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Exam (elaborations)

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

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

Institution
Radio Frequency Integrated Circuits And Systems
Course
Radio Frequency Integrated Circuits And Systems











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Written for

Institution
Radio Frequency Integrated Circuits And Systems
Course
Radio Frequency Integrated Circuits And Systems

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Uploaded on
September 13, 2025
Number of pages
312
Written in
2025/2026
Type
Exam (elaborations)
Contains
Questions & answers

Subjects

Content preview

,RadioFrequencyIntegrated
Circuits and Systems
bb bb bb




SolutionManual

Hooman Darabi
bb

,Solutions to Problem Sets b b b b b b




The selected solutions to all
bb bb b b bb b b 12 chapters problem sets are presented in this
b b b b b b b b bb bb bb



manual. The problem sets
b b b b bb b b b b depict examples of practical applications of the
b b b b b b b b b b b b



concepts described in the
b b b b b b b b bb book, more detailed analysis of some of the
b b b b b b b b b b b b b b


ideas, or in some cases
b b b b b b b b b b b b present a new concept.
b b b b bb




Note that selected problems have been
bb b b b b b b bb b b given answers already in the book.
b b b b bb b b bb

, 1 Chapter One bb



1. Using spherical coordinates, find the capacitance formed by two
b b b b b b b b b b b b b b b b



concentric spherical conducting shells of radius a, and b. What is
b b bb b b bb b b b b b b b b b b b b b b



the capacitance of a metallic marble with a diameter of 1cm in free
b b b b b b b b bb bb bb bb b b b b b b b b b b



space? Hint: let 𝑏 → ∞, thus, 𝐶
b b b b b b b b b b b b b b b b



= 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
b b b b b b




Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆.
b b bb b b b b b b b b b b b b b b bb b b



The outer surface charge density is negative, and proportionally smaller (by
b b b b bb bb bb bb bb bb bb bb bb



(𝑎/𝑏)2) to keep the total charge the same.
bb bb bb bb b b b b b b b b




-
+

+S + -
- + a
b
+
-

From Gauss’s law: bb bb




ф𝐷 ⋅ 𝑑𝑆 bb b b b b = 𝑄𝑄 bb b b = +𝜌𝑆4𝜋𝑎2
bb

𝑆
Thus, inside the sphere (𝑎
bb bb b b b b


b b≤ 𝑟 b b b b


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

Assuming a potential of 𝑉0 betw𝑎 e 1e n the 𝑎i 2nn er and
bb bb
𝜌 s2ur f a1ces, w1e have:
bb bb bb bbn nbb bb n b bn bb bb bb n bb n b b bb bb

𝑉 = − 𝜌 𝑑𝑟 𝑆 𝑎 b b bb b b b b b b

= b b


0 𝑆 ( − ) b b b b b b

2
𝑏 𝑟 𝜖 𝑎 𝑏
Thus: 𝜖
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄
𝐶 = 𝑉 = 𝜌 1 1 1 b b 1 b b b b bb bbnn bb b b



b b
𝑆
𝜖 𝑎 (𝑎 − ) 𝑎
0 2
𝑏 b b
b b



𝑏 b b

1
In the case of a metallic marble, 𝑏
bb b b → ∞,
b b 𝑎.
b b bb
= b b b b b b b b b b


and hence: 𝐶
b b b b b b Letting × bb
b b n bbn b b

36𝜋
= 4𝜋𝜀𝜀0 b b 𝜀𝜀0
−9 5
10 , and 𝑎 = 0.5𝑐𝑚,
bb b b
𝑝𝐹
b b
= 0.55𝑝𝐹.
b b b b
b b b b b b b b

9
it yields
b b b b




2. Consider the parallel plate capacitor containing two different dielectrics. Find
bb bb bb bb bb bb bb bb bb

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