Module 2 Assignment
Problem 1:
A 75 coaxial line has a current 𝑖(𝑡, 𝑧) = 1.8 𝑐𝑜𝑠(3.77 × 109 𝑡 − 18.13𝑧) 𝑚𝐴.
Determine
a.
b.
c.
d.
e.
f.
the frequency,
the phase velocity,
the wavelength,
the relative permittivity of the line,
the phasor form of the current, and
the time domain voltage on the line.
Solution:
𝜔
a. Using the general form 𝑦(𝑡, 𝑧) = 𝐴0 cos(𝜔𝑡 − βz). 𝜔 = 2𝜋𝑓; 𝑓 = 2𝜋
3.77 ∗ 109
𝑓=
≈ 600 𝑀𝐻𝑧
2𝜋
𝜔
b. Phase velocity 𝑣𝑝 = 𝛽 =
2𝜋
3.77∗109
18.13
≈ 2.08 ∗ 108 𝑚\𝑠
2𝜋
c. Wavelength 𝜆 = 𝛽 = 18.13 = 0.347𝑚
d. Relative permittivity 𝑣𝑝 =
𝑐
𝑐2
(3∗108 )
2
= 𝜖𝑟 = 𝑣 𝜇 = (208∗106 ∗1)2 = 2.08
√ 𝜖𝑟 𝜇𝑟
−𝑗18.13𝑧
𝑝 𝑟
e. Phase form of current 1.8𝑒
𝑚𝐴
f. Time domain voltage 𝑉 = 𝐼𝑅 = 75Ω ∗ 1.8 𝑐𝑜𝑠(3.77 × 109 𝑡 – 18.13𝑧)𝑚𝐴 =
0.135𝑐𝑜𝑠(3.77 × 109 𝑡 – 18.13𝑧)𝑉
Problem 2:
µ𝐻
A transmission line has the following per-unit-length parameters: 𝐿 = 0.5 𝑚 , 𝐶 =
𝑝𝐹
Ω
𝑆
200 𝑚 , 𝑅 = 4.0 𝑚 , 𝑎𝑛𝑑 𝐺 = 0.02 𝑚.
1. Calculate the
a. propagation constant and
b. characteristic impedance of this line at 800 MHz.
2. If the line is 30 cm long, what is the attenuation in dB?
3. Recalculate these quantities in the absence of loss (R = G = 0).
Solution
1. 𝑓 = 800 𝑀𝐻𝑧
a. 𝛾 = 𝛼 + 𝑗𝛽 = √(𝑅 + 𝑗𝜔𝐿)(𝐺 + 𝑗𝜔𝐶) = √(4 + 𝑗800𝜋)(0.02 + 𝑗0.32𝜋)
𝛾 = 0.54
(𝑅+𝑗𝜔𝐿)
𝑁𝑝
𝑟𝑎𝑑𝑠
+ 𝑗50.27
𝑚
𝑚
4+𝑗800𝜋
b. 𝑍0 = √ 𝐺+𝑗𝜔𝐶 = √0.02+𝑗0.32𝜋 = 49.99 + 𝑗0.46 Ω
𝑁𝑝
2. 𝛾 = 𝛼 + 𝑗𝛽; 𝛼 = 0.54 𝑚
𝛼30𝑐𝑚 = 0.54 ∗ 0.3 = 0.162
𝛼30𝑐𝑚 𝑑𝐵 = 𝛼30𝑐𝑚 ∗ 8.686 = 1.407𝑑𝐵
𝑟𝑎𝑑𝑠
3. 𝛾 = 𝑗𝛽 = 𝑗𝜔√𝐿𝐶 = 𝑗2𝜋(800 ∗ 106 )√0.5 ∗ 10−6 ∗ 200 ∗ 1−12 = 𝑗16𝜋 ≈ 𝑗50.27 𝑚
0.5∗10−6
𝐿
𝑍0 = √𝐶 = √(200∗10−12 ) = 50Ω
Problem 3:
RG-402U semirigid coaxial cable has an inner conductor diameter of 0.91 mm and a
dielectric diameter (equal to the inner diameter of the outer conductor) of 3.02 mm. Both
conductors are copper, and the dielectric material is Teflon.
1. Compute the R, L, G, and C parameters of this line at 1 GHz, and use these results to
2. find the characteristic impedance and attenuation of the line at 1 GHz.
3. Compare your results to the manufacturer’s specifications of 50Ω and 0.43 dB/m,
and discuss reasons for the difference.
Solution:
𝑆
1. Compute R, L, G, C @ 1 GHz. 𝜎 = 5.83 ∗ 107 𝑀 , 𝑅𝑠 = 0.00824, 𝑎 =
0.00302
2
0.00091
2
,𝑏 =
, 𝑡𝑎𝑛𝛿 = 0.0004, 𝜖 ′′ = 𝜖 ′ ∗ tan 𝛿 , 𝜖 ′ = 𝜖𝑟 𝜖0
𝑅
1
1
a. 𝑅 = 2𝜋𝑠 (𝑎 + 𝑏) =
𝜇
𝑏
b. 𝐿 = 2𝜋 ln 𝑎 =
c. 𝐺 =
d. 𝐶 =
2𝜋𝜔𝜖 ′′
𝑏
ln( )
𝑎
2𝜋𝜖 ′
𝑏
ln( )
𝑎
=
=
0.00824
2𝜋
4𝜋∗10−7
2𝜋
ln (
1
1
2
0.00302
2
0.00091
2
2
Ω
( 0.00091 + 0.00302 ) = 3.75 𝑚
𝑛𝐻
) = 239.91 𝑚
2𝜋∗2𝜋∗1∗109 ∗2.08∗8.854∗10−12 ∗0.0004
0.00302
2
)
ln( 0.00091
2
2𝜋∗2.08∗8.854∗10−12
0.00302
2
)
ln( 0.00091
2
= 0.24
𝑚𝑆
𝑚
𝑝𝐹
= 96.46 𝑚
2. 𝛾 = 𝛼 + 𝑗𝛽 = √(𝑅 + 𝑗𝜔𝐿)(𝐺 + 𝑗𝜔𝐶) = √(3.75 + 𝑗479.82𝜋)(0.00024 + 𝑗0.61) =
𝑁𝑝
𝑑𝐵
0.044 + 𝑗30.32, 𝛼 = 0.044 𝑚 , 𝛼𝑑𝐵 = 0.044 ∗ 8.686 = 0.38 𝑚
(𝑅 + 𝑗𝜔𝐿)
3.75 + 𝑗479.82𝜋
𝑍0 = √
=√
= 49.71 − 𝑗0.05 Ω
𝐺 + 𝑗𝜔𝐶
0.00024 + 𝑗0.61
3. My calculated impedance and attenuation came in less than what the manufacturer
specified. Some reasons for this are:
a. My calculations assume ideal materials, whereas real materials will have
different conductivities, loss tangents and dielectric constants
b. The manufacturer may have also provided data based on the conditions that
existed during the manufacturing process, which I do not account for.
Conditions like, temperature, humidity and so on.