Question 11.7: An antenna with an impedance of 40 + j30 Ω is to be matched ...

(a)

z_{L}=\frac{Z_{L}}{Z_{o}}=\frac{40+j30}{100}=0.4+j0.3

Locate z_{L} on the Smith chart as in Figure 11.24 and from this draw the s-circle so that y_{L} can be located diametrically opposite z_{L}. Thus y_{L}=1.6-j1.2. Alternatively, we may find y_{L} by using

y_{L}=\frac{Z_{o}}{Z_{L}}=\frac{100}{40+j30}=1.6+j1.2

Locate points A and B where the s-circle intersects the g = 1 circle. At A, y_{s}=-j1.04 and at B, y_{s}=+j1.04. Thus the required stub admittance is

Y_{s}=Y_{o}y_{s}=\pm j1.04\frac{1}{100}=\pm j10.4mS

Both j10.4 mS and -j10.4 mS are possible values.

(b) From Figure 11.24, we determine the distance between the load (antenna in this case) y_{L} and the stub. At A

\ell_{A}=\frac{\lambda}{2}-\frac{(62^{\circ}- -39^{\circ})\lambda}{720^{\circ}}=0.36\lambda

and at B:

\ell_{B}=\frac{(62^{\circ}–39^{\circ})\lambda}{720^{\circ}}=0.036\lambda

(c) Locate points A^{'} and B^{'} corresponding to stub admittance -j1.04 and j1.04, respectively. Determine the stub length (distance from P_{sc} to A^{'} and B^{'}):

d_{A}=\frac{88^{\circ}}{720^{\circ}}\lambda=0.1222\lambda

d_{B}=\frac{272^{\circ}}{720^{\circ}}\lambda=0.3778\lambda

Notice that d_{A}+d_{B}=0.5\lambda as expected.

(d) From Figure 11.24, s=2.7. This is the standing wave ratio on the line segment

between the stub and the load (see Figure 11.18); s = 1 to the left of the stub because the line is matched, and s=\infty along the stub because the stub is shorted.