19-1 EM Solution - Final

Electromagnetics

Final Exam Solution

Jaesang Lee

Dept. of Electrical and Computer Engineering Seoul National University

(email: jsanglee@snu.ac.kr)

Midterm 2 | Problem 1(a): Parallel-plate TR-line

(a) Derive analytically the TEM modes (i.e. E & H-fields) propagating in z-direction in an infinite parallel-plate transmission line (Assumption:

infinite in extent in x-direction, two perfectly-conducting plates are separated by distance d in y-direction and a dielectric of constitutive parameters (µ, ε) is filled in between. Hint: Start from Maxwell’s Curl equations). (40 points)

• Derivation of the TEM mode

E

_z

( ) y, z = 0 H

_z

( ) y, z = 0

⎧ ⎨

⎪

where

⎩⎪ ∂ E

∂ x = 0, ∂ H

∂ x = 0

∂E_z

∂y − ∂E_y

∂z = − jωµ^H_x !(a)

− ∂E_z

∂x + ∂E_x

∂z = − jωµ^H_y !(b)

∂E_y

∂x − ∂E_x

∂y = − jωµ^H_z !(c)

⎧

⎨

⎪⎪

⎪

⎩

⎪⎪

⎪

∂H_z

∂y + ∂H_y

∂z = jωε^E_x !(e)

− ∂H_z

∂x + ∂H_x

∂z = jωε^E_y !(f)

∂H_y

∂x − ∂H_x

∂y = jωε^E_z !(g)

⎧

⎨

⎪⎪

⎪

⎩

⎪⎪

⎪

∇ × E = − j ωµ H ∇ × H = j ωε E

- From equations (c), we know that

E

_x

( ) y, z = C ⋅ E

^x0

( ) z .

- By substituting Ex = 0 into equation (b), we get

E

_x

( d or 0, z ) = 0 → C = 0

From B.C.,

E

_t

= 0 H

_n

= 0

⎧ ⎨

⎩

at the conducting boundary (y = 0 and y = b)

∴ H

_y

( ) y, z = 0

- From equation (g), we know that and

- Equation (e) also vanishes accordingly, since Ex = Hy = 0.

H

_x

( ) y, z = H

⁰

H

^x0

( ) z ! (1)

- By differentiating equation (f) with z,

∂

²

H

_x

∂ z

²

= j ωε ∂ E

^y

∂ z = j ωε ( j ωµ H

_x

) = − ω

²

µε H

^x

⎡

⎣⎢

⎤

⎦⎥

→ d

²

H

_x

dz

²

+ ω

²

µε H

_x

= 0 ! (2)

∴ E

_x

( ) y, z = 0

(by definition) (Assumption)

equation (a)

• Derivation of the TEM mode

- By solving the ODE,

d

²

H

_x

dz

²

+ ω

²

µε H

_x

= 0 → H

_x

( ) y, z = H

0

e

^{− j}

(

^{ω µε}

)

^z

+ H

1

e

^j

(

^{ω µε}

)

^z

∴ H

_x

( ) y, z = H

⁰

H

^x0

( ) z = H

⁰

e

^{− jβz}

(∵there is no reflection!)

where

β = ω µε

- By substituting Hx back into equation (f), we get

∂ H

_x

∂ z = j ωε E

_y

→ E

_y

= 1 j ωε

∂ H

_x

∂ z = 1

j ωε ⋅ − ( j β H

₀

e

^−βz

)

∴ E

_y

( ) y, z = − µ ε H

⁰

e

^{− jβz}

= E

⁰

e

^{− jβz}

- In summary, TEM wave characterized as

E

_y

( ) y, z = E

0

e

^{− jβz}

H

_x

( ) y, z = − E η

⁰

e

^{− jβz}

⎧

⎨ ⎪

⎩ ⎪

where

β = ω µε − E

^y

H

_x

= µ

ε ! η E

_z

( ) y, z = 0

H

_z

( ) y, z = 0

⎧ ⎨

⎪

⎩⎪

and y = 0

y = d

x

y z

H

x

E

y

Midterm 2 | Problem 1(a): Parallel-plate TR-line

If boundary condition is not properly used, deduct -10 points (e.g. derivation of Ex = Hy = 0) Also, if relationship between Ey and Hx (i.e. η) is not specified, deduct -10 points.

If all derivation, although not following this approach, makes sense, give full mark.

