Issuu

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 + -

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

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

ε1

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

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

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

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

(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

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

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?

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:

vC(t)

-1

ωi(t) From the equation, the signal must be multiplied by the instantaneous frequency (𝜔𝜔𝑖𝑖 ), integrated, multiplied by 𝜔𝜔𝑖𝑖 and integrated again, and subtracted from itself. Thus, the output of the system shown above is a sinusoid whose instantaneous frequency (the input of the system) is set by 𝜔𝜔𝑖𝑖 (𝑡𝑡). 24. Show that the following circuit may be employed as an FM demodulator. Propose a proper circuitry to perform the differentiation suitable for high frequencies.

Envelope Detector

d/dt

Hint: Use the circuit below known as a balanced slope demodulator.

+ vo(t) -

f0 < f C

f0 > f C

Solution: Given the FM signal: once differentiated, we will have:

𝑣𝑣𝐶𝐶 (𝑡𝑡) = 𝐴𝐴𝐶𝐶 𝑐𝑐𝑐𝑐𝑐𝑐(� 𝜔𝜔𝑖𝑖 (𝜏𝜏) 𝑑𝑑𝑑𝑑)

𝑣𝑣𝐶𝐶′ = 𝐴𝐴𝐶𝐶 𝜔𝜔𝑖𝑖 (𝑡𝑡)𝑠𝑠𝑠𝑠𝑠𝑠(� 𝜔𝜔𝑖𝑖 (𝜏𝜏) 𝑑𝑑𝑑𝑑)

≈

Clearly, the envelope of the differentiated signal above is proportional to the instantaneous frequency. Thus, all needed is to pass 𝑣𝑣𝐶𝐶′ thought an envelope detector, and extract 𝜔𝜔𝑖𝑖 (𝑡𝑡). This is then essentially an FM demodulator. Realizing a good differentiator at high frequency may not be a trivial task. A suitable circuit to accomplish this is shown above, and whose frequency response is depicted below.

ω01

ω02

ωC

Approximate Differentiator

The circuit, known as a balanced slope detector, comprises of two LC tanks, one slightly high tuned, where as the other is slightly low tuned with respect to the carrier frequency. The FM signal is fed to each tank, performing the differentiation, after which is passed to an envelope detector as desired. From the frequency response plotted above, once the two LC tanks outputs are subtracted, around the carrier frequency (𝜔𝜔𝐶𝐶 ), the composite response may be approximated by a straight line, hence representing a differentiator (whose frequency response is ideally 𝐾𝐾𝐾𝐾). To understand better how the slope detector works, suppose the carrier frequency is 𝜔𝜔𝐶𝐶 , and that the LC tanks are tuned to 𝜔𝜔01 and 𝜔𝜔02 respectively. Furthermore, we shall assume the two tanks have equal resistor (𝑅𝑅) and capacitor (𝐶𝐶), and only the inductors are different to tune them as desired. If the tank Q is high enough (say ten or more), the composite response may be expressed as: 𝑅𝑅 𝑅𝑅 |𝑍𝑍𝑇𝑇 (𝑗𝑗𝑗𝑗)| ≈ − 2 𝜔𝜔 − 𝜔𝜔02 2 �1 + �𝜔𝜔 − 𝜔𝜔01 � �1 + ( 𝛼𝛼 ) 𝛼𝛼 1

𝐵𝐵𝐵𝐵

where 𝜔𝜔01 and 𝜔𝜔02 are the tank center frequencies, and 𝛼𝛼 = 2𝑅𝑅𝑅𝑅 = 2 . In addition, we shall set: 𝜔𝜔01 = 𝜔𝜔𝐶𝐶 + Δ𝜔𝜔 𝜔𝜔02 = 𝜔𝜔𝐶𝐶 − Δ𝜔𝜔

