Chapter 14, Problem 14.11P

(a)

To determine

The duty cycle of the radar.

Answer to Problem 14.11P

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

Explanation of Solution

Given:

Average power level, $P=20\text{W}$

  $\text{PRF}=1000\text{ pulses/s}$

Pulse width $\tau =1\mu \text{s}$

Formula used:

The duty cycle is calculated using the expression:

  $\text{duty cycle}=\frac{\text{Pulse Width}}{\text{Time period}}$

Calculation:

Period:

  $\begin{array}{l}T=\frac{1}{PRF}\\ \Rightarrow T=\frac{1\text{s}}{1000}=1.0\times {10}^{-3}\text{s}\end{array}$

Express the pulse width in seconds.

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

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

  $\begin{array}{l}\text{duty cycle}=\frac{\text{Pulse Width}}{\text{Time period}}\\ \Rightarrow \text{duty cycle}=\frac{\text{1}\text{.0}\times {\text{10}}^{-6}}{1.0\times {10}^{-3}}\\ \Rightarrow \text{duty cycle}=1.0\times {10}^{-3}\end{array}$

Conclusion:

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

(b)

To determine

The peak power of the radar.

Answer to Problem 14.11P

The peak power of the Radar is found to be 20000 W.

Explanation of Solution

Given:

Average power: $P_{\text{avg}}=20\text{ W}$

Duty cycle: $1.0\times {10}^{-3}$

Formula used:

The peak power is given by the expression,

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

Calculation:

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

  $\begin{array}{c}{P}_0=\frac{P_{\text{avg}}}{\text{Duty cycle}}\\ =\frac{20\text{ W}}{1.0\times {10}^{-3}}\\ =20000\text{ W}\end{array}$

Conclusion:

The Peak power radiated by the radar is 20000 W.