SOLUTIONS MANUAL for Radio Frequency Integrated Circuits and Systems 2nd Edition by Hooman Darabi

Radio Frequency Integrated Circuits and Systems Solution Manual Hooman Darabi

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.

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 𝑏𝑏 → ∞, thus, 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎 = 0.55𝑝𝑝𝑝𝑝.

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

+ +ρS

-

+ -

+ a + -

b

From Gauss’s law: �𝑫𝑫 ⋅ 𝑑𝑑𝑺𝑺 = 𝑄𝑄 = +𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2 𝑆𝑆

Thus, inside the sphere (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏):

𝑎𝑎2 𝒂𝒂 𝑟𝑟 2 𝒓𝒓 Assuming a potential of 𝑉𝑉0 between the inner and outer surfaces, we have: 𝑎𝑎 1 𝑎𝑎2 𝜌𝜌𝑆𝑆 1 1 𝑉𝑉0 = − � 𝜌𝜌𝑆𝑆 2 𝑑𝑑𝑑𝑑 = 𝑎𝑎2 ( − ) 𝑟𝑟 𝜖𝜖 𝑎𝑎 𝑏𝑏 𝑏𝑏 𝜖𝜖 Thus: 𝑄𝑄 𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2 4𝜋𝜋𝜋𝜋 𝐶𝐶 = = = 𝑉𝑉0 𝜌𝜌𝑆𝑆 𝑎𝑎2 (1 − 1) 1 − 1 𝜖𝜖 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏 1 In the case of a metallic marble, 𝑏𝑏 → ∞, and hence: 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎. Letting 𝜀𝜀0 = 36𝜋𝜋 × 𝑫𝑫 = 𝜌𝜌𝑆𝑆

5

10−9 , and 𝑎𝑎 = 0.5𝑐𝑐𝑐𝑐, it yields 𝐶𝐶 = 9 𝑝𝑝𝑝𝑝 = 0.55𝑝𝑝𝑝𝑝.

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

Area: A

d1

ε1

d2

ε2

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: 𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐

Accordingly:

𝜖𝜖1 𝑬𝑬𝟏𝟏 = 𝜖𝜖2 𝑬𝑬𝟐𝟐

Assuming a surface charge density of +𝜌𝜌𝑆𝑆 for the top plate, and −𝜌𝜌𝑆𝑆 for the bottom plate, the electric field (or flux has a component only in z direction, and we have: 𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐 = −𝜌𝜌𝑆𝑆 𝒂𝒂𝒛𝒛

If the potential between the top ad bottom plates is 𝑉𝑉0, based on the line integral we obtain: 𝑉𝑉0 = − �

𝑑𝑑1 +𝑑𝑑2

0

𝑑𝑑2

𝑑𝑑1 +𝑑𝑑2 −𝜌𝜌𝑆𝑆 −𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆 𝑬𝑬. 𝑑𝑑𝒛𝒛 = − � 𝑑𝑑𝑑𝑑 − � 𝑑𝑑𝑑𝑑 = 𝑑𝑑1 + 𝑑𝑑2 𝜖𝜖2 𝜖𝜖1 𝜖𝜖1 𝜖𝜖2 0 𝑑𝑑2

Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝜌𝑆𝑆 𝐴𝐴, the capacitance is found to be: 𝐶𝐶 =

𝑄𝑄 𝐴𝐴 = 𝑉𝑉0 𝑑𝑑1 + 𝑑𝑑2 𝜖𝜖1 𝜖𝜖2

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: If the conductor is perfect, opposite charges are formed on the surface, but the capacitance remains the same, that is to say, the electric fields terminate to the conductor, but are not altered. If the conductor thickness is greater than zero, but the total distance between the top and bottom plates is the same (𝑑𝑑1 + 𝑑𝑑2 ), we expect the capacitance to increase.

4. Repeat problem 2 if the dielectric boundary were placed normal to the two conducting plates as shown below.

ε1

d

A2

A1 ε2

Solution: Similar to 2, the electric flux density is in z direction, and we assume a surface charge density of +𝜌𝜌𝑆𝑆1/2 for the top plates, and −𝜌𝜌𝑆𝑆1/2 for the bottom plates. Assuming a potential of 𝑉𝑉0 between the plates, unlike 2, as 𝑫𝑫 is tangent to the surface, in general 𝑫𝑫𝟏𝟏 ≠ 𝑫𝑫𝟐𝟐 . Thus, we do not assume a uniform charge density on the plates. Furthermore, based on the line integral definition, at the boundary the tangent components of the electric field (which are in z direction) must be equal between the two dielectrics, that is: 𝑬𝑬𝟏𝟏 = 𝑬𝑬𝟐𝟐

which yields:

𝜌𝜌𝑆𝑆1 𝜌𝜌𝑆𝑆2 = 𝜖𝜖1 𝜖𝜖2

Finally, for the potential the line integral yields: 𝑉𝑉0 =

The total charge is: 𝑄𝑄 = 𝜌𝜌𝑆𝑆1 𝐴𝐴1 + 𝜌𝜌𝑆𝑆2 𝐴𝐴2

𝜌𝜌𝑆𝑆1 𝜌𝜌𝑆𝑆2 𝑑𝑑 = 𝑑𝑑 𝜖𝜖1 𝜖𝜖2

Consequently: 𝐶𝐶 =

𝑄𝑄 𝜖𝜖1 𝐴𝐴1 + 𝜖𝜖2 𝐴𝐴2 = 𝑉𝑉0 𝑑𝑑

As expected, this case turns out to be similar to two series capacitances.

5. Analogues to the capacitance, using Ohm’s law, show that the leakage conductance of an ∫ 𝐄𝐄⋅𝑑𝑑𝐒𝐒

almost perfect conductor with a non-infinite conductivity of σ is given by: 𝐺𝐺 = 𝜎𝜎 −𝑆𝑆 𝑬𝑬.𝑑𝑑𝑳𝑳. Calculate the leakage conductance of a coaxial cable with radii a and b as was used throughout the chapter.

∫

Solution: In a given conductor we have: 𝐼𝐼 = �𝐉𝐉 ⋅ 𝑑𝑑𝐒𝐒 𝑆𝑆

where 𝐉𝐉 is the current density, and by definition, for a conductor: 𝐉𝐉 = σ𝐄𝐄. According to Ohm’s law: ∫ 𝐉𝐉 ⋅ 𝑑𝑑𝐒𝐒 ∫ 𝐄𝐄 ⋅ 𝑑𝑑𝐒𝐒 𝐼𝐼 𝐺𝐺 = = 𝑆𝑆 = 𝜎𝜎 𝑆𝑆 𝑉𝑉 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳 which has a similar form as the capacitance equation: ∮ 𝑬𝑬 ⋅ 𝑑𝑑𝑺𝑺 𝑄𝑄 𝐶𝐶 = = 𝜖𝜖 𝑆𝑆 𝑉𝑉 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳 Note that the surface integral in the capacitance equation is over a closed surface. 6. Consider a very long hollow charge-free super conductor cylindrical shell with inner and outer radios of a and b, respectively. A wire with a current I is placed at the center of the cylinder. Calculate the magnetic field inside and outside considering that the magnetic field inside the shell would have to be zero. If the current I is moved away from the center but inside the shell, how the magnetic fields inside and outside would alter?

