ch5: Distributed targets

Question 1

What maybe within the sample volume?

Answer 1

• Raindrops or cloud particles

Question 2

What completely fills the radar beam?

Answer 2

• Storms and clouds
  
  • Because they are usually so large

Question 3

Place where storms and clouds don’t completely fill the radar beam

Answer 3

• 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

Question 4

The power returned to the radar will come from

Answer 4

• All the individual targets being illuminated by the radar beam wether the beam is completely filled or not

Question 5

What may misinterpret the strength of the signal?

Answer 5

• If the beam is only partially filled

Question 6

Continental clouds contain as many as

Answer 6

• 200 or more cloud droplets/cm³
• That amounts to 2x10⁸/m³

Question 7

For a radar with a 1^o antenna beam width the beam will be

Answer 7

• 1 km in diameter at a range of 57 km

Question 8

If the radar is using 1 us pulse length the effective sample volume in space will be

Answer 8

• 150 m

Question 9

The volume of the radar pulse is then illuminating (if pulse strength is 1 us)

Answer 9

• 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

Question 10

The return from meteorological targets is the combination of

Answer 10

• Billions of returns being added together

Question 11

The total backscattering cross sectional area of a meteorological target is

Answer 11

• The sum of all of the individual backscattering cross sectional area

Question 12

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

Answer 12

• There would be little time fot the raindrops to change position relative to each other or relative to the radar

Question 13

If pulses were sent nearly simultaneously

Answer 13

• The returns measured by the radar would be virtually identical

Question 14

If we waited reasonable length of time before sending a second pulse into the same point in space

Answer 14

• 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

Question 15

When sampling raindrops or other hydrometeors with radar we need to

Answer 15

• Wait long enough to allow the particles to reshuffle so a truly different arrangement can be reached

Question 16

Why should we wait long enough to allow particles to shuffle

Answer 16

• To get a good average of the true signal amplitude

Question 17

Weather echoes are

Answer 17

• 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

Question 18

Time to independence:

Answer 18

• AKA decorrelation time
• Time it takes hydrometers to rearrange them selves so the measurements are independent of one another

Question 19

Time of independence mathematical definition:

Answer 19

• Time it takes for a sample of targets to decorrelate tto a value of 0.01 from perfect coefficient

Question 20

Perfect correlation has a correlation coefficient of

Answer 20

• 1.0 (or -1.0 if there is an inverse correlation)

Question 21

Correlation coefficient of 0 means that

Answer 21

• There is no correlation at all

Question 22

Waiting for a signal to decorrelate to 0.01 means

Answer 22

• We have waited long enough that the newest sample is almost completely different than the original sample

Question 23

The decorrelation time of a sample depends on:

Answer 23

• Wavelength of the radar used
• Hydrometeors
• Turbulence within the sample volume

Question 24

Decorrelation times are shorter for

Answer 24

• 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

Question 25

What might make the sample decorelate slowly?

Answer 25

* Particles having the same size because that will make them fall together with the same terminal velocity

Question 26

What might make the sample decorrelate faster?

Answer 26

* Storm containing a wife veriety of particle sizes * There will be many different particles falling in the same volume

Question 27

The shortest decorrelation times occur when

Answer 27

* Hail and rain are in the same sample volume

Question 28

The time required for the autocorrection function to fall to a value of t= 0.01 is

Answer 28

* Aprox.t= 2 lamda to 3 lamda * T is in milliseconds and * Lamda is in centimeters

Question 29

Measured decorrelation time have ranged from

Answer 29

* 3.5 ms to nearly 30 ms

Question 30

Measured decorrelation time range depends on

Answer 30

* Storm and radar

Question 31

If we want to sample as close together in time as possible but still have independent samples, we would want to

Answer 31

* 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

Question 32

modern Doppler radars often use PRFs

Answer 32

* near 1000 Hz

Question 33

why did we use PRF near 1000 Hz and not 1000 Hz

Answer 33

* because PRF of 1000 Hz give sampling time that is too close together to have truly independent samples

Question 34

why is it necessary to average many consecutive pulses together

Answer 34

* in order to have the equivelant of just a few independent samples

Question 35

factors which can contribute to decreasing the time to independence of consecutive samples made with a radar

Answer 35

* range averaging * moving the antenna in azimuth while collecting the data (almost always done) * wind shear within the sample volume * turbulence

Question 36

the horizontal and vertical beam widths must always

Answer 36

* be in radians

Question 37

the pulse length h is:

Answer 37

* the length in space corresponding to the duration t of the transmitted pulse * AKA pulse duration

Question 38

In the equation for sample volume we used h/2 because

Answer 38

* We are interested in signals that return to the radar at precisely the same time

Question 39

How to find the total backscattering cross sectional area of targets within the radar sample volume?

Answer 39

* 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

Question 40

For sample volume we used the horizontal and vertical beam widths this assumed that

Answer 40

* 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

Question 41

2 ln (2) in the dominator of the radar pulse volume equation accounts for:

Answer 41

* The real beam shape better than the assumption

Question 42

Parameter that is related to the total backscattering cross sectional area

Answer 42

* Radar reflectivity

Question 43

Radar reflectivity:

Answer 43

* The summation is done over all individual targets in a unit volume in spave * Has units of 1/cm or cm^-1 * Intensive parameter

Question 44

…………………… is related to the backscattering cross sectional area

Answer 44

* Target size

Question 45

Reylight approximation applies if:

Answer 45

* The particles are small compared to the wavelength

Question 46

If particles are large compared to the wavelength the targets

Answer 46

* Will be in the optical region

Question 47

If particles are intermediate compared to the wavelength the targets

Answer 47

* Will be mie scatterers

Question 48

Why is Rayleigh approximation applied in meteorology?

Answer 48

* Because for most meteorological radars (wavelengths of 3 cm and larger) almost all raindrops are considered small compared to the wavelength