Lecture 2: Telegrapher Equations for Transmission Lines. Power Flow

Whites, EE 481 Lecture 2 Page 1 of 13

Lecture 2: Telegrapher Equations For Transmission Lines. Power Flow.

Microstrip is one method for making electrical connections in a microwave circuit. It is constructed with a ground plane on one side of a PCB and lands on the other:

Microstrip is an example of a transmission line, though technically it is only an approximate model for microstrip, as we will see later in this course.

Why TLs? Imagine two ICs are connected together as shown:

When the voltage at A changes state, does that new voltage appear at B instantaneously? No, of course not.

If these two points are separated by a large electrical distance, there will be a propagation delay as the change in state (electrical signal) travels to B. Not an instantaneous effect.

In microwave circuits, even distances as small as a few inches may be “far” and the propagation delay for a voltage signal to appear at another IC may be significant.

This propagation of voltage signals is modeled as a “transmission line” (TL). We will see that voltage and current can propagate along a TL as waves! Fantastic.

The transmission line model can be used to solve many, many types of high frequency problems, either exactly or approximately: Coaxial cable. Two-wire. Microstrip, stripline, coplanar waveguide, etc.

All true TLs share one common characteristic: the E and H fields are all perpendicular to the direction of propagation, which is the long axis of the geometry. These are called TEM fields for transverse electric and magnetic fields.

An excellent example of a TL is a coaxial cable. On a TL, the voltage and current vary along the structure in time t and Whites, EE 481 Lecture 2 Page 3 of 13 spatially in the z direction, as indicated in the figure below. There are no instantaneous effects.

z Fig. 1

A common circuit symbol for a TL is the two-wire (parallel) symbol to indicate any transmission line. For example, the equivalent circuit for the coaxial structure shown above is:

Analysis of Transmission Lines

As mentioned, on a TL the voltage and current vary along the structure in time (t) and in distance (z), as indicated in the figure above. There are no instantaneous effects. Whites, EE 481 Lecture 2 Page 4 of 13

i z, t

+ v z, t z - i z, t

How do we solve for v(z,t) and i(z,t)? We first need to develop the governing equations for the voltage and current, and then solve these equations.

Notice in Fig. 1 above that there is conduction current in the center conductor and outer shield of the coaxial cable, and a displacement current between these two conductors where the electric field E is varying with time. Each of these currents has an associated impedance: Conduction current impedance effects: o Resistance, R, due to losses in the conductors, o Inductance, L, due to the current in the conductors and the magnetic flux linking the current path. Displacement current impedance effects: o Conductance, G, due to losses in the dielectric between the conductors, o Capacitance, C, due to the time varying electric field between the two conductors.

To develop the governing equations for V z, t and I z, t , we will consider only a small section z of the TL. This z is so Whites, EE 481 Lecture 2 Page 5 of 13 small that the electrical effects are occurring instantaneously and we can simply use circuit theory to draw the relationships between the conduction and displacement currents. This equivalent circuit is shown below:

i z, t R z L z i z z, t + + v z, t C z G z v z z, t - -

z Fig. 2 The variables R, L, C, and G are distributed (or per-unit length, PUL) parameters with units of /m, H/m, F/m, and S/m, respectively. We will sometimes ignore losses in this course.

A finite length of TL can be constructed by cascading many, many of these subsections along the total length of the TL. In the case of a lossless TL where R = G = 0 this cascade appears as: L z L z L z L z

C z C z C z C z

z z z z

This is a general model: it applies to any TL regardless of its cross sectional shape provided the actual electromagnetic field is TEM. Whites, EE 481 Lecture 2 Page 6 of 13

However, the PUL-parameter values change depending on the specific geometry (whether it is a microstrip, stripline, two-wire, coax, or other geometry) and the construction materials.