Which is to say the two tanks are symmetrically high and low tuned around carrier frequency (𝜔𝜔𝐶𝐶 ) by an arbitrary amount of Δ𝜔𝜔, whose optimum value, we shall determine shortly. To gain more insight, let us use Taylor series to expand |𝑍𝑍𝑇𝑇 (𝑗𝑗𝑗𝑗)|: |𝑍𝑍𝑇𝑇 (𝑗𝑗𝑗𝑗)| = |𝑍𝑍𝑇𝑇 (𝑗𝑗𝑗𝑗)|′ (𝜔𝜔 − 𝜔𝜔𝐶𝐶 ) + |𝑍𝑍𝑇𝑇 (𝑗𝑗𝑗𝑗)|′′′(𝜔𝜔 − 𝜔𝜔𝐶𝐶 )3 + |𝑍𝑍𝑇𝑇 (𝑗𝑗𝑗𝑗)|′′′′′(𝜔𝜔 − 𝜔𝜔𝐶𝐶 )5 + ⋯ Note that the balanced operation results in even-order derivatives not to appear. One can verify that by choosing: 3 Δ𝜔𝜔 = � 𝛼𝛼 2

the 3rd-order derivative is eliminated, and thus: |𝑍𝑍𝑇𝑇 (𝑗𝑗𝑗𝑗)| =

4𝑅𝑅 3 𝜔𝜔 − 𝜔𝜔𝐶𝐶 64 𝜔𝜔 − 𝜔𝜔𝐶𝐶 5 � [ − � � + ⋯] 𝛼𝛼 𝛼𝛼 5 5 625

For small frequency deviations with respect to bandwidth, the second terms may be ignored, and 𝜔𝜔−𝜔𝜔𝐶𝐶

thus a linear response is obtained as desired. To be precise, for �

𝛼𝛼

� < 0.5, the second terms is

less than 0.5% and shall be comfortably ignored. The slope detector may be realized using only one LC tank based on the same principle. However, the particular circuit proposed above has all the nice properties of the balanced circuits in general. More details of the circuit operation may be found in Clarke and Hess 4.

4 K. Clarke, and D. Hess, Communications circuits: analysis and design.

3 Chapter Three 1. Find the L matrix for the following circuits:

+ v1 -

+ v2 -

Ideal

+ v1 -

+ v2 -

n1 : n 2 Solution: The voltages and currents are labeled as shown in the figure. For the circuit on the left, we can write: 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑣𝑣1 = 𝐿𝐿1 𝑖𝑖1 + 𝐿𝐿3 (𝑖𝑖1 + 𝑖𝑖2 ) = (𝐿𝐿1 + 𝐿𝐿3 ) 𝑖𝑖1 + 𝐿𝐿3 𝑖𝑖2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑣𝑣2 = 𝐿𝐿2 𝑖𝑖2 + 𝐿𝐿3 (𝑖𝑖1 + 𝑖𝑖2 ) = 𝐿𝐿3 𝑖𝑖1 + (𝐿𝐿2 + 𝐿𝐿3 ) 𝑖𝑖2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 Thus: 𝐿𝐿 + 𝐿𝐿3 𝐿𝐿3 𝐿𝐿 = � 1 � 𝐿𝐿3 𝐿𝐿2 + 𝐿𝐿3 For the circuit on right, we have: 𝑑𝑑 𝑛𝑛1 𝑣𝑣1 = 𝐿𝐿1 𝑖𝑖1 + 𝑣𝑣2 𝑑𝑑𝑑𝑑 𝑛𝑛2 𝑑𝑑 𝑛𝑛1 𝑛𝑛1 𝑑𝑑 𝑑𝑑 𝑣𝑣2 = 𝐿𝐿2 �𝑖𝑖2 + 𝑖𝑖1 � = � 𝐿𝐿2 � 𝑖𝑖1 + 𝐿𝐿2 𝑖𝑖2 𝑑𝑑𝑑𝑑 𝑛𝑛2 𝑛𝑛2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 Expressing 𝑣𝑣2 based on the currents in the 𝑣𝑣1 equation yields: 𝑑𝑑 𝑛𝑛1 𝑑𝑑 𝑛𝑛1 𝑛𝑛1 𝑑𝑑 𝑛𝑛1 𝑑𝑑 𝑣𝑣1 = 𝐿𝐿1 𝑖𝑖1 + �𝐿𝐿2 �𝑖𝑖2 + 𝑖𝑖1 �� = �𝐿𝐿1 + ( )2 𝐿𝐿2 � 𝑖𝑖1 + ( 𝐿𝐿2 ) 𝑖𝑖2 𝑑𝑑𝑑𝑑 𝑛𝑛2 𝑑𝑑𝑑𝑑 𝑛𝑛2 𝑛𝑛2 𝑑𝑑𝑑𝑑 𝑛𝑛2 𝑑𝑑𝑑𝑑 Thus: 𝑛𝑛1 𝑛𝑛1 𝐿𝐿1 + ( )2 𝐿𝐿2 𝐿𝐿 𝑛𝑛2 𝑛𝑛2 2 𝐿𝐿 = � � 𝑛𝑛1 𝐿𝐿2 𝐿𝐿2 𝑛𝑛2 Notice that for both circuits, 𝐿𝐿12 = 𝐿𝐿21 .

