Question 11.5: A load of 100 + j150 Ω is connected to a 75Ω lossless line. ...

(a) We can use the Smith chart to solve this problem. The normalized load impedance is

z_{L}=\frac{Z_{L}}{Z_{o}}=\frac{100+j150}{75}=1.33+j2

We locate this at point P on the Smith chart of Figure 11.16. At P, we obtain

|\Gamma|=\frac{OP}{OQ}=\frac{6cm}{9.1cm}=0.659

\theta_{\Gamma}=angle  POS=40^{\circ}

Hence

\Gamma=0.659\angle40^{\circ}

Check:

\Gamma=\frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}}=\frac{100+j150-75}{100+j150+75}=0.6598\angle39.94^{\circ}

(b) Draw the constant s-circle passing through P and obtain

s = 4.82

Check:

s=\frac{1+|\Gamma|}{1-|\Gamma|}=\frac{1+0.659}{1-0.659}=4.865

(c) To obtain Y_{L}, extend PO to POP^{'} and note point P^{'} where the constant s-circle meets POP^{'}. At P^{'}, obtain

y_{L}=0.228-j0.35

The load admittance is

Y_{L}=Y_{o}y_{L}=\frac{1}{75}(0.228-j0.35)=3.04-j4.67 mS

Check:

Y_{L}=\frac{1}{Z_{L}}=\frac{1}{100+j150}=3.07-j4.62 mS

(d) The 0.4\lambda corresponds to an angular movement of 0.4\times720^{\circ}=288^{\circ} on the constant s-circle. From P, we move 288° toward the generator (clockwise) on the s-circle to reach point R. At R

z_{in}=0.3+j0.63

Hence

Z_{in}=Z_{o}z_{in}=75(0.3+j0.63)=22.5+j47.25\Omega

Check:

\beta\ell=\frac{2\pi}{\lambda}(0.4\lambda)=360^{\circ}(0.4)=144^{\circ}

Z_{in}=Z_{o}\left[\frac{Z_{L}+jZ_{o}\tan\beta\ell}{Z_{o}+jZ_{L}\tan\beta\ell}\right]=\frac{75(100+j150+j75\tan144^{\circ})}{[75+j(100+j150)\tan144^{\circ}]}=54.41\angle65.25^{\circ}

or

Z_{in}=21.9+j47.6\Omega

(e) The 0.6\lambda corresponds to an angular movement of

0.6\times720^{\circ}=432^{\circ}=1  revolution +72^{\circ}

Thus, we start from P (load end), move clockwise along the s-circle 432°, or one revolution plus 72°, and reach the generator at point G. Note that to reach G from P, we have passed through point T (location of V_{min}) once and point S (location of V_{max}) twice. Thus, from the load,

1st  V_{max}  is located at      \frac{40^{\circ}}{720^{\circ}}\lambda=0.055\lambda

2nd  V_{max}  is located at      0.0555\lambda+\frac{\lambda}{2}=0.555\lambda

and the only V_{min} is located at     0.055\lambda+\lambda/4=0.3055\lambda

(f) At G (generator end)

z_{in}=1.8-j2.2

Z_{in}=75(1.8-j2.2)=135-j165\Omega

This can be checked by using eq. (11.34):

Z_{in}=Z_{o}\left[\frac{Z_{L}+jZ_{o}\tan\beta\ell}{Z_{o}+jZ_{L}\tan\beta\ell}\right]    (lossless)

where

\beta\ell=\frac{2\pi}{\lambda}(0.6\lambda)=216^{\circ}

We can see how much time and effort are saved by using the Smith chart.