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Problem: In Walter G.Carrara's book on synthetic aperture radar, the equation is presented: $\Phi(K_X, K_R) = -K_XX_t - R_B\sqrt(K_R^2 - K_X^2) +K_RR_S $ (10.30)

And this is said to come from substituting the value for $X_a$ using the Principle of Stationary Phase: $X_a = -R_BK_X/\sqrt(K_R^2 - K_X^2) + X_t $ (10.28)

into the phase of the fourier transformed signal: $\Phi(K_X, K_R) = -K_RR_t - K_XX_a + K_RR_s $ (10.29)

What I have tried: Substituting 10.28 into 10.29 as mentioned, I got: $ -K_RR_t + K_X^2R_B/\sqrt(K_R^2 - K_X^2) - K_XX_t + K_RR_s $

The following identities may be useful:

$R_B = \sqrt((Y_a - Y_t)^2 + (Z_a - Z_t)^2)$ is minimum range to scatterer

$R_t = \sqrt((Xa - X_t)^2 + R_B^2) $

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  • $\begingroup$ Why did you stop there? you have to substitute also the $X_a$ in $R_t$? $\endgroup$ – Irreducible May 30 '18 at 9:10
  • $\begingroup$ thanks! yes i had not realized that until talking to someone about it, and also got the answer just a while ago $\endgroup$ – matthew May 31 '18 at 2:24
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So if you continue the substitution you get:

$R_t = \sqrt{((-R_BK_X/\sqrt{(K_R^2 - K_X^2)} + X_t - X_t)^2 + R_B^2)} $

$R_t = \sqrt{((-R_BK_X/\sqrt{(K_R^2 - K_X^2)} )^2 + R_B^2)} $

$R_t = \sqrt{(R_B^2K_X^2/(K_R^2 - K_X^2) + R_B^2)} $

$R_t = \sqrt{(R_B^2K_X^2/(K_R^2 - K_X^2) + R_B^2 (K_R^2 - K_X^2)/(K_R^2 -K_X^2))}$

$R_t = \sqrt{(R_B^2K_R^2/(K_R^2 - K_X^2))}$

$R_t =R_BK_R/ \sqrt{(K_R^2 - K_X^2)}$

Insert that into

$ -K_RR_t + K_X^2R_B/\sqrt{(K_R^2 - K_X^2)} - K_XX_t + K_RR_s $

you get:

$ -K_R^2 R_B/ \sqrt{(K_R^2 - K_X^2)} + K_X^2R_B/\sqrt{(K_R^2 - K_X^2)} - K_XX_t + K_RR_s $

$ -(K_R^2-K_X^2) R_B/ \sqrt{(K_R^2 - K_X^2)} - K_XX_t + K_RR_s $

and there you are

$ -R_B\sqrt{(K_R^2-K_X^2)} - K_XX_t + K_RR_s $

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