Midterm 2 | Problem 1(b): Parallel-plate TR-line

(b): Q1: Derive the equations for z-dependent voltage and current on the lossless transmission line by using results obtained in (a) (Please define appropriate circuit elements per unit length that are comprised of constitutive parameters (µ, ε) and dimension of d [separation], w [width]) (45

points)

dE

_y

dz = j ωµ H

_x

! (1) dH

_x

dz = j ωε E

_y

! (2)

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

• TR line equations

- From curl equations (b) and (e), we have two ODEs as

y = 0 y = d

x

y z

H

x

E

y

- If we integrate equation (1) from y = 0 to y = d

E

_y

( ) y, z = E

⁰

e

^{− jβz}

H

_x

( ) y, z = − E η

⁰

e

^{− jβz}

⎧

⎨ ⎪

⎩ ⎪

where

β = ω µε

At y = d , ρ

_su

= − ε E

_y

J

_su

= a

_z

H

_x

⎧ ⎨

⎩⎪

At y = 0, ρ

_sl

= ε E

_y

J

_sl

= − a

_z

H

_x

⎧ ⎨

⎩⎪

w

where

L ! µ d

w (H/m)

: Inductance per unit length of parallel-plate transmission line

I z ( ) ! J

^su

( ) z w

: Total current flowing in z-direction in the upper plate

V z ( ) ! − ∫

₀^d

E

^y

dy

d

dz ∫

₀^d

E

_y

( ) y, z dy = − dV z dz ( )

⎡

⎣⎢

⎤

⎦⎥ = j ωµ H

_x

dy

0

∫

d

= j ωµ H

^x

d = j ωµ J

^su

( ) z d

⎡ ⎣⎢ ⎤

⎦⎥

→ − dV z ( )

dz = j ω µ d w

⎛ ⎝⎜ ⎞

⎠⎟ ⎡⎣ J

_su

( ) z w ⎤⎦

: Potential difference (voltage) between two plates (Jsu: surface current flowing in z-direction)

∴− dV z ( )

dz = j ω LI z ( ) ! (3)

Midterm 2 | Problem 1(b): Parallel-plate TR-line

• TR line equations

- Similarly, by integrating equation (2) from x = 0 to x = w

d

dz ∫

₀^w

H

_x

( ) y, z dx = dI z dz ( )

⎡

⎣⎢

⎤

⎦⎥ = j ωε E

_y

dy

0

∫

w

= j ωε E

^y

w

⎡ ⎣⎢ ⎤

⎦⎥

→ dI z ( )

dz = − j ω ε w d

⎛ ⎝⎜ ⎞

⎠⎟ − ⎡⎣ E

_y

d ⎤⎦

where

C ! ε w

d (F/m)

: Capacitance per unit length of parallel-plate transmission line

→ dI z ( )

dz = − j ω CV z ( ) ! (4)

− dV z ( )

dz = j ω LI z ( )

− dI z ( )

dz = j ω CV z ( )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

∴ A pair of time-harmonic transmission line equations

d

²

V z ( )

dz

²

= − ω

²

LCV z ( )

d

²

I z ( )

dz

²

= − ω

²

LCI z ( )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

If definitions of equivalent capacitance and inductance are

not made through integration, deduct -10 points per each.

d

²

V z ( )

dz

²

= − ω

²

LCV z ( )

d

²

I z ( )

dz

²

= − ω

²

LCI z ( )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

V z ( ) = V

0

e

^{− jβz}

I z ( ) = I

⁰

e

^{− jβz}

⎧ ⎨

⎪

⎩⎪

where

∵

L " µ d

w (H/m) C " ε w

d (F/m)

⎧

⎨ ⎪⎪

⎩ ⎪ β = ω LC = ω µε ⎪

(b) Q2: Find the solutions and express characteristic impedance and speed of signal propagation in terms of given parameters. (15 points)

Midterm 2 | Problem 1(b): Parallel-plate TR-line

- Characteristic Impedance (By plugging V(z) and I(z) to the paired equations)

− d

dz ( V

₀

e

^{− jβz}

) = j ω LI

⁰

e

^{− jβz}

→ j β V

₀

e

^{− jβz}

= j ω LI

₀

e

^{− jβz}

→ Z

⁰

=

V z ( )

I z ( ) = V I

⁰⁰

= C L = w d µ ε = w d η ( Ω )

- Velocity of propagation

u

_p

= ω β =

1

LC = 1

µε (m / s )

Each answer (paired solutions, characteristic impedance, velocity of propagation) takes up 5 points. Deduct -2 points if d and w are not included in Z0.

