Most accurate signal generator

[loop123]

Regarding the bottom picture: The noise is there but the sampling frequency and/or bandwidth is too low to make noise appear as jagged edges. Run an FFT on both signals (using equal record lengths and sampling frequencies).

It is the same noise in both plots. Therefore the spectrum of the noise floor is the same as well. Flat up to ~1kHz, then rolling off with 12dB/octave. The first plots adds a 50Hz sine wave to the noise, and the 2nd plot adds a 900Hz sine wave to the noise. However, time/div is different in both plots in order to fit 10 signal periods of 50Hz or 900Hz into the screen width.

[ The units of the x-axis are samples, at a sample rate of 48kSa/s, in both plots. ]

They have same noises but isnt it the 2nd plot has more clean and more easily resolvable sine waves? doesn this mean it is better to make higher frequency signal to create cleaner sine waves?

Please tell me the software you used so I can play with it.

The 2nd plot isn't more clean! It has less samples. If you can dump the samples into a file and read it into an audio processing program (like Audacity which is free), you can do an FFT analysis. But make sure record an equal number of samples for each recording if you want to compare.

gf (just wondering if a he or she), can you pls use the same sampling (used at 50Hz) at 900Hz to show the same noise as the 50Hz plot?

[gf]

The 2nd plot isn't more clean! It has less samples.

Sure, it spans a shorter time interval. But the OP wanted to know what to expect upon zoom-in.
The 2nd plot would look different, of course, if the noise would be white, and not lowpass filtered.
But without filtering, the total noise level (over the full ~20kHz audio bandwidth) would be much higher as well.

[ebastler]

gf (just wondering if a he or she), can you pls use the same sampling (used at 50Hz) at 900Hz to show the same noise as the 50Hz plot?

If you mean, show the 900 Hz signal over the same 10000 data points as used in the 50 Hz plot: You will not be able to resolve the actual 900 Hz oscillations, and will see a signal trace which is a big fat solid band, filling the plot.

[gf]

gf (just wondering if a he or she), can you pls use the same sampling (used at 50Hz) at 900Hz to show the same noise as the 50Hz plot?

With the same time scale, 180 periods of the 900Hz signal. would need to fit into the screen width. Even w/o noise, the plot becomes so dense that you hardly can see the waveform any more. You had explicitly asked how it would look if you zoom-in in order to inspect the waveform in detail.

[loop123]

gf (just wondering if a he or she), can you pls use the same sampling (used at 50Hz) at 900Hz to show the same noise as the 50Hz plot?

With the same time scale, 180 periods of the 900Hz signal. would need to fit into the screen width. Even w/o noise, the plot becomes so dense that you hardly can see the waveform any more. You had explicitly asked how it would look if you zoom-in in order to inspect the waveform in detail.

I only use and familiar with Audacity which uses time scale in horizonal and not sample format. what other popular software uses samples format so i can try them all? tnx

[ebastler]

I only use and familiar with Audacity which uses time scale in horizonal and not sample format. what other popular software uses samples format so i can try them all? tnx

It's really a simple conversion as long as a fixed sample rate is always used. 10,000 samples in 200 ms, 50 samples per ms, 1 sample per 20 µs. I am not sure where your problem with those plots lies?

[loop123]

I only use and familiar with Audacity which uses time scale in horizonal and not sample format. what other popular software uses samples format so i can try them all? tnx

It's really a simple conversion as long as a fixed sample rate is always used. 10,000 samples in 200 ms, 50 samples per ms, 1 sample per 20 µs. I am not sure where your problem with those plots lies?

If you use 40000 samples per 20µs instead of 1 sample. and you zoominto the 900Hz sine wave. You can already see the jagged edge noises, isnt it? can you change the default sampling in Audacity? what is the default sampling in Audacity per 20 µs?

[ebastler]

If you use 40000 samples per 20µs instead of 1 sample. and you zoominto the 900Hz sine wave. You can already see the jagged edge noises, isnt it?

No, not really. If you look at things on a time scale where the 900 Hz signal can be resolved (and looks smooth and sinusoidal), the noise will also look smooth and not jagged -- because its bandwidth is limited to 1000 Hz.

If you look at things on a timescale where the noise looks jagged (say the timescale used for the 50 Hz plot), the 900 Hz signal will also look "jagged": You can hardly resolve its individual oscillations, and since it has a much higher amplitude than the noise, it will fill the whole screen with trace lines.

You really seem to have a mental block here, assuming noise has to always look "noisy" -- a rapidly wiggling contribution which sits on top of your signal. But if you limit your noise spectrum to a range close to your signal frequency, the noise will indeed look similar to your signal.

[nctnico]

gf (just wondering if a he or she), can you pls use the same sampling (used at 50Hz) at 900Hz to show the same noise as the 50Hz plot?

With the same time scale, 180 periods of the 900Hz signal. would need to fit into the screen width. Even w/o noise, the plot becomes so dense that you hardly can see the waveform any more. You had explicitly asked how it would look if you zoom-in in order to inspect the waveform in detail.

I only use and familiar with Audacity which uses time scale in horizonal and not sample format. what other popular software uses samples format so i can try them all? tnx

AFAIK you can configure Audacity to show the sample numbers. There is a bit of a learning curve, but Audacity is quite versatile.

[gf]

It's really a simple conversion as long as a fixed sample rate is always used. 10,000 samples in 200 ms, 50 samples per ms, 1 sample per 20 µs.

Btw, just to clarify: I had calculated my plots at 48kSa/s (which is a commonly used audio sample rate).
1 seconds = 48000 samples, 200ms = 9600 samples, etc.

Unfortunately, I was lazy. I should have labeled the time axis in seconds (instead of samples) to avoid confusion.

[loop123]

It's really a simple conversion as long as a fixed sample rate is always used. 10,000 samples in 200 ms, 50 samples per ms, 1 sample per 20 µs.

Btw, just to clarify: I had calculated my plots at 48kSa/s (which is a commonly used audio sample rate).
1 seconds = 48000 samples, 200ms = 9600 samples, etc.

Unfortunately, I was lazy. I should have labeled the time axis in seconds (instead of samples) to avoid confusion.

Ok. I'm familiar with 48kSA/s since my Audacity uses that audio sample rate too. Let's say the audio sample rate is 2000kSA/s instead of just 48kSA/s then you can already see the jagged edge noise in the 900Hz, right?

[gf]

Ok. I'm familiar with 48kSA/s since my Audacity uses that audio sample rate too. Let's say the audio sample rate is 2000kSA/s instead of just 48kSA/s then you can already see the jagged edge noise in the 900Hz, right?

No, it is not a matter of the sample rate (48kSa/s is more than enough for this signal), and it's not a matter of sampling.
It is a matter of filtering. Feeding an analog white noise signal though a corresponding analog filter would have the same effect.

Btw, it's not difficult to do the same simulation in Audaity, too. I'm not familiar with this program, still I was able to find all the tools within a couple of minutes. Create one track and fill it with a sine wave. Create a 2nd track and fill it with white noise. Create a 3rd track by mixing the two tracks. Apply lowpass filter to the 3rd track. Done. See attachment.

[loop123]

Ok. I'm familiar with 48kSA/s since my Audacity uses that audio sample rate too. Let's say the audio sample rate is 2000kSA/s instead of just 48kSA/s then you can already see the jagged edge noise in the 900Hz, right?

No, it is not a matter of the sample rate (48kSa/s is more than enough for this signal), and it's not a matter of sampling.
It is a matter of filtering. Feeding an analog white noise signal though a corresponding analog filter would have the same effect.

Btw, it's not difficult to do the same simulation in Audaity, too. I'm not familiar with this program, still I was able to find all the tools within a couple of minutes. Create one track and fill it with a sine wave. Create a 2nd track and fill it with white noise. Create a 3rd track by mixing the two tracks. Apply lowpass filter to the 3rd track. Done. See attachment.

I didn't know Audacity can do them. I only used them to display waveforms in my amplifier. Why did you create the 3rd track. What are you trying to demonstrate so I can do more of them. Thanks.

[GigaHurts]

hey man, did you try a 555 timer?

but in all seriousness, that is quite an accuracy you are asking for not available for cheap. And if it was, you still probably don't have a cheap way of measuring it. And even if you have, you probably don't have a cheap way of calibrating it

[loop123]

Ok. I'm familiar with 48kSA/s since my Audacity uses that audio sample rate too. Let's say the audio sample rate is 2000kSA/s instead of just 48kSA/s then you can already see the jagged edge noise in the 900Hz, right?

No, it is not a matter of the sample rate (48kSa/s is more than enough for this signal), and it's not a matter of sampling.
It is a matter of filtering. Feeding an analog white noise signal though a corresponding analog filter would have the same effect.

