Chapter 14, Problem 14.28P

(a)

To determine

The duty cycle of the radar.

Answer to Problem 14.28P

The duty cycle of the radar is found to be $1.0\times {10}^{-3}$.

Explanation of Solution

Given:

  $\text{PRF}=200\text{ pulses/s}$

Pulse width $\tau =5\mu \text{s}$

Formula used:

The duty cycle is calculated using the expression,

  $\text{duty cycle}=\frac{\text{ ON time}}{\text{ Total time}}$

Calculation:

The ON time is calculated as follows:

  $\text{ON time}=\text{PRF}\times \tau$

Express the pulse width in seconds.

  $\begin{array}{c}\tau =5\mu \text{s}\times \frac{{10}^{-6}\text{s}}{1\mu \text{s }}\\ =5\times {10}^{-6}\text{s}\end{array}$

Substitute the given values in the equation and calculate the ON time.

  $\begin{array}{c}\text{ON time}=\text{PRF}\times \tau \\ =\left(200\text{ pulses}\right)\left(5\times {10}^{-6}\text{s}\right)\\ =1.0\times {10}^{-3}\text{s}\end{array}$

Calculate the duty cycle, using the given values in the formula.

  $\begin{array}{c}\text{duty cycle}=\frac{\text{ ON time}}{\text{ Total time}}\\ =\frac{1.0\times {10}^{-3}\text{s}}{1\text{ s}}\\ =1.0\times {10}^{-3}\end{array}$

Conclusion:

Thus, the duty cycle of the radar is found to be $1.0\times {10}^{-3}$.

(b)

To determine

The average power of the radar.

Answer to Problem 14.28P

The average power of the Radar is found to be 2000 W.

Explanation of Solution

Given:

Peak power

  $P_0=2\text{MW}$

Duty cycle

  $1.0\times {10}^{-3}$

Formula used:

The average power is given by the expression,

  $P_{\text{avg}}=P_0\times \text{ Duty cycle}$

Calculation:

Express the peak power $P_0$in Hz.

  $\begin{array}{c}{P}_0=2\text{MW}\times \frac{{10}^6\text{W}}{1\text{ MW}}\\ =2\times {10}^6\text{W}\end{array}$

Substitute the given values in the formula and calculate the average power.

  $\begin{array}{c}{P}_{\text{avg}}=P_0\times \text{ Duty cycle}\\ \text{=}\left(2\times {10}^6\text{W}\right)\left(1.0\times {10}^{-3}\right)\\ =2000\text{ W}\end{array}$

Conclusion:

The average power radiated by the radar is 2000 W.

(c)

To determine

The power density at 100 m.

Answer to Problem 14.28P

The power density at a distance of 100 m is found to be 26 mW/cm².

Explanation of Solution

Given:

The frequency of radiation

  $f=10\text{ GHz}$

Speed of light

  $c=3.0\times {10}^8\text{m/s}$

Diameter of the dish

  $2r=1.22\text{ m}$

Average power

  $P_{\text{avg}}=2000\text{ W}$

Distance of the point from the antenna $R=100\text{ m}$

Formula used:

If the distance of the point is nearly 4 times the diameter of the dish, then the point is in the far field. The power density $S_{\text{ff}}$at a point in the far field is given by,

  $S_{\text{ff}}=\frac{A{P}_{\text{avg}}}{{\lambda }^2{R}^2}=\frac{\pi r^2{P}_{\text{avg}}}{{\lambda }^2{R}^2}$   …….(1)

Here, A is the area of the antenna aperture and $\lambda$is the wavelength of the radiation.

The wavelength of the radiation is determined using the expression,

  $\lambda =\frac{c}{f}$   ……(2)

Calculation:

Express the frequency in Hz.

  $\begin{array}{c}f=10\text{ GHz}\times \frac{{10}^9\text{Hz}}{1\text{ GHz}}\\ =1.0\times {10}^{10}\text{Hz}\end{array}$

Determine the wavelength of the radiation by substituting the values of variables in equation (2).