I a

b

Solution: Based on Ampere’s law, for 𝑟𝑟 ≤ 𝑎𝑎 we have: � 𝑯𝑯 ⋅ 𝑑𝑑𝑳𝑳 = 𝐼𝐼

Therefore:

𝐼𝐼 𝒂𝒂 2𝜋𝜋𝜋𝜋 𝝓𝝓 For (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏), that is inside the superconductor, the magnetic field (and flux) are zero. In practice, the magnetic flux needs to be constant, so that the voltage is zero. Otherwise, there will be an infinite current induced in the superconductor. In practice however, any small change in magnetic flux will induce an infinite current, and thus, 𝑩𝑩 = 0. Furthermore, a 𝑯𝑯 =

−𝐼𝐼

surface current of 2𝜋𝜋𝜋𝜋 𝒂𝒂𝒛𝒛 flows on the inner surface,

𝐼𝐼

Outside the conductor (𝑟𝑟 ≥ 𝑏𝑏), a surface current of 2𝜋𝜋𝜋𝜋 𝒂𝒂𝒛𝒛 flows on the outer surface, and 𝐼𝐼

again, 𝑯𝑯 = 2𝜋𝜋𝜋𝜋 𝒂𝒂𝝓𝝓 .

If the current moves away from the center, 𝑩𝑩 changes for 𝑟𝑟 ≤ 𝑎𝑎, but remains the same outside the conductor. The surface current on the inner shell is not uniform anymore, but remains the same for the outer shell.

7. What is the internal inductance (per length) of a long straight wire with a circular cross 𝜇𝜇 section of radius a (use energy definition)? Answer: 8𝜋𝜋0 .

Solution: Due to symmetry, we can argue that the magnetic field has only a component in the 𝒂𝒂𝝓𝝓 direction. The current density inside the wire (𝑟𝑟 ≤ 𝑎𝑎) is: 𝜋𝜋𝑟𝑟 2 𝑟𝑟 2 𝑲𝑲 = 𝐼𝐼 2 𝒂𝒂𝒛𝒛 = 𝐼𝐼 2 𝒂𝒂𝒛𝒛 𝜋𝜋𝑎𝑎 𝑎𝑎 Accordingly, based on Ampere’s law, the magnetic field is found to be: 𝑟𝑟 2 𝐼𝐼 2 𝑟𝑟 𝑯𝑯 = 𝑎𝑎 𝒂𝒂𝝓𝝓 = 𝐼𝐼 𝒂𝒂 2𝜋𝜋𝜋𝜋 2𝜋𝜋𝑎𝑎2 𝝓𝝓 Next, we shall find the magnetic energy per unit length inside the wire: 𝜇𝜇0 𝜇𝜇0 1 𝑎𝑎 2𝜋𝜋 𝑟𝑟 2 𝜇𝜇0 2 𝟐𝟐 |𝐇𝐇| 𝑊𝑊𝐻𝐻 = � dV = � � � (𝐼𝐼 ) 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐼𝐼 2 𝑉𝑉 2 0 0 0 2𝜋𝜋𝑎𝑎2 16𝜋𝜋 1

Equating the energy to: 2 𝐿𝐿𝐼𝐼 2 , we obtain the inductance per unit length: 𝜇𝜇0 𝐿𝐿 = 8𝜋𝜋

8. Show that the DC inductance of a piece of wire with finite length l and radius r is: 𝐿𝐿 = 𝜇𝜇0 𝑙𝑙 2𝜋𝜋

2𝑙𝑙

3

(𝑙𝑙𝑙𝑙 𝑟𝑟 − 4). What is the inductance of a copper bond-wire with length of 2mm and a

diameter of 25µm (practical bonding pads in integrated circuits are typically 50×50µm2)?

Argue why traditionally, as a rule of thumb an inductance of 1nH/mm is assumed for bondwires. Solution: The inductance calculation is detailed by Rosa 1. There are two parts, the internal 𝜇𝜇 inductance, 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 , which was calculated to be 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 = 8𝜋𝜋0 𝑙𝑙 in the previous problem, and the

external inductance. As for the external inductance, let us first find the magnetic field. From the law of BiotSavart, the magnetic field at a point P normal to the paper due to an element of length 𝑑𝑑𝑑𝑑 is: 𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑑𝑑𝑑𝑑 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 4𝜋𝜋𝑅𝑅 2 4𝜋𝜋(𝑥𝑥 2 + (𝑦𝑦 − 𝑏𝑏)2 )3/2 where 𝐼𝐼 is the wire current uniformly distributed, and the rest of the parameters are shown in the figure below.

Wire dx dy θ

R

l

y

P

b x The magnetic field at P due to the entire length of the wire is then: 𝑙𝑙 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼 𝑙𝑙 − 𝑏𝑏 𝑏𝑏 = ( + ) 𝐻𝐻 = � 2 2 3/2 4𝜋𝜋𝜋𝜋 �𝑥𝑥 2 + (𝑙𝑙 − 𝑏𝑏)2 √𝑥𝑥 2 + 𝑏𝑏 2 0 4𝜋𝜋(𝑥𝑥 + (𝑦𝑦 − 𝑏𝑏) ) 𝐼𝐼

If the integral were to be taken from −∞ to +∞, the field would be 2𝜋𝜋𝜋𝜋 as we calculated before for a piece of wire with infinite length. To find the inductance, we calculate the magnetic flux as follows: 𝜙𝜙 = �𝐁𝐁 ⋅ 𝑑𝑑𝐒𝐒 = 𝑆𝑆

𝜇𝜇0 𝐼𝐼 ∞ 𝑙𝑙 𝑙𝑙 − 𝑏𝑏 𝑏𝑏 � � ( + ) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 4𝜋𝜋 𝑥𝑥=𝑟𝑟 𝑏𝑏=0 𝑥𝑥�𝑥𝑥 2 + (𝑙𝑙 − 𝑏𝑏)2 𝑥𝑥√𝑥𝑥 2 + 𝑏𝑏 2

1 Edward B. Rosa, Bulletin of the Bureau of Standards, vol. 4, no.2 , pp 301-305, 1907.

which is found to be: 𝜇𝜇0 𝐼𝐼 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 √𝑟𝑟 2 + 𝑙𝑙 2 𝑙𝑙[𝑙𝑙𝑙𝑙 + − ] 2𝜋𝜋 𝑟𝑟 𝑙𝑙 𝑙𝑙 From this the external inductance is: 𝜙𝜙 =

𝜇𝜇0 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 √𝑟𝑟 2 + 𝑙𝑙 2 𝑙𝑙[𝑙𝑙𝑙𝑙 + − ] 2𝜋𝜋 𝑟𝑟 𝑙𝑙 𝑙𝑙 And the total inductance would be: 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 =

𝜇𝜇0 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 1 √𝑟𝑟 2 + 𝑙𝑙 2 𝑙𝑙[𝑙𝑙𝑙𝑙 + + − ] 2𝜋𝜋 𝑟𝑟 𝑙𝑙 4 𝑙𝑙 For 𝑟𝑟 ≪ 𝑙𝑙, the inductance is roughly: 𝜇𝜇0 2𝑙𝑙 3 𝐿𝐿 ≈ 𝑙𝑙(𝑙𝑙𝑙𝑙 − ) 2𝜋𝜋 𝑟𝑟 4 For typical values of 𝑟𝑟 = 12.5𝜇𝜇𝜇𝜇 , and 𝑙𝑙 = 2𝑚𝑚𝑚𝑚, the inductance is found to be about 𝐿𝐿 = 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 + 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 =

2𝑙𝑙

2.01nH. Given the logarithmic nature of the term 𝑙𝑙𝑙𝑙 𝑟𝑟 , as a rule of thumb we assign an inductance of about 1nH/mm for a piece of wire. For the reference, a 1mm long wire inductance is 0.87nH.

9. In Faraday’s experiment, assume the switch has a resistance of R, and the two coils are identical with an inductance of L. The battery voltage is VBAT. Find the time-varying current in the coil. Assuming the iron toroid has a large permeability, find the magnetic flux in the second coil and estimate the emf read by the galvanometer.

Solution: We use the following circuit model to obtain the current in the primary:

VBAT

i1

L

i2

emf

R VBAT t

Setting 𝑡𝑡 = 0 at the instant of switch closing, the primary inductor current is readily found to be: 𝑉𝑉𝐵𝐵𝐵𝐵𝐵𝐵 𝑖𝑖1 = (1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 ) 𝑅𝑅 where 𝜏𝜏 = 𝐿𝐿/𝑅𝑅 is the time constant. The secondary current (𝑖𝑖2 ) is equal to this, and from that the flux in the secondary is: 𝑉𝑉𝐵𝐵𝐵𝐵𝐵𝐵 𝜙𝜙2 = 𝐿𝐿𝐿𝐿2 = 𝐿𝐿 (1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 ) 𝑅𝑅 Thus, the voltage read by the galvanometer is: 𝑑𝑑𝜙𝜙2 𝑒𝑒𝑒𝑒𝑒𝑒 = = 𝑉𝑉𝐵𝐵𝐵𝐵𝐵𝐵 𝑒𝑒 −𝑡𝑡/𝜏𝜏 𝑑𝑑𝑑𝑑 Indicating a voltage spike of 𝑉𝑉𝐵𝐵𝐵𝐵𝐵𝐵 , decaying eventually to zero, as shown in the figure above.

10. Consider a series RLC circuit below where the inductor has an initial current of I0. Solve the circuit differential equation, and find the inductor current. What are the total energies stored in the inductor, and dissipated in the resistor over time?

I0

L

C

R

Solution: The differential equation describing the circuit is: 𝜕𝜕 2 𝑖𝑖𝐿𝐿 𝑅𝑅 𝜕𝜕𝑖𝑖𝐿𝐿 1 + + 𝑖𝑖 = 0 2 𝜕𝜕𝑡𝑡 𝐿𝐿 𝜕𝜕𝜕𝜕 𝐿𝐿𝐿𝐿 𝐿𝐿

The equation may be solved readily using our findings for the parallel circuit, and the duality: 𝑖𝑖𝐿𝐿 (𝑡𝑡) = 𝐼𝐼0

𝜔𝜔0 −𝛼𝛼𝛼𝛼 𝑒𝑒 cos(𝜔𝜔𝑑𝑑 𝑡𝑡 + 𝜙𝜙) 𝜔𝜔𝑑𝑑

𝑣𝑣𝐶𝐶 (𝑡𝑡) =

𝐼𝐼0 −𝛼𝛼𝛼𝛼 𝑒𝑒 sin 𝜔𝜔𝑑𝑑 𝑡𝑡 𝐶𝐶𝜔𝜔𝑑𝑑

where 𝛼𝛼 = 𝑅𝑅/2𝐿𝐿, and the rest of the parameters have been already defined for the parallel circuit. Additionally, we have:

1 𝑊𝑊𝑇𝑇 (𝑡𝑡) = 𝑊𝑊𝑐𝑐 (𝑡𝑡) + 𝑊𝑊𝐿𝐿 (𝑡𝑡) ≈ 𝐿𝐿𝐼𝐼0 2 𝑒𝑒 −2𝛼𝛼𝛼𝛼 2

11. For the circuit below, the inductor 𝐿𝐿1 has an initial stored current of 𝐼𝐼0 . The switch is closed at 𝑡𝑡 = 0. Find the final current of the inductors at 𝑡𝑡 = ∞. Answer: 𝑖𝑖𝐿𝐿1 (∞) = −𝑖𝑖𝐿𝐿2 (∞) = 𝐿𝐿1 𝐼𝐼 𝐿𝐿1 +𝐿𝐿2 0

t=0

I0

L1

L2

R

Solution: At the instant of switch closing (𝑡𝑡 = 0), the inductors current cannot change. Thus: 𝑖𝑖𝐿𝐿1 (0+ ) = 𝐼𝐼0 𝑖𝑖𝐿𝐿2 (0+ ) = 0 Consequently, according to KCL, the first inductor initial current must entirely go through the resistor at 𝑡𝑡 = 0+ . This leads to a sudden jump in the resistor voltage, which eventually decays to zero, as the final voltage across the inductors must be zero. Hence, we can write the resistor voltage (𝑣𝑣(𝑡𝑡)) as: 𝑣𝑣(𝑡𝑡) = −𝑅𝑅𝐼𝐼0 𝑒𝑒 −𝑡𝑡/𝜏𝜏 where 𝜏𝜏 =

𝐿𝐿1 𝐿𝐿2 𝐿𝐿1 +𝐿𝐿2

𝑅𝑅

. This waveform is shown below.

