Solution Manual for Radio Frequency Integrated Circuits and Systems 2nd Edition by Hooman Darabi , ISBN: 9781107194755 |All Chapters Verified| Guide A+

,RadioFrequencyIntegrated
Circuits and Systems
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SolutionManual

Hooman Darabi
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,Solutions to Problem Sets p o p o p o




The selected solutions to all
po po p o po p o 12 chapters problem sets are presented in this
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manual. The problem sets
p o p o po p o p o depict examples of practical applications of the
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concepts described in the
p o p o p o p o po book, more detailed analysis of some of the
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ideas, or in some cases
p o p o p o p o p o p o present a new concept.
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Note that selected problems have been
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, 1 Chapter One po



1. Using spherical coordinates, find the capacitance formed by two
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concentric spherical conducting shells of radius a, and b. What is
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the capacitance of a metallic marble with a diameter of 1cm in free
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space? Hint: let 𝑏 → ∞, thus, 𝐶
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= 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
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Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆.
p o po p o p o p o p o p o p o p o po p o



The outer surface charge density is negative, and proportionally smaller (by
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(𝑎/𝑏)2) to keep the total charge the same.
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-
+

+S + -
- + a
b
+
-

From Gauss’s law: po po




ф𝐷 ⋅ 𝑑𝑆 po p o p o = 𝑄𝑄 po p o = +𝜌𝑆4𝜋𝑎2
po

𝑆
Thus, inside the sphere (𝑎
po po p o p o


p o≤ 𝑟 p o p o


≤ 𝑏): p o
𝑎2
𝐷 = 𝜌𝑆 𝑎𝑟 p o
𝑟2 o uter
p o p o p o
p o

Assuming a potential of 𝑉0 betw𝑎 e 1e n the 𝑎i2nn er and
po po
𝜌 s2ur f a1ces, w1e have:
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𝑉 = − 𝜌 𝑑𝑟 𝑆 𝑎 p o po p o p o p o

= p o


0 𝑆 ( − ) p o p o p o

2
𝑏 𝑟 𝜖 𝑎 𝑏
Thus: 𝜖
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄
𝐶 = 𝑉 = 𝜌 1 1 1 p o 1 p o p o po ponn po p o


−
p o
𝑆
𝜖 𝑎 (𝑎 − ) 𝑎
0 2
𝑏 p o
p o



𝑏 p o

1
In the case of a metallic marble, 𝑏
po p o → ∞,
p o 𝑎.
p o po
= p o p o p o p o p o


and hence: 𝐶
p o p o p o Letting × po
p o n pon p o

36𝜋
= 4𝜋𝜀𝜀0 p o 𝜀𝜀0
−9 5
10 , and 𝑎 = 0.5𝑐𝑚,
po p o
𝑝𝐹
p o
= 0.55𝑝𝐹.
p o p o
p o p o p o p o

9
it yields
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2. Consider the parallel plate capacitor containing two different dielectrics. Find
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