Week 5

Question 1

Targets are detected when

Answer 1

The signal out of envelope detector crosses threshold

Question 2

Detection criterion

Answer 2

Established using N received pulses from target

Question 3

Analog radar displays

Answer 3

A-scope and PPI use persistence on tube to integrate signals

PPI tube persistence&raquo_space; pulse repetition interval

Returns from target contribute to brightness of spot on screen

Analog integration - no automatic detection

A-scope, PPI display requires a human observer (read range and angle from PPI)

Question 4

Analog displays - observers

Answer 4

Can interpret signals

Get fatigued when staring at screen

Can fail to notice targets

Can misread range or angle

Question 5

Automatic target detection

Answer 5

Removes observer as a detector

Threshold circuit deletes all targets below threshold

Has no intelligence, does not get fatigued, always reports range and angle correctly

Used by modern radar systems (present processed info to observer for interpretation, i.e., removes noise from targets and screen and labels target with additional info)

Question 6

Air traffic control radar labels target on screen with

Answer 6

ID, altitude

Question 7

Range gating

Answer 7

Dividing up received signal in time

A sample of the receiver output at an instant of time

Integration cannot be carried out without range gates

Question 8

Natural time step for range gating

Answer 8

τ, the pulse width (corresponds to a step in range equal to ΔR the range resolution)

Question 9

How does range gating work

Answer 9

Output of receiver is sampled at discrete instants of time after each pulse

Integration proceeds by adding up the samples returned by the pulses for each delay time

Threshold then applied to decide whether each range gate contains a target

Range dimension of the radar display will be divided into segments (cells)

For each cell a decision will be made as to whether a target is present - the cell might ‘light up’

There will be no ability to resolve multiple targets within a cell

Question 10

M out of N detector

Answer 10

Signal pulses appear in same range gate, noise pulses appear randomly in time

Require M detections in N tries to declare that there is a target

In each range gate try N times (usually number of pulses per target)

1 for detection, 0 for no detection

Threshold at M detections

Question 11

M out of N detector - Pfa

Answer 11

Pfa: Probability that a noise spike occurs in the same range gate on successive radar transmission falls rapidly with increase in N

Pd: probability that a target pulse appears in the same range gate on successive radar transmission falls slowly with increase in N

Probability of N noise pulse crossing threshold in the same range gate

Question 12

m-out-of-n tries

Answer 12

Cost of improvement is an extended dwell time in order to conduct test N times

Improvement to Pd and Pfa

With M out of N detection we can meet spec with a smaller radar (can meet spec with a lower single-pulse S/N with M out of N processing than with single-pulse detection

Question 13

Processing gain

Answer 13

Reduction in the S/N from M out of N detection required to meet spec

Question 14

Target RCS Fluctuations

Answer 14

Targets typically have dimensions much larger than wavelength (ships, aircraft, people, except for raindrops where D

Question 15

Two scatterer case

Answer 15

Equal RCS

Electric fields of reflected waves from target 1 and 2 can:

• add in phase (E1+E2=2E)
• add in anti phase (E1-E2=0)

Received power is proportional to E^2

Received power varies between 0 and 4P

Question 16

Meaning of RCS for complex targets

Answer 16

RCS not single-valued

Because its RCS is the average value of σ over its PDF, the variability of σ needs to be handled statistically

Question 17

Statistical models for RCS fluctuation

Answer 17

Case 0 - no fluctuating σ

2 PDFs x 2 fluctuation rates = 4 models (SW1, SW2, SW3, SW4)

Choice depends on the nature of the target

Question 18

2 types of PDFs

Answer 18

Rayleigh and 4th degree chi-square

Question 19

Two types of fluctuations

Answer 19

Fast, with σ varying from pulse to pulse

Slow, with σ varying from scan to scan (dwell to dwell)

Question 20

SW1

Answer 20

Rayleigh, slow

Most commonly applied

Higher detection probabilities require larger increases in (S/N) - curve diverges

Question 21

SW3

Answer 21

Chi-square, degree 4; slow

Question 22

SW2

Answer 22

Rayleigh, fast

Question 23

SW4

Answer 23

Chi-square, degree 4; fast

Question 24

Targets made up of many small scatterers

Answer 24

SW1, SW2

Question 25

Target made up of a combination of one large scatterer and many small scatterers

Answer 25

SW3, SW4

Question 26

Finding (S/N) with Swerling Targets

Answer 26

Find the (S/N) needed to meet spec with a non fluctuating model Decide which swerling model best fits target Read off additional S/N needed for the Swerling target Add additional S/N to that needed for SW0 target