2. Find the transfer function and input impedance of the double-tuned circuit shown below.

+ vIN -

i1 iS

i2 L C

vOUT -

ZIN Solution: For the coupled inductor, given the currents and voltages labeled above, we can write: 𝑗𝑗𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗𝑗𝑗 𝐼𝐼1 𝑉𝑉 �� 𝐼𝐼𝐼𝐼 � = � 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 𝑗𝑗𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗𝑗𝑗 𝐼𝐼2 𝑀𝑀

Or, solving for the currents, knowing that 𝑘𝑘 = 𝐿𝐿 we arrive at:

1 (𝑉𝑉 − 𝑘𝑘𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 ) 𝑗𝑗𝑗𝑗𝑗𝑗(1 − 𝑘𝑘 2 ) 𝐼𝐼𝐼𝐼 1 𝐼𝐼2 = (−𝑘𝑘𝑉𝑉𝐼𝐼𝐼𝐼 + 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 ) 𝑗𝑗𝑗𝑗𝑗𝑗(1 − 𝑘𝑘 2 ) 𝐼𝐼1 =

Furthermore, a KCL at the input yields: 𝐼𝐼𝑠𝑠 = 𝐺𝐺𝐺𝐺𝐼𝐼𝐼𝐼 + 𝑗𝑗𝐶𝐶𝐶𝐶𝐶𝐶𝐼𝐼𝐼𝐼 + 𝐼𝐼1 where 𝐺𝐺 = 1/𝑅𝑅. Also a KCL at the output gives: 𝐺𝐺𝐺𝐺𝑂𝑂𝑂𝑂𝑂𝑂 + 𝑗𝑗𝐶𝐶𝐶𝐶𝐶𝐶𝑂𝑂𝑂𝑂𝑂𝑂 + 𝐼𝐼2 = 0 Replacing 𝐼𝐼1 and 𝐼𝐼2 with their voltage equivalent in the two equations above, we will have: 1 𝑘𝑘 �𝐺𝐺 + 𝑗𝑗𝑗𝑗𝑗𝑗 + � 𝑉𝑉𝐼𝐼𝐼𝐼 − 𝑉𝑉 = 𝐼𝐼𝑠𝑠 2 𝑗𝑗𝑗𝑗𝑗𝑗(1 − 𝑘𝑘 ) 𝑗𝑗𝑗𝑗𝑗𝑗(1 − 𝑘𝑘 2 ) 𝑂𝑂𝑂𝑂𝑂𝑂 −𝑘𝑘 1 𝑉𝑉𝐼𝐼𝐼𝐼 + �𝐺𝐺 + 𝑗𝑗𝑗𝑗𝑗𝑗 + � 𝑉𝑉 =0 2 𝑗𝑗𝑗𝑗𝑗𝑗(1 − 𝑘𝑘 ) 𝑗𝑗𝑗𝑗𝑗𝑗(1 − 𝑘𝑘 2 ) 𝑂𝑂𝑂𝑂𝑂𝑂 We may directly solve the equations above for 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 , but given the symmetry of the circuit, we take a different approach, which is easier and more intuitive. Let us define: 𝑉𝑉𝐼𝐼𝐼𝐼 + 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 𝑉𝑉1 = 2 𝑉𝑉𝐼𝐼𝐼𝐼 − 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 𝑉𝑉2 = 2 By adding and subtracting the main two equations, we will have: 1 �𝐺𝐺 + 𝑗𝑗𝑗𝑗𝑗𝑗 + � 𝑉𝑉 = 𝐼𝐼𝑠𝑠 /2 𝑗𝑗𝑗𝑗𝑗𝑗(1 + 𝑘𝑘) 1 1 �𝐺𝐺 + 𝑗𝑗𝑗𝑗𝑗𝑗 + � 𝑉𝑉 = 𝐼𝐼𝑠𝑠 /2 𝑗𝑗𝑗𝑗𝑗𝑗(1 − 𝑘𝑘) 2 Thus:

1 1 𝑉𝑉1 = 𝑅𝑅𝐼𝐼𝑠𝑠 𝜔𝜔 𝜔𝜔 2 1 + 𝑗𝑗𝑄𝑄1 (𝜔𝜔 − 𝜔𝜔1 ) 1 1 1 𝑉𝑉2 = 𝑅𝑅𝐼𝐼𝑠𝑠 𝜔𝜔 𝜔𝜔 2 1 + 𝑗𝑗𝑄𝑄2 (𝜔𝜔 − 𝜔𝜔2 ) 2 where, as we defined in Chapter 3, 1 𝜔𝜔1/2 = �𝐿𝐿𝐿𝐿(1 ± 𝑘𝑘) 𝑄𝑄1/2 = 𝑅𝑅𝑅𝑅𝜔𝜔1/2 Thus, we finally arrive at: 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 𝑅𝑅 1 1 = [ − ] 2 1 + 𝑗𝑗𝑄𝑄 � 𝜔𝜔 − 𝜔𝜔1 � 1 + 𝑗𝑗𝑄𝑄2 ( 𝜔𝜔 − 𝜔𝜔2 ) 𝐼𝐼𝑠𝑠 1 𝜔𝜔 𝜔𝜔2 𝜔𝜔 𝜔𝜔 1 And: 𝑉𝑉𝐼𝐼𝐼𝐼 𝑅𝑅 1 1 = [ + ] 𝑍𝑍𝐼𝐼𝐼𝐼 = 𝜔𝜔 𝜔𝜔 𝐼𝐼𝑠𝑠 2 1 + 𝑗𝑗𝑄𝑄 � − 1 � 1 + 𝑗𝑗𝑄𝑄2 ( 𝜔𝜔 − 𝜔𝜔2 ) 1 𝜔𝜔 𝜔𝜔2 𝜔𝜔 𝜔𝜔 1 Which is the result we presented in Chapter 3 example. 3. Using parallel-series transformation, find the values of L and C to match the 250Ω amplifier input resistance to 50Ω.

RIN

Matching Network Solution: Using parallel to series conversion, the parallel 𝑅𝑅𝐼𝐼𝐼𝐼 and 𝐶𝐶 may be represented by a series resistance of: 𝑋𝑋𝐶𝐶 2 𝑅𝑅𝐼𝐼𝐼𝐼 𝑅𝑅𝐼𝐼𝐼𝐼 2 + 𝑋𝑋𝐶𝐶 2 And a reactance of: 𝑅𝑅𝐼𝐼𝐼𝐼 2 −𝑋𝑋𝐶𝐶 𝑅𝑅𝐼𝐼𝐼𝐼 2 + 𝑋𝑋𝐶𝐶 2 where 𝑋𝑋𝐶𝐶 is the capacitance impedance. To match to 𝑅𝑅𝑠𝑠 = 50Ω, we shall set:

𝑅𝑅𝐼𝐼𝐼𝐼

which results in:

𝑋𝑋𝐶𝐶 2

𝑅𝑅𝐼𝐼𝐼𝐼 2 + 𝑋𝑋𝐶𝐶 2

= 𝑅𝑅𝑠𝑠

𝑅𝑅𝑠𝑠 𝑋𝑋𝐶𝐶 = 𝑅𝑅𝐼𝐼𝐼𝐼 � = 125Ω 𝑅𝑅𝐼𝐼𝐼𝐼 − 𝑅𝑅𝑠𝑠

With 𝑋𝑋𝐶𝐶 = 125Ω, the corresponding series capacitive reactance is: 𝑅𝑅𝐼𝐼𝐼𝐼 2 −𝑋𝑋𝐶𝐶 = −100Ω 𝑅𝑅𝐼𝐼𝐼𝐼 2 + 𝑋𝑋𝐶𝐶 2 Accordingly, the series inductor reactance must be: 𝑋𝑋𝐿𝐿 = 100Ω.

