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A radio signal is sampled at $f_{s}=5512$ Hz, and $n=2^{14}=16384$ points are used in the FFT to display it as a spectrogram. The frequency resolution is $\frac{f_{s}}{n}=0.33642578125$ Hz/point, displayed in the spectrogram containing $\frac{n}{2}=8192$ pixels (Nyquist). Assuming there is no DC offset, the first pixel displayed is $f_{min}=0$ Hz, the last corresponds to $f_{max}=\frac{f_{s}}{2}-\frac{f_{s}}{n}=2755.66357421875$ Hz. So far, I understand. But, what's the precise bandwidth? Is this $\frac{f_{s}}{2}=2756$ Hz, or rather $f_{max}-f_{min}=2755.66357421875$ Hz?

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  • $\begingroup$ a radio signal has frequency components much hither than 2756 Hz (which is your Nyquist frequency). $\endgroup$ – robert bristow-johnson Dec 24 '19 at 14:19
  • $\begingroup$ @robertbristow-johnson To eliminate that distraction, just put a bandpass filter around whatever frequency the "radio signal" is at, and assume the sampler has sufficient analog front-end bandwidth. The sampler can be either real or complex too, and if complex we can use an optional Hilbert transform on the bandpass signal prior to sampling to avoid a 3 dB Noise Figure penalty. $\endgroup$ – Dan Boschen Dec 27 '19 at 14:11
  • $\begingroup$ well, without that " optional Hilbert transform[er]", sampling a bandpass signal at sub-sample rate is sometimes problematic. personally, if i were to design a "software radio", i would, in the analog domain, bump it down to an intermediate frequency (IF) and sample the IF properly at more than twice the highest IF frequency. then do all of this Hilbert and complex signal processing all in the mind of the CPU or DSP chip. $\endgroup$ – robert bristow-johnson Dec 27 '19 at 16:05
  • $\begingroup$ @robertbristow-johnson sure or another reasonable (and often implemented option) is to bump it down to a lower IF within the performance range of the ADC and then under-sample that with more reasonable ratios. $\endgroup$ – Dan Boschen Dec 27 '19 at 16:38
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This question seems to highlight some of the perils of translating concepts between the continuous domain and a discrete one. My understanding is that bandwidth is defined in the continuous domain as the difference between the upper and lower band limits. Since it is the interval length, whether the interval is open or closed doesn't matter. Open or closed means if the endpoints are considered part of the interval. Since we are always saying "up to the Nyquist", considering the endpoints to be open seems logical.

The extra twist in this is your stipulation that the DC component is zero. This would seem to narrow the bandwidth at the lower end, but to what? In the continuous case, it would be an infinitesimal amount.

Can this properly be translated to the discrete case you describe? I don't think so. There may be a common convention for doing so that I am unaware of. Perhaps your approach, but then you should strike off the gap at both ends, since you have excluded DC.

Upon some reflection, I am going to disagree with MBaz again, and say that $f_s/2$ is the theoretical accurate answer based on the continuous definition of bandwidth.

I will agree that these are purely theoretical considerations with negligible practical value, but then again, so are a lot of exam questions.

You should probably strike the word "radio", it stopped RB-J from giving a good answer. Probably a better answer than what I just gave.


$$ H_k(\omega) = \sum_{n=0}^{N-1}W_N^{-kn}e^{-jn\omega} $$

$$ W_N^{-kn} = \left( e^{-j \frac{2\pi}{N}} \right) ^{-kn} = e^{j kn \frac{2\pi}{N}} $$

$$ \begin{aligned} H_k(\omega) &= \sum_{n=0}^{N-1}e^{j kn \frac{2\pi}{N}}e^{-jn\omega} \\ &= \sum_{n=0}^{N-1} e^{-jn ( -k \frac{2\pi}{N} + \omega) } \\ &= \sum_{n=0}^{N-1} \left( e^{-j ( -k \frac{2\pi}{N} + \omega )} \right)^n \\ &= \frac{1 - \left( e^{-j ( -k \frac{2\pi}{N} + \omega )} \right)^N }{ 1 - e^{-j ( -k \frac{2\pi}{N} + \omega )} } \\ &= \frac{1 - e^{-j ( -k 2\pi + \omega N )} }{ 1 - e^{-j ( -k \frac{2\pi}{N} + \omega )} } \\ &= \frac{1 - e^{-j \omega N } }{ 1 - e^{-j ( -k \frac{2\pi}{N} + \omega )} } \\ &= \frac{e^{-j \frac{\omega N}{2} }}{e^{-j ( \frac{-k\pi}{N} + \frac{\omega}{2} )}} \cdot \frac{ e^{j \frac{\omega N}{2} } - e^{-j \frac{\omega N}{2} } }{ e^{j ( \frac{-k\pi}{N} + \frac{\omega}{2} )} - e^{-j ( \frac{-k\pi}{N} + \frac{\omega}{2} )} } \\ &= e^{j \left( \frac{-k\pi}{N} - \frac{\omega (N-1)}{2} \right) } \cdot \frac{ \sin \left( \frac{\omega N}{2} \right) }{ \sin \left( \frac{-k\pi}{N} + \frac{\omega}{2} \right) } \\ \end{aligned} $$

