,RadioFrequencyIntegrated
n n
Circuits and Systems
nn nn nn
Solution Manual
nn
Hooman Darabi
nn
,Solutions to Problem Sets nn nn nn
The selected solutions to all 12 chapters problem sets are presented in this manual. The
nn nn nn nn nn nn nn nn nn nn nn nn nn nn
problem sets depict examples of practical applications of the concepts described in the
nn nn nn nn nn nn nn nn nn nn nn nn nn
book, more detailed analysis of some of the ideas, or in some cases present a new
nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn
concept.
nn
Note that selected problems have been given answers already in the book.
nn nn nn nn nn nn nn nn nn nn nn
, 1 Chapter One nn
1. Using spherical coordinates, find the capacitance formed by two concentric
nn nn nn nn nn nn nn nn nn
spherical conducting shells of radius a, and b. What is the capacitance of a
nn nn nn nn nn nn nn nn nn nn nn nn nn nn
metallic marble with a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶
nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn
= 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
n n nn nn nn
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer
nn nn nn nn nn nn nn nn nn nn nn nn nn
surface charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the
nn nn nn nn nn nn nn nn nn nn nn nn nn
total charge the same.
nn nn nn nn
-
+
+S - + a + -
b
+
-
From Gauss’s law:nn nn
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
nn nn nn nn nn nn
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟
nn nn nn nn nn nn
≤ 𝑏):
nn nn
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 nn nn
𝑟 nn
nn
Assuming a potential of 𝑉0 between
𝑎 1 the inner
2 and 2 1
outer surfaces, 1 have:
we
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
nn nn nn nn nn nn nn nn n n nn nn nn nn nn
= −
n n
𝑉 nn
nn
n n nn nn n n nn n n
0 𝑆 ( − ) nn nn n n n n n
n
2
𝑏 𝑟 𝜖 𝑎 𝑏 nn
Thus: 𝜖 nn
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 n n n n
𝐶 = 𝑉 =
𝜌 𝑆 21 1 1 1 nn nn nn
n n
n n
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 nn nn
nn nn
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 𝑎. Letting =
nn nn nn nn nn nn nn nn nn nn nn nn nn
n n n n nn
nn= 4𝜋𝜀𝜀0 nn 𝜀𝜀0 × nn
nn
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝑝𝐹 = 0.55𝑝𝐹. nn
nn nn 𝐶 = nn nn nn nn n n nn n n
nn nn
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the
nn nn nn nn nn nn nn nn nn nn
total capacitance as a function of the parameters shown in the figure.
nn nn nn nn nn nn nn nn nn nn nn nn
n n
Circuits and Systems
nn nn nn
Solution Manual
nn
Hooman Darabi
nn
,Solutions to Problem Sets nn nn nn
The selected solutions to all 12 chapters problem sets are presented in this manual. The
nn nn nn nn nn nn nn nn nn nn nn nn nn nn
problem sets depict examples of practical applications of the concepts described in the
nn nn nn nn nn nn nn nn nn nn nn nn nn
book, more detailed analysis of some of the ideas, or in some cases present a new
nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn
concept.
nn
Note that selected problems have been given answers already in the book.
nn nn nn nn nn nn nn nn nn nn nn
, 1 Chapter One nn
1. Using spherical coordinates, find the capacitance formed by two concentric
nn nn nn nn nn nn nn nn nn
spherical conducting shells of radius a, and b. What is the capacitance of a
nn nn nn nn nn nn nn nn nn nn nn nn nn nn
metallic marble with a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶
nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn nn
= 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
n n nn nn nn
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer
nn nn nn nn nn nn nn nn nn nn nn nn nn
surface charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the
nn nn nn nn nn nn nn nn nn nn nn nn nn
total charge the same.
nn nn nn nn
-
+
+S - + a + -
b
+
-
From Gauss’s law:nn nn
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
nn nn nn nn nn nn
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟
nn nn nn nn nn nn
≤ 𝑏):
nn nn
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 nn nn
𝑟 nn
nn
Assuming a potential of 𝑉0 between
𝑎 1 the inner
2 and 2 1
outer surfaces, 1 have:
we
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
nn nn nn nn nn nn nn nn n n nn nn nn nn nn
= −
n n
𝑉 nn
nn
n n nn nn n n nn n n
0 𝑆 ( − ) nn nn n n n n n
n
2
𝑏 𝑟 𝜖 𝑎 𝑏 nn
Thus: 𝜖 nn
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 n n n n
𝐶 = 𝑉 =
𝜌 𝑆 21 1 1 1 nn nn nn
n n
n n
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 nn nn
nn nn
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 𝑎. Letting =
nn nn nn nn nn nn nn nn nn nn nn nn nn
n n n n nn
nn= 4𝜋𝜀𝜀0 nn 𝜀𝜀0 × nn
nn
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝑝𝐹 = 0.55𝑝𝐹. nn
nn nn 𝐶 = nn nn nn nn n n nn n n
nn nn
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the
nn nn nn nn nn nn nn nn nn nn
total capacitance as a function of the parameters shown in the figure.
nn nn nn nn nn nn nn nn nn nn nn nn