4. Repeat the problem by using Smith Chart.

Solution: The calculations are shown on the chart below. We start with a resistance of 250Ω, or normalized resistance of 5 shown as the point P1.

x3=0.4 r=1

P3 P2 r2=0.2

P1 P4 x4=-2

For ease of calculations, we convert that to a conductance of 0.2 shown as point P2, and perform the calculations in the admittance domain. To match to 50Ω, we must move P2 on the mirrored version of r=1 circle as highlighted in the chart above. This corresponds to point P3, which is capacitive (note we are dealing with admittance). The other solution on the circle is inductive and hence invalid. On the chart we can read a value of x=0.4 for the point P3. Accordingly, the amount of capacitance needed to move P2 to P3 must be: 0.4 𝑋𝑋𝐶𝐶 −1 = 50

Which leads to 𝑋𝑋𝐶𝐶 = 125Ω, as calculated in the previous problem. When converted back to impedance, we arrive at point P4, which clearly resides on r=1 circle, and whose normalized reactance is read to be -2 on the chart. Consequently, we shall add and inductance value of 𝑋𝑋𝐿𝐿 = 2 × 50 = 100Ω to match to 50Ω.

5. Using parallel-series transformation, design an LC matching network to match an amplifier with an input impedance of 20Ω||1pF to 50Ω.

20Ω

1pF

-80Ω

-80×202/(202+802)=-4.7Ω

20×802/(202+802)=19Ω

Solution: We shall first represent the amplifier with a series equivalent circuit as shown below. The steps are obvious, and depicted in the figure. We assume the frequency of operation is 2GHz, and hence the capacitance impedance will be -80Ω.

XL’=XL-4.7

The series inductor from the matching network, along with the series capacitance from the amplifier itself may be represented by a net reactance of 𝑋𝑋𝐿𝐿′ as follows: 𝑋𝑋𝐿𝐿′ = 𝑋𝑋𝐿𝐿 − 4.7Ω Converting the series combination of 𝑋𝑋𝐿𝐿′ and the amplifier resistance (which is 19Ω), we arrive at a new parallel resistance of: 192 + 𝑋𝑋𝐿𝐿 ′2 19 Which is set to 50Ω to create the desired impedance for matching. This results in 𝑋𝑋𝐿𝐿′ = 24Ω, or 𝑋𝑋𝐿𝐿 = 28.7Ω. From this, the shunt reactance is:

(192+XL’2)/19

19Ω

-4.7Ω

(192+XL’2)/XL’

Next, we will add the matching network as shown below.

192 + 𝑋𝑋𝐿𝐿 ′2 = 39Ω 𝑋𝑋𝐿𝐿 ′

And thus, 𝑋𝑋𝐶𝐶 = −39Ω

6. Plot on Smith chart the impedance of a series RLC circuit over frequency. Solution: The impedance of the series RLC circuit is: 1 𝑍𝑍 = 𝑅𝑅 + 𝑗𝑗𝑗𝑗𝑗𝑗 + = 𝑅𝑅 + 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗𝑗𝑗 And the reactance is: 𝐿𝐿(𝜔𝜔2 − 𝜔𝜔0 2 ) 𝑋𝑋 = 𝜔𝜔 1 where 𝜔𝜔0 = is the resonance frequency. The real part of the impedance is frequency √𝐿𝐿𝐿𝐿

𝑅𝑅

independent, and hence the impedance resides on the constant 𝑟𝑟 = 𝑅𝑅 circle as shown below. 0

Above the resonance, the reactive part is negative, and the impedance moves on the constant 𝑟𝑟 circle, but in the bottom half of the chart. On the other hand, below resonance, the impedance is inductive and it moves on the top half of the constant 𝑟𝑟 circle.

ω<ω0 ω=ω0 ω>ω0

ω0

At 𝜔𝜔0 , the impedance is purely resistive, and is shown above on the chart. At DC or infinity, it resides on the open circuit point at the far right of the chart. 7. Consider the following two-port system terminated to admittance 𝑌𝑌𝐿𝐿 in the second port and to a voltage source 𝑣𝑣𝑠𝑠 with output admittance 𝑌𝑌𝑠𝑠 in the first port. (a) Prove that:

𝑌𝑌12 𝑌𝑌21 𝑌𝑌22 + 𝑌𝑌𝐿𝐿 𝑌𝑌12 𝑌𝑌21 𝑌𝑌𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑌𝑌22 − 𝑌𝑌11 + 𝑌𝑌𝑆𝑆 𝑉𝑉𝐿𝐿 𝑌𝑌𝑆𝑆 𝑌𝑌21 𝐴𝐴𝑣𝑣 = = − (𝑌𝑌𝑆𝑆 + 𝑌𝑌11 )(𝑌𝑌𝐿𝐿 + 𝑌𝑌22 ) − 𝑌𝑌12 𝑌𝑌21 𝑉𝑉𝑠𝑠 (b) Repeat part (a) and calculate 𝑍𝑍𝑖𝑖𝑖𝑖 , 𝑍𝑍𝑜𝑜𝑜𝑜𝑜𝑜 and 𝐴𝐴𝑣𝑣 in terms of Z parameters. 𝑌𝑌𝑖𝑖𝑖𝑖 = 𝑌𝑌11 −

[Y]

YIN

YOUT

Solution: Assuming voltage and current phasors of 𝑉𝑉1, 𝑉𝑉2, 𝐼𝐼1 , and 𝐼𝐼2 at the two ports, we have: 𝐼𝐼1 = 𝑌𝑌𝑠𝑠 (𝑉𝑉𝑠𝑠 − 𝑉𝑉1 ) 𝐼𝐼2 + 𝑌𝑌𝐿𝐿 𝑉𝑉2 = 0 Also from Y matrix definition: 𝐼𝐼1 = 𝑌𝑌11 𝑉𝑉1 + 𝑌𝑌12 𝑉𝑉2 𝐼𝐼2 = 𝑌𝑌21 𝑉𝑉1 + 𝑌𝑌22 𝑉𝑉2 Accordingly: (𝑌𝑌𝐿𝐿 + 𝑌𝑌22 )𝑉𝑉2 + 𝑌𝑌21 𝑉𝑉1 = 0 Hence: −𝑌𝑌21 𝑉𝑉1 𝐼𝐼1 = 𝑌𝑌11 𝑉𝑉1 + 𝑌𝑌12 𝑉𝑉2 = 𝑌𝑌11 𝑉𝑉1 + 𝑌𝑌12 𝑌𝑌𝐿𝐿 + 𝑌𝑌22 which leads to the input admittance: 𝐼𝐼1 𝑌𝑌12 𝑌𝑌21 𝑌𝑌𝐼𝐼𝐼𝐼 = = 𝑌𝑌11 − 𝑉𝑉1 𝑌𝑌22 + 𝑌𝑌𝐿𝐿 For the output admittance, by definition the source voltage must be shorted. Thus: 𝐼𝐼1 = −𝑌𝑌𝑠𝑠 𝑉𝑉1 , which leads to: 𝐼𝐼1 = 𝑌𝑌11 𝑉𝑉1 + 𝑌𝑌12 𝑉𝑉2 = −𝑌𝑌𝑠𝑠 𝑉𝑉1 Or: 𝑌𝑌12 𝑉𝑉2 𝑉𝑉1 = − 𝑌𝑌11 + 𝑌𝑌𝑠𝑠 Thus: −𝑌𝑌12 𝑉𝑉2 𝐼𝐼2 = 𝑌𝑌21 𝑉𝑉1 + 𝑌𝑌22 𝑉𝑉2 = 𝑌𝑌21 + 𝑌𝑌22 𝑉𝑉2 𝑌𝑌11 + 𝑌𝑌𝑠𝑠 which leads to the output admittance:

𝐼𝐼2 𝑌𝑌12 𝑌𝑌21 = 𝑌𝑌22 − 𝑉𝑉2 𝑌𝑌11 + 𝑌𝑌𝑠𝑠 Knowing the input admittance, the output admittance may have been deduced by inspection and given the symmetry of the circuit. To calculate the voltage gain, we have: 𝐼𝐼1 = 𝑌𝑌𝑠𝑠 (𝑉𝑉𝑠𝑠 − 𝑉𝑉1 ) = 𝑌𝑌11 𝑉𝑉1 + 𝑌𝑌12 𝑉𝑉2 (𝑌𝑌𝐿𝐿 + 𝑌𝑌22 )𝑉𝑉2 + 𝑌𝑌21 𝑉𝑉1 = 0 Thus: −(𝑌𝑌𝐿𝐿 + 𝑌𝑌22 ) 𝑌𝑌𝑠𝑠 𝑉𝑉𝑠𝑠 = (𝑌𝑌𝑠𝑠 + 𝑌𝑌11 )𝑉𝑉1 + 𝑌𝑌12 𝑉𝑉2 = (𝑌𝑌𝑠𝑠 + 𝑌𝑌11 ) 𝑉𝑉2 + 𝑌𝑌12 𝑉𝑉2 𝑌𝑌21 which leads to the voltage gain: 𝑉𝑉2 𝑌𝑌𝑆𝑆 𝑌𝑌21 𝐴𝐴𝑣𝑣 = = − (𝑌𝑌𝑆𝑆 + 𝑌𝑌11 )(𝑌𝑌𝐿𝐿 + 𝑌𝑌22 ) − 𝑌𝑌12 𝑌𝑌21 𝑉𝑉𝑠𝑠 Note that clearly: 𝑉𝑉𝐿𝐿 = 𝑉𝑉2 . Similar procedure leads to very similar equations expressed in terms of impedances: 𝑍𝑍12 𝑍𝑍21 𝑍𝑍𝑖𝑖𝑖𝑖 = 𝑍𝑍11 − 𝑍𝑍22 + 𝑍𝑍𝐿𝐿 𝑍𝑍12 𝑍𝑍21 𝑍𝑍𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑍𝑍22 − 𝑍𝑍11 + 𝑍𝑍𝑆𝑆 𝑍𝑍𝐿𝐿 𝑍𝑍21 𝐴𝐴𝑣𝑣 = − (𝑍𝑍𝑆𝑆 + 𝑍𝑍11 )(𝑍𝑍𝐿𝐿 + 𝑍𝑍22 ) − 𝑍𝑍12 𝑍𝑍21 This is expected from duality. 𝑌𝑌𝑂𝑂𝑂𝑂𝑂𝑂 =

8. In problem 7, prove that the condition of 𝑌𝑌𝑠𝑠 + 𝑌𝑌𝑖𝑖𝑖𝑖 = 0 or 𝑌𝑌𝐿𝐿 + 𝑌𝑌𝑜𝑜𝑜𝑜𝑜𝑜 = 0 for the boundary between stable and unstable regions both lead to (𝑌𝑌𝑠𝑠 + 𝑌𝑌11 )(𝑌𝑌𝐿𝐿 + 𝑌𝑌22 ) − 𝑌𝑌12 𝑌𝑌21 = 0, which is equivalent to the condition that 𝐴𝐴𝑣𝑣 approaches infinity.

Solution: The condition 𝑌𝑌𝑠𝑠 + 𝑌𝑌𝑖𝑖𝑖𝑖 = 0 for instance leads to: 𝑌𝑌12 𝑌𝑌21 𝑌𝑌𝑠𝑠 + 𝑌𝑌𝑖𝑖𝑖𝑖 = 𝑌𝑌𝑠𝑠 + 𝑌𝑌11 − =0 𝑌𝑌22 + 𝑌𝑌𝐿𝐿 Thus: (𝑌𝑌𝑠𝑠 + 𝑌𝑌11 )(𝑌𝑌𝐿𝐿 + 𝑌𝑌22 ) − 𝑌𝑌12 𝑌𝑌21 = 0 Exact same equation is deduced by setting 𝑌𝑌𝐿𝐿 + 𝑌𝑌𝑜𝑜𝑜𝑜𝑜𝑜 to zero. In either condition the denominator of 𝐴𝐴𝑣𝑣 approaches zero, or 𝐴𝐴𝑣𝑣 approaches infinity.

9. In problem 7, when the input and output are simultaneously conjugate matched the power gain becomes maximum. This would happen if 𝑌𝑌𝑠𝑠 and 𝑌𝑌𝐿𝐿 are found such that the following two equations are met: 𝑌𝑌12 𝑌𝑌21 𝑌𝑌𝑖𝑖𝑖𝑖 = 𝑌𝑌11 − = 𝑌𝑌𝑠𝑠∗ 𝑌𝑌22 + 𝑌𝑌𝐿𝐿

𝑌𝑌12 𝑌𝑌21 = 𝑌𝑌𝐿𝐿∗ 𝑌𝑌11 + 𝑌𝑌𝑠𝑠 Such 𝑌𝑌𝑠𝑠 and 𝑌𝑌𝐿𝐿 are called 𝑌𝑌𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 and 𝑌𝑌𝐿𝐿,𝑜𝑜𝑜𝑜𝑜𝑜 . (a) Prove that 𝑌𝑌𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 and 𝑌𝑌𝐿𝐿,𝑜𝑜𝑜𝑜𝑜𝑜 are given by: 𝑌𝑌𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑌𝑌22 −

