I'm trying to reconcile the fact that decimation without low-pass filtering does not modify the ratio of signal to interference power in the time domain with the fact that, in the frequency domain, interfering signals can alias on top of the signal of interest. I'm thinking about something here in the wrong way but I'm not sure what.

Say I have a signal $x[k]$ made up of a signal of interest $s[k]$ plus an additive interfering signal $n[k]$, so $x[k]=s[k]+n[k]$. Say both signals are bandlimited with non-intersecting bands, and I'm sampling fast enough to separate those bands. The first gap in my understanding is how to characterize the "SNR" in the time domain vs. the frequency domain. "SNR" might not be the right term here, I'm just thinking about in general how well I can discriminate between the two signals.

In the time domain I can find the average power of $s$ and the average power of $n$ and say those are directly competing at all samples, but in the frequency domain, assuming no leakage between bands, the "SNR" is infinite. I know that typically one could find the "time-domain SNR" from the spectrum by something like an integration of the signal energies over all frequencies, but there seems to be something to the fact that, no matter how small of an "SNR" that would give me, I could still separate the signals in frequency and pull out the signal of interest. So what is the metric that really matters in this case?

If I then decimate $x[k]$ such that the bands of $s$ and $n$ overlap, the "SNR" in the frequency domain seems to have been reduced, since now there's signal and interference energy competing in the same frequency bins and I can no longer easily separate out $s$ simply by looking at the spectrum. However, in the time domain, assuming the statistics of $s$ and $n$ are constant, there is apparently no change in the ratio of their average powers. The amount of signal and interference energy competing in a sample is the same.

So, what is the concise reason that I would need to filter before decimating? Is it simply that I can no longer separate the signals using a bandpass filter or other spectral analysis? How is the loss of discriminatory ability characterized in the time domain samples?

For example, in the image below $s$ is a complex tone (just plotting real parts) at 13 Hz with amplitude 1 and $n$ is a tone at 113 Hz with amplitude 1. In both time-domain plots you can clearly see $s$ with some higher-frequency interference. In the frequency-domain plots, after decimation, $n$ has aliased on top of $s$ and they're not separable. If anything, I'm tempted to think that in the time-domain plot $s$ has become more visible.

enter image description here

Here is the code I used:

Fs = 500; % sample rate
N = 1000; % number of samples
t = 0:1/Fs:(N-1)/Fs; % sampling times

fs = 13; % signal of interest frequency
fn = 113; % interfering signal frequency

s = exp(1i*2*pi*fs*t); % signal of interest
n = exp(1i*2*pi*(fn*t+rand)); % interference
x = s + n; % observed signal

% Decimated signals
dec = 5;
s_dec = s(1:dec:end);
n_dec = n(1:dec:end);
x_dec = x(1:dec:end);
N_dec = numel(s_dec);

% Frequency domain
fax = (-floor(N/2):ceil(N/2)-1)*Fs/N;
fax_dec = (-floor(N_dec/2):ceil(N_dec/2)-1)*Fs/N_dec;

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  • $\begingroup$ By "decimation", you mean down-sampling? Of course,, if you do any resampling of a digital signal, you must LPF it before resampling because any surviving images will potentially fall into your baseband and become aliases. This LPFing is what happens when you use a sinc function (or similar) in your interpolation kernel. $\endgroup$ Jan 23 at 1:38
  • $\begingroup$ And I don't know what you mean by "and I'm sampling fast enough to separate those bands." You need to sample fast enough to separate the baseband from the first image of the baseband. Then you can worry about what filter you need to separate $s[k]$ from $n[k]$. $\endgroup$ Jan 23 at 1:41
  • $\begingroup$ By "decimation" I mean selecting every Mth sample. $\endgroup$
    – Roger
    Jan 23 at 18:29
  • $\begingroup$ By "and I'm sampling fast enough to separate those bands." I mean $s$ occupies some bandwidth [a, b] and $n$ occupies some bandwidth [c, d] with c > b and Fs > d $\endgroup$
    – Roger
    Jan 23 at 18:32
  • $\begingroup$ Fs > 2d , correct? $\endgroup$ Jan 23 at 20:12

1 Answer 1


From a radio receiver perspective, the SNR of concern would be the difference between the signal and noise of the samples used for demodulation decisions. With proper design, this is done after optimized filtering that removes much of the noise that is not within the occupied bandwidth of the signal of interest.

