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L-section Network<br />
(1) complex ZL to real Z0 matching<br />
Z0<br />
jX<br />
admittance<br />
(a)<br />
jB<br />
5-4<br />
Z L<br />
Z0<br />
jB<br />
jX<br />
admittance<br />
(b)<br />
How to determine jX & jB ? Let zL = ZL / Zo = (RL + jXL) / Zo = r + jx<br />
1. Analytic Solutions or 2. <strong>Smith</strong> <strong>Chart</strong> Solution<br />
(1) if RL Z0 ( z 1)<br />
[zL is inside the (1 + jx) circle ] => choose (a) why?<br />
1<br />
for impedance matching (to Z0) => jX <br />
Z0<br />
jB 1( RL jX L)<br />
<br />
2 2<br />
B(<br />
XRL XLZ ) R Z<br />
X<br />
0 L 0 L RL<br />
Z0<br />
RL<br />
X L Z0RL<br />
B <br />
<br />
=> <br />
2 2<br />
X(<br />
1 BXL) BZ0RL X<br />
RL<br />
X L<br />
L <br />
X<br />
( 1 B)<br />
( X LZ0<br />
RL<br />
) ( Z0<br />
B RL<br />
)<br />
(2) if RL Z0 ( z 1)<br />
[zL is outside the (1 + jx) circle ] => choose (b) why?<br />
1 1<br />
for impedance matching (to Z0) => jB <br />
<br />
RL j( X XL) Z0<br />
BZ0(<br />
X X L) Z0 RL<br />
<br />
X RL(<br />
Z0<br />
RL<br />
) X L<br />
<br />
=> <br />
(<br />
X XL) BZ0RL B ( Z R ) R Z<br />
r x<br />
r < 1<br />
r > 1<br />
0<br />
L<br />
L<br />
(1+jx) circle<br />
2011-12 H.-R. Chuang EE NCKU<br />
zL<br />
Z L / Zo<br />
r jx<br />
L<br />
( Z L Zo<br />
) /( Z L Zo<br />
) |<br />
L<br />
| <br />
0<br />
Z L
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