EE2025 Engineering Electromagnetics: July-Nov 2019

Tutorial 6: Waveguides
1. A section of a rectangular waveguide of cross-section 2 cm × 1.5 cm is to be used as a delay line
in a radar at 10 GHz. What should be the length of the section to realize a delay of 10 nsec ?

Solution:
Cutoff frequency of T E 10 ,

c 3 × 108
fc = = = 7.5GHz
2a 2 × 2 × 10−2
Cutoff frequency of the next higher order mode T E 01 (Since, a < 2b can’t be T E 20 ),

0 c 3 × 108
fc = = = 10GHz
2b 2 × 1.5 × 10−2
Thus only possible mode of transmission is T E 10
Group velocity
s r
fc 7.5 2
vg = c × 1 − ( )2 = 3 × 108 × 1−( ) = 1.984 × 108 ms−1
fo 10

Length for a delay of 10 ns,

l = vg × 10 × 10−9 = 1.984m = 198.4cm

2. (a) Derive an expression for the conductive loss in a rectangular waveguide for the fundamental
mode.
(b) In a 5 cm × 3 cm waveguide the T E10 mode is propagating at 10 GHz. The total power
carried by the waveguide is 10 W. If the conductivity of the waveguide walls is 2 ×107 f/m, find
the attenuation constant of the waveguide in dB/m.

Solution:

(a) Refer to section 6.8.2 of R.K.Shevgaonkar’s ”Electromagnetic waves”.

(b) Conductive loss in T E10 mode in a rectangular waveguide is given by,

Rs (1 + 2b (fc /f )2 )
αc = p a (1)
ηb 1 − (fc /f )2

q q
ωµ 2π×10×109 ×4π×10−7
Rs = 2σ = 2×2×107
= 0.044 Ω
Considering dielectric as air,
v 3×108
fc = 2a = 2×0.05 = 3 GHz
√
η = 377 × 1 − 0.32 = 359.63Ω
0.06
0.044(1+ 0.05 (0.3)2 )
αc = √
359.63×0.03 1−0.32
= 4.74 × 10−3 N p/m = 0.041 dB/m
3. In an air-filled rectangular waveguide with a = 2.286 cm and b = 1.016 cm, the y component of
the TE mode is given by

Ey = sin(2πx/a) cos(3πy/b) sin(10π × 1010 t − βz)V /m

Find
(a) the operating mode
(b) the propagation constant γ
(c) the intrinsic impedance η
(d) Find the dielectric loss αd of the mode when the waveguide is filled with water( Complex
permittivity of water at 50 GHz is 0 (16.40 − j26.31))

Solution: (a) m = 2, n= 3, so the mode is T E23

(b) γ = jβ s  2
fc
β = β1 1− , f = 50GHz
f
r
1 mπ 2  nπ 2
fc = √ + = 46.19GHz
2π µ a b
β = 400.9rad/m, γ = j400.9m−1

(c) Intrinsic impedance for the TE mode

η1
η=r  2 = 984.6Ω
1 − ffc

(d)
00 σ
 =
ω
σ = 2π × 50 × 109 × 8.85 × 10−12 × 26.31 = 73f/m
ση 0
αd = p
2 1 − (fc /f )2
r
0 µ
η = = 93Ω

αd = 8.865 × 103 N p/m

4. A frequency of 9.5 GHz is used to excite all possible modes in a hollow rectangular waveguide of
dimensions 3 cm × 2 cm. The length of the waveguide is 100 m. Find the difference between time
of arrivals of the fastest mode and the slowest mode.

Solution:
Given, Operating frequency = 9.5 GHz, a=3 cm and b=2 cm
length of the waveguide=100 m
q
cutoff frequency, fc = 2π√1µ0 0 ( mπ 2 + ( nπ )2 = 15 × 109 ( m2 ) + ( n2 )
p
a ) b 9 4
q
2 2 2 2
For, f > fc ⇒ 9.5 × 109 > 15 × 109 ( m9 ) + ( n4 ) ⇒ ( m9 ) + ( n4 ) < 0.4

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 Therefore, possible operating modes are T E10 , T E01 , T E11 and T M11 .
q q
Group velocity of fastest mode (T E11 ) = c 1 − ( ffc )2 = c 1 − ( 9.5
9 2
) = 2.55 × 108 m/s
q q
Group velocity of slowest mode (T E10 ) = c 1 − ( ffc )2 = c 1 − ( 9.5
5 2
) = 0.96 × 108 m/s
100 100
Difference of time of arrival of slowest and fastest mode = ( 0.96×108 ) − ( 2.55×108 ) = 0.649 µs

5. A 240 degree phase shift is produced by a 4 GHz signal when traveling along a dielectric filled
waveguide 3 cm long. If the cutoff frequency of the waveguide when air-filled is 10 GHz, calculate
the relative permittivity of the dielectric.

Solution: Cut-off frequency in air-filled waveguide = 10GHz

Cut-off frequency in dielectric waveguide fc = 10GHz
√
r
◦
Given phase shift βl = 240 and l = 0.03m
β = 400π
9 rad/m
r
√  2
ω r
β= c 1 − ffc
Given f = 4GHz, substituting the values in above equation,
r = 9.0046

6. For a square waveguide, show that attenuation αc is minimum for T E10 mode when f = 2.962fc

Solution: In a rectangular waveguide, for TE10 mode

  2

1 b fc
+
αc = 2R 2 a f
p
s r  , where Rs = πµf /σ
ηb  2
fc
1− f

For square waveguide a = b, we get

 
  2
1 fc
+
2Rs  r2 f
αc = ηa  2

1− ffc

dαc
For the frequency at which αc is minimum, df =0
Solving the equation, we get f 4 − 9fc2 f 2 + 2fc4 = 0
substituting f = kfc , in the above equation we obtain
k 4 − 9k 2 + 2 = 0 ⇒ r2 − 9r + 2 = 0 ; (r = k 2 )
solving the quadratic equation for r we get r = k 2 = 8.772, 0.228 ⇒ k ≈ 2.962, 0.477.
Even though we get two solutions (numerically), k < 1 is not a physically realizable solution
because for the operating frequency f < fc the mode of interest itself is not supported by the
waveguide.
Therefore, for minimum attenuation f = 2.962fc

7. A 4 cm square waveguide is filled with a dielectric with complex permittivity c = 16o (1 − j10−4 )
and is excited with the T M21 mode. If the waveguide operates at 10 % above the cutoff frequency,
calculate attenuation αd . How far can the wave travel down the guide before its magnitude is
reduced by 20 % ?

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Solution:
σ
c = 0 − j00 =  − j (2)
ω
Comparing this with

c = 16o (1 − j10−4 ) = 16o − j16o × 10−4 (3)

σ
 = 16o , = 16o × 104 (4)
ω
√
For T M21 mode, (with c0 = c/ r )
1
c0 m2 n2 2

fc = + 2 = 2.096 GHz (5)
2 a2 b

f = 1.1fc = 2.306 GHz (6)

10−9
σ = 16o ω × 10−4 = 16 × 2π × 4.6123 × 109 × × 10−4 = 2.05 × 10−4 (7)
36π
r
0 µ
η = = 30π (8)


ση 0
αd = p = 0.0320 N p/m (9)
2 1 − (fc /f )2

Eo e−αd z = 0.8Eo → z = 6.97m (10)

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