Transmission Line Equations

To develop the governing equation for v z, t , apply KVL in Fig. 2 above (ignoring losses) i z, t v z,, t L z v z z t (2.1a),(1) t Similarly, for the current i z, t apply KCL at the node v z z, t i z,, t C z i z z t (2.1b),(2) t Then:

1. Divide (1) by z : v z z,,, t v z t i z t L (3) z t In the limit as z 0, the term on the LHS in (3) is the forward difference definition of derivative. Hence, v z,, t i z t L (2.2a),(4) z t

2. Divide (2) by z : Whites, EE 481 Lecture 2 Page 7 of 13

i z z,, t i z t v z z0 , t C (5) z t Again, in the limit as z 0 the term on the LHS is the forward difference definition of derivative. Hence, i z,, t v z t C (2.2b),(6) z t

Equations (4) and (6) are a pair of coupled first order partial differential equations (PDEs) for v z, t and i z, t . These two equations are called the telegrapher equations or the transmission line equations.

Recap: We expect that v and i are not constant along microwave circuit interconnects. Rather, (4) and (6) dictate how v and i vary along the TL at all times.

TL Wave Equations

We will now combine (4) and (6) in a special way to form two equations, each a function of v or i only.

To do this, take of (4) and of (6): z t 2v z,, t 2 i z t (4) : L (7) z z2 z t Whites, EE 481 Lecture 2 Page 8 of 13

2i z,, t 2 v z t (6) : C (8) t t z t 2 Substituting (8) into (7) gives: 2v z,, t 2 v z t LC (9) z2 t 2 Similarly, 2i z,, t 2 i z t LC (10) z2 t 2

Equations (9) and (10) are the governing equations for the z and t dependence of v and i. These are very special equations. In fact, they are wave equations for v and i!

We will define the (phase) velocity of these waveforms as 1 v [m/s] (2.16) p LC so that (9) becomes 2v z,, t1 2 v z t 2 2 2 (11) z vp t

Voltage Wave Equation Solutions

There are two general solutions to (11): 1. v z, t v t z (12) vp Whites, EE 481 Lecture 2 Page 9 of 13

v is any twice-differentiable function that contains t, z, and

vp in the form of the argument shown. It can be verified that (12) is a solution to (11) by substituting (12) into (11) and showing that the LHS equals the RHS.

Equation (12) represents a wave traveling in the +z direction

with speed vp 1/ LC m/s.

To see this, consider the example below with vp = 1 m/s:

At t = 1 s, focus on the peak located at z = 1.5 m. Then, z 1.5 s t 11 0.5 vp v p The argument s stays constant for varying t and z. Therefore, at t = 2 s, for example, then

2 z s0.5 t 1 vp Therefore, z 2.5 m So the peak has now moved to position z = 2.5 m at t = 2 s. Whites, EE 481 Lecture 2 Page 10 of 13

Likewise, every point on this function moves the same distance (1 m) in this time (1 s). This is called wave motion.

The speed of this movement is z 1 m m 1 v t 1 s s p

2. v z, t v t z (13) vp This is the second general solution to (11). This function v

represents a wave moving in the -z direction with speed vp.

The complete solution to the wave equation (11) is the sum of (12) and (13) v z, t v tz v t z (14) vp v p v and v can be any suitably differentiable functions, but with arguments as shown.

Current Wave Equation Solutions

A similar analysis can be performed for current waves on the TL. The governing equation for i z, t is 2i z,, t1 2 i z t 2 2 2 (15) z vp t Whites, EE 481 Lecture 2 Page 11 of 13

The complete general solution to this current wave equation can be determined in a manner similar to the voltage as i z, t i tz i t z (16) vp v p

+z wave - z wave

Furthermore, the function i can be related to the function v and i can be related to v .

For example, substituting i t z and v t z into (6), vp vp differentiating then integrating gives 1 i Cv vp or

i vp Cv (17) But, 1 C v C C p LC L We will define L Z [ ] (2.13),(18) 0 C as the characteristic impedance of the transmission line. (Note that in some texts, Z0 is denoted as Rc , the characteristic resistance of the TL).

With (18), (17) can be written as Whites, EE 481 Lecture 2 Page 12 of 13

z v t v i t z p (19) vp Z0 Similarly, it can be shown that z v t v i t z p (20) vp Z0 The minus sign results since the current is in the -z direction.

Finally, substituting (19) and (20) into (16) gives 1 1 i z, t v tz v t z (21) vp v p ZZ0 0 This equation as well as (14) v z, t v tz v t z (22) vp v p are the general wave solutions for v and i on a transmission line.

Power Flow

These voltage and current waves transport power along the TL. The power flow carried by the forward wave p z, t is v2 z, t p z,,, t v z t i z t (23) Z0 which is positive indicating power flows in the +z direction. Similarly, the power flow of the reverse wave is Whites, EE 481 Lecture 2 Page 13 of 13

v2 z, t p z,,, t v z t i z t (24) Z0 which is negative indicating power flows in the -z direction. Whites, EE 481/581 Lecture 3 Page 1 of 9

Lecture 3: Phasor Wave Solutions to the Telegrapher Equations. Termination of TLs.

We will continue our TL review by considering the steady state response of TLs to sinusoidal excitation.

Consider the following TL in the sinusoidal steady state:

l Zs

+ Vs Z0, vp ZL - z

We derived in the previous lecture the wave equations for the voltage and current as 2v z,, t1 2 v z t 2 2 2 (1) z vp t 2i z,, t1 2 i z t and 2 2 2 (2) z vp t

For sinusoidal steady state, we will employ the phasor representation of the voltage and current as v z, t e V z e j t (3) i z, t e I z e j t (4) where V z and I z are spatial phasor functions.

Substituting (3) into (1) gives 2 2 d V z 1 2 2 2j V z 2 V z (5) dz vp v p We define LC [rad/m] (2.12a),(6) as the phase constant for reasons that will be apparent shortly. (L and C are the usual TL per-unit-length parameters.)

2 2 2 From (6) LC 2 vp Substituting this into (5) gives d2 V z 2V z 0 (7) dz2 Similarly, from (4) and (2) we can derive d2 I z 2I z 0 (8) dz 2 Equations (7) and (8) are the wave equations for V and I in the frequency domain (i.e., the phasor domain).

The solutions to these two second-order ordinary differential equations are j z j z V z Vo e V o e (2.14a),(9) j z j z and I z Io e I o e (10)

VVIIo,,, o o o are complex constants. Whites, EE 481/581 Lecture 3 Page 3 of 9

We can confirm the correctness of these two solutions by direct j z substitution into (7) and (8). For example, substituting Vo e from (9) into (7) gives 2 j z 2 ? Vo j e V z 0 2j z 2 j z ? or Vo e V o e 0 j z which is indeed true. Therefore, Vo e in (9) is a valid solution to (7).

The constants Io and Io in (10) can be expressed in terms of Vo and Vo . In particular, it can be shown that

Vo Io (11) Z0

Vo and Io (12) Z0 If we substitute (11) and (12) into (10) we find that VV I zo ej z o e j z (2.14b),(13) ZZ0 0

j z j z and V z Vo e V o e (2.14a),(14) Both of these equations should be committed to memory. They are the general form of phasor voltages and currents on transmission lines. Whites, EE 481/581 Lecture 3 Page 4 of 9

The first terms in (13) and (14) are the phasor representation of waves propagating in the +z direction along the TL. The second terms in both equations represent waves propagating in the -z direction.

Discussion

As stated above, the first terms in (13) and (14) are the phasor representation of waves traveling in the +z direction. To see this, convert the first term in (14) to the time domain: j z j t j j t z vzt, eVeeo eVee o

Vocos t z V o cos t z

z Vo cos t vp We can clearly see in this last result that we have a function

of time with argument t z/ vp . From our previous discussions with TLs we recognize that this is a wave that is

propagating in the +z direction with speed vp.

j z j z Similarly, we can show that Vo e (and Io e ) are waves propagating in the -z direction. Whites, EE 481/581 Lecture 3 Page 5 of 9

Generality of TL Theory

It was mentioned in the last lecture that transmission lines could be used to model the voltage and current waves on any structure supporting only TEM waves.

What changes from structure to structure are the values for L, C, R, and G, as shown below in Table 2.1 for three common TEM structures:

Consequently, the values of Z0, vp, (and ) generally all change from one TL to another. The numerical values can also be changed within a type of TL by varying the dimensions and construction materials. Whites, EE 481/581 Lecture 3 Page 6 of 9

Termination of Transmission Lines

We will now consider the termination of TLs that are excited by sinusoidal steady state sources.

Adding terminations produces reflections so that the total voltage and current anywhere on the TL are sums of forward and reverse propagating waves. From (13) and (14), the voltage and current on the TL will have the form j z j z V z Vo e V o e (15) VV and I zo ej z o e j z (16) ZZ0 0

The “lumped load” ZL that terminates the TL is considered a boundary condition for the voltage and current in (15) and (16):

V z0 I z 0 ZL (17)

Therefore, we can solve for Vo in terms of Vo by applying this boundary condition as:

from (15): V z0 Vo V o 1 from (16): I z0 Vo V o Z0 Whites, EE 481/581 Lecture 3 Page 7 of 9

Forming the ratio of these quantities gives

V 0 VVo o Z0 ZL IVV0 o o

Solving for VVo/ o , and defining this ratio as the voltage reflection coefficient at the load (z = 0), we find VZZ o L 0 (2.35),(18) L VZZ oz 0 L 0

Note that, in general, L is a complex number since ZL is complex.

Example N3.1: For an open-circuit load on the TL shown above, compute the load reflection coefficient and sketch the voltage and current magnitude on the TL.

For an open circuit load ZL (i.e., an extremely large impedance), so that from (18) L 1. With this value of L , then from (15) and (16) the solutions for V z and I z are

j zVo j z V z Vo e e2 V o cos z (2.46a),(19) Vo

V V j2 V and I zo ej z o e j z o sin z (2.46b),(20) ZVZ0o 0

L Whites, EE 481/581 Lecture 3 Page 8 of 9

These two equations (19) and (20) are not traveling waves. So, where has the traveling wave behavior in V z and I z gone? The interference between the incident and reflected waves produces standing waves, such as these.

V z and I z are shown here for the open-circuit load:

/2 between |voltage| and V z |current| maxima or minima

2 Vo

z I z

2 Vo

Open-circuit Z0 , load

3 0 z 4 2 4

Phasor voltage and current magnitudes vary noticeably along TLs provided the TL length is greater than about 0.05 or so.

Remember, though, that we are plotting the magnitude of phasor voltages and currents. The voltage and current oscillate Whites, EE 481/581 Lecture 3 Page 9 of 9 as functions of time with amplitudes equal to V z and I z , respectively, at each point along the TL.

|V(z)|

Zs z

+ Vs -

t t t Whites, EE 481/581 Lecture 4 Page 1 of 12

Lecture 4: TL Input Impedance, Time Average Power, Return and Insertion Losses. VSWR.

Example N4.1: Determine an expression for the voltage at the input to the TL assuming Rs = Z0:

l Rs

+ +

Vs Vg Z0, - - z

Zin z = 0

To calculate the input voltage Vg , we’ll first determine the effective impedance seen at the TL input terminals seen looking towards the load at z = 0. This is called the input impedance

Zin .

Forming the ratio of (19) and (20) from the previous lecture gives V l2 V cos l Z o jZcot l [ ] inI l j2 V 0 o sin l Z0 In other words, the input impedance is purely reactive

ZinjX in where X inZ 0 cot l (2.46c) A plot of this reactance is shown in Fig. 2.8c of the text.

Because of the impedance transformation properties of TLs, this input impedance Zin will generally not equal Z L . Remarkably, Zin can assume any value of reactance from to depending on the length l of the TL.

An equivalent circuit can now be constructed at the input to the

TL by using Rs and Zin as

Using voltage division,

Zin jZ0 cot l VVVg s s ZinZ 0 jZ 0cot l Z 0

This circuit voltage Vg is also the voltage on the TL at z l . That is, from (19) in the previous lecture

V z l2 Vo cos l

Since Vg V z l , we can equate these two voltages giving

jZ0 cot l 2Vo cos l V s jZ0cot l Z 0 More often than not, expressions of this type are used to determine Vo in terms of Vs , Rs , and Z L . We’ll see more on this topic in Lecture 5. Whites, EE 481/581 Lecture 4 Page 3 of 12

Input Impedance of a Transmission Line

In problems like the one in the last example, it is helpful to have an analytical expression for the input impedance of an arbitrarily terminated TL.