In this graphic I show an example of two waveforms in two different channels, along with an overall noise density $N_o$ that is a distribution as Watts/Hz. The signal power is also distributed over an occupied bandwidth such as signal S1 below with bandwidth $B$ Hz. The total noise that would be considered for SNR of signal S1 is the noise within the occupied bandwidth $B$, so $BN_o$ Watts. given $N_o$ as a flat density the noise further away from S1 will be easily reduced to insignificant levels with a properly designed "Matched Filter", but also assumes receiver filtering has sufficiently reduced S2 from otherwise being an interference degradation to SNR.

two signals

An easy way to see this more intuitively and experimentally is to create samples of Additive White Gaussian Noise (AWGN) of variance $\sigma^2$ and sampling rate $f_s$ and add that to a mean DC level $\sqrt{E}$ as used to transmit one bit as a "1". (similarly a level of $-\sqrt{E}$ would be used to transmit one bit as a "0"). The total noise power prior to any further processing is the variance of the AWGN, and the total signal power is the DC level squared such that the SNR in dB is given by $10Log_{10}(E/\sigma^2)$. However if we used $N$ successive samples to transmit each bit, the data rate would be less, and therefore the occupied bandwidth will be less, but the receiver could then average those $N$ samples prior to deciding if a 1 or 0 was transmitted. Averaging as a filter reduces the noise, and specifically the variance of the noise as AWGN after a moving average of $N$ samples would go down by $N$. Such averaging would also reduce the power of signals at higher frequencies (other channels), but for non-white noise additional specific filtering can be more effective. A moving average is a low pass filter with an equivalent noise bandwidth of $f_s/N$ where $N$ is the sample duration of the average, so the decrease in noise is consistent with the selection through filtering of a portion of the total noise.

So what we see from this exercise is similar to what is demonstrated in the graphic above (our waveform here is directly at baseband but could be centered on any other frequency within our sampled bandwidth): the AWGN that was generated is a white (equal over all frequencies) noise density and prior to any averaging (filtering) the total noise power matched the variance for the noise as created. The moving average over the duration of the bit is the optimum Matched Filter for such a simple waveform in the presence of AWGN and reduced the overall noise accordingly. The SNR is typically given by the both the signal and the noise within the occupied bandwidth of the signal, as we have the opportunity with proper design to filter out noise that occupies other frequencies. In reality other noise will get in and this all goes into the detailed design of the receiver and the interference and co-channel environment in which in needs to operate.

Decimation and Filtering

As far as filtering prior to down-sampling (decimation is the combination of a filter and down-sampler), the concise reason is to not degrade SNR due to folding of the broadband noise floor (or other signals if also in band). It is the signals (which can and do include the digital noise floor) that would fold directly into our primary band of interest ($B$ in my graphic above) that would be of concern. This assumes maximizing SNR, receiver sensitivity, dynamic range etc is a prime concern as it would typically be in most receiver designs.

This is identically the same reason why we need anti-alias filtering prior to sampling. See this recent answer describing what occurs. Even when there are no other signals present, there will always be an elevated noise floor (unless it has been previously filtered), and not including any anti-alias filtering will cause all the noise from the alias frequency regions to fold onto the primary signal band of interest, elevating the noise floor. In the analog system this would be the amplified received signal from the antenna. In the digital system this would be samples of that noise combined with the quantization noise floor of the A/D. (Ideally assuming we want maximum sensitivity, it is the samples of the received noise that would dominate). When we sample from Analog to Digital, we are down-sampling from an infinite sampling rate to a finite sampling rate. When we decimate, we down-sample from a higher sampling rate to a lower rate. All the same concepts of aliasing apply in that anti-alias filtering is required prior to A/D sampling, and prior to digital down-sampling.

In the OP's final example, a 13 Hz and (synchronized) 113 Hz combine such that when down-sampled the aliases constructively add. Similarly, if we had the exact same modulated waveform centered at 13 Hz and again at 113 Hz, the result would be the same waveform with constructive interference. Offsetting the frequencies slightly would be more realistic in showing the effect of two independent waveforms.


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