• Attenuation in the parallel-plate transmission lines caused by…

- (1) Lossy dielectric (σ ≠ 0)

- (2) Imperfectly conducting walls (σc ≠ ∞)

α = α

_d

+ α

_c

= σ

2 η + 1 d

π f ε σ

_c

y

z

G

Ru

Rl

• Conductance (G) between two conductors per unit length

<Circuit representation>

∵ RC = − E ⋅ d l

∫

L

σ E ⋅ d s

" ∫

S

⋅

ε E ⋅ d s

" ∫

S

− E ⋅ d l

∫

L

=

ε

σ = C G G = C σ

ε

(from right)

= ε w

d ⋅ σ

ε = σ w

d (S/m)

where σ is the conductivity of the dielectric Δz

(c) Q1: Assuming the transmission line is lossy (i.e. the conducting wall has finite conductivity and the dielectric has nonzero conductivity), build the “distributed-element” equivalent circuit model for infinitesimally short segment (Δz) of the line and explain what each element represents. (40 points)

Midterm 2 | Problem 1(c): Parallel-plate TR-line

* No need for derivation of this equation 15 points given for conductance derivation

Midterm 2 | Problem 1(c): Parallel-plate TR-line

a

_y

p

_σ

c

[W/m

²

] = 1

2 Re ( a

_z

E

_z

× a

_x

H

_x^*

) = 1 2 Re ( a

^y

J

^{s 2}

Z

^s

) ! (1)

Z

_s

= E

_z

H

_x

= E

_z

J

_su

= η

_c

! (2)

(= Intrinsic impedance of the plate)

η

_c

= R

_s

+ jX

_s

= ( 1 + j ) π σ f µ

^c

c

( Ω ) ! (3)

(Refer to lecture note 3-2) - Time-average power dissipated on unit surface [W/m²] due to Ez

J

_su

= a

_z

H

_x

= a

_z

σ

_c

E

_z

(A/m) J

_sl

= − J

_su

“Surface” current (Js) (∵f↑ → skin depth↓)

“Surface” impedance (Zs) of the plate

E

_z

= J

_su

Z

_s

and H

_x

= J

_su

! (4)

- From eqn. (2), we get

p

_σ

c

= 1

2 Re ( J

_s ²

Z

_s

) = 1 2 J

^{s 2}

R

^s

(W/m

²

)

- By plugging eqn. (4) and (3) into eqn. (1), we get

- Power dissipated per unit length [W/m] through the plate of width w

P

_σ

c

= wp

_σ

c

= 1

2 w J

_s ²

R

_s

(W/m) = 1

2 wJ

_s ²

R

_s

w

⎛ ⎝⎜ ⎞

⎠⎟ = 1

2 I

²

R

_s

w

⎛ ⎝⎜ ⎞

⎠⎟ ∴ R = R

_u

+ R

_l

= 2 R

_s

w

⎛ ⎝⎜ ⎞

⎠⎟ = 2 w

π f µ

_c

σ

_c

( Ω / m )

25 points given for resistance derivation

- Deduct -5 points if 2 in the final form missing - Only 15 points given if only answer is given

Midterm 2 | Problem 1(c): Parallel-plate TR-line

y

z

Δz

GΔz

RΔz LΔz

CΔz

µ d

L w (H / m )

2 w

π f µ

_c

σ

_c

R ( Ω / m )

G σ w

d (S / m )

C ε w

d (F / m )

Parameter Formula Unit

Midterm 2 | Problem 1(c): Parallel-plate TR-line

(c) Q2: Discuss whether there is any difference in the E & H-fields in the lossy transmission line compared to those in the lossless counterpart.

(15 points)

H

x

E

y

E

z

• Resistance (R) along the conductors per unit length

- In actual cases, conductivity of the plate is finite (σc ≠ ∞)

∴ small, yet non-vanishing longitudinal field (Ez) “induced!” (Hx → Jsu → Ez!) - R obtained by relationship between power loss at the surface vs. surface current

w

Δz

J

su

(c) Q3: Under what conditions do we use a “distributed-element” model? (20 points)

- When used?

‣ Very, very long TR-line (>240 km): AC voltage and current at one location different from those at other (∵signal speed ≠ ∞)

‣ At very high frequency: physical dimension of circuit ~ wavelength of electrical signal

‣ For these cases, wires or lines are not perfect conductors and their impedance matters (represented by R, L, C, G)

‣ c.f) Lumped-element model (R, L, C, G NOT depending on length and concentrated at singular points) e.g. regular circuit we use Deduct -5 points if finite conductivity is not mentioned

Deduct -5 points if description how surface current is induced (due to magnetic field) is missing.