-I0L1L2/(L1+L2)

0

t

τ=L1L2/R(L1+L2)

-RI0 v(t)

The total area under the resistor voltage is:

∞

� 𝑣𝑣(𝑡𝑡)𝑑𝑑𝑑𝑑 = −𝐼𝐼0 0+

𝐿𝐿1 𝐿𝐿2 𝐿𝐿1 + 𝐿𝐿2

The final current of the inductors may be found then: 1 ∞ 𝐿𝐿1 (∞) 𝑖𝑖𝐿𝐿1 = � 𝑣𝑣(𝑡𝑡)𝑑𝑑𝑑𝑑 + 𝑖𝑖𝐿𝐿1 (0+ ) = +𝐼𝐼0 𝐿𝐿1 0+ 𝐿𝐿1 + 𝐿𝐿2 And: 1 ∞ 𝐿𝐿1 𝑖𝑖𝐿𝐿2 (∞) = � 𝑣𝑣(𝑡𝑡)𝑑𝑑𝑑𝑑 + 𝑖𝑖𝐿𝐿2 (0+ ) = −𝐼𝐼0 𝐿𝐿2 0+ 𝐿𝐿1 + 𝐿𝐿2 This shows that the initial energy stored in 𝐿𝐿1 will not be entirely dissipated in the resistor. In 𝐿𝐿

1 , independent of the resistor value, loops in the two fact, a constant current of 𝐼𝐼0 𝐿𝐿 +𝐿𝐿 1

inductors.

2

12. For the problem above argue intuitively how the final current of the two inductors look with respect to each other. Find the total energy dissipated in the resistor, and from that find the final energy and the currents of the two inductors. Is there a condition that leads to the initial energy of the inductor 𝐿𝐿1 completely dissipated, leading to zero final current? Answer: ∞ (−𝑅𝑅𝐼𝐼0 𝑒𝑒 −𝑡𝑡/𝜏𝜏 )2

Resistor energy: 𝐸𝐸𝑅𝑅 = ∫0

𝑅𝑅

1

𝐿𝐿 𝐿𝐿

𝑑𝑑𝑑𝑑 = 2 (𝐿𝐿 1+𝐿𝐿2 )𝐼𝐼0 2 . 1

2

Solution: As the resistor steady state voltage, and consequently current must be zero, we expect the inductors final current to be the same, but in opposite direction. The final energy of inductors must be equal to the initial energy, less the amount dissipated in the resistor. So all needed is to find the energy dissipated in the resistor. This can be readily done given the resistor voltage obtained in the previous problem: ∞ ∞ |𝑣𝑣(𝑡𝑡)|2 𝜏𝜏 1 𝐿𝐿1 𝐿𝐿2 2 𝐸𝐸𝑅𝑅 = � 𝑑𝑑𝑑𝑑 = 𝑅𝑅𝐼𝐼0 � 𝑒𝑒 −2𝑡𝑡/𝜏𝜏 𝑑𝑑𝑑𝑑 = 𝑅𝑅𝐼𝐼0 2 = ( )𝐼𝐼0 2 𝑅𝑅 2 2 𝐿𝐿1 + 𝐿𝐿2 0 0 The energy dissipated in the resistor is independent of its value. Consequently: 1 1 𝐿𝐿1 𝐿𝐿2 1 𝐿𝐿1 2 𝑊𝑊𝐿𝐿1 (∞) + 𝑊𝑊𝐿𝐿2 (∞) = 𝐿𝐿1 𝐼𝐼0 2 − � � 𝐼𝐼0 2 = 𝐼𝐼 2 2 𝐿𝐿1 + 𝐿𝐿2 2 𝐿𝐿1 + 𝐿𝐿2 0 2 1

Since 𝑊𝑊𝐿𝐿1 (∞) + 𝑊𝑊𝐿𝐿2 (∞) = 2 (𝐿𝐿1 + 𝐿𝐿2 )𝐼𝐼𝐿𝐿1/2 (∞)2, the inductors final current is readily 𝐿𝐿

1 obtained to be ±𝐼𝐼0 𝐿𝐿 +𝐿𝐿 , in agreement with the previous problem. 1

2

Clearly, the energy dissipated in the resistor is always smaller than the inductor initial energy (unless 𝐿𝐿2 = ∞, that is to say if 𝐿𝐿2 is removed). Thus, there always exists a non-zero final current looping in the inductors.

13. Suppose an LC tank used in a voltage-controlled oscillator (VCO) consists of a switchable capacitance of 𝐶𝐶𝐹𝐹 , and a varactor with nominal capacitance of 𝐶𝐶(𝑣𝑣). We define the VCO gain 𝜕𝜕𝜔𝜔

as: 𝐾𝐾𝑉𝑉𝑉𝑉𝑉𝑉 = 𝜕𝜕𝜕𝜕 , where V is the varactor voltage. Show that 𝐾𝐾𝑉𝑉𝑉𝑉𝑉𝑉 varies with frequency cubed

(𝜔𝜔0 3 ) as the switchable capacitance changes the nominal frequency of oscillation, 𝜔𝜔0 .

Solution: Assuming an inductance of 𝐿𝐿 for the tank, the frequency of oscillation is: 1 1 𝜔𝜔0 = = (𝐶𝐶𝐹𝐹 + 𝐶𝐶(𝑣𝑣))−1/2 �𝐿𝐿(𝐶𝐶𝐹𝐹 + 𝐶𝐶(𝑣𝑣)) √𝐿𝐿 According to VCO gain definition: 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 −1/2 𝐾𝐾𝑉𝑉𝑉𝑉𝑉𝑉 = = = (𝐶𝐶𝐹𝐹 + 𝐶𝐶(𝑣𝑣))−3/2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 √𝐿𝐿 After rearranging, we obtain: −𝐿𝐿 𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕 (𝐿𝐿(𝐶𝐶𝐹𝐹 + 𝐶𝐶(𝑣𝑣)))−3/2 = − 𝐿𝐿𝜔𝜔0 3 𝐾𝐾𝑉𝑉𝑉𝑉𝑉𝑉 = 2 𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 Given that the varactor voltage characteristics is known and fixed, the equation indicates 𝐾𝐾𝑉𝑉𝑉𝑉𝑉𝑉 dependence on frequency cubed.