𝑌𝑌12 𝑌𝑌21 + |𝑌𝑌12 𝑌𝑌21 |(𝐾𝐾 + √𝐾𝐾 2 − 1) 2𝑅𝑅𝑅𝑅(𝑌𝑌22 ) 𝑌𝑌12 𝑌𝑌21 + |𝑌𝑌12 𝑌𝑌21 |(𝐾𝐾 + √𝐾𝐾 2 − 1) 𝑌𝑌𝐿𝐿,𝑜𝑜𝑜𝑜𝑜𝑜 = 2𝑅𝑅𝑅𝑅(𝑌𝑌11 ) 𝑌𝑌𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 =

Where 𝐾𝐾 is equal to: 𝐾𝐾 =

2𝑅𝑅𝑅𝑅(𝑌𝑌11 )𝑅𝑅𝑅𝑅(𝑌𝑌22 )−𝑅𝑅𝑅𝑅(𝑌𝑌12 𝑌𝑌21 ) |𝑌𝑌12 𝑌𝑌21 |

𝑌𝑌

(b) Prove that under the above optimum condition the power gain is equal to: 𝐺𝐺𝑝𝑝 = �𝑌𝑌21 � (𝐾𝐾 − 12

√𝐾𝐾 2 − 1), which also proves that for a reciprocal network (𝑌𝑌21 = 𝑌𝑌12 ) the power gain is less than unity. As a special case, an LTI passive network cannot amplify the power. Solution: Solving the set of equations:

𝑌𝑌12 𝑌𝑌21 ⎧𝑌𝑌𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 ∗ = 𝑌𝑌11 − ⎪ 𝑌𝑌22 + 𝑌𝑌𝐿𝐿,𝑜𝑜𝑜𝑜𝑜𝑜 𝑌𝑌12 𝑌𝑌21 ⎨𝑌𝑌 ∗ ⎪ 𝐿𝐿,𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑌𝑌22 − 𝑌𝑌 + 𝑌𝑌 11 𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 ⎩ leads to the answer. The math is too difficult however, and instead, we take a more intuitive approach. We shall first find the power gain and the available gain of the circuit. By definition: 𝑃𝑃𝐿𝐿 𝑉𝑉2 2 𝑅𝑅𝑅𝑅[𝑌𝑌𝐿𝐿 ] 𝐺𝐺𝑝𝑝 = =� � 𝑃𝑃𝐼𝐼𝐼𝐼 𝑉𝑉1 𝑅𝑅𝑅𝑅[𝑌𝑌𝐼𝐼𝐼𝐼 ] From the previous problem, we have: 𝑌𝑌12 𝑌𝑌21 𝑌𝑌𝐼𝐼𝐼𝐼 = 𝑌𝑌11 − 𝑌𝑌22 + 𝑌𝑌𝐿𝐿 (𝑌𝑌𝐿𝐿 + 𝑌𝑌22 )𝑉𝑉2 + 𝑌𝑌21 𝑉𝑉1 = 0 Thus: 𝑌𝑌21 2 𝑅𝑅𝑅𝑅[𝑌𝑌𝐿𝐿 ] 𝐺𝐺𝑝𝑝 = � � 𝑌𝑌𝐿𝐿 + 𝑌𝑌22 𝑅𝑅𝑅𝑅[𝑌𝑌𝐼𝐼𝐼𝐼 ] Interestingly, the power gain is only a function of 𝑌𝑌𝐿𝐿 (and not 𝑌𝑌𝑠𝑠 ). Thus, assuming 𝑌𝑌𝐿𝐿 = 𝐺𝐺𝐿𝐿 + 𝑗𝑗𝐵𝐵𝐿𝐿 , we can argue that the optimum condition occurs by maximizing the power gain. That is by setting: 𝜕𝜕𝜕𝜕𝑝𝑝 =0 𝜕𝜕𝐺𝐺𝐿𝐿 𝜕𝜕𝜕𝜕𝑝𝑝 =0 𝜕𝜕𝐵𝐵𝐿𝐿 Similarly, the available gain is: