I've created both single tone signal and a LFM chirp signal with same pulse duration and cross correlated both of them with their own(matched filter). Peak points of both results were same as expected which is the total energy of input pulse. I've read about SNR gain with signal compression in radar equations and chirp is an example of compressed signal. Where does this SNR gain come from? Does it simply mean matched filter? If so pulse compression is no different than a single tone as far as I understand(range resolution is the only difference).

Edit: Equation 7.4 states that output peak SNR to input SNR is proportional to both bandwidth and pulse length. It looks at the output to input SNR ratio which is affected by the input noise because of bandwidth used at the input. Does this mean a single tone signal has a better input SNR because of less bandwidth? Then what is the paper trying to say by comparing output SNR to input SNR if peak point is the total energy of my pulses on both compressed and single tone after a matched filter?

Edit2: After some thinking I've concluded it as follows: No matter what the signal modulation or bandwidth, peak SNR at matched filter output is same for input pulses of same energy. However, input noise power is proportional to the bandwidth of signal which gets filtered on matched filter and that's what we call matched filter gain. Correct me if I'm wrong.

  • $\begingroup$ Actually, input SNR is a function of signal bandwidth. Signal power is same for all cases, but higher the signal compression(bandwidth) higher the input noise which is going to get filtered in matched filter anyways. $\endgroup$ – menderft Oct 26 '16 at 19:06
  • $\begingroup$ Noise is not same, more the bandwidth, you get noises from more frequencies. $\endgroup$ – menderft Oct 26 '16 at 19:44
  • $\begingroup$ If we assume that we are only passing the band of interest which is equal to the bandwidth, then dont we have more noise as we let a greater band enter the matched filter? $\endgroup$ – menderft Oct 26 '16 at 20:43
  • $\begingroup$ Lets talk on an example as it may make the things easier. I am sampling with an ADC where fs=100Mhz. My band of interest is from 25Mhz to 50Mhz. Shouldn't I filter out the band from 0 to 25MHz before going into matched filter? $\endgroup$ – menderft Oct 26 '16 at 20:50
  • $\begingroup$ Yeah, now I have a 25Mhz band of noise at the input matched filter. If I've had a band of interest from 37.5 to 50Mhz and I filter out from 0 to 37.5MHz, now I have a 12.5MHz band of noise at the input of matched filter. $\endgroup$ – menderft Oct 26 '16 at 20:55

It's often said that pulse compression gives you a gain proportional to the time-bandwidth product (otherwise known as the pulse compression ratio, or $PCR$). This is a really misleading statement, and it had me confused enough to sit down and think about it for awhile. I thought I'd share some of my findings that I pieced together from both reading the internet and from my own DSP simulations.

The claim

Before looking into this, I had assumed that, if I take two pulses of equal amplitude $A_0 = \sqrt{P_0}$ and equal pulsewidth $\tau$, one of them a simple sine (no frequency modulation) and the other an LFM chirp, then the matched filter output $SNR$ of the chirp would be larger than that of the sine, precisely by a factor of $PCR$. I had reasoned that they both have equal energy $(E = P_0 \tau)$, but the chirp has a much narrower peak width than the sine $(\delta t = \tau/PCR)$, which would result in a larger peak height, inversely proportional to keep the total energy fixed: $$E = (P_0) (\tau) = (P_0 PCR)(\frac{\tau}{PCR}) $$

This is wrong, yet for some reason it's stated exactly this way on Wikipedia's article on the subject. Someone should fix this?

What actually happens

It turns out that both signals have the same peak power, despite the fact that the chirp's energy is squeezed into a smaller width. Therefore, we conclude that the total energy of the matched filter output of an LFM is actually smaller!

At first this looks like a problem. Moreover, you're not going to do any better with any other filter, because matched filters are optimal. But as has been noted already, the same peak power will be observed regardless of the waveform. Improved $SNR$ isn't the advantage of an LFM. The real benefit is the resolution. With the same $\tau$, I can resolve echoes that are less than $\tau$ apart, without losing $SNR$. That's pretty awesome, it turns out.

How to actually observe "Pulse Compression Gain"

If you really want to see pulse compression gain, then you've gotta compare something different. Indeed, the subject of pulse compression is usually introduced by explaining the tension between time resolution and $SNR$, before pulse compression. With a simple sine, you need to have a longer $\tau$ to improve $SNR$ (more total energy), but then you have the problem that you can't resolve echoes as well as you did before. If you want to resolve echoes better, you have to either crank up your peak power or sacrifice $SNR$.

As I mentioned above, pulse compression allows you to resolve shorter time-scales while keeping $SNR$ fixed. Alternatively, you could try comparing two signals with equal bandwidth: a simple sine wave with pulsewidth $\tau$ and an LFM chirp with pulsewidth $\alpha \tau$. Both pulses have the same resolving capability when $\alpha = PCR$ of the chirp. But the chirp will a larger peak power, and thus larger $SNR$, by a factor of precisely the $PCR$. This is simply because you have a longer pulse, not because more energy has been "squeezed" into a smaller time bin.


You are correct that the peak output will be the same regardless of pulse shape for pulses with the same energy. This is a tautology, because the peak of the autocorrelation function is identically equal to the signal energy.

What is important to note here is that SNR (signal to noise ratio) gain requires noise to actually be present. The SNR gain achieved will vary widely by pulse shape, because they will have different frequency responses.

All that is guaranteed with a matched filter is that the SNR will be maximized over all possible receive filters for a given transmit filter.


SNR gain or comes from matched filter, independent of bandwidth. But resolution is directly proportional to bandwidth. SNR gain is proportional to pulse length.

  • $\begingroup$ Edited my question. $\endgroup$ – menderft Oct 26 '16 at 7:01

No matter what the signal modulation or bandwidth, peak SNR at matched filter output is same for input pulses of same energy.

You're almost there... what is "peak energy" anyway? The important quantity is peak power, and it is indeed increased by the factor of PCR at the matched filter output.

if I take two pulses of equal amplitude A0=P0−−√ and equal pulsewidth τ, one of them a simple sine (no frequency modulation) and the other an LFM chirp, then the matched filter output SNR of the chirp would be larger than that of the sine, precisely by a factor of PCR.

The ratio of Signal power to Noise power is larger by a factor of PCR, even though the ratio of Signal energy to Noise energy is unchanged. In fact the latter goes down slightly because the filter is imperfect and both attenuates the signal energy and adds thermal noise energy. Clearly we can't make more energy out of thin air using math. What we can do is improve the detectability of the same amount of energy.

For example a 10 microsecond long coded input pulse exits the pulse compression filter as 1 microsecond long temporally compressed pulse, thus (ideally) Signal power has increased by a factor of 10 because its duration has been decreased by a factor of 10 while still having the same amount of energy. While Noise power has not been correlated in the filter and so has not been multiplied whatsoever by the filter. Decreasing the length of the pulse by a factor of 10 increased the power of the pulse by a factor of 10 (power SNR is boosted by 10 dB).

Implemented electronically, one can say that receivers measure voltage. Detection is enabled by the boost in the ratio of Signal voltage (increased by 10 dB thanks to pulse compression) to Noise voltage (unchanged by pulse compression). However, it's also possible to perform mathematical pulse compression using digital signal processing after analog to digital conversion of the--still 'long' in the temporal sense--received pulse.


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