14. Find the Q of parallel RC, RL circuit shown below. Show that the overall Q can be expressed 1

1

1

by: 𝑄𝑄 = 𝑄𝑄 + 𝑄𝑄 assuming the inductor and capacitor are high-Q. 𝐿𝐿

𝐶𝐶

L RC

C rL

Solution: If high-Q, the inductor equivalent parallel resistor is (see Chapter 3 also and seriesparallel transformation): 𝐿𝐿𝜔𝜔0 𝑅𝑅𝑃𝑃𝑃𝑃 ≈ 𝑄𝑄𝐿𝐿 2 𝑟𝑟𝐿𝐿 = 𝑄𝑄𝐿𝐿 𝑟𝑟 = 𝑄𝑄𝐿𝐿 𝐿𝐿𝜔𝜔0 𝑟𝑟𝐿𝐿 𝐿𝐿 And the total Q is: 𝑅𝑅𝐶𝐶 ||𝑅𝑅𝑃𝑃𝑃𝑃 𝑄𝑄 = 𝐿𝐿𝜔𝜔0 which can be rearranged as:

1

1 1 1 = 𝐿𝐿𝜔𝜔0 ( + ) 𝑅𝑅𝐶𝐶 𝑄𝑄𝐿𝐿 𝐿𝐿𝜔𝜔0 𝑄𝑄

Since at resonance 𝐿𝐿𝜔𝜔0 = 𝐶𝐶𝜔𝜔 , we have: 0

1 1 1 1 1 = + = + 𝑄𝑄 𝑅𝑅𝐶𝐶 𝐶𝐶𝜔𝜔0 𝑄𝑄𝐿𝐿 𝑄𝑄𝐶𝐶 𝑄𝑄𝐿𝐿

15. For an RLC one-port with an input impedance of 𝑍𝑍(𝑗𝑗𝑗𝑗), the quality factor is sometimes defined as: |𝐼𝐼𝐼𝐼[𝑍𝑍]| 𝑄𝑄 = 𝑅𝑅𝑅𝑅[𝑍𝑍] Using the energy definition of the Q, and the concept of complex power justify this

1

definition. Find a similar equation based on the admittance of the one-port: 𝑌𝑌(𝑗𝑗𝑗𝑗) = 𝑍𝑍(𝑗𝑗𝑗𝑗).

Discuss how this definition works for a series (or parallel) RLC circuit.

Solution: For the one-port shown below, according to Telegen’s theorem (although discussed in basic circuit theory, we shall prove Telegen’s theorem in Chapter 4 problem sets), we have: 1 1 𝑃𝑃 = 𝑍𝑍|𝐼𝐼|2 = � 𝑍𝑍𝑚𝑚 |𝐼𝐼𝑚𝑚 |2 2 2 𝑚𝑚

where 𝑃𝑃 is the complex average power derived to the one-port, and 𝑚𝑚 denotes an arbitrary branch with its corresponding impedance and peak current phasor.

+ V -

I

One-Port

Z(s) The equation may be broken into sum of all resistive, capacitive, and magnetic powers of all branches as follows: 1 1 1 1 |𝐼𝐼 |2 𝑃𝑃 = � 𝑅𝑅𝑖𝑖 |𝐼𝐼𝑖𝑖 |2 + � 𝑗𝑗𝑗𝑗𝑗𝑗𝑘𝑘 |𝐼𝐼𝑘𝑘 |2 + � 2 2 2 𝑗𝑗𝜔𝜔𝜔𝜔𝑛𝑛 𝑛𝑛 which can be expressed as:

𝑖𝑖

𝑘𝑘

𝑛𝑛

1 1 1 𝑃𝑃 = � 𝑅𝑅𝑖𝑖 |𝐼𝐼𝑖𝑖 |2 + 2𝑗𝑗𝑗𝑗[� 𝐿𝐿𝑘𝑘 |𝐼𝐼𝑘𝑘 |2 − � 𝐶𝐶𝑛𝑛 |𝑉𝑉𝑛𝑛 |2 ] 2 4 4

𝑖𝑖 𝑘𝑘 𝑛𝑛 1 2 Considering that 2 𝑅𝑅𝑖𝑖 |𝐼𝐼𝑖𝑖 | represents the average power dissipated in resistor, and 4 𝐿𝐿𝑘𝑘 |𝐼𝐼𝑘𝑘 |2 1

1

and 4 𝐶𝐶𝑛𝑛 |𝑉𝑉𝑛𝑛 |2 represent the average energy in inductor and capacitor respectively, we can rewrite:

This yields:

Thus:

1 𝑍𝑍|𝐼𝐼|2 = 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 + 2𝑗𝑗𝑗𝑗(𝐸𝐸𝑀𝑀 − 𝐸𝐸𝐶𝐶 ) 2 𝑍𝑍 =

2𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 + 4𝑗𝑗𝑗𝑗(𝐸𝐸𝑀𝑀 − 𝐸𝐸𝐶𝐶 ) |𝐼𝐼|2

|𝐼𝐼𝐼𝐼[𝑍𝑍]| 2|𝐸𝐸𝑀𝑀 − 𝐸𝐸𝐶𝐶 | = 𝜔𝜔 𝑅𝑅𝑅𝑅[𝑍𝑍] 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 The result obtained, though close, is somewhat different from the Q definition we showed earlier in the chapter. There is an additional factor of two, and that what appears in the numerator is the difference between the magnetic and capacitive energies. In the case of a parallel RLC circuit at resonance, this particular definition yields a quality factor of zero! However, this definition proves to be much more useful for RL or RC circuits. For instance, 1

1

in a series RL circuit, we have: 𝐸𝐸𝑀𝑀 = 4 𝐿𝐿|𝐼𝐼|2, 𝐸𝐸𝐶𝐶 = 0, and 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = 2 𝑅𝑅|𝐼𝐼|2, which results in: 𝐿𝐿𝐿𝐿 𝑅𝑅 as expected. When characterizing integrated inductors for instance, this definition is more commonly used. Note that for an integrated inductor at resonance, the Q will be zero according to the above definition, consistent with the plots we showed earlier. Similarly, we can argue: |𝐼𝐼𝐼𝐼[𝑌𝑌]| 𝑄𝑄 = 𝑅𝑅𝑅𝑅[𝑌𝑌] which is a more handy equation to characterize parallel RL or RC circuits. 𝑄𝑄 =

𝑧𝑧

𝑧𝑧

16. Show that the solution of the general form: 𝑉𝑉(𝑧𝑧, 𝑡𝑡) = 𝑓𝑓1 �𝑡𝑡 − 𝜈𝜈� + 𝑓𝑓2 �𝑡𝑡 + 𝜈𝜈� satisfies the transmission line equation. Find the proper value of velocity ν, to satisfy the equation. 𝑧𝑧

Solution: Let us define: 𝑢𝑢 = 𝑡𝑡 − 𝜈𝜈. We have:

Similarly:

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 −1 𝜕𝜕𝑓𝑓1 1 𝜕𝜕𝑓𝑓2 = = + 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜈𝜈 𝜕𝜕𝜕𝜕 𝜈𝜈 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑉𝑉 1 𝜕𝜕 2 𝑓𝑓1 1 𝜕𝜕 2 𝑓𝑓2 1 𝜕𝜕 2 𝑓𝑓1 𝜕𝜕 2 𝑓𝑓2 = + = [ + ] 𝜕𝜕𝑧𝑧 2 𝜈𝜈 2 𝜕𝜕𝑢𝑢2 𝜈𝜈 2 𝜕𝜕𝑢𝑢2 𝜈𝜈 2 𝜕𝜕𝑢𝑢2 𝜕𝜕𝑢𝑢2