Btw, it's not difficult to do the same simulation in Audaity, too. I'm not familiar with this program, still I was able to find all the tools within a couple of minutes. Create one track and fill it with a sine wave. Create a 2nd track and fill it with white noise. Create a 3rd track by mixing the two tracks. Apply lowpass filter to the 3rd track. Done. See attachment.

Ok I tried Audacity generator and mixing. I first created white noise in 1st track. then apply 1000kHz filter to the white noise. In the first screenshot I added a 50Hz sine wave in 2nd track and mix in 3rd track. There is noise.



In the following image I added a 900Hz sine wave in 2nd track with the same white noise 1000kHz filtered in first track and mix in 3rd track. The jagged edge is not seen.



If the above methods are correct. Can we just say that in the 900Hz.  There is simply less noise per sine wave compared to the 50Hz in first image that is why there is no jagged edge in the 3rd track?

[Anthocyanina]

Ok. I'm familiar with 48kSA/s since my Audacity uses that audio sample rate too. Let's say the audio sample rate is 2000kSA/s instead of just 48kSA/s then you can already see the jagged edge noise in the 900Hz, right?

No, it is not a matter of the sample rate (48kSa/s is more than enough for this signal), and it's not a matter of sampling.
It is a matter of filtering. Feeding an analog white noise signal though a corresponding analog filter would have the same effect.

Btw, it's not difficult to do the same simulation in Audaity, too. I'm not familiar with this program, still I was able to find all the tools within a couple of minutes. Create one track and fill it with a sine wave. Create a 2nd track and fill it with white noise. Create a 3rd track by mixing the two tracks. Apply lowpass filter to the 3rd track. Done. See attachment.

Ok I tried Audacity generator and mixing. I first created white noise in 1st track. then apply 1000kHz filter to the white noise. In the first screenshot I added a 50Hz sine wave in 2nd track and mix in 3rd track. There is noise.

(Attachment Link)

In the following image I added a 900Hz sine wave in 2nd track with the same white noise 1000kHz filtered in first track and mix in 3rd track. The jagged edge is not seen.

(Attachment Link)

If the above methods are correct. Can we just say that in the 900Hz.  There is simply less noise per sine wave compared to the 50Hz in first image that is why there is no jagged edge in the 3rd track?

there is the same amount/magnitude/presence of noise, it just looks different. you can do an FFT to confirm that, then increase the noise amplitude, and do another FFT and see how the noise floor rises equally on both 50 hz and 900 hz if the applied noise is the same

[loop123]

Ok. I'm familiar with 48kSA/s since my Audacity uses that audio sample rate too. Let's say the audio sample rate is 2000kSA/s instead of just 48kSA/s then you can already see the jagged edge noise in the 900Hz, right?

No, it is not a matter of the sample rate (48kSa/s is more than enough for this signal), and it's not a matter of sampling.
It is a matter of filtering. Feeding an analog white noise signal though a corresponding analog filter would have the same effect.

Btw, it's not difficult to do the same simulation in Audaity, too. I'm not familiar with this program, still I was able to find all the tools within a couple of minutes. Create one track and fill it with a sine wave. Create a 2nd track and fill it with white noise. Create a 3rd track by mixing the two tracks. Apply lowpass filter to the 3rd track. Done. See attachment.

Ok I tried Audacity generator and mixing. I first created white noise in 1st track. then apply 1000kHz filter to the white noise. In the first screenshot I added a 50Hz sine wave in 2nd track and mix in 3rd track. There is noise.

(Attachment Link)

In the following image I added a 900Hz sine wave in 2nd track with the same white noise 1000kHz filtered in first track and mix in 3rd track. The jagged edge is not seen.

(Attachment Link)

If the above methods are correct. Can we just say that in the 900Hz.  There is simply less noise per sine wave compared to the 50Hz in first image that is why there is no jagged edge in the 3rd track?

there is the same amount/magnitude/presence of noise, it just looks different. you can do an FFT to confirm that, then increase the noise amplitude, and do another FFT and see how the noise floor rises equally on both 50 hz and 900 hz if the applied noise is the same

There is the same amount of noise given the same time. But one 900Hz sine wave can form superposition with the noise at smaller scale that is why the 900Hz sine wave has no jagged edge. Is this simple interpretation correct?

If wrong. Then is it like the quantum vacuum where no matter how smaller scale you try to sample it, the noise is the same because the quantum vacuum is random and lorentz invariant and background independent?

[radiolistener]

Then is it like the quantum vacuum where no matter how smaller scale you try to sample it, the noise is the same because the quantum vacuum is random and lorentz invariant and background independent?

