Introduction to Radar Basics - MATLAB Simulations for Radar Systems Design - page 72

Solution:

From Eq. (1B.18) we have

F 2 –

G 1

1 F 3 –

G 1 G 2

1 –

G 1 G 2 G 3

F 4 1

F = + --------------- + --------------- + ---------------------

1

G 1 G 2 G 3 G 4 F 1 F 2 F 3 F 4

1 dB 6 dB 6 dB

– 1 dB 20 dB – 8 dB 60 dB 10 dB

100 10

0.7943 0.1585

10 6

1.2589 3.9811 3.9811

It follows that

3.9811 –

0.7943

1 10 – 1 3.9811 – 1

F = 1.2589 + ------------------------- + ------------------------------- + --------------------------------------------------- = 5.3629

100 × 0.7943 0.158 × 100 × 0.7943

F = 10 log ( 5.3628 ) = 7.294 dB