𝜕𝜕 2 𝑉𝑉 𝜕𝜕 2 𝑓𝑓1 𝜕𝜕 2 𝑓𝑓2 = + 𝜕𝜕𝑡𝑡 2 𝜕𝜕𝑢𝑢2 𝜕𝜕𝑢𝑢2

Thus:

𝜕𝜕 2 𝑉𝑉 1 𝜕𝜕 2 𝑉𝑉 = 𝜕𝜕𝑧𝑧 2 𝜈𝜈 2 𝜕𝜕𝑡𝑡 2 which is the lossless transmission line differential equation. 17. Design a 4nH single-layer spiral inductor assuming the inductance may be approximated by: 𝐿𝐿 = 𝜇𝜇0 𝑁𝑁 2 𝑟𝑟, where N is the number of turns, and r is the spiral radius. Assume a metal sheet resistance of 10mΩ/□ , and constrain the design to an area of 200×200µm2, with an inner diameter of greater than 150µm. The spacing between the metals is 5µm. The goal is to maximize the Q given the constrains. Neglect the skin effect and other high frequency factors, and find the optimum Q. Solution: The inductor layout is depicted below. Although we showed that the integrated inductance value is a more complex function of the parameters, we shall use 𝐿𝐿 = 𝜇𝜇0 𝑁𝑁 2 𝑟𝑟 solely as an exercise here.

l

150µm

W

S

Since the inductor inner and outer radii are given, the inductance only depends on the number of turns, 𝑁𝑁. To optimize Q, then we need to find the resistance as a function of 𝑁𝑁 as well. For an arbitrary leg belonging to the nth turn we can write: 𝑙𝑙𝑛𝑛 = 𝑙𝑙 − 2(𝑛𝑛 − 1)(𝑊𝑊 + 𝑆𝑆) where 𝑙𝑙 = 200𝜇𝜇𝜇𝜇 is the outer leg length, and 𝑙𝑙𝑛𝑛 is the nth leg length. For the last leg, given an inner radius of 150µm, we can establish the following constraint:

200𝜇𝜇𝜇𝜇 − 2𝑁𝑁𝑁𝑁 − 2(𝑁𝑁 − 1)𝑆𝑆 = 150𝜇𝜇𝜇𝜇 Since 𝑆𝑆 = 5𝜇𝜇𝜇𝜇 is given, we have: 𝑁𝑁𝑁𝑁 + (𝑁𝑁 − 1)5 = 25𝜇𝜇𝜇𝜇 Or: 30𝜇𝜇𝜇𝜇 𝑊𝑊 = − 5𝜇𝜇𝜇𝜇 𝑁𝑁 Calculating the total length (and ignoring the edges double counting), we have: 𝑁𝑁

𝑙𝑙 𝑇𝑇 = 4 � 𝑙𝑙𝑛𝑛 = 4𝑁𝑁𝑁𝑁 − 4𝑁𝑁(𝑁𝑁 − 1)(𝑊𝑊 + 𝑆𝑆) 𝑛𝑛=1

Replacing 𝑊𝑊 as a function of 𝑁𝑁, we arrive at: 𝑙𝑙 𝑇𝑇 = 𝑁𝑁680𝜇𝜇𝜇𝜇 + 120𝜇𝜇𝜇𝜇 The total DC resistance as a result is: 𝑁𝑁680 + 120 𝑁𝑁680 + 120 𝑅𝑅 = 𝑅𝑅□ = 𝑅𝑅□ 30 𝑊𝑊 𝑁𝑁 − 5 We can see that the DC resistance also roughly increases with 𝑁𝑁 2 , same as the inductance. The quality factor is readily found to be: 𝐿𝐿𝐿𝐿 𝜇𝜇0 𝑟𝑟𝑟𝑟 𝜇𝜇0 𝑟𝑟𝑟𝑟 𝑁𝑁(6 − 𝑁𝑁) 𝑁𝑁 2 𝑄𝑄 = = = 𝑅𝑅 𝑅𝑅□ 𝑁𝑁680 + 120 𝑅𝑅□ 136𝑁𝑁 + 24 30 𝑁𝑁 − 5

which expresses 𝑄𝑄 as a function of 𝑁𝑁 as the only variable. Taking the derivative, the 𝑄𝑄 is maximized for 𝑁𝑁 = 1.2. So it appears that a single-turn design is the optimum choice. The corresponding line width would be then: 25µm, and the total DC resistance is 0.32Ω. Approximating 𝑟𝑟 ≈ 100√2𝜇𝜇𝜇𝜇, the DC inductance is 0.18nH, and the quality factor is 14.1 at 4GHz.

2 Chapter Two 1. 2. Using Fourier transform basic definition, prove the Parseval’s energy theorem: ∞

∞

� |𝑥𝑥(𝑡𝑡)|2 𝑑𝑑𝑑𝑑 = � |𝑋𝑋(𝑓𝑓)|2 𝑑𝑑𝑑𝑑 −∞

−∞

Solution: The proof readily follows from the basic Fourier integral definition. We have: �

∞

|𝑥𝑥(𝑡𝑡)|2

−∞

∞

𝑑𝑑𝑑𝑑 = � 𝑥𝑥(𝑡𝑡)𝑥𝑥 −∞

∗ (𝑡𝑡)𝑑𝑑𝑑𝑑

After rearranging the integral, we will have: ∞

∞

∞

∞

∞

= � 𝑥𝑥(𝑡𝑡)[� 𝑋𝑋(𝑓𝑓)𝑒𝑒 𝑗𝑗2𝜋𝜋𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑 ]∗ 𝑑𝑑𝑑𝑑 −∞

−∞

∞

� |𝑥𝑥(𝑡𝑡)|2 𝑑𝑑𝑑𝑑 = � 𝑋𝑋 ∗ (𝑓𝑓) �� 𝑥𝑥(𝑡𝑡)𝑒𝑒 −𝑗𝑗2𝜋𝜋𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑� 𝑑𝑑𝑑𝑑 = � |𝑋𝑋(𝑓𝑓)|2 𝑑𝑑𝑑𝑑 −∞

−∞

−∞ ∞ Note that the term inside the bracket: ∫−∞ 𝑥𝑥(𝑡𝑡)𝑒𝑒 −𝑗𝑗2𝜋𝜋𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑 is 𝑋𝑋(𝑓𝑓).

