Practical_Antenna_Handbook_0071639586
C h a p t e r 4 : T r a n s m i s s i o n L i n e s a n d I m p e d a n c e M a t c h i n g 125 • Similarly, suppose we disconnect the load completely, so that the transmission line is looking into an open circuit. In this case, Z L = ∞ and I L = 0. But if the current in the advancing wave is to be continuous, there must be an equal and opposite current flowing at the end of the transmission line to make the net load current zero. We call this new current I REF , and it must be exactly equal in magnitude to I FWD but of opposite polarity through the load. In other words, I REF = –I FWD . The minus sign simply means the reflected current has the same amplitude as the forward current, but it is directed in the opposite direction—i.e., back toward the source. If we could stand on the junction of the line and the load, we would see two waves, moving in opposite directions, but since the currents are 180 degrees out of phase with each other, we see no net current in either direction. Because there are both forward and reflected waves at the junction of the line and the load, V and I there are the sum of the forward and reflected voltages and currents, respectively. Therefore: V I TOTAL Z L = (4.19) TOTAL Z L VFWD + VREF = (4.20) I + I FWD REF where V FWD = incident (i.e., forward) voltage V REF = reflected voltage I FWD = incident current I REF = reflected current Using Ohm’s law and algebraic substitution to define the currents at the boundary in terms of the voltage components and the characteristic impedance of the line: V Z FWD I FWD = (4.21) 0 and V = – REF (4.22) I REF Z0 (The minus sign in Eq. [4.22] indicates that I REF is in the opposite direction.) The two expressions for current (Eqs. [4.21] and [4.22]) may be substituted into Eqs. (4.19) and (4.20) to yield
126 P a r t I I : F u n d a m e n t a l s Z L V = V Z FWD FWD 0 + V V – Z REF REF 0 (4.23) Repeating Eq. (4.18), which defined the reflection coefficient G as the ratio of reflected voltage to incident voltage: G V REF = V FWD (4.24) Rearranging Eq. (4.23) to combine all terms involving V REF separately from those involving V FWD and substituting into Eq. (4.24) gives ZL − Z0 Γ = (4.25) Z + Z Example 4.4 A 50-Ω transmission line is connected to a 30-Ω resistive load. Calculate the reflection coefficient G. L 0 Solution 0 G Z Z L – = ZL + Z0 (50 Ω) – (30 Ω) = (50 Ω) + (30 Ω) = 20 80 = 0.25 Example 4.5 In Example 4.4, the incident (forward) voltage is 3 V rms. Calculate the reflected voltage. Solution If G = V V REF FWD , then V REF = ΓV FWD = (0.25)(3V) = 0.75V (4.26)
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126 P a r t I I : F u n d a m e n t a l s<br />
Z<br />
L<br />
V<br />
=<br />
V<br />
Z<br />
FWD<br />
FWD<br />
0<br />
+ V<br />
V<br />
–<br />
Z<br />
REF<br />
REF<br />
0<br />
(4.23)<br />
Repeating Eq. (4.18), which defined the reflection coefficient G as the ratio of reflected<br />
voltage to incident voltage:<br />
G V REF<br />
=<br />
V<br />
FWD<br />
(4.24)<br />
Rearranging Eq. (4.23) to combine all terms involving V REF separately from those involving<br />
V FWD and substituting into Eq. (4.24) gives<br />
ZL<br />
− Z0<br />
Γ = (4.25)<br />
Z + Z<br />
Example 4.4 A 50-Ω transmission line is connected to a 30-Ω resistive load. Calculate<br />
the reflection coefficient G.<br />
L<br />
0<br />
Solution<br />
0<br />
G Z Z L<br />
–<br />
=<br />
ZL<br />
+ Z0<br />
(50 Ω) – (30 Ω)<br />
=<br />
(50 Ω) + (30 Ω)<br />
= 20<br />
80<br />
= 0.25<br />
<br />
Example 4.5 In Example 4.4, the incident (forward) voltage is 3 V rms. Calculate the<br />
reflected voltage.<br />
Solution If<br />
G =<br />
V<br />
V<br />
REF<br />
FWD<br />
,<br />
then<br />
<br />
V<br />
REF<br />
= ΓV<br />
FWD<br />
= (0.25)(3V)<br />
= 0.75V<br />
(4.26)
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