C sin (nφ) + D cos (nφ)] e−jβzz

])Eφ =

−jβ2

0 − β2z

(βz

ρ

∂Ez

∂φ

)= A

−jβ2

0 − β2z

(βz

ρ

[Jn (βcρ) [nC cos (nφ)− nD′ sin (nφ)] e−jβzz

])Hρ =

jβ2

0 − β2z

(ωε

ρ

∂Ez

∂φ

)=A

jβ2

0 − β2z

(ωε

ρ

[Jn (βcρ) [nC cos (nφ)− nD′ sin (nφ)] e−jβzz

])Hφ =

−jβ2

0 − β2z

(ωε

∂Ez

∂ρ

)= A

−jβ2

0 − β2z

(ωε[

βcJ′n (βcρ) [C sin (nφ) + D cos (nφ)] e−jβzz])

where Ez = AJn (βcρ) [C sin (nφ) + D cos (nφ)] e−jβzz.

Cutoff propagation constant can be derived from the above equations by using boundary conditions.Since Eφ is tangential to the boundary of the cylindrical waveguide, boundary conditions are

{Eφ = 0, when ρ = a . (23)

So, boundary condition for TM modes is

Jn (βca) = 0. (24)


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S. R. Zinka [email protected] May … · Helmholtz Wave EquationTE, TM, and TEM Modes – RectRectangular WaveguideTE, TM, and TEM Modes – CylCylindrical Waveguide