10 points given per each case (i.e. very long line or very high frequency)

(d) Derive general transmission line equations from the circuit obtained in (c) (Hint: Use Kirchhoff’s current and voltage laws). (40 points)

Midterm 2 | Problem 1(d): Parallel-plate TR-line

y

z

Δz

GΔz

RΔz LΔz

CΔz

Series components

• R (Ω/m): finite conductivity of the plates (σc)

• L (H/m): H-field in the wire and self-inductance Shunt components

• C (F/m): Two conductors + dielectric

• G (S/m): non-zero conductivity of dielectric (σ)

v(z,t) v(z+Δz,t)

i(z,t) Node i(z+Δz,t)

Loop

Δz +

-

- Kirchoff’s voltage law (around loop L)

− v( z, t ) + R Δ z ⋅ i ( z, t ) + L Δ z ∂ i (z, t )

∂ t + v( z + Δ z, t ) = 0 → − v( z + Δ z, t ) − v( z, t )

Δ z ! − ∂ v( z, t )

∂ z = Ri( z, t ) + L ∂ i ( z, t )

∂ t

<Infinitesimal TR line element> <Equivalent circuit>

Δ z → 0

- Kirchoff’s current law (to node N)

i ( z, t ) − G Δ z ⋅ v(z + Δ z, t ) − C Δ z ∂ v( z + Δ z, t )

∂ t − i ( z + Δ z, t ) = 0 → − i ( z + Δ z, t ) − i ( z, t )

Δ z ! − ∂ i( z, t )

∂ z = Gv( z, t ) + C ∂ v( z, t )

∂ t

•General transmission-line equations

: A pair of 1st-order PDEs in v(z,t) and i(z,t)

− ∂ v( z, t )

∂ z = Ri( z, t ) + L ∂ i ( z, t )

∂ t

− ∂ i ( z, t )

∂ z = Gv(z, t ) + C ∂ v( z, t )

∂ t

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

v( z, t ) = Re ⎡⎣ V ( z )e

^jωt

⎤⎦

i ( z, t ) = Re ⎡⎣ I ( z )e

^jωt

⎤⎦

⎧ ⎨

⎪

⎩⎪

since

,

− dV ( z )

dz = ( R + j ω L ) I ( z )

− dI ( z )

dz = ( G + j ω C ) V ( z )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

Time-harmonic TR-line equation

Midterm 2 | Problem 1(d): Parallel-plate TR-line

• Wave characteristics on an infinite TR line

- To solve for V(z) and I(z), coupled time-harmonic TR equations combined as

− d

²

V ( z )

dz

²

= ( R + j ω L ) dI ( z )

dz = − ( R + j ω L ) ( G + j ω C ) V ( z )

− d

²

I ( z )

dz

²

= ( G + j ω C ) dV ( z )

dz = − ( G + j ω C ) ( R + j ω L ) I ( z )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

(Double derivative)

d

²

V ( z )

dz

²

= γ

²

V ( z ) d

²

I ( z )

dz

²

= γ

²

I ( z )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

where

γ = α + j β

= ( R + j ω L ) ( G + j ω C )

If Kirchhoff’s laws are correctly applied and the final forms are correct, full mark should be given.

If the constant (R + jωL) or (G + jωC) is not right (including sign), deduct -5 points per each.

If only the final forms are given, only 20 points will be provided.

(e) Q1: Is the lossy transmission line dispersive? (10 points)

Midterm 2 | Problem 1(e): Parallel-plate TR-line

γ = ( R + j ω L ) ( G + j ω C ) = ( RG − ω

²

LC ) + j ω ( LG + RC )

= ⎡⎣ κ ω ( ) e

^{jΘ( )ω}

⎤⎦

¹²

= κ ω ( ) e

^{jΘ( )2ω}

- Propagation constant (γ)

= κ ω ( ) cos Θ ( ) ω

2 + j sin Θ ( ) ω 2

⎛

⎝⎜

⎞

⎠⎟ = α ω ( ) + j β ω ( )

u

_p

= ω

β ω ( ) = κ ω ( ) ω sin Θ ( ) ω

2

- Phase velocity

∴Phase velocity varies with frequency, indicating that the signal of different frequency within the lossy TR-line travels at different

velocity. Thus, yes, the lossy TR-line is a dispersive system.

- Characteristic impedance (Z0)

Z

₀

= V z ( )

I z ( ) = V I

0⁰

= R + j ω L

G + j ω C = ℜ ( ) ω + j Λ ( ) ω

Q2: Are voltage and current in such system in-phase or out-of-phase? (10 points)

∴Characteristic impedance is defined as the ratio of amplitude of voltage and current on the TR-line which has an imaginary term accounting for phase shift in case that TR-line is lossy. Thus,

voltage and current in such a lossy TR-line is out-of-phase.

For problem 1(e) Q1-Q2, if only answers are given without explanation, 5 points given per each.

For Q3, full mark will be provided if two facts are stated with explanation. Without explanation, 5 points given per each fact.