  Are you kidding?   

[Anthocyanina]

Ok. I'm familiar with 48kSA/s since my Audacity uses that audio sample rate too. Let's say the audio sample rate is 2000kSA/s instead of just 48kSA/s then you can already see the jagged edge noise in the 900Hz, right?

No, it is not a matter of the sample rate (48kSa/s is more than enough for this signal), and it's not a matter of sampling.
It is a matter of filtering. Feeding an analog white noise signal though a corresponding analog filter would have the same effect.

Btw, it's not difficult to do the same simulation in Audaity, too. I'm not familiar with this program, still I was able to find all the tools within a couple of minutes. Create one track and fill it with a sine wave. Create a 2nd track and fill it with white noise. Create a 3rd track by mixing the two tracks. Apply lowpass filter to the 3rd track. Done. See attachment.

Ok I tried Audacity generator and mixing. I first created white noise in 1st track. then apply 1000kHz filter to the white noise. In the first screenshot I added a 50Hz sine wave in 2nd track and mix in 3rd track. There is noise.

(Attachment Link)

In the following image I added a 900Hz sine wave in 2nd track with the same white noise 1000kHz filtered in first track and mix in 3rd track. The jagged edge is not seen.

(Attachment Link)

If the above methods are correct. Can we just say that in the 900Hz.  There is simply less noise per sine wave compared to the 50Hz in first image that is why there is no jagged edge in the 3rd track?

there is the same amount/magnitude/presence of noise, it just looks different. you can do an FFT to confirm that, then increase the noise amplitude, and do another FFT and see how the noise floor rises equally on both 50 hz and 900 hz if the applied noise is the same

There is the same amount of noise given the same time. But one 900Hz sine wave can form superposition with the noise at smaller scale that is why the 900Hz sine wave has no jagged edge. Is this simple interpretation correct?

If wrong. Then is it like the quantum vacuum where no matter how smaller scale you try to sample it, the noise is the same because the quantum vacuum is random and lorentz invariant and background independent?

if your noise contains every frequency between 1 and 1000hz, your low frequency signal has what you call jagged edges because the signal's period is too long compared to most of the noise, so the higher frequency noise will look "jagged" because, let's say the 1khz component of the noise, it goes up and down in a much shorter time than your 50hz signal. at 900hz, the period of the signal is much shorter than that of most of the components of the noise, and very close to that of the higher frequency components of the noise, so as your signal goes up and down, the higher frequency noise goes up and down at about the same rate, so there is a visually different effect if you observe the captured waveform, and the lower frequency noise is also there, but its period is so much longer that it is barely "visible" if you observe a single, or a few periods of your high frequency signal.

do this in audacity, generate a 50hz signal, without noise, and generate a 900hz signal, also without noise, then add them.
if you zoom out, you will see the 50hz signal with the 900hz making it look fuzzy, but if you zoom in, you will see the 900hz signal slowly being offset from the center at a rate of 50hz. both signals are equally present in that sum output, it just looks different depending on how zoomed in or out you look at it.

[loop123]

if your noise contains every frequency between 1 and 1000hz, your low frequency signal has what you call jagged edges because the signal's period is too long compared to most of the noise, so the higher frequency noise will look "jagged" because, let's say the 1khz component of the noise, it goes up and down in a much shorter time than your 50hz signal. at 900hz, the period of the signal is much shorter than that of most of the components of the noise, and very close to that of the higher frequency components of the noise, so as your signal goes up and down, the higher frequency noise goes up and down at about the same rate, so there is a visually different effect if you observe the captured waveform, and the lower frequency noise is also there, but its period is so much longer that it is barely "visible" if you observe a single, or a few periods of your high frequency signal.

do this in audacity, generate a 50hz signal, without noise, and generate a 900hz signal, also without noise, then add them.
if you zoom out, you will see the 50hz signal with the 900hz making it look fuzzy, but if you zoom in, you will see the 900hz signal slowly being offset from the center at a rate of 50hz. both signals are equally present in that sum output, it just looks different depending on how zoomed in or out you look at it.

Thanks guys. The appearances and explanations of the noises at 50Hz and 900Hz are very clear now.

