Waveforms SDK Correctly Start and Stop Analog Out

[timmolter]

Hi,

I think I may have either found a bug or I'm doing something wrong. What I want to do is start and stop analog out. Here's how I'm doing it:

Start

FDwfAnalogOutNodeEnableSet(idxChannel, true);
FDwfAnalogOutNodeFunctionSet(idxChannel, waveform.getId());
FDwfAnalogOutNodeFrequencySet(idxChannel, frequency);
FDwfAnalogOutNodeAmplitudeSet(idxChannel, amplitude);
FDwfAnalogOutNodeOffsetSet(idxChannel, offset);
FDwfAnalogOutNodeSymmetrySet(idxChannel, dutyCycle);
return FDwfAnalogOutConfigure(idxChannel, true);

Stop

// FDwfAnalogOutNodeOffsetSet(idxChannel, 0);
return FDwfAnalogOutConfigure(idxChannel, false);

Notice that `FDwfAnalogOutNodeOffsetSet` is commented out. In the commented out case, the waveforms does stop (I have an additional oscilloscope looking at it), but the DC offset remains on. To get the offset to go back to zero, I need to comment that line in to manually set it to zero before stopping the analog out waveform. 

So my question is, why do we need to manually set the offset before stopping the wave? Why is there a lingering DC output on that channel when stopping the waveform? Or am I not stopping the waveform correctly using `FDwfAnalogOutConfigure`?

Thanks in advance!

 

[attila]

Hello,

The offset is not changed on start/stop because this has pretty long settling time.

Applying offset on start/stop it would cause similar effects: