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 135 Example 4.6 A transmission line is connected to a mismatched load. Calculate both the VSWR and the VSWR decibel equivalent if the reflection coefficient G is 0.25. Solution G ( a) VSWR = 1+ 1– G 1+ 0.25 = 1– 0.25 = 1.25 0.75 = 1.67 :1 ( b)VSWR = 20 log (VSWR) dB = (20) (log 1.67) = (20)( 0.22) = 4.4 dB SWR on a transmission line is important for several reasons. • The higher the SWR, the higher the maximum voltages on a transmission line for a given power applied or delivered to the load. High SWR can cause unexpected destruction of the line and any electronics equipment attached to it. • Many of today’s solid-state transmitters, transceivers, and amplifiers reduce RF output or shut down completely in the presence of high SWR. • The reflected wave on a transmission line represents energy “rejected” by the load. For a specified power delivered to the antenna, all components must be “supersized” for a line with a high SWR. • A transmission line operating with high SWR will exhibit greater loss per unit length than its specified loss for a matched load impedance. Mismatch (VSWR) Losses The power reflected from a mismatched load represents a potential loss and can have profound implications, depending on the installation. For example, one result might be a slight loss of signal strength at a distant point from an antenna. A more serious problem can result in the destruction of components in the output stage of a transmitter. The latter problem so plagued early solid-state transmitters that designers opted to include shutdown circuitry to sense high VSWR and turn down output power proportionately—a practice that remains to this day. In microwave measurements, VSWR on the transmission lines (that interconnect devices under test, instruments, and signal sources) can cause erroneous readings— and invalid measurements.
136 P a r t I I : F u n d a m e n t a l s Determination of VSWR losses must take into account two VSWR situations. Figure 4.9 shows a transmission line of impedance Z 0 interconnecting a load impedance Z L to a source with output impedance Z S . There is a potential for impedance mismatch at both ends of the line. When one end of the line is matched (either Z S or Z L ), the mismatch loss (ML) caused by SWR at the mismatched end is ⎡ ⎛ ⎞ ⎤ ML = – 10 log ⎢1– ⎜ SWR – 1 2 ⎟ ⎥ ⎣ ⎝SWR + 1⎠ ⎦ (4.37) which from Eq. (4.35) is 2 ML = – 10 log (1– G ) (4.38) Example 4.7 A coaxial transmission line with a characteristic impedance of 50 Ω is connected to the 50-Ω output (Z 0 ) of a signal generator, and also to a 20W load impedance Z L . Calculate the mismatch loss. Solution (a) First find the VSWR: VSWR VSWR = Z / Z= Z / Z 0 L = (50 Ω) = /(50 (20 Ω) / (20 Ω) = 2.5:1= 2.5:1 0 L (b) Mismatch loss: ⎡ ⎛ ⎞ ⎤ ML = – 10 log ⎢1– ⎜ SWR – 1 2 ⎟ ⎥ ⎣ ⎝SWR + 1⎠ ⎦ ⎡ ⎛ ⎞ ⎤ = – 10 log ⎢1– ⎜ 2.5 – 1 2 ⎟ ⎥ ⎣ ⎝ 2.5 + 1⎠ ⎦ ⎡ ⎛ ⎞ ⎤ = – 10 log ⎢1– ⎜ 1.5 2 ⎟ ⎥ ⎣ ⎝3.5⎠ ⎦ ⎡ 2 = – 10 log ⎣1– (0.43) ⎤ ⎦ = – 10 log [ 1– 0.185] = – 10 log [ 0.815] = (–10) (–0.089) = 0.89
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136 P a r t I I : F u n d a m e n t a l s<br />
Determination of VSWR losses must take into account two VSWR situations. Figure<br />
4.9 shows a transmission line of impedance Z 0 interconnecting a load impedance Z L to<br />
a source with output impedance Z S . There is a potential for impedance mismatch at both<br />
ends of the line.<br />
When one end of the line is matched (either Z S or Z L ), the mismatch loss (ML) caused<br />
by SWR at the mismatched end is<br />
⎡ ⎛ ⎞ ⎤<br />
ML = – 10 log ⎢1– ⎜<br />
SWR – 1<br />
2<br />
⎟ ⎥<br />
⎣ ⎝SWR + 1⎠<br />
⎦<br />
(4.37)<br />
which from Eq. (4.35) is<br />
2<br />
ML = – 10 log (1– G )<br />
(4.38)<br />
Example 4.7 A coaxial transmission line with a characteristic impedance of 50 Ω is connected<br />
to the 50-Ω output (Z 0 ) of a signal generator, and also to a 20W load impedance<br />
Z L . Calculate the mismatch loss.<br />
Solution<br />
(a) First find the VSWR:<br />
VSWR VSWR = Z / Z= Z / Z<br />
0<br />
L<br />
= (50 Ω) = /(50 (20 Ω) / (20 Ω)<br />
= 2.5:1= 2.5:1<br />
0<br />
L<br />
(b) Mismatch loss:<br />
⎡ ⎛ ⎞ ⎤<br />
ML = – 10 log ⎢1– ⎜<br />
SWR – 1<br />
2<br />
⎟ ⎥<br />
⎣ ⎝SWR + 1⎠<br />
⎦<br />
⎡ ⎛ ⎞ ⎤<br />
= – 10 log ⎢1– ⎜<br />
2.5 – 1<br />
2<br />
⎟ ⎥<br />
⎣ ⎝ 2.5 + 1⎠<br />
⎦<br />
⎡ ⎛ ⎞ ⎤<br />
= – 10 log ⎢1– ⎜<br />
1.5<br />
2<br />
⎟ ⎥<br />
⎣ ⎝3.5⎠<br />
⎦<br />
⎡<br />
2<br />
= – 10 log ⎣1– (0.43) ⎤<br />
⎦<br />
= – 10 log [ 1– 0.185]<br />
= – 10 log [ 0.815]<br />
= (–10) (–0.089)<br />
= 0.89