There is another topic that I've been thinking for over a month. Can you share any software that can demonstrate the power of digital filter especially oversampling that can apply lowpass or bandpass filter with brick wall frequency response. My BMA-200 that uses the AMP01 has 2 order Butterworth filter with -12dB/octave response while my gtec USBamp has no amplifier but only ADC that directly maps the microvolt signal at +- 250mV. I'd like to know if it's oversampling (ability to do moving averages) capability can make me see signal that I can't with the BMA (between 1 to 2400Hz), and worth spending $3000 just to get the software to run it. Here is the spec of the USBamp. Pls share sofware to demonstrate the power of oversampling via digital filter. How much can it beat 2 order Butterworth filter with its almost brick wall response?  Member "gf" told me not make another thread so I post this question here.

Remember that a 2nd order Butterworth filter with less vertical low-pass edge only adds about 7.5% noise compared to the digital brick-wall filter. So does digital filter oversampling only remove the noise above the cut-off or can it also make the noise less below the cutoff? Please share software so I can try to see the process. Thanks.


https://www.gtec.at/product/gusbamp-research/

"
Each of the 16 analog to digital converters operates at 2.4576 MHz. Oversampling 64 times yields the internal sampling rate of 38,400 Hz (per channel and for all channels!). In addition, a powerful floating point Digital Signal Processor performs oversampling and real-time filtering of the biosignal data (between 0 Hz – 2,400 Hz). Therefore, a typical sampling frequency of 256 Hz yields an oversampling rate of 9,600. This results in a very high signal to noise ratio, which is especially critical when recording evoked potentials (EP) in the EEG or identifying small amplitude changes in high-resolution ECG recordings. You are measuring far below the noise-range of conventional amplifiers."

[loop123]

for your original 200$ budget, i would get an owon HDS242S.  it's a very reasonable oscilloscope, multimeter and generator for the price, here's the owon's generator in yellow, vs a keysight 33212a in blue, both set for the same frequency and amplitude of 1.8vpp. you can see the measured Vpp and frequency for both channels to the right of the waveforms.

keep in mind that the output of both generators is going to a high impedance load (the analog discovery's 1Mohm inputs) and will be halved if you connect them to a 50 ohm load. the owon can output 5vpp to high impedance loads and 2.5vpp to 50 ohm loads, so if that doesn't work for you, then yeah, this won't work, but within that voltage range, the generator of the owon is pretty reasonable for the frequency range you want, and you also get an oscilloscope and multimeter.

You said "the owon can output 5vpp to high impedance loads and 2.5vpp to 50 ohm loads". I need the 5vpp voltage  bec it would be at least 1.76V rms for the E1DA. How does it calibrate 5vpp for 1Mohm load and 2.5vpp for 50 ohm load?  What if my load or amplifier has arbitrary input impedance like 10,000 Megaohm in the case of my BMA-200 (see below) or 640 Ohm for my E1DA ADC? how do you compute the voltage output for those impedances? what if I will use a 1Mega ohm resistor in parallel to the output of the Owon HDS242S to be sure the load would be 1Mega ohm for unknown input impedance like my USBamp?  I think the Owon with all in one function would be more useful. The RF attenuator can wait.

[gf]

Member "gf" told me not make another thread so I post this question here.

I rather meant that all this stuff would better fit into your "Audio ADC" thread which already did exist before. These questions are not related to the topic "Most accurate signal generator".

There is another topic that I've been thinking for over a month. Can you share any software that can demonstrate the power of digital filter especially oversampling that can apply lowpass or bandpass filter with brick wall frequency response. My BMA-200 that uses the AMP01 has 2 order Butterworth filter with -12dB/octave response while my gtec USBamp has no amplifier but only ADC that directly maps the microvolt signal at +- 250mV. I'd like to know if it's oversampling (ability to do moving averages) capability can make me see signal that I can't with the BMA (between 1 to 2400Hz), and worth spending $3000 just to get the software to run it. Here is the spec of the USBamp. Pls share sofware to demonstrate the power of oversampling via digital filter. How much can it beat 2 order Butterworth filter with its almost brick wall response?"

Remember that a 2nd order Butterworth filter with less vertical low-pass edge only adds about 7.5% noise compared to the digital brick-wall filter.

You answer the question yourself. The equivalent noise bandwidth of a 2nd order Butterworth is about 1.08 x cut-off frequency, so the difference is not large.

If your Butterworth filter is not digital, but resides in the analog frontend, then additional noise may be introduced in the signal path between the filter and the ADC (including ADC noise). Then an additional digital filter may help if you don't need to full 0...sample_rate/2 bandwidth. [ But don't overreact before you have evidence that this is really relevant in your setup -- it could also be negligible. ]

So does digital filter oversampling only remove the noise above the cut-off...

