Planar waveguides are an important type of waveguide that can be used in integrated circuits to connect microwave circuit elements. Examples include strip lines, microstrip lines, coplanar waveguides, and slotted lines. When designing a planar waveguide circuit, its characteristic impedance, dispersion, phase velocity, attenuation, and other parameters must be determined. Commercial software tools can help with designing waveguides to meet specifications for impedance and phase shift. Planar waveguides find applications in microwave integrated circuits, filters, phase shifters, and feeding energy from generators to antennas.

1. Planar waveguides
Planar waveguides are an important subclass of
waveguides (transmission lines)
Planar waveguides can be used in integrated
circuits to connect the various microwave circuit
elements
Examples are shown in the next three slides
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2. Applications of planar
waveguides
Also planar transmission lines can be used to
feed microwave energy from a microwave
generator to an antenna as shown
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3. Applications of planar
waveguides
Planar waveguides can be used to
interconnect the various parts of a
microwave amplifiers
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4. Applications of planar
waveguides
Planar wave guides also can be used to
design filters and phase shifters
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Filter Phase shifter

5. Types of planar wave guides
There are many types of planar waveguides
available. Examples are
 Strip line
 Microstrip line
 Coplanar waveguide
 Slotted lines
These waveguides support TEM, TE and TM
wave propagations
However only TEM will be considered
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6. Planar waveguide design
parameters
When designning a given microwave circuit, it is
desired to know the characteristic impedance,
dispersion curves, phase velocity, phase delay,
capacitance, inductance and the attenuation per unit
length of the waveguide
Computing these parameters can be performed by
solving the Helmholtz equation
However, the computation can be greatly simplified
by the use of commercial microwave design tools
such as CST, AWR microwave office, Agilent ADS,
and HFSS
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7. Strip-line
The strip line is composed from a thin conducting
strip of width 𝑊 centered between two conducting
ground planes of separation 𝑏
The region between the conducting planes is filled
with a dielectric material whose relative permitivity
is 𝜖 𝑟 as shown
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8. Strip line
The strip line can be considered as a planar
coaxial cable
Exact solution for the electric field can be
obtained by a method known as conformal
mapping
However this avoided in practice and an
empirical formulas can be used to design the
strip line for specific characteristic impedance
𝑍0, phase velocity 𝑣 𝑝, propagation constant 𝛽
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9. Formulas for Propagation Constant,
Characteristic Impedance
The phase velocity can be given by
𝑣 𝑝 =
1
𝜇𝜖
=
𝑐
𝜖 𝑟
The propagation constant is given by
𝛽 =
𝜔
𝑣 𝑝
= 𝜔 𝜇0 𝜖0 𝜖 𝑟 = 𝜖 𝑟 𝑘0
The characteristic impedance can be approximated by
𝑍0 =
30𝜋
𝜖 𝑟
𝑏
𝑊𝑒 + 0.441𝑏
Where 𝑊𝑒 is the effective width of the center conductor
which is approximated by
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10. Strip line width for specific 𝑍0
The strip line width that can be used to design a
strip line with a specific characteristic impedance
can be determined from the following equation
𝑥 =
30𝜋
𝜖 𝑟 𝑍0
− 0.441
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11. Attenuation due to conductor
loss
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12. Example
Find the width for a 50 Ω copper stripline
conductor with 𝑏 = 0.32 𝑐𝑚 and 𝜖 𝑟 = 2.2. If the
dielectric loss tangent is 𝑡𝑎𝑛𝛿 = 0.001 and the
operating frequency is 10 GHz, calculate the
attenuation in 𝑑𝐵/𝜆 . Assume a conductor
thickness of 𝑡 = 0.01 mm
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13. Solution
To find the strip line width we can rely on the
equations in slide 10
𝜖 𝑟 𝑍0 = 2.2 50 = 74.2 < 120
Therefore
𝑥 =
30𝜋
𝜖 𝑟
− 0.441 = 0.83
We can use the equation
𝑊
𝑏
= 𝑥 ⇒ 𝑊 = 𝑏𝑥 = 0.32 × 0.83 = 0.266 𝑐𝑚
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14. Solution
The total attenuation is the attenuation due to
the conductor loss plus the attenuation due to
dielectric material, therefore
𝛼 = 𝛼 𝑐 + 𝛼 𝑑
𝛼 𝑑 =
𝑘𝑡𝑎𝑛𝛿
2
But
𝑘 = 𝜔 𝜇𝜖 =
2𝜋𝑓 𝜖 𝑟
𝑐
= 310.6
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15. Solution
Now 𝛼 𝑑 =
310.6×0.001
2
= 0.155 𝑁𝑝/𝑚
The conductor loss can be determined from
The surface resistance for the copper is
𝑅 𝑠
𝜔𝜇
2𝜎
= 0.026 Ω 𝑎𝑡 10 𝐺𝐻𝑧
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16. Solution
𝛼 𝑐 = 2.6 ×
10−3
𝑅 𝑠 𝜖 𝑟 𝑍0 𝐴
30𝜋(𝑏 − 𝑡)
= 0.122 𝑁𝑝/𝑚
The total attenuation is
𝛼 = 𝛼 𝑑 + 𝛼 𝑐 = 0.277
𝑁𝑝
𝑚
𝛼 𝑑𝐵 = 20 log10 𝑒 𝛼 = 2.41 𝑑𝐵/𝑚
The wavelength at 10 𝐺𝐻𝑧 is
𝜆 =
𝑐
𝜖 𝑟 𝑓
= 2.02 𝑐𝑚
The attenuation in terms of wavelength is
𝛼 𝑑𝐵 = 2.41 × 0.0202 = 0.049 𝑑𝐵/𝜆
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17. Microstrip line
The microstrip line is one of the most popular
type of transmission lines
The microstrip line is composed from a signal
conductor of width 𝑊 and ground plane
separated by a dielectric material as shown
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18. Phase velocity, propagation
constant
The phase velocity on microstrip line is given by
𝑣 𝑝 =
𝑐
𝜖 𝑒
The propagation constant is given by
𝛽 = 𝜔 𝜇0 𝜖0 𝜖 𝑒 = 𝑘0 𝜖 𝑒
Where 𝜖 𝑒 is the effective dielectric constant 1 <
𝜖 𝑒 < 𝜖 𝑟
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19. Effective dielectric constant 𝜖 𝑒
The effective dielectric constant can be
approximated by
𝜖 𝑒 =
𝜖 𝑟 + 1
2
+
𝜖 𝑟 − 1
2
1
1 + 12𝑑/𝑊
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20. Characteristic impedance
analysis formula
If the dimensions of the microstrip are known, then
the characteristic impedance can be determined
from
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21. Characteristic impedance
Synthesis formula
If the microstrip is designed to have a specific
characteristic impedance, then the ratio of its width
𝑊 to the dielectric thickness 𝑑 can be determined
form
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22. Microstrip attenuation
The attenuations for quasi TEM microstrip are
determined from the following two equations
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23. Example microstrip design
Design a microstrip line on a 0.5 mm alumina
substrate ( 𝜖 𝑟 = 9.9, 𝑡𝑎𝑛 𝛿 = 0.001 ) for a 50 Ω
characteristic impedance.
Find the length of this line required to produce a
phase delay of 270° at 10 GHz, and compute the
total loss on this line, assuming copper
conductors.
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24. Solution
If we assume that
𝑊
𝑑
< 2, the width 𝑊 can be
determined from
𝑊
𝑑
=
8𝑒 𝐴
𝑒2𝐴 − 2
𝐴 =
𝑍0
60
𝜖 𝑟 + 1
2
+
𝜖 𝑟 − 1
𝜖 𝑟 − 1
0.23 +
0.11
𝜖 𝑟
= 2.142
𝑊/𝑑 = 0.9654 which is less than 2
The required width of microstrip line is 𝑊 =
0.9654𝑑 = 0.483 𝑚𝑚
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25. Solution
The length of the line that can produce a phase
shift of 𝜙 = 270° can be determined from the
electrical length
𝜖 𝑒 = 6.665
𝜙 = 𝛽𝑙 = 𝜖 𝑒 𝑘0 𝑙 = 270
𝑘0 =
2𝜋𝑓
𝑐
= 209.4 𝑚−1
𝑙 =
270
𝜋
180
𝜖 𝑒 𝑘0
= 8.72 𝑚𝑚
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26. Solution
By solving for 𝛼 𝑑 and 𝛼 𝑐 , we have a total
attenuation of 𝛼 = 𝛼 𝑑 + 𝛼 𝑐 = 0.022 + 0.094 =
0.116 𝑑𝐵𝑚/𝑐𝑚
The attenuation within the length of the phase
shifter is given by
𝑎𝑡𝑡𝑒𝑛𝑢𝑎𝑡𝑖𝑜𝑛 = 𝛼𝑑𝐵𝑐𝑚 × 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒
𝑎𝑡𝑡𝑒𝑛𝑢𝑎𝑡𝑖𝑜𝑛 = 0.116 × 0.872 = 0.101 𝑑𝐵
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27. Coplanar waveguide
The coplanar waveguide is composed
from a central conductor with two ground
planes as shown below
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28. Coplanar waveguide
The impedance of the coplanar waveguide can
be determined the width of the central conductor
𝑊 and the spacing from the ground planes 𝑆
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29. Commercial software
The dimensions of planar wave guide can
be adjusted for specific 𝑍0, 𝑣 𝑝, 𝛽, 𝑎𝑛𝑑 𝜙 by
using CAD tool such on line tools or
Linecalc in ADS
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30. Linecalc
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You can find the line calculator program
under tools, then LineCalc