As we saw in the last lecture, the voltage and current everywhere on a homogeneous TL are j z j z V z Vo e V o e (2.34a),(1) VV and I zo ej z o e j z (2.34b),(2) ZZ0 0 We can readily construct an input impedance expression for a TL of length l by dividing (1) and (2) for some arbitrary load reflection coefficient L at z = 0:

j lVo j l j l j l V l Vo e e V o e L e (3) Vo

VVVoj l o j l o j l j l I l e e eL e (4) ZVZ0o 0 such that V ej l e j l j2 l V l o L 1 Le ZZin 0 j2 l (2.43) I lVo j l j l 1 L e eL e Z0

Substituting for L and simplifying gives Whites, EE 481/581 Lecture 4 Page 4 of 12

ZL jZ0 tan l ZZin 0 [ ] (2.44),(5) Z0 jZL tan l This is the input impedance for a lossless TL of length l and characteristic impedance Z0 with an arbitrary load ZL.

Three special cases are:

1. With an open circuit load (Z L ), (5) yields

ZinjZ 0 cot l [ ] (2.46c),(6) as we derived in the last lecture.

2. With a short circuit load (Z L 0), (5) yields

ZinjZ 0 tan l [ ] (2.45c),(7) A plot of this input reactance is shown in Fig. 2.6c.

3. With the resistive load Z L Z0 , (5) yields

ZinZ 0 [ ]

The input impedance is Z0 regardless of the length of the TL.

All of these last three expressions should be committed to memory. You will use them often in microwave circuits.

Note that both input impedances (6) and (7) are purely reactive, which is expected since neither type can dissipate energy, assuming lossless TLs. Whites, EE 481/581 Lecture 4 Page 5 of 12

Time Average Power Flow on TLs

A hugely important part of microwave engineering is delivering signal power to a load. Examples include efficiently delivering power from a source to an antenna, or maximizing the power delivered from a filter to an amplifier.

Often, the “power” we are ultimately concerned with is the time average power Pav, expressed as 1 P z e V z I z (8) av 2 This expression is similar to that used in circuit analysis.

Substituting V(z) and I(z) from (1) and (2) into (8) gives 2 V 1 o *j 2 l j 2 l 2 Pav z e1 LLL e e (9) 2 Z0 Notice that the second and third terms are conjugates so that j2 l j 2 l j 2 l LLLe e j2 m e The real part of this sum is zero. Consequently, (9) simplifies to 2 1 Vo 2 Pav 1 L [W] (2.37),(10) 2 Z0 Since this power is not a function of z (true for a lossless and homogeneous TL), a z-dependence is no longer indicated for

Pav. Whites, EE 481/581 Lecture 4 Page 6 of 12

It is important to reiterate that we’re assuming a lossless TL throughout this analysis. These results are not valid for lossy TLs.

Equation (10) is very illuminating. It shows that the total time average power delivered to a load is equal to the incident time 2 average power VZo 2 0 minus the reflected time average 2 2 power VZo 2 0 .

The relative reflected time average power from an arbitrary load 2 on a lossless TL is the ratio of the two terms in (10) = L .

From (10) we see that if the load is entirely reactive so that

L 1, then Pav 0 and no time average power is delivered to the load, as expected. For all other passive loads, Pav 0.

The relative time average power that is not delivered to the load can be considered a “loss” since the signal from the generator was intended to be completely transported – not returned to the generator.

This return loss (RL) is defined as 2 RL 10log10L 20log 10 L dB (2.38),(11) The two extremes for return loss with a passive load are:

1. A matched load where L 0 and RL dB (no reflected power), and Whites, EE 481/581 Lecture 4 Page 7 of 12

2. A reactive load where L 1 and RL 0 dB (all power reflected).

Transmission Coefficient and Insertion Loss

Insertion loss is a term closely related to return loss. Consider a junction of two semi-infinite TLs as shown in Fig. 2.9:

Z0, 0 Z1, 1

z z = 0

We’ll arbitrarily assume that a voltage wave is incident from z < 0. From (1), the total voltages in the two regions are:

j0 z j0 z V z Vo e e z 0 (2.50a),(12)

j1 z and V z V1 e z 0 (13)

In these expressions, Vo is the complex amplitude of the incident voltage wave and V1 is the complex amplitude of the transmitted voltage wave. There is no reflection on the right- hand TL so there is only the outgoing term.

