SME
Remote Sensing · Problem type
Radar azimuth resolution, real and synthetic aperture
Use this for “A C-band radar (5.6 cm) 850 km from the ground has a 10 m antenna. Compare the real-aperture and SAR azimuth resolution.”
Chart
A log scale: each grid line up is 10 times the one below.
This is an example. Type your own values to start a new problem.
λ
Wavelength
example
R
Range
example
L
Antenna length
example · m
θ
Beamwidth
worked out (not typed)
δᵣₑₐₗ
Real-aperture resolution
calculated
δSAR
SAR resolution
calculated · m
k
Times finer with SAR
worked out (not typed)
Formulas
θ
=
λ
L
0.0056
=
0.056
10
δᵣₑₐₗ
=
θ
×
R
4,760 = 0.0056 × 850,000
δSAR
=
L
2
5
=
10
2
k
=
δᵣₑₐₗ
δSAR
952
=
4,760
5
Formulas are worked in m, rad. The step-by-step shows the conversions.
Assumptions
•
The beam spreads along the track at an angle of about λ ÷ L radians, so its footprint there is that angle times the range.
•
Two targets side by side along the track are told apart only when they are not in the beam together.
•
A synthetic aperture radar (SAR) adds up the echoes of many pulses into one long synthetic antenna, so its resolution is L ÷ 2 at any range.
Step-by-step
WE KNOW
λ = 5.6 cm, R = 850 km, L = 10 m
FIND
beamwidth (θ), real-aperture resolution (δᵣₑₐₗ), SAR resolution (δSAR), times finer with SAR (k)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m, rad).
λ = 5.6 cm = 0.056 m (1 m = 100 cm)
R = 850 km = 850,000 m (1 km = 1000 m)
Step 2 · Find beamwidth (θ)
Use θ = λ ÷ L
The longer the antenna in wavelengths, the narrower its beam: the wavelength over the length is the spread in radians.
θ
=
0.056
10
θ
=
0.0056
rad
Step 3 · Find real-aperture resolution (δᵣₑₐₗ)
Use δᵣₑₐₗ = θ × R
The beam’s width on the ground along the track is the spread in radians times the range; targets closer together than this share the beam.
δᵣₑₐₗ = 0.0056 × 850,000
δᵣₑₐₗ = 4,760 m
Step 4 · Find SAR resolution (δSAR)
Use δSAR = L ÷ 2
The synthetic antenna grows with the range just as the beam does, so the two cancel and half the real antenna’s length is left.
δSAR
=
10
2
δSAR = 5 m
Step 5 · Find times finer with SAR (k)
Use k = δᵣₑₐₗ ÷ δSAR
How many times smaller the SAR’s footprint is than the real beam’s.
k
=
4,760
5
k
=
952
Step 6 · Convert the answers
Change each answer back to the unit you chose.
θ = 0.0056 rad = 5.6 mrad (1 rad = 1000 mrad)
Check
Put the numbers back in. Both sides should match.
0.0056
=
0.056
10
✓
4,760 = 0.0056 × 850,000 ✓
5
=
10
2
✓
952
=
4,760
5
✓
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