Midterm 2 | Problem 1(e): Parallel-plate TR-line

• At very high frequency (R << ωL, G << ωC)

Q3: How do these characteristics change as frequency varies? (20 points)

γ = ( R + j ω L ) ( G + j ω C ) = j ω L 1 + R

j ω L

⎛

⎝⎜

⎞

⎠⎟ j ω C 1 + G j ω C

⎛

⎝⎜

⎞

⎠⎟ ! j ω LC 1 + R

2 j ω L

⎛

⎝⎜

⎞

⎠⎟ 1 + G

2 j ω C

⎛

⎝⎜

⎞

⎠⎟ ∵ lim

x→0

( 1 + x )

ⁿ

" ( 1 + nx )

! j ω LC 1 + 1 2 j ω

R

L + G C

⎛ ⎝⎜ ⎞

⎠⎟ −

1 4 ω

²

RG LC

⎛

⎝⎜

⎞

⎠⎟

Binomial approx.

∴ α ! 1

2 R C

L + G L C

⎛

⎝⎜

⎞

⎠⎟ ,

β ! ω LC

(“nearly” a linear function of ω) (Neglected at high frequency)

- Phase velocity:

Z

₀

= R + j ω L

G + j ω C =

j ω L

j ω C 1 +

R j ω L

⎛

⎝⎜

⎞

⎠⎟

1/2

1 + G j ω C

⎛

⎝⎜

⎞

⎠⎟

−1/2

! L

C 1 + R 2 j ω L

⎛

⎝⎜

⎞

⎠⎟ 1 − G

2 j ω C

⎛

⎝⎜

⎞

⎠⎟

! L

C 1 + 1 2 j ω

R

L − G C

⎛ ⎝⎜ ⎞

⎠⎟ −

1 4 ω

²

RG LC

⎡

⎣⎢

⎤

⎦⎥

(Almost zero phase shift between E and H)

- Characteristic impedance:

u

_p

= ω β !

1

LC

(Approximately constant)

∴ R

₀

! L

C , X

₀

! − L

C

1 2 ω

R

L − G C

⎛ ⎝⎜ ⎞

⎠⎟ ! 0

∴At very high frequency, phase velocity becomes constant and characteristic impedance becomes purely resistive, indicating the TR-line becomes a dispersion-less system and its voltage and current are approximately in-phase.

(f) Q1: Derive the condition under which the lossy transmission line becomes distortion-less. Q2: In such system, are voltage and current in- phase or out-of-phase? Please explain. (40 points)

Midterm 2 | Problem 1(f): Parallel-plate TR-line

From characteristic impedance (Z0),

Z

₀

= R + j ω L

G + j ω C =

R + j ω L

( ) ( G − j ω C )

G + j ω C

( ) ( G − j ω C ) = RG + ω

2

LC + j ω ( LG − RC )

G

²

+ ω

²

C

²

To have reactance to be zero,

LG − RC = 0 → ∴ R

L = G C γ = ( R + j ω L ) ( G + j ω C ) = C

L ( R + j ω L ) u

^p

= ω β = LC 1

Z

₀

= R + j ω L

G + j ω C =

L C

Non-dispersive (i.e. Distortionless)

No phase shift (in-phase)

One should understand duality between in-phase propagation and distortionless condition. If one

derives the relation from Z0 and apply that to γ, full mark given. Answers without derivation 20 points given.

Find the input impedance of the finite-length transmission line at a distance z’ from a load when the load is (a) purely resistive, (b) open-circuit, (c) short-circuit, (d) length is an integer multiple of quarter wave and (e) half wave (Denote the characteristic impedance of the line as Z0). (50 points)

Midterm 2 | Problem 2

Z ( ) z ′ = Z

0

Z

_L

+ Z

₀

tanh γ z ′

Z

₀

+ Z

_L

tanh γ z ′ ( ) Ω

(a) When the load is purely resistive (ZL = RL)

Z ( ) z ′ = Z

0

R

_L

+ Z

₀

tanh γ z ′ Z

₀

+ R

_L

tanh γ z ′

(b) When the load is open-circuit (ZL → ∞)

Z ( ) z ′ = lim

Z_L→∞

Z

₀

Z

_L

+ Z

₀

tanh γ z ′

Z

₀

+ Z

_L

tanh γ z ′ = Z

₀

tanh γ z ′

(c) When the load is short-circuit (ZL → 0)

Z ( ) z ′ = lim

Z_L→0

Z

₀

Z

_L

+ Z

₀

tanh γ z ′

Z

₀

+ Z

_L

tanh γ z ′ = Z

₀

tanh γ z ′

* For (d) and (e), please assume lossless TR-line (i.e., γ = jβ) (d) length is an “odd” integer multiple of quarter-wavelength

Z ( ) z ′ = Z

⁰

R

^L

+ Z

⁰

tanh γ z ′

Z

₀

+ R

_L

tanh γ z ′ l = ( 2n − 1 ) λ

4 , ( n = 1, 2, 3, ! )

