Hi Satya,
You can build a simple voltage divider by using resistors in the circuit measuring the voltage of the 0 to 10v signal source. There are several web sites with illustrations of the circuit available...for example, you can visit http://www.aikenamps.com/VoltageDividerRule.htm for an illustration of the circuit. In your case, since you want to divide your voltage by 2 (where 10v becomes 5v), you would want to select your resistance values of R1 and R2 to be the same. You would want to choose a resistance value high enough so that it does not cause a load on your signal. 10k ohms is usually a good number. So applying the formula Vout=Vin*R1/(R1+R2), you would get Vout=Vin*10k/(10k+10k), or Vout=Vin*1/2. The result
is that you are dividing your input voltage by 2. So to connect the circuit, the output from your signal source will go in one end of a 10k ohm resistor, and the other end of the resistor will go to the input of your DAQ device (SCXI 1200, ACH0 for example). For the second resistor, you would connect one end up to ACH0 and then to the negative end (or ground) of your signal source. Then you can compensate for your voltage divider in software by multiplying the measured signal value by 2. So if the DAQ device measures an input of 2.50v, then the software should multiply that number by 2 to indicate that a 5.00v signal is seen at the signal source.
One thing to keep in mind is by doing this, you are actually decreasing the resolution by 2. This may be ok in your situation, depending on how much resolution you need to have on your signal source. Since the SCXI 1200 has a 12-bit A/D converter, you are essentially only using 11-bits to characterize your signal, which would mean the
smallest detectable voltage change from your signal source is 0.0049 volts (10/2^11).
Hope this helps!
Wilbur Shen
National Instruments