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Doppler Radar Relative Velocities

  1. Jun 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A radar at rest emits microwaves at a frequency of 1.63 GHz onto a truck moving away from the radar. There is a frequency difference between the received and sent frequencies of 281 Hz. What is the velocity of the truck?

    2. Relevant equations
    f(obs)=f(sent)*[(v+/-observer velocity)/(v+/-source velocity)]


    3. The attempt at a solution

    (1.63e9 Hz + 281 Hz) = (1.63e9 Hz)*((343 m/s - X)/(343 - 0))
    (((1.63e9 Hz + 281 Hz)*343) / 1.63e9 Hz) - 343 m/s = X
    X=5.91e-5 m/s

    but because that answer was ridiculous, I redid the question with 3e8 m/s as my V), which gave X=49.88 m/s, but that was wrong too.
     
  2. jcsd
  3. Jun 20, 2008 #2

    Tom Mattson

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    Science Advisor
    Gold Member

    It certainly is. That's because you're using the speed of sound. Radar waves are electromagnetic radiation!

    That's better. Can you please show your calculation so we can check it?
     
  4. Jun 20, 2008 #3
    (1.63e9 Hz + 281 Hz) = (1.63e9 Hz)*((3e8 m/s - X)/(3e8 - 0))
    (((1.63e9 Hz + 281 Hz)*3e8) / 1.63e9 Hz) - 3e8 m/s = X
    X=51.7 m/s

    Ah, that makes sense, but that isn't the right answer either.
     
  5. May 10, 2009 #4
    Your equation looks wrong.
    A travelling wave in air is about the same as empty space -- the error introduced will be very small.
    The gun can be modeled as a CW source of 1.63GHz radiation sending out a wave according to:
    Computing a radian frequency: ω = 2ָ*π*f and c = 3x10**8 m/s
    The E field intensity will go like:
    E(generate) = cos( ω * t - ω/c * x )
    The E-field will be canceled at the reflector and the reverse wave to cancel it, will need to have the same frequency AT the location of the moving target. Set that location to x = v*t and solve for when:

    E(cancel) = cos( ω1 * t + ω1/c * v*t )
    E(generate) = cos( ω * t - ω/c * v*t )

    are the same frequency.

    So, ω1 * t ( 1 + v/c ) = ω * t ( 1 - v/c )
    This solves to: ω1 = ω * ( 1 - v/c )/ ( 1 + v/c ) = ω * ( c - v )/ ( c + v )
    The difference in frequency is: ω * ( c - v - (c+v) / (c+v)) = ω*( -2v / (c + v) )
    Solve for v. I get a reasonable answer between 50 and 60mph. 20 to 30 m/s.
     
    Last edited: May 10, 2009
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