We will define the voltage transmission coefficient T as

j1 z V1 e T z 0 (14) V e j0 z o z 0 Whites, EE 481/581 Lecture 4 Page 8 of 12 so that (13) can be written as

V1 j1 z j 1 z V z Vo e V o Te z 0 (2.50b),(15) Vo

At the junction of these TLs, the two boundary conditions are that the voltage and current are each continuous across the junction. Equating voltages (12) and (15) at z = 0 gives T 1 (2.51),(16) The reflection coefficient for this junction of two TLs is Z Z 1 0 (17) Z1Z 0 Substituting (17) into (16) and simplifying gives 2Z T 1 (2.51),(18) Z1Z 0

This transmission coefficient between two “ports” in a microwave circuit is often expressed in decibels as the insertion loss, IL: 2 IL 10log10TT 20log 10 dB (2.52),(19) The two extremes for insertion loss in a passive circuit are: 1. A matched junction where 0, so that T 1 and IL 0 dB (all power transmitted), and 2. A completely reflecting junction where 1, so that T 0 and IL dB (no power transmitted). Whites, EE 481/581 Lecture 4 Page 9 of 12

Voltage Standing Wave Ratio

As we’ve seen, there is generally some amount of reflection of voltage and current waves from discontinuities and loads attached to a TL.

To help quantify the amount of interference that exists on a TL, we define the voltage standing wave ratio (VSWR) as V z VSWR max (2.41),(20) V z min where V z and V z are the maximum and minimum max min voltage magnitudes, respectively, found anywhere on a long TL.

As shown in the text, we can determine expressions for these quantities. Specifically, V z V 1 (2.40a),(21) max o L and V z V 1 (2.40b),(22) min o L

Substituting these into the definition of VSWR in (20) gives 1 VSWR= L (2.41),(23) 1 L From this expression, we can definitely see that VSWR is intimately related to the amount of reflection at the load

(through L ) and the subsequent interference on the TL. Whites, EE 481/581 Lecture 4 Page 10 of 12

Special cases:

1. If Z L 0 (short-circuit load) then L 1. Consequently,

L 1 VSWR ,

2. If Z L (open-circuit load) then L 1. Consequently,

L 1 VSWR ,

3. If Z L Z0 (matched load) then L 0. Consequently,

L 0 VSWR 1. Regardless of the load, 1 VSWR . Whites, EE 481/581 Lecture 4 Page 11 of 12

Example N4.2: Compute the VSWR and return loss for the TL shown below. Plot the magnitude of the phasor voltage from z = 0 to z = -7 cm. From this plot, confirm the value of VSWR that you computed earlier. Whites, EE 481/581 Lecture 4 Page 12 of 12 Whites, EE 481 Lecture 5 Page 1 of 9

Lecture 5: Generator and Load Mismatches on TLs.

Up to this point, we have focused primarily on terminated transmission lines that lacked a specific excitation. That is, the TL was semi-infinite and terminated by a load impedance.

In this lecture, we’ll complete our review of TLs by adding a voltage source together with an arbitrary load (Fig. 2.19):

This TL model is very useful and applicable to a wide range of practical engineering situations.

Quantities of interest in such problems include the input impedance (for matching purposes) and signal power delivered to the load.

We will first consider the computation of the latter quantity assuming the TL is lossless.

Proceeding, the voltage on the TL is expressed by

j z j z V z Vo e V o e j z j z or V z Vo e L e (2.69),(1)

ZL Z0 where L (2.68),(2) ZL Z0

We’ll assume that the physical properties of the TL, the source and the load are known. This leaves the complex constant Vo as the only unknown quantity in (1).

Generally speaking, we compute Vo by applying the boundary condition at the TL input. (Recall that we have already applied boundary conditions at the load.) This is accomplished by applying voltage division at the input:

Zin VVin g (3) ZZin g

Observe that Vin is an electrical circuit quantity.