−∞

3. In the ladder structure shown below, show that the transfer function is of the form: 𝑣𝑣𝑂𝑂𝑂𝑂𝑂𝑂 (𝑠𝑠) 1 = 𝑖𝑖𝐼𝐼𝐼𝐼 (𝑠𝑠) 𝑎𝑎𝑛𝑛 𝑠𝑠 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑠𝑠 𝑛𝑛−1 + ⋯ + 𝑎𝑎1 𝑠𝑠 + 𝑎𝑎0 Where n is the number of reactive components, and 𝑎𝑎𝑛𝑛 is their product. …

iIN

+ vOUT -

Solution: Considering how the current is divided within the ladder, To find 𝑎𝑎𝑛𝑛 intuitively, as we discussed in the chapter, at high frequencies, with the capacitors being short, and the inductors open, we can monitor the current dividing in any given branch as follows: For an arbitrary part of the ladder comprising of the shunt capacitance 𝐶𝐶𝑖𝑖 , and the series inductance 𝐿𝐿𝑖𝑖 , the total impedance at the right side of the capacitance is mostly dominated by the inductance, and is roughly 𝐿𝐿𝑖𝑖 𝑠𝑠. Consequently, the current entering the branch is divided by: 1 1 𝐶𝐶𝑖𝑖 𝑠𝑠 ≈ ≈ 1 𝐿𝐿𝑖𝑖 𝐶𝐶𝑖𝑖 𝑠𝑠 2 + 𝐿𝐿 𝑠𝑠 𝑛𝑛𝑛𝑛 𝐶𝐶𝑖𝑖 𝑠𝑠 When all the branches considered, the output current, which determines the output voltage appearing on the resistor, is of the form:

𝑛𝑛/2

� 𝑖𝑖=1

1 1 = 2 𝐿𝐿𝑖𝑖 𝐶𝐶𝑖𝑖 𝑠𝑠 [∏ 𝐿𝐿𝑖𝑖 𝐶𝐶𝑖𝑖 ]𝑠𝑠 𝑛𝑛

assuming 𝑛𝑛 is even (we have equal number of capacitances and inductances). The case of 𝑛𝑛 being odd is very similar. 1

Since at high frequencies only the term 𝑎𝑎 𝑠𝑠𝑛𝑛 of the transfer function is dominant, we 𝑛𝑛

conclude:

𝑛𝑛

𝑎𝑎𝑛𝑛 = � 𝐿𝐿𝑖𝑖 𝐶𝐶𝑖𝑖 1

which is the proof.

4. The input impedance of the circuit shown below is expressed as: 𝑁𝑁(𝑠𝑠) 𝑎𝑎𝑛𝑛 𝑠𝑠 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑠𝑠 𝑛𝑛−1 + ⋯ + 𝑎𝑎1 𝑠𝑠 + 𝑎𝑎0 𝑍𝑍(𝑠𝑠) = = 𝐷𝐷(𝑠𝑠) 𝑏𝑏𝑚𝑚 𝑠𝑠 𝑚𝑚 + 𝑏𝑏𝑚𝑚−1 𝑠𝑠 𝑚𝑚−1 + ⋯ + 𝑏𝑏1 𝑠𝑠 Where |𝑛𝑛 − 𝑚𝑚| ≤ 1.

C

Z2(s)

Z(s)

Show that the value of the series input capacitor is: 𝜕𝜕 𝐷𝐷(𝑠𝑠)� 𝜕𝜕𝜕𝜕 𝑠𝑠=0 𝐶𝐶 = 𝑁𝑁(0)

Solution: We can rewrite 𝑍𝑍(𝑠𝑠) as: 𝑎𝑎𝑛𝑛 𝑠𝑠 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑠𝑠 𝑛𝑛−1 + ⋯ + 𝑎𝑎1 𝑠𝑠 + 𝑎𝑎0 𝑍𝑍(𝑠𝑠) = 𝑠𝑠(𝑏𝑏𝑚𝑚 𝑠𝑠 𝑚𝑚−1 + 𝑏𝑏𝑚𝑚−1 𝑠𝑠 𝑚𝑚−1 + ⋯ + 𝑏𝑏1 ) Using partial fraction expansion (note 𝑍𝑍(𝑠𝑠) has a pool at 𝑠𝑠 = 0): 𝐾𝐾 𝑎𝑎′𝑛𝑛−1 𝑠𝑠 𝑛𝑛−1 + 𝑎𝑎′𝑛𝑛−2 𝑠𝑠 𝑛𝑛−2 + ⋯ + 𝑎𝑎′1 𝑍𝑍(𝑠𝑠) = + 𝑠𝑠 𝑏𝑏𝑚𝑚 𝑠𝑠 𝑚𝑚−1 + 𝑏𝑏𝑚𝑚−1 𝑠𝑠 𝑚𝑚−1 + ⋯ + 𝑏𝑏1 Clearly: 𝑎𝑎0 𝑁𝑁(0) 𝐾𝐾 = 𝑠𝑠𝑠𝑠(𝑠𝑠)|𝑠𝑠=0 = = 𝜕𝜕 𝑏𝑏1 𝐷𝐷(𝑠𝑠)� 𝜕𝜕𝜕𝜕 𝑠𝑠=0 1

It is evident that 𝑍𝑍(𝑠𝑠) consists of a capacitance of 𝐾𝐾 in series with another one-port. Hence:

𝜕𝜕 𝐷𝐷(𝑠𝑠)� 1 𝜕𝜕𝜕𝜕 𝑠𝑠=0 𝐶𝐶 = = 𝐾𝐾 𝑁𝑁(0) The condition |𝑛𝑛 − 𝑚𝑚| ≤ 1 arises from the fact that for 𝑍𝑍(𝑠𝑠) to be realizable by a lumped RLCM network, it needs to be a positive real function, that is: 𝑅𝑅𝑅𝑅[𝑍𝑍(𝑠𝑠)] ≥ 0 when 𝑅𝑅𝑅𝑅[𝑠𝑠] ≥ 0. The proof is given by Brune, and is a sufficient and necessary condition that must be satisfied when synthesizing the circuits. It follows that in a positive real function, the difference between degrees of the numerator and denominator cannot exceed one. For more details, see Dimopoulos 2 or Temes 3. 5. A one-port has the input impedance: 2𝑠𝑠 3 + 2𝑠𝑠 + 1 𝑠𝑠 2 + 1 Using a similar approach as problem 3, find the value of the series input inductance, and synthesize the rest of the circuit. 𝑍𝑍(𝑠𝑠) =

Solution: We may write:

2𝑠𝑠(𝑠𝑠 2 + 1) + 1 1 𝑍𝑍(𝑠𝑠) = = 2𝑠𝑠 + 𝑠𝑠 2 + 1 𝑠𝑠 2 + 1

Accordingly, the one-port may be synthesized though a 2H inductor, in series with a parallel 1

LC circuit with 1H inductance and 1F capacitance (to realize the 𝑠𝑠2 +1 portion).

2H

1F

1H

Z(s) The synthesized circuit is depicted above.

2 H. Dimopoulos, Analog filters, Theory, design, and synthesis, Springer 2011.

3 G. Temes, J. Laparta, Introduction to circuit synthesis and design, McGraw Hill, 1977.

6. Prove that the Butterworth function: |𝐻𝐻(𝑓𝑓)| = derivatives are equal to zero at 𝑓𝑓 = 0.

1

𝑓𝑓 𝐵𝐵

�1+( )2𝑛𝑛

is maximally flat, that is, its first n

Solution: All we need to show is that the first n derivatives of the transfer function is equal to zero. For simplicity, we assume 𝐵𝐵 = 1 as it does not impact the derivatives. Forming: 1 = (1 + 𝑓𝑓 2𝑛𝑛 )−1/2 𝐺𝐺(𝑓𝑓) = 2𝑛𝑛 �1 + 𝑓𝑓 We have: 𝐺𝐺′(𝑓𝑓) = −𝑛𝑛𝑓𝑓 2𝑛𝑛−1 (1 + 𝑓𝑓 2𝑛𝑛 )−3/2 Which is zero for 𝑓𝑓 = 0. The second derivative is: 3

𝐺𝐺 ′′ (𝑓𝑓) = −𝑛𝑛(2𝑛𝑛 − 1)𝑓𝑓 2𝑛𝑛−2 (1 + 𝑓𝑓 2𝑛𝑛 )−2 + 3𝑛𝑛2 𝑓𝑓 4𝑛𝑛−2 (1 + 𝑓𝑓 2𝑛𝑛 )−5/2 Which is also zero for 𝑓𝑓 = 0. The rest of the derivatives may be done very easily as well. For instance, for 𝑛𝑛 = 1, we can see that 𝐺𝐺′(𝑓𝑓) = −𝑓𝑓(1 + 𝑓𝑓 2 )−3/2 And clearly the second derivative is not zero anymore.

7. Show that all the poles of the normalized Butterworth filter (B = 1) lie on unity circle in the s-plane. Discuss the pole locations for even and odd values of n. Hint: Prove: 𝐻𝐻(𝑠𝑠)𝐻𝐻(−𝑠𝑠) = 1

� 1+(−𝑠𝑠2 )𝑛𝑛

𝑠𝑠=𝑗𝑗𝑗𝑗

2𝑘𝑘+1

𝜋𝜋

. Show that the roots of 𝐻𝐻(𝑠𝑠) are: 𝑠𝑠𝑘𝑘 = 𝑒𝑒 𝑗𝑗( 2𝑛𝑛 𝜋𝜋+ 2 ) , where 𝑘𝑘 = 0, 1, ⋯ , 𝑛𝑛 − 1.

Solution: Since |𝐻𝐻(𝑓𝑓)| =

1

𝑓𝑓 𝐵𝐵

�1+( )2𝑛𝑛

To find the roots, we shall set:

, it follows:

𝐻𝐻(𝑠𝑠)𝐻𝐻(−𝑠𝑠) =

1 � 1 + (−𝑠𝑠 2 )𝑛𝑛 𝑠𝑠=𝑗𝑗𝑗𝑗

1 + (−𝑠𝑠 2 )𝑛𝑛 = 0

It follows:

(−𝑠𝑠 2 )𝑛𝑛 = −1 = 𝑒𝑒 𝑗𝑗(2𝑘𝑘+1)𝜋𝜋 where 𝑘𝑘 = 0, 1, ⋯ , 𝑛𝑛 − 1. Consequently: −𝑠𝑠 2 = 𝑒𝑒

Thus:

(2𝑘𝑘+1)𝜋𝜋 𝜋𝜋 + ) 2 2𝑛𝑛

Which leads to: 𝑠𝑠 = 𝑒𝑒 𝑗𝑗(

𝑠𝑠 2 = 𝑒𝑒

𝑗𝑗(2𝑘𝑘+1)𝜋𝜋 𝑛𝑛

𝑗𝑗(2𝑘𝑘+1)𝜋𝜋 𝑛𝑛 𝑒𝑒 𝑗𝑗𝑗𝑗

. Clearly, all the roots are on the unity circuit as |𝑠𝑠| = 1.

8. For the following active gyrator, the capacitor C is voltage-dependent given by 𝐶𝐶 = 𝑓𝑓(𝑣𝑣), 1

1

prove that the effective inductor is current-dependent with value given by 𝐿𝐿 = 𝑔𝑔𝑔𝑔2 𝑓𝑓(𝑔𝑔𝑔𝑔 𝑖𝑖𝐼𝐼𝐼𝐼 ).

-gm

gm

iIN

+ v -

Solution: Assuming an input voltage of 𝑣𝑣𝐼𝐼𝐼𝐼 , we have: 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑔𝑔𝑚𝑚 𝑣𝑣𝐼𝐼𝐼𝐼 = 𝐶𝐶 = 𝑓𝑓(𝑣𝑣) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 Furthermore: 𝑖𝑖𝐼𝐼𝐼𝐼 + (−𝑔𝑔𝑚𝑚 𝑣𝑣) = 0 Thus: 1 𝑖𝑖𝐼𝐼𝐼𝐼 𝑑𝑑 𝑖𝑖𝐼𝐼𝐼𝐼 𝑣𝑣𝐼𝐼𝐼𝐼 = 𝑓𝑓( ) ( ) 𝑔𝑔𝑚𝑚 𝑔𝑔𝑚𝑚 𝑑𝑑𝑑𝑑 𝑔𝑔𝑚𝑚 Or: 1 𝑖𝑖𝐼𝐼𝐼𝐼 𝑑𝑑𝑖𝑖𝐼𝐼𝐼𝐼 𝑣𝑣𝐼𝐼𝐼𝐼 = [ 2 𝑓𝑓 � �] 𝑔𝑔𝑚𝑚 𝑑𝑑𝑑𝑑 𝑔𝑔𝑚𝑚

C = f(v)

1

𝑖𝑖

which indicates a nonlinear inductance at the input, whose value is: 𝑔𝑔 2 𝑓𝑓(𝑔𝑔𝐼𝐼𝐼𝐼 ). 𝑚𝑚

𝑚𝑚

9. A transfer function 𝐻𝐻(𝑓𝑓) is plotted below, which has a unity magnitude and phase of −𝛼𝛼 and +𝛼𝛼 for positive and negative frequencies, respectively. Prove that the impulse response is 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

given by the following expression: ℎ(𝑡𝑡) = 𝜋𝜋𝜋𝜋 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 × 𝛿𝛿(𝑡𝑡).