Radio Frequency Integrated Circuits and Systems – Hooman Darabi | Complete Solution Manual (Chapters 1–12) with Verified Answers and Rationales

Solution Manual for Radio Frequency Integrated Circuits And Systems

Solution Manual

By Hooman Darabi

All Chapters (1-12) Q&A Verified With Rationales| A+ Assuared

ISBN-13: 978-0521190794




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,Solutions to Problem Sets



The selected solutions to all 12 chapters problem sets are presented in this manual. The problem
sets depict examples of practical applications of the concepts described in the book, more detailed
analysis of some of the ideas, or in some cases present a new concept.

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




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,Chapter One: RF Components
1. Using spherical coordinates, find the capacitance formed by two concentric spherical
conducting shells of radius a, and b. What is the capacitance of a metallic marble with a
diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.

Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface
charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge
the same.


-
+

+S - + a + -

b
+
-


From Gauss’s law:
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏):
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟
𝑟
Assuming a potential of 𝑉0 between the inner and outer surfaces, we have:
𝑎 2 𝜌 1 1
𝑉0 = − � 𝜌𝑆 𝑎 𝑑𝑟 = 𝑆 𝑎2( − )
1

𝑏 𝑟2 𝜖 𝑎 𝑏
Thus: 𝜖
𝜌𝑆4𝜋𝑎2 4𝜋𝜖
𝑄𝑄
𝐶=𝑉 = =
𝜌 1 1 1 1
0 𝑆 −
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 𝑏
2
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀0 𝑎. Letting 𝜀𝜀0 = ×
36𝜋
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 5 𝑝𝐹 = 0.55𝑝𝐹.
9


2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
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, capacitance as a function of the parameters shown in the figure.




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