Question about Nyquist sampling - filtering of high image copies

Suppose I sample an audio signal with a sampling frequency > twice the highest audio frequency.

ie, the frequency transform of the sampled signal looks like the image on top here, where there is no overlap of copies:

Also, suppose that all the high image copies are outside the audible range. Now is there any need for a brickwall filter in this situation? I'd think it was unnecessary since the audible part of the original and sampled signal are identical. As a specific example, say we sampled a signal at 96 kHz. Now is there a need for any filtering to remove high image copies? Thanks.

• in a sense, your finite hearing range becomes the lowpass filter that filters out the images. Mar 5, 2023 at 7:50
• @robertbristow-johnson, yes. That's what I was thinking also. But I'm wondering if there are some unforeseen side-effects to having those images there. Mar 5, 2023 at 8:37
• BTW, that graphic was done by me circa 2 decades ago, back when I could haunt Wikipedia under my own name. Mar 5, 2023 at 18:00
• @robertbristow-johnson, very nice graphic. Mar 5, 2023 at 18:05
• My version of the Nyquist-Shannon sampling theorem was a lot better, for pedagogical purposes than what that article became. Now I am banned from Wikipedia, so I can only edit anonymously. Mar 5, 2023 at 18:18

In your scenario where an audio analog signal having nonzero spectral components from zero Hz to, say, 15 kHz, is sampled at a rate of 96 kHz the separation between the blue and green curves (in your linked web page picture) will be larger than the separation shown in the picture.

To answer your question of: "I'm wondering if there are some unforeseen side-effects to having those images there.", the answer is "No". And I say "No" because there is no spectral overlap of the sampled x[n] signal's blue and green curves, thus no "aliasing" errors have been caused by sampling the original analog signal at 96 kHz.

Now if you were to perform some processing on the discrete x[n] sampled signal that results in some sort of frequency translation (such AM modulation, or discrete signal decimation) causing overlap of the blue and green curves then you'll be introducing errors in your new "processed" discrete signal. The bottom line here is well-known; For any discrete lowpass signal the sample rate must be greater than twice the highest-frequency spectral component.

On that linked web page the page's author wrote: "A brick-wall low-pass filter, H(f), removes the images, leaves the original spectrum, X(f), and recovers the original signal from its samples." That sentence, without further explanation, is super misleading! There is NO digital filter that will eliminate the spectral images (the green curves) from a discrete signal sequence.ㅤ

• //"That sentence, without further explanation, is super misleading!"// Really?? How is the sentence, as it is, misleading? Mar 5, 2023 at 18:11
• //" There is NO digital filter that will eliminate the spectral images (the green curves) from a discrete signal sequence."// Where does that sentence say anything about a digital filter? Even so, as long as we're willing to put in enough MIPS and tolerate enough throughput delay, in upsampling, a digital filter can get arbitrarily close to that ideal brick wall and we audio interpolationists do that with a variety of tools firpm(), firls(), kaiser().*sinc(). Mar 5, 2023 at 18:12
• All discrete signal sequences have spectral "replica6tions" centered at integer multiples of the sequence's sample rate (ie., the green spectral curves). What I'm saying is: There is NO digital signal processing that we can perform on a discrete sequence that will eliminate (attenuate) those spectral replications. Mar 5, 2023 at 18:53
• Richard, every discrete-time system has repeated images. But you can still using DSP, emulate a continuous-time filter with a continuous-time finite impulse response. A simple idea that I have always had, but got published by Smith and Gossett is a polyphase FIR interpolation with a very high upsample ratio (I think $r=512$ is about right) and then linear interpolation remaining to make the impulse response continuous-time. Mar 5, 2023 at 23:55
• Locked by system until edited, didn't feel compelled enough to edit or remark it. But I'll remark it now - or more. @OlliNiemitalo Mar 12, 2023 at 13:23