Radar cross section for a weather radar.

Discussion in 'Homework Help' started by lam58, Jan 20, 2016.

  1. lam58

    Thread Starter Member

    Jan 3, 2014
    Hi, so I'm stuck on a tutorial question regarding a weather radar. The question is as follow:

    Q) A weather radar used for measuring rainfall operates at 5.65GHz. The antenna has a gain of 45dBi and the transmitter emits 1.1us duration of pulse 250kW power. The pulse repetition period is 1ms.

    i) Assuming a receiver noise figure of 1.8 dB what is the minimum cross section target that can be detected at a range of 100km;
    ii) Suppose linear FM pulse compression is used, with a bandwidth of 10MHz and assuming the pulse duration remains the same what is the minimum detectable cross section in this case.
    iii) What is the range resolution in each case.

    My attempt:

    So the target is a clutter target (or whatever it is called), therefore I use the following radar equation:

     \frac{P_r}{N} = \frac{P_t G^2 \lambda^2 \tau \sigma_b}{4\pi^3 R^4 KTB}

    So if the radar isn't modulated, it is safe to assume (or so I've been told) that the pulse width, \tau, is matched with bandwidth, B, and they cancel out leaving:

     \frac{P_r}{N} = \frac{P_t G^2 \lambda^2  \sigma_b}{4\pi^3 R^4 KT}

    To find the RCS I rearranged the above to give:
     \sigma_b = \frac{4\pi^3 R^4 KT}{P_t G^2 \lambda^2}

    To find wavelength:
    \lambda = \frac{c}{f} = \frac{3x10^3}{5.65x10^9} = 0.0531

    If noise figure = 1.8 dB, noise factor = 1.51 Watts = KTB and bandwidth is 1/pulse width = 1/1.1us = 909.1 kHz. This implies that:

     KT = \frac{1.51}{909.1x10^3} = 1.661x10^{-6}

    Gain is 48dBi = 19.95 kW. R = distance = 100 km, Pt = 250 kW.

    Plugging in all the numbers gives me a RCS value of 73418 metres squared. Which seems a little high, even for RCS. Although it amounts to 48.65dB.

    For part ii) I use a similar approach, however now bandwidth and pulse width are not matched. Pulse width duration is still 1.1us but the new bandwidth is 10MHz. So I have to now use both pulse width and KTB i.e.

     \sigma_b = \frac{4\pi^3 R^4 KTB}{P_t G^2 \lambda^2 \tau}

    KTB = 1.661x10^{-6} \times 10 MHz = 16.61
    \tau = 1.1\mu s

    Which gives me a RCS of  7.25x10^{17}
    or 178.6 dB, which is surely wrong.

    What am I doing wrong?
  2. sailorjoe


    Jun 4, 2013
    This is interesting, and I'll have to get back up to speed on radar equations to really help. The only thing I noticed is that you rearranged your equation to solve for RCS, but the max RCS of a sphere is σmax = π ·R2. Not sure of the radius of a raindrop, but I don't expect the raindrops to match your rearranged equation. There's probably data on raindrop size, and you could try using that in the non-rearranged equation and see what happens. In the meantime, I'll study this some more and see if I can help.