Chapter 10, Problem 10.20P

(a) Determine s on the transmission line of Figure 10.32. Note that the dielectric is air. (b) Find the input impedance. (c) If $\omega L=10\Omega$, find I_s. (d) What value of L will produce a maximum value for |I_s| at $\omega =1$ Grad/s? For this value of L, calculate the average power (e) supplied by the source; (f) delivered to $Z_L=40+j30\Omega$.

Figure 10.32 See Problem 10.20.

To determine

(a)

The value ofs on the transmission line.

Answer to Problem 10.20P

The value of son the transmission line is 2.

Explanation of Solution

Given:

   $Z_L=\text{ (40+30j)}$

$Z_o\text{=}\text{50}\Omega$

   $l=2.7\lambda$

The given figure is shown below.

Concept Used:

The term s is calculated by

$s=\frac{1+\left|\Gamma \right|}{1-\left|\Gamma \right|}$

Calculation:

The reflection coefficient is

$\begin{array}{l}\Gamma =\frac{40+j30-50}{40+j30+50}\\ =\frac{(-10)+j(30)}{(90)+j(30)}\\ =j0.3333\end{array}$

The magnitude of the reflection coefficient is, $\left|\Gamma \right|e^{j\Gamma }=0.3333e^{j(\pi /2)}$

The standing wave ratio is calculated as

$s=\frac{1+\left|\Gamma \right|}{1-\left|\Gamma \right|}$

$s=\frac{1+0.3333}{1-0.333}=1.998\approx \text{2}$

Conclusion:

The value of s in the transmission line is 2.

To determine

(b)

The input impedance.

Answer to Problem 10.20P

The input impedance of the transmission line is $61.82-j37.493$.

Explanation of Solution

Given:

   $Z_L=\text{ (40+30j)}$

$Z_o\text{=}\text{50}\Omega$

   $l=2.7\lambda$

The given figure is shown below.

Concept Used:

The input impedance is calculated by

$Z_{in}=Z_0\frac{Z_L\cos (\beta l)+j{Z}_0\sin (\beta l)}{Z_0\cos (\beta l)+j{Z}_L\sin (\beta l)}$

Calculation:

The input impedance of the transmission line is calculated as

$Z_{in}=Z_0\frac{Z_L\cos (\beta l)+j{Z}_0\sin (\beta l)}{Z_0\cos (\beta l)+j{Z}_L\sin (\beta l)}$

Let $\frac{2\pi }{\lambda }$ be $\beta \text{.}$

   $\begin{array}{l}{\text{Z}}_{in}=Z_0\left(\frac{Z_L\cos \left(\frac{2\pi }{\lambda }\right)\left(2.7\lambda \right)+j{Z}_0\sin \left(\frac{2\pi }{\lambda }\right)}{Z_0\cos \left(\frac{2\pi }{\lambda }\right)\left(2.7\lambda \right)+j{Z}_L\sin \left(\frac{2\pi }{\lambda }\right)}\right)\\ =Z_0\left(\frac{Z_0\cos \left(2\pi \left(2.7\right)\right)+j{Z}_0\sin \left(2\pi \left(2.7\right)\right)}{Z_0\cos \left(2\pi \left(2.7\right)\right)+j{Z}_L\sin \left(2\pi \left(2.7\right)\right)}\right)\\ =Z_0\left(\frac{Z_L\left(-0.3090\right)+j{Z}_0\left(-0.95\right)}{Z_0\left(-0.3090\right)+j{Z}_L\left(-0.95\right)}\right)\\ =Z_0\left(\frac{Z_L\left(0.3090\right)+j{Z}_0\left(0.95\right)}{Z_0\left(0.3090\right)+j{Z}_L\left(0.95\right)}\right)\\ Z_{in}=Z_0\left(\frac{Z_L\left(0.3090\right)+j{Z}_0\left(0.95\right)}{Z_0\left(0.3090\right)+j{Z}_L\left(0.95\right)}\right)\end{array}$

   $\begin{array}{l}{Z}_{in}=50\left(\frac{\left(40+j30\right)\left(0.3090\right)+j50\left(0.95\right)}{50\left(0.3090\right)+j\left(40+30j\right)\left(0.95\right)}\right)\\ Z_{in}=50\left(\frac{12.36+j9.27+j47.5}{15.45+j\left(38+28.5j\right)}\right)\\ Z_{in}\text{ =}50\left(\frac{12.36+j56.77}{15.45+j\left(38+28.5j\right)}\right)\\ Z_{in}=50\left(\frac{12.36+j56.77}{15.45+j38-28.5}\right)=50\left(\frac{12.36+j56.77}{-13.05+j38}\right)\\ Z_{in}=61.82-j37.493\end{array}$

Conclusion:

The input impedance of the transmission line is $61.82-j37.493$.

To determine

(c)

The source current $I_s$.

Answer to Problem 10.20P

The source current is $I_s=1.098+j0.369\text{A}$.