However, at the input to the TL, voltage must be continuous from the generator to the TL. This implies that Vin must also equal V ( z l ) on the TL.

Proceeding, then from (1) at the input j l j l V z l Vo e L e (4) Equating (3) and (4) to enforce the boundary condition at the TL input we find Whites, EE 481 Lecture 5 Page 3 of 9

j l j l Zin Vo e L e V g ZZin g 1 Zin j l j l or Vo V g e L e [V] (2.70),(5) ZZin g

Maximum Power

Because the TL is lossless, the time average power Pav delivered to the input of the TL must equal the time average power delivered to the load. Therefore, * 1* 1 Vin Pav e[] V in I in e V in * 2 2 Zin 2 Vin 1 or Pav e * [W] (2.74),(6) 2 Zin (Note that there is an error in (2.74) in your text.)

Now, substituting (3) into (6) gives 2 2 V g Zin 1 Pav e * (2.74),(7) 2 ZinZZg in

If we define ZinR in jX in and Z g Rg jX g , (7) becomes 2 Vg R P in (2.75),(8) av 2 2 2 RRXXing in g Whites, EE 481 Lecture 5 Page 4 of 9

Employing this last result, we’ll consider three special cases for

Pav in an effort to maximize this quantity. We will assume that

Z g is both nonzero and fixed:

(1.) Load is matched to the TL: Z L Z0 .

From (2), L 0 in this situation, which also implies that

ZinZ 0 . [This should be intuitive. If not, see (2.43).]

Consequently, from (8) with RZin 0 and X in 0: 2 Vg Z P 0 (2.76),(9) av,1 2 2 2 ZRX0 g g

(2.) Generator is matched to an arbitrarily loaded TL:

Zin Z g and L 0.

Specific values for l , Z0 , and ZL would need to be

chosen so that Zin Zg . Then from (8) and with RRin g

and X in X g : 2 V R P g g av,2 2 2 2 RRXXg g g g 2 Vg Rg or Pav,2 2 2 (2.78),(10) 8 RXg g Whites, EE 481 Lecture 5 Page 5 of 9

(3.) Maximum power transfer theorem applied at the TL * input: Zin Z g . (2.80),(11)

In this situation, RRin g , X in X g (conjugate match),

and L 0, so that from (8) 2 V R P g g av,3 2 2 2 RRXXg g g g 2 Vg or Pav,3 (2.81),(12) 8Rg Previous EE 322 students should recognize this as the maximum available source power.

Now, which of these three situations (9), (10), or (12) provides the most time average power delivered to the load?

Clearly, Pav,3P av,2 (equal when X g 0).

It can be shown that Pav,3P av,1.

Therefore, Pav,3 is the largest.

In conclusion, to transfer maximum time average power to a load, we generally need to conjugate match the generator impedance to the TL input impedance.

Note that maintaining a low VSWR ( L 0) doesn’t necessarily guarantee maximum Pav , though it could. (When?) Whites, EE 481 Lecture 5 Page 6 of 9

Efficiency

We haven’t said anything about efficiency yet. That is, what percentage of the source power is delivered to the load?

With Z gZZ L 0 , the load and the source are both matched to the TL. However, only one half of the source power is delivered to the load so the efficiency is 50%. For a matched line, that’s as good as it gets.

One way to increase efficiency is to decrease Rg (from Z0) and conjugate match the source to the TL input. The line may no longer be matched. Nevertheless, the power from multiple reflections can add in phase to increase the time average signal power delivered to the load. Whites, EE 481 Lecture 5 Page 7 of 9

Example N5.1: For the TL shown, determine the VSWR on the TL and the time-averaged power delivered to the load. Whites, EE 481 Lecture 5 Page 8 of 9 Whites, EE 481 Lecture 5 Page 9 of 9

That’s the long way to solve this problem. There’s even an easier way to calculate the time-averaged power delivered to the load. Since for this problem the source is matched to the TL characteristic impedance, then from Homework #3 2 Vg 2 Pav 1 (14) 8Z0 such that 1002 1 2 P 1 22.2 W av 8 50 3 as computed above.