  $\begin{array}{c}\lambda =\frac{c}{f}\\ =\frac{3.0\times {10}^8\text{m/s}}{1.0\times {10}^{10}\text{Hz}}\\ =0.03\text{ m}\\ =3\text{ cm}\end{array}$

Express the radius R in cm.

  $\begin{array}{c}r=\frac{1.22\text{ m}}{2}\\ =0.61\text{ m}\\ =\text{61 cm}\end{array}$

Express the power in mW.

  $\begin{array}{c}{P}_{\text{avg}}=2000\text{ W}\times \frac{{10}^3\text{ mW}}{1\text{ W}}\\ =2.0\times {10}^6\text{mW}\end{array}$

Substitute the values of variables in equation (1) and calculate the power density.

  $\begin{array}{c}{S}_{\text{ff}}=\frac{\pi r^2{P}_{\text{avg}}}{{\lambda }^2{R}^2}\\ =\frac{\left(3.14\right){\left(\text{61 cm}\right)}^2\left(2.0\times {10}^6\text{mW}\right)}{{\left(3\text{ cm}\right)}^2{\left(100\times 100\text{ cm}\right)}^2}\\ =25.96{\text{ mW/cm}}^2\\ =26{\text{ mW/cm}}^2\end{array}$

Conclusion:

Thus, the power density at a distance of 100 m is found to be 26 mW/cm².

(d)

To determine

If a person standing in the field of the Radar at a distance of 100 m from it for an hour is within the permissible limits of exposure according to OSHA standards.

Answer to Problem 14.28P

The person standing in the field of the Radar at a distance of 100 m from it for an hour is within the permissible limits of exposure according to OSHA standards.

Explanation of Solution

Given:

The frequency of radiation

  $f=10\text{ GHz}$

Beam width

  $\theta =2.5°$

Rotational frequency

  $n=4\text{ rpm}$

Power density at a distance 100 m

  $S_{\text{ff}}=26{\text{ mW/cm}}^2$

Permissible exposure limit by OSHA standards: 10 mW/cm²averaged over 6 min.

Calculation:

Determine the time period of rotation using the expression,

  $T=\frac{1}{n}$

Substitute the given value of n.

  $\begin{array}{c}T=\frac{1}{n}\\ =\left(\frac{1}{4}\right)\left(60\right)\text{ s}\\ \text{=12 s}\end{array}$

For every rotation, the time of exposure t is given by,

  $t=\frac{\theta }{360}T$

Substitute the given values in the expression.

  $\begin{array}{c}t=\frac{\theta }{360°}T\\ =\left(\frac{2.5°}{360°}\right)\left(12\text{ s}\right)\\ =8.33\times {10}^{-2}\text{s}\end{array}$

The number of rotations made in 1 h is given by

  $\begin{array}{c}N=4\text{rpm}\times \text{60 min}\\ \text{=240}\end{array}$

Total exposure time is given by,

  $\tau =Nt$

Substitute the given values in the expression.

  $\begin{array}{c}\tau =Nt\\ =\left(240\right)\left(8.33\times {10}^{-2}\text{s}\right)\\ =20\text{ s}\end{array}$

According to OSHA standards, the permissible level of exposure is 10 mW/cm²averaged over 6 min. This means that the allowed energy density levels are,

  $\begin{array}{c}{E}_{\text{allowed}}=\left(10{\text{ mW/cm}}^2\right)\left(6\times 60\text{ s}\right)\\ =3600{\text{ mJ/cm}}^2\end{array}$

The energy density the person is exposed in 1 h is given by,

  $E=S_{\text{ff}}\tau$

Substitute the given values in the expression.

  $\begin{array}{c}E=S_{\text{ff}}\tau \\ =\left(26{\text{ mW/cm}}^2\right)\left(20\text{ s}\right)\\ =520{\text{ mJ/cm}}^2\end{array}$

Conclusion:

Thus, the person standing in the field of the Radar at a distance of 100 m from it for an hour is within the permissible limits of exposure according to OSHA standards.