High Resolution Tactical Synthetic Aperture Radar (TSAR) - MATLAB Simulations for Radar Systems Design - page 593

Denote their angular displacement as , and let be the minimum
Doppler spread between the two scatterers such that they will appear in two
distinct Doppler filters. Using the same methodology that led to Eq. (12.20) we
get

∆θ ∆ f d min

2 v
λ

------ ∆θ

∆ f d min = sin β k

(12.23)

where β k is the elevation angle corresponding to the kth range bin.

The bandwidth of the individual Doppler filters must be equal to the inverse of
the coherent integration interval T ob (i.e., ∆ f d min = 1 ⁄ T ob ). It follows that

λ
2 vT ob

∆θ = ----------------------------

(12.24)

sin β k

Substituting L for vT ob yields

λ

∆θ = --------------------

(12.25)

2 L sin β k

Therefore, the SAR azimuth resolution (within the kth range bin) is

λ

--------------------

∆ A g = ∆θ R k = R k (12.26)

2 L sin β k

Note that when β k = 90° , Eq. (12.26) is identical to Eq. (12.10).

12.4. SAR Radar Equation

The single pulse radar equation was derived in Chapter 1, and is repeated
here as Eq. (12.27),

P t G 2 λ 2 σ
4( 3 R 4 kT 0 BL Loss

SNR = ---------------------------------------------

(12.27)

where: is peak power; is antenna gain; is wavelength; is radar cross
section; is radar slant range to the range bin; is BoltzmanÓs constant;
is receiver noise temperature; is receiver bandwidth; and is radar
losses. The radar cross section is a function of the radar resolution cell and ter-
rain reflectivity. More precisely,

P t G λ σ

R k kth k

T 0 B L Loss

σ 0 ∆ A g c τ σσ 0 ∆ R g

-----

= ∆ A g = sec ψ g

(12.28)

2