ch5: Distributed targets Flashcards
What maybe within the sample volume?
- Raindrops or cloud particles
What completely fills the radar beam?
- Storms and clouds
- Because they are usually so large
Place where storms and clouds don’t completely fill the radar beam
- Along the boundary of a storm
- Because the radar beam will be moving from no echo to echo and vice versa
- Near the top or bottom of a storm
- The beam will be partially in and partially out of echo
The power returned to the radar will come from
- All the individual targets being illuminated by the radar beam wether the beam is completely filled or not
What may misinterpret the strength of the signal?
- If the beam is only partially filled
Continental clouds contain as many as
- 200 or more cloud droplets/cm3
- That amounts to 2x108/m3
For a radar with a 1o antenna beam width the beam will be
- 1 km in diameter at a range of 57 km
If the radar is using 1 us pulse length the effective sample volume in space will be
- 150 m
The volume of the radar pulse is then illuminating (if pulse strength is 1 us)
- More than 2x106 cloud droplets simultaneously
- The number of precipitation sized particles is lower than this
- Typical rain will have on the order of a few to a few hundred raindrops per cubic meter
- Thus there might be something like 109 to 1012 raindrops in a single radar sample volume
- This is still a very large number of particles
The return from meteorological targets is the combination of
- Billions of returns being added together
The total backscattering cross sectional area of a meteorological target is
- The sum of all of the individual backscattering cross sectional area
If we send a pulse of radar energy into a storm and get an echo back then send a second pulse into the storm immediately after the first
- There would be little time fot the raindrops to change position relative to each other or relative to the radar
If pulses were sent nearly simultaneously
- The returns measured by the radar would be virtually identical
If we waited reasonable length of time before sending a second pulse into the same point in space
- The arrangement of particles bring sampled by the radar might be different
- Between these two limits is a region of interest and importance for radar
When sampling raindrops or other hydrometeors with radar we need to
- Wait long enough to allow the particles to reshuffle so a truly different arrangement can be reached
Why should we wait long enough to allow particles to shuffle
- To get a good average of the true signal amplitude
Weather echoes are
- Constantly changing
- A single instantaneous measurement might not be a good measure of the true signal strength
- By averaging several samples together we get a better measurement of a storm intensity
Time to independence:
- AKA decorrelation time
- Time it takes hydrometers to rearrange them selves so the measurements are independent of one another
Time of independence mathematical definition:
- Time it takes for a sample of targets to decorrelate tto a value of 0.01 from perfect coefficient
Perfect correlation has a correlation coefficient of
- 1.0 (or -1.0 if there is an inverse correlation)
Correlation coefficient of 0 means that
- There is no correlation at all
Waiting for a signal to decorrelate to 0.01 means
- We have waited long enough that the newest sample is almost completely different than the original sample
The decorrelation time of a sample depends on:
- Wavelength of the radar used
- Hydrometeors
- Turbulence within the sample volume
Decorrelation times are shorter for
- Shorter wavelength radars
- When a short wavelength is used particles do not have to travel as far to change position significantly relative to the radar
What might make the sample decorelate slowly?
- Particles having the same size because that will make them fall together with the same terminal velocity
What might make the sample decorrelate faster?
- Storm containing a wife veriety of particle sizes
- There will be many different particles falling in the same volume
The shortest decorrelation times occur when
- Hail and rain are in the same sample volume
The time required for the autocorrection function to fall to a value of t= 0.01 is
- Aprox.t= 2 lamda to 3 lamda
- T is in milliseconds and
- Lamda is in centimeters
Measured decorrelation time have ranged from
- 3.5 ms to nearly 30 ms
Measured decorrelation time range depends on
- Storm and radar
If we want to sample as close together in time as possible but still have independent samples, we would want to
- sample at a rate given approximately by 10t0.01.
- this suggests we should sample at rates on the order of 3 to 30 times a second (3-30 Hz)
- most radars sample at rates much higher than this
- this suggests we should sample at rates on the order of 3 to 30 times a second (3-30 Hz)
modern Doppler radars often use PRFs
- near 1000 Hz
why did we use PRF near 1000 Hz and not 1000 Hz
- because PRF of 1000 Hz give sampling time that is too close together to have truly independent samples
why is it necessary to average many consecutive pulses together
- in order to have the equivelant of just a few independent samples
factors which can contribute to decreasing the time to independence of consecutive samples made with a radar
- range averaging
- moving the antenna in azimuth while collecting the data (almost always done)
- wind shear within the sample volume
- turbulence
the horizontal and vertical beam widths must always
- be in radians
the pulse length h is:
- the length in space corresponding to the duration t of the transmitted pulse
- AKA pulse duration
In the equation for sample volume we used h/2 because
- We are interested in signals that return to the radar at precisely the same time
How to find the total backscattering cross sectional area of targets within the radar sample volume?
- Determine the backscattering cross sectional area of a unit volume and multiply this by the total sample volume
- Where the summation is over all of the individual backscattering cross-sectional areas in a unit volume
For sample volume we used the horizontal and vertical beam widths this assumed that
- All of the energy in the radars transmitted pulse is contained within the half power beam widths
- Real radars don’t have such nicely behaved beam patterns
2 ln (2) in the dominator of the radar pulse volume equation accounts for:
- The real beam shape better than the assumption
Parameter that is related to the total backscattering cross sectional area
- Radar reflectivity
Radar reflectivity:
- The summation is done over all individual targets in a unit volume in spave
- Has units of 1/cm or cm^-1
- Intensive parameter
…………………… is related to the backscattering cross sectional area
- Target size
Reylight approximation applies if:
- The particles are small compared to the wavelength
If particles are large compared to the wavelength the targets
- Will be in the optical region
If particles are intermediate compared to the wavelength the targets
- Will be mie scatterers
Why is Rayleigh approximation applied in meteorology?
- Because for most meteorological radars (wavelengths of 3 cm and larger) almost all raindrops are considered small compared to the wavelength