For similar math, see my blog article DFT Bin Value Formulas for Pure Complex Tones. The section called "As N Gets Large" covers the relation between "alias" sinc, aka Dirichlet Kernel, and the normalized sinc function.

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  • $\begingroup$ @Dan Boschen No, that isn't what I meant. The issue has to do with open or closed endpoints on the intervals and how they should be dealt with within the DFT bins. When the lower band limit is zero (DC), normally one would say that DC is part of the range, so a closed point, but the Nyquist is usually considered an open point (i.e. not part of the range). The "no DC offset" seems to imply the DC point is also open. Now, when you consider the intervals at the bin level, does this exclude the bins (or half of them as you suggest the bin widths are partitioned)? $\endgroup$ – Cedron Dawg Dec 27 '19 at 4:11
  • $\begingroup$ We both seem to think not. This means that N becomes irrelevant, hence this is strictly a continuous definition. $\endgroup$ – Cedron Dawg Dec 27 '19 at 4:13
  • $\begingroup$ Ah! I see what you meant. Sorry, yes I would agree that N is irrelevant. Even at bin level, simply consider the frequency response of the first bin which I will call a Sinc function for brevity. To make it waveform independent consider white noise as a signal since it has equal power at all frequencies (or just consider the frequency responses alone as we would when we consider the bandwidth of any process-- but if you have to consider any signal, white noise would be consistent). This is why I showed the frequency responses as I think that helps explain a lot re that end point. $\endgroup$ – Dan Boschen Dec 27 '19 at 4:22
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If the signal is complex, the theoretical unique maximum bandwidth of the DFT process is $f_s$ and if the signal is real this would be $f_s/2$.

However if you processed your signal to remove its DC offset (rather than it being a sampled zero-mean signal) then this is a high pass filter which will reduce the theoretical bandwidth of the composite DC removal + DFT process accordingly. See this blog post by Rick Lyons that explains this clearly as the filtering analogy to subtracting the mean: https://www.dsprelated.com/showarticle/58.php. What reduction in bandwidth occurs is specific to your DC removal technique.

Assuming you are not filtering your signal in any way, but the signal is zero-mean then the maximum theoretical unique bandwidth of that signal in the DFT would be $f_S$ if complex or $f_s/2$ if real. Here is one way to help explain that:

Each bin in the DFT, if not further windowed, has an equivalent noise bandwidth that is 1 bin wide centered on the bin center frequency (bin number 0 is centered on DC, bin 1 is centered on $F_s/N$, bin 2 on $2F_s/N$ etc). "Equivalent noise bandwidth" is the bandwidth of a theoretical brick-wall filter that would have the same output power given a white noise input. Thus the bandwidth of each bin is $F_s/N$.

Further, in this simpler case that the input is not windowed in the time domain prior to taking the FFT, the sum of the total power out of each bin is equal to the total power of the input waveform (Parseval's theorem) if we scale the DFT magnitudes by $1/\sqrt{N}$.

Since there are N total bins, and the equivalent noise bandwidth of each bin is one bin wide, then the total equivalent noise bandwidth accounting for all bins in the DFT is $F_s$.

This assumes a complex signal, which can have a unique bandwidth up to $F_s$. If the signal is real then half the spectrum must be the complex conjugate of the other half and therefore in that case can only have a unique bandwidth of $F_s/2$. With the realities of aliasing and achievable filtering, the achieved available bandwidth must be less (35-40% is common, but depends entirely on the complexity of the filtering you will allow).

