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
We’ll assume that the physical properties of the TL, the source and the load are known. This leaves the complex constant V0+ as the only unknown quantity in (1). Generally speaking, we compute V0++ 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:
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
Equating (3) and (4) to enforce the boundary condition at the TL input we find
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,
Now, substituting (3) into (6) gives
If we define Zin = Rin + jXin and Zg = R g+ jX g, (7) becomes
Employing this last result, we’ll consider three special cases for Pav in an effort to maximize this quantity. We will assume that Zg is both nonzero and fixed: (1.) Load is matched to the TL: ZL = Z0 . From (2), TL =0 in this situation, which also implies that Zin = Z0 . [This should be intuitive. If not, see (2.43).] Consequently, from (8) with Rin = Z0 and Xin = 0:
(2.) Generator is matched to an arbitrarily loaded TL: Zin = Zg and TL = 0.
Specific values for ? l , Z0 , and ZL would need to be chosen so that Zin = Zg . Then from (8) and with Rin = Rg and Xin = Xg :
(3.) Maximum power transfer theorem applied at the TL input: * Zin = Zg . In this situation, R in= R g, Xin = Xg (conjugate match), And TL ? 0, so that from (8)
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, Pav3 £ Pav2 (equal when Xg = 0).
• It can be shown that Pav3 > Pav1 Therefore, Pav3 is the largest. In conclusion, to transfer maximum time average power to a load, we need to conjugate match the generator impedance to the TL input impedance. Note that maintaining a low VSWR ( TL » 1 ) doesn’t necessarily guarantee maximum Pav , though it could. (When?)
We haven’t said anything about efficiency yet. That is, what percentage of the source power is delivered to the load? With Zg = ZL = Z 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.
Example N5.1: For the TL shown, determine the VSWR on the TL and the time-averaged power delivered to the load.