Assume a monostatic radar is transmitting a pulsed waveform. The pulse has been modulated by a Barker code with$\ N$ chips or bits. The waveform chip rate is $\tau_c $, the radar receiver's detection bandwidth is $\beta_c = 1/\tau_c$, and the waveform's pulsewidth is $\ N*\tau_c$.

I've heard conflicting opinions about a signal processing gain,$\ P_g$, due to compressing the pulse that is proportional to the waveform's time-bandwidth (or pulse compression ratio (PCR)) product (or equivalently the product of the uncompressed pulsewidth $\tau_p$ and the compressed bandwidth $\beta_c$) of the pulse: $\ P_g = \tau_p*\beta_c=N\tau_c*\beta_c$

Assuming that you start with the power form of the radar range equation: Power form of SNR Radar Range Equation

Is it appropriate to add a$\ P_g$ term to the power form of the SNR radar range equation?

SNR plus processing gain

(Pulse compression allows a radar to transmit a longer pulse to get more energy on the target while simultaneously achieving the range resolution of a much shorter pulse.)

FWIW, I have the same question about the energy form of the radar range SNR equation:

Energy form of the radar range snr equation

My question is similar to this older question but more specific

  • $\begingroup$ Yes. If you search for 'radar equation pulse compression,' you will find derivations that include the SNR gain from pulse compression. $\endgroup$ Feb 9, 2018 at 18:47

1 Answer 1


The answer is yes but one has to specify $B_n$ properly to avoide possible confusions. In case if one uses a pulse compression, the bandwidth through which the receiver collects the noise will normally be $B_n = \beta_c$. Then, the "new" signal-to-noise ratio should be written as:

$SNR = \dfrac{P_TG_TG_R\lambda^2\sigma{P_g}}{(4\pi)^3R^4(kT_{sys}\beta_c)} = \dfrac{P_TG_TG_R\lambda^2\sigma{\tau_p\beta_c}}{(4\pi)^3R^4(kT_{sys}\beta_c)} = \dfrac{P_TG_TG_R\lambda^2\sigma{\tau_p}}{(4\pi)^3R^4(kT_{sys})}$

As you see, the signal-to-noise ratio remained the same as for the pulse without modulation (for which $B_n = 1/\tau_p$).

So, the pulse compression helps to increase the range resolution but the signal-to-noise ratio remains the same as for the unmodulated pulse. This holds true if you do not change $\tau_p$ and the peak power $P_T$ when you modulate your pulse.

The confusion (and as you mentioned "conflicting opinions") comes from the fact that it is quite often forgotten that if you increase the bandwidth of your pulse then you also increase the bandwidth through which the receiver collects the noise since radar's receiver is matched to the pulse and therefore has a bandwidth $\beta_c$.

Your last equation is correct. But since you put $F_n$ - the noise figure- then it should be specified that this is the signal-to-noise raito at the output of the receiver that has a noise figure $F_n$.


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