# Does oversampling improve processing gain?

I am used to Time-Bandwidth product describing the processing gain or improvement in SNR for a spread spectrum signal using a matched filter.

One could view the bandwidth $B$ as the sample rate (complex) and the time $T$ is the integration time and thus there are $TB$ samples. Thus, the matched filter produces a gain of $10\log_{10}N$, with $N$ being the number of samples.

So with all of this in mind, suppose a bandpass filter removes all noise power except for the signal bandwidth, thus the noise power is $kTB$. The signal power is the amplitude $A$ squared, $A^2$. If we over sample the signal by some factor $M$, then I want to say that the processing gain is now $MTB$ because there are $M$ more samples.

I also have worked some theory of this as well as a MATLAB script and all is indicating that it is true. Am I missing something? Why not over sample signals to provide more processing gain using a longer matched filter?

• Yes, Process Gain =10log((Fs/2)/BW). However, it increases the complexity, power consumption and cost. – msm Sep 2 '16 at 7:44
• What you say is certainly true with respect to quantization noise. For a given ADC, over sampling, low-pass filter and decimating will certainly improve your signal to quantization noise ratio. – Pedro_Uno Sep 16 '19 at 13:26
• If an ADC is sufficiently dithered, you can say that quantization noise is evenly spread over the Nyquist range of the ADC. Increasing the sampling rate spreads more of that energy into spectrum that you will filter out with your matched filter. The rule of thumb is that for each increase in sampling rate by 4 you get 6dB increase in SNR. Note that often this increase is not as great as what you might get from using a better ADC. – Pedro_Uno Sep 16 '19 at 13:33

You need to define what you mean by SNR and processing gain, because Marcus Muller and Qasim Chaudhari can both be considered correct depending on your definitions of SNR and processing gain. A processing gain is usually taken to be the SNR at the output of a system divided by the SNR at the input of a system. If you consider the SNR to be the signal power within the band of interest divided by the noise power within the band of interest, then there is no processing gain as you increase the sampling rate. This assumes an ideal (sinc) filter. In fact, without a perfect matched filter and perfect stopband attenuation (equal to zero), there is a processing loss caused by the out-of-band noise aliasing back into the spectrum after downsampling to the symbol rate. Although in all practical cases, this is very small and not really worth talking about. In other words, there is a fundamental limit to the SNR you can achieve by oversampling in this case. Without oversampling, it is likely that you will have poor filtering, and out-of-band noise may leak into the spectrum of interest (degrading it).

Now consider the SNR to be defined as the signal power divided by the total noise power contained in the samples (i.e. across the larger Nyquist bandwidth). In this case, you can reduce the noise level by filtering the excess noise that was added out-of-band. In other words, you will get a processing gain that is equal to the sampling rate divided by the true bandwidth of your signal (one over the baud interval for single carrier pulse shaped systems). This is often called a processing gain in ADC literature for example. Notice that if in your original samples, the matched filter design is good enough to reject out-of-band noise to negligible levels then there is no benefit in letting more noise into the system simply to reject it in order to obtain a high processing gain. It is probably worth noting that oversampling does result in a higher order matched filter which typically would result in a better stopband attenuation (making the excess noise less pronounced). This effect is very subtle and will never allow you to achieve a better SNR than you could by using a perfect matched filter (stopband attenuation = 0) at a lower sampling rate.

Shortly put: Your revelation is true, the more you oversample, the lower the noisyness of your observed signal becomes.

Think about it this way: at any sampling rate, white noise in any two samples is uncorrelated, whilst the signal being oversamples is very correlated across samples. Now, if you low-pass filter the noisy signal, the uncorrelated noise should "cancel" on average, whereas the signal "sums up". So, yes, there is a gain proportional to the oversampling ratio.

Why not over sample signals to provide more processing gain using a longer matched filter?

Technical feasibility. In fact, this is quite often done (you should read up on "delta sigma converters", which is an oversampling technique), and it's even especially often applied to SDR. For example (and I know I always cite these devices, sorry), a Ettus USRP X310 does sample with a (typically fixed) rate of 200 MS/s complex. As a user, you can get (nearly) any integer fraction of that rate, let's say 200MS/s / 16 = 12.5MS/s. The oversampling will "shape" the white noise, and due to the USRP decimating with "good" low-pass filters internally, you get an improved SNR. Linearity considerations show that it doesn't make a difference whether you apply your matched filter to the 12.5 MS/s stream or the 200 MS/s stream, assuming the bandwidth of your signal < 12.5 MHz.

However, at some point, things get hard. Your PC will have a really hard time dealing with 200 MS/s at all (which is why the decimation has to happen on the USRP itself, not after sending the samples to the PC), and that's not even accounting for the fact that you want to apply a filter to that. ADCs that are even faster than that can be bought, yes, but a) they get really expensive and b) you pay for higher sampling rate with reduced sample bit depth, and hence, higher quantization noise.

• there's been a downvote, which is democratically very fine, but if the person downvoting me would like to leave a comment on what they don't like about my answer, I'd have the chance to address that. – Marcus Müller Oct 3 '16 at 23:41
• could you explain your point b), why do we have higher quantization noise ? According to this dsp.stackexchange.com/questions/31899/…, which seems correct, we should have lower quantization noise at ADC. Thanks. @Marcus – AlexTP Apr 22 '17 at 17:21
• @AlexTP ah, that was badly expressed by me! Technologically, you "pay" for higher sampling rate (at the same price/power consumption/complexity...) by only being able to buy lower-resolution ADCs. Does that explain the point? – Marcus Müller Apr 22 '17 at 17:45
• yeah, I understand now, economical​ stuff. Thanks :) – AlexTP Apr 22 '17 at 17:48
• @AlexTP not even necessarily economical. Admittedly, if you have deep pockets, you can nowadays buy gigasample ADCs, but you simply can't buy a 16 bit 10 GHz ADC. It doesn't exist (to my knowledge). – Marcus Müller Apr 22 '17 at 17:49

Have a look at this paper: "Performance analysis of an all-digital BPSK direct-sequence spread-spectrum IF receiver architecture" by B.-Y. Chung, et al. In the second paragraph of the second page, it states that upsampling by 4 provides 6dB of processing gain. So your revelation is true in a sense. But this argument is true if the sampled noise values are uncorrelated. For example, suppose that you have a signal with BW=20MH, and your symbol rate is Rsym=5Msym/sec. Sampling rate of Rsamp=20MSample/sec would give you 6dB (4 times) of processing gain, by averaging out the noise samples within each symbol. But the thing is, what happen if the noise BW is less than 20MHz. In this case, averaging will not reduce noise power as much as 6dB. The effective reduction in noise power due to upsampling is BW/Rsym, where BW is the noise bandwidth of your system, and Rsym is the symbol rate(5Msym/sec in our case example).