Question 27

Typical values of additional S/N needed to meet the Pd spec with a fluctuating target

Answer 27

SW1 or 2: 7-8dB | SW3 or 4: 3-4dB

Question 28

Impact of target fluctuations on pulse integration

Answer 28

Coherent integration requires ‘coherence’ in the pulse-to-pulse returns (the returns need to be highly correlated in phase) Fast fluctuations of SW2 and SW4 targets defeat coherent integration of pulse returns Does not affect non-coherent integration

Question 29

Efficiency of envelope detection

Answer 29

The output SNR from an envelope detector is less than the input S/N Efficiency of detector gets worse as (S/N)in approaches 1 or 0 dB

Question 30

When noise voltage > signal voltage

Answer 30

Noise ‘captures’ the detector Does not allow the single pulse S/N for non-coherent integration to fall below +3dB at threshold (does not apply to coherent detection)

Question 31

range ambiguity

Answer 31

PRF corresponds to a max unambiguous range R = Run + x with x < Run, R’ = x = R-Run; real target can have any value {R} = mRun + x; correct target is R=R’=x Echo from the first pulse is associated with the second pulse (R>Run) All ranges > Run are aliased to the measurement interval {0, Run} {R} < Rmax if targets do not occur beyond Rmax Only if Rmax < Run have we eliminated the range ambiguity

Question 32

Resolving range ambiguity

Answer 32

Using two PRFs can resolve the range ambiguity that arise with individual PRFs mRun1 = nRun2 If new Run is too far away, there is no possibility of a target being present of or of it having a large enough RCS to be detected

Question 33

Up to what range can we avoid ambiguity by combining observations with two PRFs

Answer 33

First occurrence of mRun1 = nRun2 = Run_eff All target ranges beyond this range will aliase to nearer ranges Avoid choosing PRFs with common multiples

Question 34

The atmosphere as a propagation medium includes the effects of

Answer 34

``` Absorption Refraction Volumetric scattering Turbulence Called atmospherics ```

Question 35

In order of increasing altitude, the temperature

Answer 35

False through the troposphere (0-10km) Rises through the stratosphere (10-50km) Falls through the mesosphere (50-90km) Riss through the thermosphere (above 90km)

Question 36

Structure of the atmosphere

Answer 36

Each layer has an upper boundary where the temp reaches a minimum (tropopause, mesopause) or maximum (stratopause) At a pause the vertical gradient in temp is zero (dT/dz = 0)

Question 37

Refraction in the atmosphere

Answer 37

Refraction in the troposphere generally bends radar signal downwards Causes errors in knowing target’s position but extends the range of radar systems Obeys Snell’s Law applied to the index of refraction Fairly predictable but can produce bizarre anomalies and what amount to mirages

Question 38

Snell’s Law

Answer 38

nsin(i) | n - index of refraction, i - angle of incidence

Question 39

Effect of earth’s surface

Answer 39

Effectively blocks propagation of E/M waves Plane of the horizon defines two regions

Question 40

For a target in the line-of-sight region

Answer 40

Transmissions have direct access

Question 41

For a target in the shadow region

Answer 41

Transmissions only have access via a diffraction or refraction mechanism

Question 42

Distance to horizon

Answer 42

Microwave radars are line of sight devices Some bending of rays occur due to refraction and targets can be detected beyond the physical horizon

Question 43

The ionosphere and refraction

Answer 43

Ionosphere begins about 80km altitude and extends through a peak at about 300 km to about 1000 km Has electrical properties because it has ions and electrons Refraction in the ionosphere is strong on HF frequencies and can cause signals to reflect and propagate downwards Basis for over-the-horizon (OTH) radar

Question 44

What reveals the ionosphere

Answer 44

Airglow

Question 45

How can HF signal skip to great distances

Answer 45

By ‘bouncing’ off the ionosphere and Earth’s surface

Question 46

Atmospheric attenuation and absorption

Answer 46

An E/M wave loses energy from the intended direction of propagation by absorption and scattering (amounts to loss)

Question 47

Absorption is largely due to

Answer 47

Oxygen and water vapor

Question 48

Scattering is due to

Answer 48

Particles that deflect wave energy away from the intended direction of propagation

Question 49

Major factors in attenuation

Answer 49

Water content and temperature (highly variable)

Question 50

Clear air and water vapor attenuation

Answer 50

Attenuation in clear air varies strongly with frequency above 10 GHz -significant peaks in attenuation occur within absorption bands that are due to molecular oxygen and water vapor Water vapor produces a minor absorption band between 20 and 24 GHz (k-band) -oxygen produces a major observation band that peaks near 60 GHz at almost 20 dB/km