31. Starting the line calculator
If you press the start LineCalc item, you may obtain a
window as shown below
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32. Entering the dielectric parameters
You can enter the
dielectric (substrate)
parameters in the
following section of
the LineCalc window
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33. Computing
𝑍0, 𝑝ℎ𝑎𝑠𝑒 𝑠ℎ𝑖𝑓𝑡 𝜙 , 𝜖 𝑒 (Analysis)
Enter the width and length of the transmission line and
click on the analyze window, the LineCalc program will
compute 𝑍0, 𝜙, 𝜖 𝑒 as illustrated by
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34. Computing the 𝑊 and 𝐿 for a
specific 𝑍0 and 𝜙 (Synthesis)
Enter the value of the desired 𝑍0 and the phase shift in
the section provided, press synthesize, the program will
compute the width and length of the transmission line
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35. Comparison of Common Transmission
Lines and Waveguides
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36. 36Falah Mohammed
Microwave Frequency Bands (taken from wikipedia)
Letter Designation Frequency range
L band 1 to 2 GHz
S band 2 to 4 GHz
C band 4 to 8 GHz
X band 8 to 12 GHz
Ku band 12 to 18 GHz
K band 18 to 26.5 GHz
Ka band 26.5 to 40 GHz
Q band 33 to 50 GHz
U band 40 to 60 GHz
V band 50 to 75 GHz
E band 60 to 90 GHz
W band 75 to 110 GHz
F band 90 to 140 GHz
D band 110 to 170 GHz