Simple explanation of Coherent integration radar

Question

I have a physics background, and I'm reading some physics data analysis papers where they keep throwing around the term coherent integration. I've done the google search, but the best answer I could come to was in coherent integation you, add voltages, and with in-coherent intergration you add power.

Now this may make more sense to a person with signal processing background, but I can't figure it out.

My specific papers, are about coherently integrating $n$ number of days of time-series data to find a weak signal.

Answer 1

Yes, that is about it: coherent integration = voltages (currents) are added, noncoherent integration = powers are added.

For example, say the transmitted radar pulse is $$p(t)cos(\omega t) \text{ where } p(t)=1, \text {when } |t|<T_p/2 \text { and } 0 \text { otherwise} $$. Now the radar receiver gets this pulse with some unknown phase and amplitude as $Ap(t-\tau)cos(\omega t +\phi) + \nu'_c cos(\omega t) + \nu'_s sin(\omega t)$ where $\nu'_c, \nu'_s$ are iid normal processes representing the received noise . Now the receiver mixes this with its own local oscillator $cos(\omega t)$ and $sin(\omega t)$ and gets the baseband IQ pair over the $T_p$ pulse length as $$I=Acos(\phi)+\nu_c$$ and $$Q=Asin(\phi)+\nu_s$$ after having properly delayed the terms by $\tau$ and $Var[\nu_c]=Var[\nu_s]=\sigma^2$. These values are stored. Having transmitted $M$ such pulses the stored values are summed, in radar parlance "integrated", the result being $$ \sum {I} = MAcos(\phi) + \sum \nu_c$$ where $Var{[\sum \nu_c]} = MVar[\nu_c]=M\sigma ^2$ and similar for the Q term. The power in the signal is $S=(MAcos(phi))^2+(MAsin(phi))^2=(MA)^2$ while in the noise is $M\sigma^2 + M\sigma^2 = 2M\sigma^2$. Therefore the resulting SNR after integration is: $$ \rho_M =\frac {(MA)^2}{2M \sigma^2}= M \rho _1$$