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I have just started work with radars. I read that incoming signal in case of Doppler radar is frequency shifted. On adding it to original signal we get Doppler shift. But we don't do it directly. First we make I component (by mixing incoming signal with LO signal) then Q signal by mixing original signal with phase shifted LO then do I + jQ. This I understand is to determine the direction of motion either towards or away from Radar. I am attaching excerpt from book. Can anyone please explain how this one signal in each waveform becomes zero?

From Micro Doppler Effect in Radar by Victor C. Chen

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Keep in mind that the value of fD can be positive or negative. The sign of fD has no effect on the I signal (cos(x) = cos(-x)), but it reverses the sign of the Q signal (sin(x) = -sin(-x)).

The 90° phase shifter delays the Q signal — its output is Q90(t) = -sin(2πfDt - π/2) = cos(2πfDt) when fD is positive, but it becomes -cos(2πfDt) when fD is negative.

Therefore, when fD is positive, the upper adder outputs cos(2πfDt) - cos(2πfDt) = 0, but when fD is negative, it outputs cos(2πfDt) - (-cos(2πfDt)) = 2 cos(2πfDt).

Similarly, when fD is positive, the lower adder outputs cos(2πfDt) + cos(2πfDt) = 2 cos(2πfDt), but when fD is negative, it outputs cos(2πfDt) + (-cos(2πfDt)) = 0.

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  • \$\begingroup\$ If we look like this -sin(2*pifdt- pi/2)=sin(pi/2 - 2*pifdt). And sin(pi/2-a) equals cosa no matter what sign of a is. (Here a would be 2*pifdt). So in either case output is same. I am thinking like this. Any comments? \$\endgroup\$ – Aashish Sharma Oct 6 '14 at 5:29
  • \$\begingroup\$ You're still ignoring the fact that the sign of the Q signal changes with the sign of \$f_D\$, before you apply the processing shown in the diagram. \$\endgroup\$ – Dave Tweed Oct 6 '14 at 11:10
  • \$\begingroup\$ Thanks a ton. This was bugging me a lot. Now got hold of it. \$\endgroup\$ – Aashish Sharma Oct 7 '14 at 9:12

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