Below is a simple example of a 4 point DFT showing the Frequency Response of each bin. For example when a time domain signal is a fixed exponential frequency centered on bin 1, all the other bins will have 0 output as their response is 0 where they cross the bin 1 frequency. Likewise if a time domain signal is a fixed exponential frequency centered on any other bin. When a signal has a frequency that is in between bins, all the bins will have a non-zero output (this is "spectral leakage"). This shows the actual frewquency response, and the equivalent noise bandwidth which is given by the area under those responses is exactly 1 bin wide for each. Further as N approaches infinity, each of those responses approach a Sinc function (shifted in frequency to be centered on each bin). What results here are "aliased" Sinc functions, with an exact expression given as:

$$H_k(\omega) = \sum_{n=0}^{N-1}W_N^{-kn}e^{-jn\omega}$$

Where:

$\omega$ is the continuous radian frequency from 0 to $2\pi$

$k$ is the DFT bin number, $0$ to $N-1$

$W_N$ is the Nth root of unity = $e^{-j2\pi/N}$ as computed in the DFT


Frequency Responses Showing Bandwidth of Each DFT Bin

DFT


Consider this additional description that may help some of the thought processes by using theoretical extremes:

Consider a Sinc function time domain waveform that extends from $-\infty$ to $+\infty$: This signal has no DC value and a continuous "brickwall" spectrum in frequency with bandwidth B (or B/2 if you only consider the positive frequency axis for real signals).

Next consider this signal to be sampled, as shown in the graphic below at exactly $f_s = B$. If the time domain signal extends from $-\infty$ to $+\infty$, even after sampled the frequency spectrum will be unchanged (continuous). This is the Discrete Time Fourier Transform.

enter image description here

This is the only possible signal either sampled or continuous time domain with a "theoretical" bandwidth of B if you define bandwidth as a brickwall response. But note in this case that even though there is no DC, the spectrum as it passes over DC is no different than any other location within B. (As "DC" is infinitely thin). With finite sampling durations (and hence the DFT) with this particular waveform we can only approach this theoretical limit, we will not have a brick wall response and therefore we would need to carefully define "bandwidth" to provide a theoretically specific result, but I think it helps to illuminate the considerations with bandwidth of zero-mean signals- that from a theoretical viewpoint the fact that the signal is zero-mean does not impact its bandwidth.

If the waveform being received has no DC offset or if the DC offset was removed by subtracting the mean are two very different things. DC removal by subtracting the mean is a process that has bandwidth. If you pre-process your waveform by subtracting the mean then you are filtering it with a high-pass filter and reducing its bandwidth (removing DC as well as other low frequency components). This is an important distinction.

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  • $\begingroup$ Aren't radio waves real and polarized? Never mind the OP's frequency range error. If I am not mistaken, your chart pertains to the complex case. You kept editing, but your $H_k(\omega)$ definition is now simply the DFT of $e^{-jn\omega}$. I will show the closed form solution as a followup. Still, I don't think you are getting to the heart of the question. The OP seems to think you exclude the Nyquist bin, then further stipulates "no DC offset", wondering (as I truly still am) whether that has an impact on bandwidth calculation. $\endgroup$ – Cedron Dawg Dec 27 '19 at 2:59
  • $\begingroup$ $H_k(\omega)$ is a continuous function of $\omega$, so it's not the DFT. This is the closed form frequency response of a DFT bin. This is the bandwidth of a sampled radio signal- so it can be real or complex depending on the way that signal is received. If you have a zero mean signal because you have removed the mean from the signal you sampled would be very different than if you sample a zero mean (noise) signal. My explanation above helps explain the difference: pure DC has 0 bandwidth! $\endgroup$ – Dan Boschen Dec 27 '19 at 3:24
  • $\begingroup$ If you sample a random distributed stationary ergodic signal where each sample has been selected from a zero mean process, and take the DFT of that sampled waveform, the first bin will have the same noise distribution as any other bin! (Which you can experimentally confirm by repeating the process and evaluating the standard deviation of each of the bins). If you actually subtract the mean to make it zero, you are effectively nulling the first DFT bin, but now you have modified the waveform. Therefore, the bandwidth is $F_s$ or $F_s/2$ depending on if the signal is complex or real. $\endgroup$ – Dan Boschen Dec 27 '19 at 3:33
  • $\begingroup$ Any pure tone will have a zero signal bandwidth, no? So, unless we are talking about the DFT's bandwidth, the answer is signal dependent. I don't think this was the OP's intent. Your conclusion is no different than mine, but I don't see any attempt to bring it into the discrete domain. You split the continuous domain into pieces and then reassemble them. This seems to confirm my hunch that this is strictly a continuous domain definition. $\endgroup$ – Cedron Dawg Dec 27 '19 at 3:40
  • $\begingroup$ Yes if you complex conjugate $W_N$ and change the independent variable from a continuous variable $\omega$ to the discrete variable k you do get the DFT- not sure of your point though? $\endgroup$ – Dan Boschen Dec 27 '19 at 3:42

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