TESTBANKS BY TESTBANKSNERD
Solution Manual Radio Frequency Integrated Circuits and Systems
by Hooman Darabi
,TESTBANKS BY TESTBANKSNERD
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.
,TESTBANKS BY TESTBANKSNERD
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
𝐷 = 𝜌𝑆
𝑎𝑟
𝑟2
Assuming a potential of 𝑉0 between the inner and outer surfaces, we have:
𝑎 2 1 1
𝑉0 = − � 1 𝜌𝑆 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎2( − )
𝑏 𝜖
𝑟2 𝜖 𝑎 𝑏
Thus:
𝑄𝑄 𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝐶=𝑉 =
𝜌𝑆 2 1 1 1 1
𝑎 ( − ) 𝑎−𝑏
0
𝜖 𝑎 𝑏
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 𝑎. Letting 𝜀𝜀0 = ×
36𝜋
4𝜋𝜀𝜀0
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 capacitance as a function of the parameters shown in the figure.
, TESTBANKS BY TESTBANKSNERD
Area: A
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal
component of the electric flux density has to be equal in each dielectric. That is:
𝐷1 = 𝐷𝟐𝟐
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate, the
electric field (or flux has a component only in z direction, and we have:
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
𝑑1+𝑑2 𝑑2
−𝜌𝑆 𝑑1+𝑑2 −𝜌
𝑆 𝜌𝑆 𝜌𝑆
𝑉0 = − � 𝐸. 𝑑𝑧 = − � 𝜖 𝑑𝑧 − � 𝜖 𝑑𝑧 = 𝜖 𝑑1 + 𝜖 𝑑2
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
𝑄𝑄 𝐴
𝐶= 𝑉 =𝑑 𝑑
0 1+ 2
𝜖1 𝜖 2
which is analogous to two parallel capacitors.
3. What would be the capacitance of the structure in problem 2 if there were a third
conductor with zero thickness at the interface of the dielectrics? How would the
electric field lines look? How does the capacitance change if the spacing between the
top and bottom plates are kept the same, but the conductor thickness is not zero?
Solution Manual Radio Frequency Integrated Circuits and Systems
by Hooman Darabi
,TESTBANKS BY TESTBANKSNERD
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.
,TESTBANKS BY TESTBANKSNERD
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
𝐷 = 𝜌𝑆
𝑎𝑟
𝑟2
Assuming a potential of 𝑉0 between the inner and outer surfaces, we have:
𝑎 2 1 1
𝑉0 = − � 1 𝜌𝑆 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎2( − )
𝑏 𝜖
𝑟2 𝜖 𝑎 𝑏
Thus:
𝑄𝑄 𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝐶=𝑉 =
𝜌𝑆 2 1 1 1 1
𝑎 ( − ) 𝑎−𝑏
0
𝜖 𝑎 𝑏
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 𝑎. Letting 𝜀𝜀0 = ×
36𝜋
4𝜋𝜀𝜀0
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 capacitance as a function of the parameters shown in the figure.
, TESTBANKS BY TESTBANKSNERD
Area: A
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal
component of the electric flux density has to be equal in each dielectric. That is:
𝐷1 = 𝐷𝟐𝟐
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate, the
electric field (or flux has a component only in z direction, and we have:
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
𝑑1+𝑑2 𝑑2
−𝜌𝑆 𝑑1+𝑑2 −𝜌
𝑆 𝜌𝑆 𝜌𝑆
𝑉0 = − � 𝐸. 𝑑𝑧 = − � 𝜖 𝑑𝑧 − � 𝜖 𝑑𝑧 = 𝜖 𝑑1 + 𝜖 𝑑2
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
𝑄𝑄 𝐴
𝐶= 𝑉 =𝑑 𝑑
0 1+ 2
𝜖1 𝜖 2
which is analogous to two parallel capacitors.
3. What would be the capacitance of the structure in problem 2 if there were a third
conductor with zero thickness at the interface of the dielectrics? How would the
electric field lines look? How does the capacitance change if the spacing between the
top and bottom plates are kept the same, but the conductor thickness is not zero?