Explanation of Solution

Given:

   $Z_L=\text{ (40+30j)}$

$Z_o\text{=}\text{50}\Omega$

   $l=2.7\lambda$

Calculation:

The source current is calculated by

$I_s=\frac{V_S}{Z_{in}+20+j\omega L}$

Let $V_s=100\text{V}$

$Z_{in}$ = $\text{61}\text{.82}-j37.493$

$\begin{array}{c}{I}_s=\frac{V_S}{Z_{in}+20+j\omega L}\\ I_s=\frac{100}{61.82-j37.493+20+j10}\\ =\frac{100}{81.82-j27.493}\\ =1.098+j0.369A\end{array}$

Conclusion:

Thus, the source current is $I_s=1.098+j0.369\text{A}$.

To determine

(d)

The value of L which produces maximum value for $I_s$.

Answer to Problem 10.20P

The value of L which produces maximum value for $I_s$ , $L=0.037493\mu \text{H}$.

Explanation of Solution

Given:

   $Z_L=\text{ (40+30j)}$

$Z_o\text{=}\text{50}\Omega$

   $l=2.7\lambda$

$\omega =1$

The given circuit is shown below.

Concept Used:

The maximum value of L is calculated by

   $I_s=\frac{V_s}{Z_{in}+20+j\omega L}$

Calculation:

   $\begin{array}{l}{I}_s=\frac{V_s}{Z_{in}+20+j\omega L}\\ I_s=\frac{100}{61.82-j37.493+20+j({10}^9)L}\\ =\frac{100}{61.82-j37.493+20+j({10}^9)L}\\ =\frac{100}{81.82+j\left(({10}^9)L-37.493\right)}\end{array}$

The magnitude of the source current is,

   $I_s=\frac{100}{\sqrt{{81.82}^2+{\left(\left({10}^9\right)L-37.493\right)}^2}}$

Differentiating with respect to L, $\begin{array}{l}\frac{d{l}_s}{dL}=\frac{d}{dL}\left(\frac{100}{\sqrt{{81.82}^2+{\left(\left({10}^9\right)L-37.493\right)}^2}}\right)\\ \frac{d{l}_s}{dL}=0-100\frac{1}{2\sqrt{{81.82}^2+{\left(L-37.493\right)}^2}}(2){10}^9\left({10}^{9}L-37.493\right)\left(1\right)\\ \frac{d{l}_s}{dL}=-{10}^{11}\frac{\left(\left({10}^9\right)L-37.493\right)}{\sqrt{{81.82}^2+{\left(L-37.493\right)}^2}}\end{array}$

$\begin{array}{l}\frac{d{l}_s}{dL}=0\\ -{10}^{11}\frac{\left({10}^{9}L-37.493\right)}{\sqrt{{81.82}^2+{\left(L-37.493\right)}^2}}=0\\ \frac{\left({10}^{9}L-37.493\right)}{\sqrt{{81.82}^2+{\left(L-37.493\right)}^2}}=0\\ \left({10}^{9}L-37.493\right)=0\\ L=37.493\times {10}^{-9}H=\text{0}\text{.037493}\mu \text{H}\end{array}$

Conclusion:

The value of L which produces maximum value for $I_s$ , $L=0.037493\mu \text{H}$.

To determine

(e)

The average power delivered by the source.

Answer to Problem 10.20P

The average power delivered by the source is, $\text{74}\text{.5W}$.

Explanation of Solution

Given:

   $Z_L=\text{ (40+30j)}$

$Z_o\text{=}\text{50}\Omega$

   $l=2.7\lambda$

$\omega =1$

The given circuit is shown below.

Concept Used:

The average power is calculated by

   $P_{in}=\frac{1}{2}{\left(I_S\right)}^2\left(Z_{in}\right)$

Calculation:

$\begin{array}{l}{I}_s=\frac{V_S}{Z_{in}+20+j\omega L}\\ I_s=\frac{100}{61.82-j37.493+20+j10}=\frac{100}{81.82-j27.493}=1.098+j0.369A\end{array}$

Considering the real part only

Average power is calculated as

   $\begin{array}{l}{P}_{in}=\frac{1}{2}{\left(I_S\right)}^2\left(Z_{in}\right)\\ P_{in}=\frac{1}{2}{\left(1.098\right)}^2\left(61.82\right)=74.5\text{ W}\end{array}$

Conclusion:

Thus, the average power delivered by the source is, $\text{74}\text{.5W}$.

To determine

(f)

Average power delivered to Z_L.

Answer to Problem 10.20P

The average power delivered to the load is $40.996\text{W}$.

Explanation of Solution

Given:

${\text{Z}}_L\text{= (40+30}j\text{)}$

${\text{Z}}_o\text{=50}\Omega$

   $l=2.7\lambda$

$\omega =1$

Concept Used:

The average power delivered is calculated by

$P_{av,L}=P_{av}\left(1-{\left|\Gamma \right|}^2\right)$

Calculation:

Thus, the total power delivered to the load is $\begin{array}{l}{P}_{av,L}=P_{av}\left(1-{\left|\Gamma \right|}^2\right)\\ =46.0064\left(1-{\left|j0.33\right|}^2\right)=46.0064\left(0.8911\right)=40.996342\text{W}\end{array}$

Conclusion:

Thus, the average power delivered to the load is $40.996\text{W}$.