∴ lim

tan( )β^{l →±∞}

Z

_i

= lim

tan( )β^{l →±∞}

R

₀

Z

_L

+ jR

₀

tan β l R

₀

+ jZ

_L

tan β l =

R

₀²

Z

_L

( ) Ω β l = 2 π

λ ⋅ ( 2n − 1 ) λ

4 = ( 2 n − 1 ) π

2 → tan β l = tan 2 ( n − 1 ) π

2

⎡

⎣⎢

⎤

⎦⎥ → ±∞

(e) length is an integer multiple of half-wavelength

β l = 2 π λ ⋅

n λ 2

⎛ ⎝⎜ ⎞

⎠⎟ = n π → tan β l = 0

∴ lim

tan( )βl ^→0

Z

_i

= lim

tan( )βl ^→0

R

₀

Z

_L

+ jR

₀

tan β l

R

₀

+ jZ

_L

tan β l = Z

^L

( ) Ω

5 points

10 points

10 points

10 points

10 points If all correct, additional 5 points provided.

Q1. Provide analytical derivation of the standing wave ratio (SWR) for the lossless transmission line based on its definition. (35 points)

Midterm 2 | Problem 3

V ( ) z ′ = I

^L

2 ⎡⎣ ( Z

_L

+ Z

₀

) e

^{γ z′}

+ ( Z

^L

− Z

⁰

) e

^{−γ z′}

⎤⎦

- Lossless TR-line → γ = jβ and Z0 = R0

→ V ( ) z ′ = I

^L

2 ( Z

_L

+ R

₀

) e

^jβz′

⎡⎣ 1 + Γ e

^{− j2βz′}

⎤⎦

V ( ) z ′ = I

^L

2 ( Z

_L

+ Z

₀

) e

^{γ z′}

⎡⎣ 1 + Γ e

^{−2γ z′}

⎤⎦

= I

_L

2 ( Z

_L

+ R

₀

) e

^jβz′

⎡⎣ 1 + Γ e

^{j(θΓ−2βz′)}

⎤⎦

∵ γ = j β and Z

₀

= R

₀

( )

∵ Γ = Γ e

^{− jθΓ}

( )

∵ Γ " Z

_L

− Z

₀

Z

_L

+ Z

₀

⎛

⎝⎜

⎞

⎠⎟

max V ( ) z ′ = V

max

= I

_L

2 ⋅ Z

_L

+ R

₀

⋅ ( 1 + Γ )

min V ( ) z ′ = V

^min

= I

^L

2 ⋅ Z

_L

+ R

₀

⋅ ( 1 − Γ )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

- Voltage for finite-length TR-line

SWR or S ! V

_max

V

_min

= 1 + Γ

1 − Γ Γ = S − 1 S + 1

If only answer is there, 15 points given.

Q2. Discuss physical significance of SWR (e.g. when it does occur, how it is used, determining factors and so on). (15 points)

Midterm 2 | Problem 3

-

SWR appears when the voltage and current form standing waves in the TR-line, indicating that the TR-line is finite in length and thus there exists reflected wave (Γ > 0). [5 points. “Standing wave caused by incident + reflected waves” should be mentioned]

-

SWR depends on the reflection coefficient and thus a load value (ZL). [5 points]

-

SWR is used for determining the arbitrary resistive load (RL). [5 points]

The first and second statements MUST BE included. Any other description that makes sense, additional 5 points given.

The SWR on a 50 (Ω)-lossless transmission line terminated in an unknown load impedance is found to be 3. The distance between successive voltage maxima is 20 (cm), and the first maximum is located at 5 (cm) from the load.

(a) Determine Γ and ZL. (40 points)

Midterm 2 | Problem 4

- Step 1) Express ZL in terms of R0 and voltage reflection coefficient Γ

Z

_L

= V ( ) z ′

I ( ) z ′

_z′=0

= R

⁰

1 + Γ e

jθ_Γ

1 − Γ e

^jθΓ

- Step 2) By applying condition below, we can find θΓ for the first voltage maxima

θ

_Γ

− 2 β d = − 2 n π

_n=0

→ θ

_Γ

= 2 β d

^Here,

β = 2 π

λ

^where

λ

2 = 20 (cm)

Distance between successive voltage maxima

→ θ

_Γ

= 2 × 5 π × 0.05 = 0.5 π (rad)

= 2 π

0.4 = 5 π (rad/m)

- Step 3) By measuring S, we can get |Γ| as

Γ = S − 1

S + 1 = 1

2 ∴ Z

_L

= R

₀

1 + Γ e

^jθΓ

1 − Γ e

^jθΓ

= 50 1 + j 0.5

1 − j 0.5 = 30 + j 40 ( Ω )

