I have a small function generator: http://www.velleman.co.uk/contents/en-uk/p371_hpg1.html

And using the following device for DAQ: http://www.mccdaq.com/PDFs/manuals/USB-1208LS.pdf

I hook up the generator output directly to the daq board. At 1Hz output I see he following plot(sampling rate is 500Hz):

enter image description here

It doesn't look like a nice square wave. When I increase the freq. it gets a little better.

What can be the problem?

enter image description here

  • 2
    \$\begingroup\$ Hah, you've got your 'scope set for AC coupling. (oops or AC coupling into the DAC perhaps) \$\endgroup\$ Commented Dec 1, 2015 at 19:52
  • \$\begingroup\$ how can I fix this since Im using a DAQ board not a scope? And why does it happen? \$\endgroup\$
    – user16307
    Commented Dec 1, 2015 at 19:53
  • 1
    \$\begingroup\$ Are you AC coupling somewhere? It's possible it is in the ground line. Is the DAC single ended? \$\endgroup\$ Commented Dec 1, 2015 at 19:57
  • \$\begingroup\$ yes it is single ended. other channels are not plugged \$\endgroup\$
    – user16307
    Commented Dec 1, 2015 at 19:58
  • 1
    \$\begingroup\$ I don't want you to blow up your DAC, but try hooking the sig gen ground to the DAC ground. (maybe with ~1k R's in series to limit current?) \$\endgroup\$ Commented Dec 1, 2015 at 20:47

1 Answer 1


Yes, the schematic of the signal generator (page 8) clearly shows that it is AC-coupled.

IC4 is the output attenuator, and IC3B is the output driver. C4, C20 and C21 are the coupling between them. In conjunction with R7, they give you a time constant of roughly 1 second, which explains the "droop" you see on a 1-Hz square wave.

EDIT: The issue here is that the DAC (IC2) is single-ended — it can only produce positive voltages. In order to get an output that swings both positive and negative, the AC coupling is used to remove the DC bias.

If your application does not require negative output voltages, you could simply short out the coupling capacitors by installing a temporary wire across one of them.

If you do still want bipolar output voltages, but a much lower cutoff frequency, Google for the phrase "dc offset servo circuit". The basic concept is that you use an active low-pass filter to isolate the DC component, and then subtract this from the original signal in the output stage.

The point is, that it's much easier (i.e., more compact) to create a low-pass filter with a low cutoff frequency than it is to create a high-pass filter with the same cutoff frequency. For example, a 1 MΩ resistor and a 100 µF capacitor will get you down to the milliHertz range.

  • \$\begingroup\$ why would they do that. so is this device useless for 1Hz??? \$\endgroup\$
    – user16307
    Commented Dec 2, 2015 at 12:32
  • \$\begingroup\$ It's probably primarily intended for audio testing, in which a high-pass cutoff frequency of 0.16 Hz is more than adequate. \$\endgroup\$
    – Dave Tweed
    Commented Dec 2, 2015 at 12:37
  • \$\begingroup\$ is it possible to correct it for me? \$\endgroup\$
    – user16307
    Commented Dec 2, 2015 at 12:59
  • \$\begingroup\$ See edit above. \$\endgroup\$
    – Dave Tweed
    Commented Dec 2, 2015 at 13:57
  • \$\begingroup\$ they sell this device without mentioning you will never get a nice square wave for very low freq. \$\endgroup\$
    – user16307
    Commented Dec 2, 2015 at 14:00

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