Answer to Question #164718 in Electrical Engineering for mohammed

Question

A lossless TL with if the voltage a cross a load connected to lossless TL is 20 Vrms

and reflection coefficient is equal to KR = 0.7270°.

1. Find VSWR (S) for this TL.

2. Find the voltage and current scales.

3. Find the voltage and current at d=0, 1/8, 1/4, 31/8 and 1/2 using Crank diagram

and tabulate the results, including the distances and the values for maximum and

minimum voltage, current, i.e. Emax and Emin at dmax and dmin, respectively.

4. From Crank diagram confirm that S has the same value found theoretically

already in (1) above.

5. Use Crank diagram to evaluate the transmitted power.

6. Draw clearly the SWR pattern from d=0 to d=1/2 showing the maximum

and minimum positions in your plot.

7. If the load is changed to Zr= ZO how would the SWR pattern look like?

Why? Draw it clearly.

Answer 1

1.Reflection coefficient:

$\Gamma(z)=\Gamma_Le^{j2\beta z}=\Gamma_Le^{j\theta}$

we have: $\theta=0.7270\degree$

$\Gamma(z)=\frac{V_0^-}{V_0^+}e^{j2\beta z}$

$V_0^+=V_0^-=20\ V$

$VSWR=\frac{1+|\Gamma_L|}{1-|\Gamma_L|}$

$|\Gamma_L|=1\implies VSWR=\infin$

2.

$|V(z)|=|V_0^+||1+\Gamma(z)|$

$|1+\Gamma(z)|=\sqrt{(1+cos\theta)^2+sin^2\theta}=2$

$|V(z)|=20\cdot2=40\ V$

$|I(z)|=\frac{|V_0^+|}{Z_0}|1-\Gamma(z)|$

$|1-\Gamma(z)|=\sqrt{(1-cos\theta)^2+sin^2\theta}=0.01$

$|I(z)|=20\cdot0.01/Z_0=0.2/Z_0\ A$

3.

for $d=0$ :

$V(z)=V_0^+=20\ V$

$I(z)=V_0^+/Z_0=20/Z_0\ V$

for $d=1/8$ :

$|1+\Gamma(z)|=\sqrt{(1+cos(\theta/8))^2+sin^2(\theta/8)}=2$

$|V(z)|=20\cdot2=40\ V$

$|1-\Gamma(z)|=\sqrt{(1-cos(\theta/8))^2+sin^2(\theta/8)}=0.002$

$|I(z)|=20\cdot0.002/Z_0=0.04/Z_0\ A$

for $d=1/4$ :

$|1+\Gamma(z)|=\sqrt{(1+cos(\theta/4))^2+sin^2(\theta/4)}=2$

$|V(z)|=20\cdot2=40\ V$

$|1-\Gamma(z)|=\sqrt{(1-cos(\theta/4))^2+sin^2(\theta/4)}=0.003$

$|I(z)|=20\cdot0.003/Z_0=0.06/Z_0\ A$

for $d=31/8$ :

$|1+\Gamma(z)|=\sqrt{(1+cos(31\theta/8))^2+sin^2(31\theta/8)}=2$

$|V(z)|=20\cdot2=40\ V$

$|1-\Gamma(z)|=\sqrt{(1-cos(31\theta/8))^2+sin^2(31\theta/8)}=0.05$

$|I(z)|=20\cdot0.05/Z_0=1/Z_0\ A$

for $d=1/2$ :

$|1+\Gamma(z)|=\sqrt{(1+cos(\theta/2))^2+sin^2(\theta/2)}=2$

$|V(z)|=20\cdot2=40\ V$

$|1-\Gamma(z)|=\sqrt{(1-cos(\theta/2))^2+sin^2(\theta/2)}=0.006$

$|I(z)|=20\cdot0.006/Z_0=0.12/Z_0\ A$

$V_{max}=40\ V$ at $d=1/8,1/4,1/2,31/8$

$V_{min}=20\ V$ at $d=0$

$I_{max}=20/Z_0\ A$ at $d=0$

$I_{min}=0.006/Z_0\ A$ at $d=1/2$

4.

5.

Power equals area of rectangle ABCD on Crank diagram.

$P=|V(z)||I(z)|=40\cdot0.2/Z_0=8/Z_0\ W$

6.For voltage:

$SWR=\frac{|V(z)|{max}}{|V(z)|_d}$

where $|V(z)|_d$ is voltage for different $d$ .

$d=0$ : $SWR=1$ , $d=1/8,1/4,1/2$ : $SWR=2$

For current:

$SWR=\frac{|I(z)|{max}}{|I(z)|_d}$

where $|I(z)|_d$ is current for different $d$ .

$d=0$ : $SWR=1$, $d=1/8$ : $SWR=20/0.04=500$

$d=1/4$ : $SWR=20/0.06=333$

$d=1/2$ : $SWR=20/0.006=3333$

7.If the load is changed to $Z_L=Z_0$ :

$\Gamma_L=\frac{Z_l-Z_0}{Z_L+Z_0}=0$

$VSWR=1$

$V(z)=V_0^+=20\ V$

$I(z)=V_0^+/Z_0=20/Z_0\ V$

SWR=1

for all values of $d$