Give 40 points if all procedures included

(b) Find the equivalent length (lm) and terminating resistance (Rm), for two cases where Rm > 50 (Ω) and Rm < 50 (Ω), such that input impedance is equal to ZL. (60 points)

Midterm 2 | Problem 4

- Case 1: RM > 50 (Ω) → resistive termination with voltage maxima

d = 5 cm

ZL

lm

lM

R

_M

= SR

₀

= 3 × 50 = 150 ( ) Ω

l

_M

+ d = λ

2 → l

_M

= λ

2 − d = 0.2 − 0.05 = 0.15 (m)

R

_m

= R

₀

S = 50

3 = 16.7 ( ) Ω l

_m

+ d = λ

4 → l

_m

= λ

4 − d = 0.1 − 0.05 = 0.05 (m)

- Case 1: Rm > 50 (Ω) → resistive termination with voltage minima

30 points per each case given

(a) A d-c voltage V0 is applied at t = 0 directly to the input terminals on an open-circuited lossless line of length l. Sketch voltage and current waves on the line for the following time intervals: 0 < t < T, T < t < 2T, 2T < t < 3T, 3T < t < 4T, t > 4T where T is time required for the signal to transverse the entire line. (100 points)

Midterm 2 | Problem 5

10 points per each graph (8 total, 80 points) 15 points for t > 4T

5 points for voltage & current values (V1+, I1+, …)

(b) As shown below, a lossless transmission line with a characteristic resistance R0 is terminated at z = l with a capacitor CL. A d-c voltage V0 is applied to the line at z = 0 through a series resistance R0. When the switch is closed at t = 0, plot iL(t), vL(t) and the amplitude of the reflected

wave V1–(t) and v(z, t1) where T < t1 < 2T, here t1 is moment when the wave reflected at the load end is in the process of travelling toward the source. (100 points)

Midterm 2 | Problem 5

- Condition

‣ TR-line terminated with an capacitor load (CL)

‣ Internal (or series) impedance Rg = R0

‣ Voltage initially sent down to TR-line:

V

₁⁺

= R

₀

R

_g

+ R

₀

V

₀

= V

₀

2

- At t = T (= u/l)

‣ V1+ reached at z = l, reflected by capacitor (ΓL)

‣ V1– = ΓLV1+ generated and travel in -z direction

v

_L

( ) t = V

¹⁺

+ V

¹⁻

( ) t ! ( ) 1

- At z = l after reflection (i.e. t ≥ T)

- Equivalently,

i

_L

( ) t = C

L

dv

_L

( ) t

dt ! ( ) 3 i

_L

( ) t = V

¹⁺

R

₀

− V

₁⁻

( ) t

R

₀

→ R

₀

i

_L

( ) t = V

1

+

− V

₁⁻

( ) t ! ( ) 2

- By eliminating V1-(t) by combining eqns. (1) and (2), we have

v

_L

( ) t + R

0

i

_L

( ) t = 2V

1

+

! ( ) 4

- By substituting eqn. (3) into eqn. (4), we have

dv

_L

( ) t

dt + R

₀

v

_L

( ) t

R

₀

C

_L

= 2 V

₁⁺

R

₀

C

_L

, ( t ≥ T )

Midterm 2 | Problem 5

dv

_L

( ) t

dt + R

₀

v

_L

( ) t

R

₀

C

_L

= 2 V

₁⁺

R

₀

C

_L

, ( t ≥ T )

- IVP for first-order differential equation

and

v

_L

( ) T = 0

- By applying Laplacian operator,

sV s ( ) − v

^L

( ) T + V s ( )

R

₀

C

_L

= 2 V

₁⁺

R

₀

C

_L

→ V s ( ) = 2 V

¹⁺

R

₀

L

_L

1

s s ( + 1 R

₀

C

_L

) = 2 V

¹⁺

⎡ ⎣ ⎢ 1 s − s + 1 1 R

⁰

C

^L

⎤ ⎦ ⎥

- By applying inverse Laplacian transform to V(s),

v

_L

( ) t = 2V

¹⁺

⎡ 1 − e

^{−(Rt0−TCL)}

⎣ ⎢

⎢

⎤

⎦ ⎥

⎥ i

_L

( ) t = 2 V

¹⁺

R

₀

1 − e

⁻

t−T

( )

R₀C_L

⎡

⎣ ⎢

⎢

⎤

⎦ ⎥

⎥ V

₁⁻

( ) t = 2 V

¹⁺

1

2 − e

⁻

t−T

( )

R₀C_L

⎡

⎣ ⎢

⎢

⎤

⎦ ⎥

⎥

30 points for correct derivation of vL, iL, V1-

15 points per each graph (4 total, 60 points) 10 points additional if all correct

