Page 1 Solution Manual for Radio Frequency
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
RADIO FREQUENCY
INTERGRATED CIRCUITS
SYSTEMS SOLUTION
MANUAL BY HOOMAN
DARABI LATEST UPDATE.
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
,Page 2 Solution Manual for Radio Frequency
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
Radio Frequency Integrated
Circuits and Systems
Solution Manual
Hooman Darabi
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
,Page 3 Solution Manual for Radio Frequency
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
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.
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
, Page 4 Solution Manual for Radio Frequency
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
1 Chapter One
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
𝑎𝑟 𝐷 = 𝜌𝑆
𝑟2
Assuming a potential of 𝑉0 between the inner and outer surfaces, we have:
𝑎 2 1 1
𝑉0 = − � 1 𝜌𝑆 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎 ( − )
2
𝑏 𝜖
𝑟2 𝜖 𝑎 𝑏
Thus:
𝑄𝑄 𝜌𝑆4𝜋𝑎2
𝐶 =𝑉 = = 4𝜋𝜖
𝜌𝑆 2 1 1 1 1
−
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 𝑏
0
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 = 𝑝𝐹 = 0.55𝑝𝐹.
5
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
capacitance as a function of the parameters shown in the figure.
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
RADIO FREQUENCY
INTERGRATED CIRCUITS
SYSTEMS SOLUTION
MANUAL BY HOOMAN
DARABI LATEST UPDATE.
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
,Page 2 Solution Manual for Radio Frequency
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
Radio Frequency Integrated
Circuits and Systems
Solution Manual
Hooman Darabi
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
,Page 3 Solution Manual for Radio Frequency
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
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.
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
, Page 4 Solution Manual for Radio Frequency
Solution Manual
Integrated
for Circuits
Radio Frequency
and Systems
Integrated
by Hooman
Circuits
Darabi
and Systems by Hooman Darabi .pdf
1 Chapter One
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
𝑎𝑟 𝐷 = 𝜌𝑆
𝑟2
Assuming a potential of 𝑉0 between the inner and outer surfaces, we have:
𝑎 2 1 1
𝑉0 = − � 1 𝜌𝑆 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎 ( − )
2
𝑏 𝜖
𝑟2 𝜖 𝑎 𝑏
Thus:
𝑄𝑄 𝜌𝑆4𝜋𝑎2
𝐶 =𝑉 = = 4𝜋𝜖
𝜌𝑆 2 1 1 1 1
−
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 𝑏
0
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 = 𝑝𝐹 = 0.55𝑝𝐹.
5
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
capacitance as a function of the parameters shown in the figure.
-2026 Solution Manual for Radio Frequency
Solution
Integrated
ManualCircuits
for Radio
and
Frequency
Systems Integrated
by HoomanCircuits
Darabi and Systems by Hooman Darabi
