error using coupling property for analog input in LabVIEW DAQ

[NIST]

I am trying to use the coupling property of LabVIEW 7.1 to apply AC coupling to an input signal from a function generator. I am using the DAQ max icons to create the channel as well as start, read, stop, etc. in my labVIEW program and am using the DAQCard-6062E. card for data aquisition. i am trying to use a property node to apply the AC coupling and am getting an error message that says that 'the requested value is not a supported value for this property'. can you provide any guidence?

[Dennis_Knutson]

You get the unsupported error because the daq card you're using does not support AC coupling. It is DC coupling only.

[af.ra]

I have a S series DAQ device and I faced the same problem. AC coupling is not supported by my device, therefore , I cant use the coupling property node and I realy need to have a AC coupled signal.

here is the signal I'm acquiring and the second picture is the way I want it to be looked like. Does anybody have any suggestions for me?

[Dennis_Knutson]

What size capacitor have you tried? A quick Google search for capacitive coupling found some sites that will calculate it based on a few simple parameters.

[af.ra]

thank you for your response.

I did not try using any capacitor. I dont know how to do that.

By using a capacitor, do you mean something like that?

[af.ra]

does anybody know how and where I should insert a capacitor in series with a signal to ac couple it with my daq device? does anybody have done that before?

[johnsold]

The way you show in the image you posted will provide AC coupling. You probably also need a resistor in parallel with the input to the DAQ device so the you can be sure there is a DC path to allow charging and discharging of the capacitor.  That resistor will likely determine the input impedance of your measuring system. Select the RC time constant to provide the low frequency cutoff you want for your signals.

Lynn

[af.ra]

thank you so much.

can you explain more how I should usethe resistor in parallel with device? Do you mean to insert it in paralle with the capacitor?

So, I think I can calculate the size of the capacitor by this equation:

where:

f = desirable cuttoff frequency

R= resistance of the resistor which would be the input impedance

right?

Does it matter what resistor to use? or I just need to balance this equation?

[johnsold]

The resistor is not in parallel with the capacitor. If it is connected that way, it defeats the purpose of AC coupling by providing the parallel path for DC. The resistor is connected from AI+ to AI-.

The formula you have is correct.  The input impedance of your DAQ device is specified as 10^11 ohms in parallel with 100 pF.With the resistor connected as described above, the resistance is in parallel with the input impedance of the DAQ device. Commonly available resistors have much lower resistance than the input impedance of the DAQ device so the effective input impedance will essentially be that of the resistor. Suppose you choose a resistor of 1 megohm. And suppose that you want fc to be <= 10 Hz. Then from the formula, C >= 16 nF.  A standard value would be 22 nF.

Lynn

[af.ra]

Thank you so much. it really helped.