Midterm 2 | Problem 6

(a) Discuss the working principle and advantage / disadvantage (if any) of quarter-wave transformer, single-stub and double-stub connections to the transmission line for impedance matching. (60 points)

• Quarter-wave transformer (20 points)

- Working principle [15 points]

‣ TR-line of characteristic impedance R0’ serially extended by λ/4 to the main line of R0 and terminated with load RL

‣ For impedance matching, R0’ should satisfy the condition:

- Disadvantage [5 points]

‣Cannot be used for complex-valued (or arbitrary) load

• Single-stub matching (20 points)

- Working principle [5 points]

‣ Open- or short-circuited line section connected in parallel with main TR-line at an appropriate distance from the load

‣ Such extended line has same characteristic impedance (R0) as the main line

‣ For impedance matching, the following condition should be satisfied:

- Advantage [5 points]

‣ It can achieve impedance matching for complex-valued load - Disadvantage [10 points]

‣ Distance of stub from load should change as frequency of signal varies, which is difficult from a mechanical perspective

′

R

₀

= R

_L

R

₀

y

_s

! − jb

_B

! ( ) 1 y

_B

= 1 + jb

_B

! ( ) 2

⎧ ⎨

⎪

⎩⎪

Midterm 2 | Problem 6

• Double-stub matching (20 points)

- Working principle [10 points]

‣ Two short-circuited stubs attached at fixed locations and apart by d0. d0 can be arbitrarily chosen.

‣ Same condition should be satisfied as that for single-stub matching

- Advantage [10 points]

‣ Only need to adjust their lengths lA and lB for matching with ZL without changing their locations

y

_s

! − jb

_B

! ( ) 1 y

_B

= 1 + jb

_B

! ( ) 2

⎧ ⎨

⎪

⎩⎪

As long as the statements marked blue included, points will be given.

Midterm 2 | Problem 6

(b) SWR of a 50 (Ω)-lossless transmission line is measured to be 3 for signal of wavelength = 0.4 (m). The first voltage minimum appears at zm’ = 0.05 (m) away from the load end. In that case, find Γ, ZL, lm and Rm by using Smith Chart. (120 points) [Also, briefly describe the key steps in

using the Smith Chart for this problem].

• Procedure

(1) On positive real-axis, PM represents r = S = 3.0 (= RL/R0) (2) Then, we have circle of radius |Γ| = 0.5 (θΓ yet unknown)

(3) Intersection between negative real-axis and the circle (4) : Pm [Γ < 0 → RL < R0] → Voltage minima at z’ = 0

(see slide 13-2)

(5) To find load impedance, move from Pm along perimeter by z’m/λ = 0.05/0.4 = 0.125 [in the CCW direction.]

(6) PL represents reflection coefficient → Γ = –j0.5 (7) At PL, Read r = 0.6, x = 0.8 → zL = 0.6 + j0.8 (8) Thus, ZL = R0 · zL = 30 – j40 (Ω)

∵ Γ = R

_L

− R

₀

R

_L

+ R

₀

= 0.5

⎛

⎝⎜

⎞

⎠⎟

If Smith Chart is well used according to above procedures, full mark given.

Midterm 2 | Problem 6

(c) A 50 (Ω)-lossless transmission line is connected to a load impedance of ZL = 35 – j47.5 (Ω). Find the position and length of a short-circuited stub required to match the line by using Smith Chart. (120 points) [Also, briefly describe the key steps in using the Smith Chart for this

problem].

- Find normalized zL (=ZL/R0 = 0.7 – j0.95) on impedance chart.

(1) Rotate zL by 180º to convert it to yL (Point PL). The chart now becomes admittance chart. We start from here.

(2) Draw |Γ|-circle passing through PL.

(3) Find two intersecting points between |Γ|-circle and (g = 1)-circle (PB1 and PB2) that yield yB1 = 1 + j1.2 and yB2 = 1 – j1.2 (satisfying condition 1)

(4) Determine d1 and d2 from angles between [OPL and OPB1] and between [OPL

and OPB2] in CW direction.

(5) Read the angle values for ys1 = –j1.2 and ys2 = j1.2 (at points Ps1 and Ps2)

(Satisfying condition 2). Determine lB1 and lB2 from angles between [OPsc and OPs1] and between [OPsc and OPs2].

d

₁

= ( 0.168 − 0.109 ) λ = 0.059 λ

d

₂

= ( 0.332 − 0.109 ) λ = 0.223 λ

⎧ ⎨

⎪

⎩⎪

l

_B1

= ( 0.361 − 0.250 ) λ = 0.111 λ

l

_B2

= ( 0.139 + 0.250 ) λ = 0.389 λ

⎧ ⎨

⎪

⎩⎪

^{45 points}

45 points 30 points