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Correction of Frequency in Arrival Direction Detection

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Advanced Member level 5
Oct 21, 2006
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Carrier frequency of receiving signal is same as local frequency of receiver for rader system, if we ignore doppler shift.

In this case, we can evaluate arrival direction by using music

However if a target generates own carrier which is slightly different from local frequecy of receiver, I don't think we can apply above algorithm directly.
Here I have to correct frequency difference.

Am I correct ?

I think correction is not required for angular information. For phase monopulse receiver both antennas will receive same transmitter frequency, DOA may be extracted using simple FFT (which may be replaced by MUSIC)

For example:
receiver frequency F
transmitter frequency F+1MHz
antenna 1 mixer output -> ADC -> FFT -> found peak at 1MHz, phase phi1
antenna 2 mixer output -> ADC -> FFT -> found peak at 1MHz, phase phi2
finding DOA: (phi2-phi1)*k depending on antenna spacing, angle will be mirrored if transmitter frequency is Ftr = F - 1MHz

Or you meant some other case?

Maybe this will interest you, something about correction:

It is a frequency compensate actually, since frequency deviation of 1MHz is evaluated.

E[(ftx-frx)*time] is not zero.
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One of my 1st designs was replacing ground based Radar with Doppler using OCXO +PLL to 100kHz sub-carrier FM pilot tone on Tx, then received and mixed with same OCXO ( using PLL to scale to same 10MHz) to get change in RANGE and azimuth shift with 2 antenna. It worked very well since my design was stable to 1e-10 for rocket harsh environment of 100g shock, 50g accel, and 15g vibe.

Range is phase of 1 cycle at speed of light and velocity is d (phi) /dt so counting cycles determines total range.

I used an analog sawtooth mixer so the result was an analog linear phase shift with range. It worked very well and the error frequency was nulled before launch. Circa 1976

The resulting out waveform was an exact duplicate of the sawtooth due to my method of precise very fast S&H such that if each transition of the sawtooth were used to accumulate instead , you would see the precise parabolic path of the rocket as perceived by the apparent tangent distance called "Range" and this can see the peak apparent altitude ( just after Apogee) and thus the mixed output sawtooth slows to a zero ramp 0 f. or DC then changes direction or slope coming down to earth. In our case Apogee was over 500 miles for plasma research.
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E[exp(j*2*pi*(ftx-frx)*time)] is not zero.

What is E[] ?

My current understanding is that direction may be found regardless of transmitter frequency deviation.
How is your transmitter signal is different from Doppler signal reflected from moving object?

transmitter with frequency F+F(t), F(t) unknown
receiver with LO frequency F
antenna 1: F(t) phase shift phi1
antenna 2: F(t) phase shift phi2
antenna 3: F(t) phase shift phi3
antenna 4...

DOA is found by analyzing phi1,phi2,phi3 using FFT/music/etc.

it seems that I am missing something here

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