Chapter 11, Problem 11.6P

A uniform plane wave has electric field E₈ = (Ey₀ay - E_za₃)e ^-axe ^-jBx V/m. The intrinsic impedance of the medium is given as $\eta$

= $\eta$

where ^ is a constant phase, (a) Describe the wave polarization and state the direction of propagation, (b) Find H_s. (c) Find s(x, t) and H(x, f). (d) Find < S > in W/m². (e) Find the time-average power in watts that is intercepted by an antenna of rectangular cross-section, having width -w and height h, suspended parallel to the yz plane, and at a distance d from the wave source.

To determine

(a)

The wave polarization and the direction of propagation.

Explanation of Solution

Given:

Electric Field, $E_s=(E_{y0}{a}_y-E_{z0}{a}_z)e^{-\alpha x}{e}^{-j\beta x}\text{V/m}$.

Intrinsic Impedance of the medium is $\eta =|\eta |e^{j\varphi }.$

Wave polarization describes the direction as well as magnitude of EM field of wave. When the plane is vibrating, it is said to be wave polarization. The electromagnetic wave vibrates at right angles. Mostly, wave vibrates in one plane like vibrating up and down or vibrating sideways. This polarization is called linear polarization.

If the wave is linearly polarised in the y - z plane, then it propagates in the forward x direction (as seen from the $e^{-j\beta x}$ factor in the given electric field equation).

To determine

(b)

The value of $H_s$

Answer to Problem 11.6P

The value of $H_s=\left[\frac{E_{y0}}{\left|\eta \right|}{a}_z+\frac{E_{z0}}{\left|\eta \right|}{a}_y\right]e^{-\alpha x}{e}^{-j\varphi }{e}^{-j\beta x}\text{A/m}\text{.}$

Explanation of Solution

Given:

Electric Field $E_s=(E_{y0}{a}_y-E_{z0}{a}_z)e^{-\alpha x}{e}^{-j\beta x}\text{V/m}\text{.}$

Intrinsic Impedance of the medium is $\eta =|\eta |e^{j\varphi }.$

Calculation:

When E_S is crossed with H_s, the resultant vector will be in positive x -direction.

   $E_s=(E_{y0}{a}_y-E_{z0}{a}_z)e^{-\alpha x}{e}^{-j\beta x}\text{V/m}$

The value of H_s is found by changing the y -component to z -component and z -component to negative y -component.

   $\begin{array}{l}{H}_s=\left[\frac{E_y}{\eta }{a}_z+\frac{E_z}{\eta }{a}_y\right]\\ H_s=\left[\frac{E_{y0}}{\left|\eta \right|}{a}_z+\frac{E_{z0}}{\left|\eta \right|}{a}_y\right]e^{-\alpha x}{e}^{-j\varphi }{e}^{-j\beta x}\text{A/m}\end{array}$

Conclusion:

The value of $H_s=\left[\frac{E_{y0}}{\left|\eta \right|}{a}_z+\frac{E_{z0}}{\left|\eta \right|}{a}_y\right]e^{-\alpha x}{e}^{-j\varphi }{e}^{-j\beta x}\text{A/m}\text{.}$

To determine

(c)

The value of $\epsilon (x,t)$ and $H(x,t)$.

Answer to Problem 11.6P

The value of $\epsilon (x,t)$ and $H(x,t)$ are

   $\epsilon (x,t)=[E_{y0}{a}_y-E_{z0}{a}_z]e^{-\alpha x}\cos (\omega t-\beta x)$

   $H(x,t)=[E_{y0}{a}_z+E_{z0}{a}_y]e^{-\alpha x}\cos (\omega t-\beta x-\varphi )$

Explanation of Solution

Given:

Electric Field is $E_s=(E_{y0}{a}_y-E_{z0}{a}_z)e^{-\alpha x}{e}^{-j\beta x}V/m.$

Intrinsic impedance of the medium is $\eta =|\eta |e^{j\varphi }.$

Calculation:

$\epsilon (x,t)$ is found by taking the real part of E_s

   $\begin{array}{l}\epsilon (x,t)=\Re e\left\{E_s{e}^{j\omega t}\right\}\\ \epsilon (x,t)=[E_{y0}{a}_y-E_{z0}{a}_z]e^{-\alpha x}\cos (\omega t-\beta x)\end{array}$

$H(x,t)$ is found by taking the real part of H_s.

   $\begin{array}{l}H(x,t)=\Re e\left\{H_s{e}^{j\omega t}\right\}\\ H(x,t)=[E_{y0}{a}_z+E_{z0}{a}_y]e^{-\alpha x}\cos (\omega t-\beta x-\varphi )\end{array}$

Conclusion:

Therefore, the value of $\epsilon (x,t)$ and $H(x,t)$ are

   $\epsilon (x,t)=[E_{y0}{a}_y-E_{z0}{a}_z]e^{-\alpha x}\cos (\omega t-\beta x)$

   $H(x,t)=[E_{y0}{a}_z+E_{z0}{a}_y]e^{-\alpha x}\cos (\omega t-\beta x-\varphi )$

To determine

(d)

The value of $<S>{\text{in W/m}}^{\text{2}}$.

Answer to Problem 11.6P

The value of $<S>=\frac{1}{2}(E_{y0}^2+E_{z0}^2\cos \varphi a_x{\text{)W/m}}^{\text{2}}$.

Explanation of Solution

Given:

Electric field is $E_s=(E_{y0}{a}_y-E_{z0}{a}_z)e^{-\alpha x}{e}^{-j\beta x}\text{V/m}$.

Intrinsic Impedance of the medium is $\eta =|\eta |e^{j\varphi }$.

Calculation:

The value of $<S>$ is calculated as:

   $\begin{array}{l}<S>=\frac{1}{2}\Re e\{E_s\times H_s^{*}\}\\ <S>=\frac{1}{2}\Re e\{\left(E_{y0}{a}_y-E_{z0}{a}_z)e^{-\alpha x}{e}^{-j\beta x}\right)\times \left(\left[\frac{E_{y0}}{|\eta }{a}_z+\frac{E_{z0}}{|\eta }{a}_y\right]e^{-\alpha x}{e}^{-j\varphi }{e}^{-j\beta x}\right)\}\\ <S>=\frac{1}{2}(E_{y0}^2+E_{z0}^2\cos \varphi a_x{\text{)W/m}}^{\text{2}}\end{array}$

Conclusion:

The value of $<S>=\frac{1}{2}(E_{y0}^2+E_{z0}^2\cos \varphi )a_x{\text{W/m}}^{\text{2}}$.

To determine

(e)

Time average power in watts.

Answer to Problem 11.6P

Time average power is $P=\frac{1}{2}(wh)(E_{y0}^2+E_{z0}^2)e^{-2\alpha d}\cos \varphi \text{W}\text{.}$

Explanation of Solution

Given:

Electric field is $E_s=(E_{y0}{a}_y-E_{z0}{a}_z)e^{-\alpha x}{e}^{-j\beta x}\text{V/m}\text{.}$

Intrinsic impedance of the medium is $\eta =|\eta |e^{j\varphi }.$

The width of antenna is w and height of antenna is h. It is located at distance of d from the wave source.

Calculation:

Time average power is calculated as

   $\begin{array}{l}P={\displaystyle \int {\displaystyle \int {}_{{}_{plate}}}<S>\cdot dS}\\ P=|<S>|{}_{x=d}\times area\\ P=\left(\frac{1}{2}(E_{y0}^2+E_{z0}^2\cos \varphi a_x\right)\times area\\ P=\frac{1}{2}(wh)(E_{y0}^2+E_{z0}^2)e^{-2\alpha d}\cos \varphi \text{W}\end{array}$

Conclusion:

Time average power is $P=\frac{1}{2}(wh)(E_{y0}^2+E_{z0}^2)e^{-2\alpha d}\cos \varphi \text{W}$.