RADIO FREQUENCY INTER
gc gc
GRATED CIRCUITS SYSTE
gc gc
MS SOLUTION MANUAL BY
gc gc gc gc
HOOMAN DARABI LATEST
gc gc gc
UPDATE.
,RadioFrequencyIntegrated
gc gc gc
Circuits and Systems
gc gc
SolutionManual
gc
Hooman Darabi
gc
,Solutions to Problem Sets gc gc gc
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem se
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
ts depict examples of practical applications of the concepts described in the book, more detailed
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
analysis of some of the ideas, or in some cases present a new concept.
gc gc gc gc gc gc gc gc gc gc gc gc gc
Note that selected problems have been given answers already in the book.
gc gc gc gc gc gc gc gc gc gc gc
, 1 Chapter One gc
1. Using spherical coordinates, find the capacitance formed by two concentric spherical c
gc gc gc gc gc gc gc gc gc gc gc
onducting shells of radius a, and b. What is the capacitance of a metallic marble with a di
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
ameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
gc gc gc gc gc gc gc gc gc gc gc gc g c gc gc gc
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge t
gc gc gc gc gc gc gc gc gc gc gc gc gc gc
he same.
gc
-
+
+S - + a + -
b
+
-
From Gauss’s law:
gc gc
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
gc gc gc gc gc gc
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏)
gc gc gc gc g c gc gc gc
: 𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟
𝑟
gc gc g c
gc
Assuming a potential of 𝑉0 between𝑎 the inner
𝑎 2 1
2 and outer surfaces, we1have:
𝑑𝑟 = 𝜌𝑆 𝑎
gc
1 gc gc gc gc ggc
c gc g c gc gc gc gc gc
𝑉 =−
g c
𝜌 gc
g c
g c gc gc g c gc g c
0 𝑆 ( − ) gc gcg c g c g
c
𝑏 g c g c 𝑟2 𝜖 𝑎 𝑏
Thus: 𝜖
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 g c g c
𝐶 =𝑉 = 1 1 1 1
𝜌
g c gc gc
g c
g c
𝑆
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 −𝑏
0 2 gc
gc
gc gc
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀 𝑎. Letting 𝜀𝜀0 =
gc gc gc gc gc gc gc gc gc gc gc gc g c gc gc gc
g c g c g c g c ×
36𝜋
0
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
gc
g c gc
gc gc g c gc gc gc g c g c g c
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
gc gc gc gc gc gc gc gc gc gc gc gc
capacitance as a function of the parameters shown in the figure.
gc gc gc gc gc gc gc gc gc gc
gc gc
GRATED CIRCUITS SYSTE
gc gc
MS SOLUTION MANUAL BY
gc gc gc gc
HOOMAN DARABI LATEST
gc gc gc
UPDATE.
,RadioFrequencyIntegrated
gc gc gc
Circuits and Systems
gc gc
SolutionManual
gc
Hooman Darabi
gc
,Solutions to Problem Sets gc gc gc
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem se
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
ts depict examples of practical applications of the concepts described in the book, more detailed
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
analysis of some of the ideas, or in some cases present a new concept.
gc gc gc gc gc gc gc gc gc gc gc gc gc
Note that selected problems have been given answers already in the book.
gc gc gc gc gc gc gc gc gc gc gc
, 1 Chapter One gc
1. Using spherical coordinates, find the capacitance formed by two concentric spherical c
gc gc gc gc gc gc gc gc gc gc gc
onducting shells of radius a, and b. What is the capacitance of a metallic marble with a di
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
ameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
gc gc gc gc gc gc gc gc gc gc gc gc g c gc gc gc
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface
gc gc gc gc gc gc gc gc gc gc gc gc gc gc gc
charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge t
gc gc gc gc gc gc gc gc gc gc gc gc gc gc
he same.
gc
-
+
+S - + a + -
b
+
-
From Gauss’s law:
gc gc
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
gc gc gc gc gc gc
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏)
gc gc gc gc g c gc gc gc
: 𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟
𝑟
gc gc g c
gc
Assuming a potential of 𝑉0 between𝑎 the inner
𝑎 2 1
2 and outer surfaces, we1have:
𝑑𝑟 = 𝜌𝑆 𝑎
gc
1 gc gc gc gc ggc
c gc g c gc gc gc gc gc
𝑉 =−
g c
𝜌 gc
g c
g c gc gc g c gc g c
0 𝑆 ( − ) gc gcg c g c g
c
𝑏 g c g c 𝑟2 𝜖 𝑎 𝑏
Thus: 𝜖
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 g c g c
𝐶 =𝑉 = 1 1 1 1
𝜌
g c gc gc
g c
g c
𝑆
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 −𝑏
0 2 gc
gc
gc gc
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀 𝑎. Letting 𝜀𝜀0 =
gc gc gc gc gc gc gc gc gc gc gc gc g c gc gc gc
g c g c g c g c ×
36𝜋
0
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
gc
g c gc
gc gc g c gc gc gc g c g c g c
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
gc gc gc gc gc gc gc gc gc gc gc gc
capacitance as a function of the parameters shown in the figure.
gc gc gc gc gc gc gc gc gc gc