Any filter removes (or better say attenuates) frequencies outside its passband. A filter cannot distinguish useful signal and noise, but only frequencies. Consequently, without attenuating components of the useful signal as well, a filter can only eliminate/reduce noise in a frequency band which is not covered by the useful signal.

https://www.gtec.at/product/gusbamp-research/

"Each of the 16 analog to digital converters operates at 2.4576 MHz. Oversampling 64 times yields the internal sampling rate of 38,400 Hz (per channel and for all channels!). In addition, a powerful floating point Digital Signal Processor performs oversampling and real-time filtering of the biosignal data (between 0 Hz – 2,400 Hz). Therefore, a typical sampling frequency of 256 Hz yields an oversampling rate of 9,600. This results in a very high signal to noise ratio, which is especially critical when recording evoked potentials (EP) in the EEG or identifying small amplitude changes in high-resolution ECG recordings. You are measuring far below the noise-range of conventional amplifiers."

Any delta-sigma ADC is based on oversampling and subsequent decimation to the output sample rate (which implies digital filtering), and that includes all today's audio ADCs. That's not special, but that's simply how a delta-sigma ADC works. And the mentioned "typical sampling frequency of 256 Hz" is not suitable for you, as it would limit the bandwidth to 128Hz (theoretically, and in practice even less, say 100...120Hz), but you want 1000Hz bandwidth for the useful signal. The low sample rate is not relevantt anyway for the noise consideration, but only filtering matters, and filtering can be done at a higher sample rate, too [although it requires more processing resources then].

[gf]

for your original 200$ budget, i would get an owon HDS242S.  it's a very reasonable oscilloscope, multimeter and generator for the price, here's the owon's generator in yellow, vs a keysight 33212a in blue, both set for the same frequency and amplitude of 1.8vpp. you can see the measured Vpp and frequency for both channels to the right of the waveforms.

keep in mind that the output of both generators is going to a high impedance load (the analog discovery's 1Mohm inputs) and will be halved if you connect them to a 50 ohm load. the owon can output 5vpp to high impedance loads and 2.5vpp to 50 ohm loads, so if that doesn't work for you, then yeah, this won't work, but within that voltage range, the generator of the owon is pretty reasonable for the frequency range you want, and you also get an oscilloscope and multimeter.

You said "the owon can output 5vpp to high impedance loads and 2.5vpp to 50 ohm loads". I need the 5vpp voltage  bec it would be at least 1.76V rms for the E1DA. How does it calibrate 5vpp for 1Mohm load and 2.5vpp for 50 ohm load?  What if my load or amplifier has arbitrary input impedance like 10,000 Megaohm in the case of my BMA-200 (see below) or 640 Ohm for my E1DA ADC? how do you compute the voltage output for those impedances?

Consider the generator as a voltage source (which supplies the no-load voltage up to 5Vpp) in series with a 50 Ohm resistor. The 50 Ohm source resistance and the load resistance form a voltage divider. Example: If the load is 50 Ohm, then you get a 50 Ohm : 50 Ohm divider, and the voltage across the load is (50 / (50 + 50)) times the generator's no-load voltage. For 640 Ohm load, the voltage across the load will be (640 / (50 + 640)) times the no-load voltage.

what if I will use a 1Mega ohm resistor in parallel to the output of the Owon HDS242S to be sure the load would be 1Mega ohm for unknown input impedance like my USBamp?

Any load greater than 10 kOhms can be considered to be virtually "no load" and the error will be less than 0.5%. So don't worry when you connect high impedance loads.

[loop123]

Is the Owon HDS242S the best 3 in 1 there is (Oscilloscope, Multimeter, Signal Generator combined) in the $200 price range? Any other alternative? Once I get it. I'll use it for the next 20 years.  My multimeter is 20 years old. My pocket Oscilloscope f-NiRSi DSO-TC3 is not working, and I need Signal generator for occasional calibration.   Can its oscilloscope mode be even better than my E1DA Cosmos ADC + Audacity combo or inferior? My amp output is between 0.3V to 2V peak to peak.

[Aldo22]

Is the Owon HDS242S the best 3 in 1 there is (...) in the $200 price range?

Do you mean USD or CAD?

The HDS242S does not cost USD 200. https://www.aliexpress.com/item/1005006068203110.html
Whether something is "the best" depends on what you need it for.
For example, it has a 14bit / 8 kpts signal generator (good), but only max 2.5Vpp (maybe a bit low).

Once I get it. I'll use it for the next 20 years

Are you sure about that? A lot happens in 20 years...