Question

The frequency sweep is 383-438 Mhz with a period of 15 microseconds. The first return occurs at 429 MHz. What is the range? (answer in meters to one decimal point)

Answer 1

Answer:

The range is 1881.8 m.

Explanation:

Given that,

Time $t=15\times10^{-6}\sec$

Frequency range $\Delta f= f_{f}-f_{i}$

$\Delta f= 429-383$

$\Delta f=46\ MHz$

The value of $\dfrac{df}{dt}$

$\dfrac{df}{dt}=\dfrac{438-383}{15\times10^{-6}}$

We need to calculate the range

Using formula of range

$R=\dfrac{c\Delta f}{2\times\dfrac{df}{dt}}$

Put the value into the formula

$R=\dfrac{3\times10^{8}\times46}{2\times\dfrac{438-383}{15\times10^{-6}}}$

$R=1881.8\ m$

Hence, The range is 1881.8 m.