Electronic – output amplitude of Direct Digital Sythesizers as AD9850

ddsfunction generatorsignal

By looking at AD9850 DDS or other similar parts made by Analog Devices Company it is obvious that the output signal amplitude changes by frequency. This can be shown as x= pi* fout / fClock and amplitude changes by Sin[x]/x.
By using this formula we should see an amplitude drop less that 20% between 1Hz to 50 MHz but this is not the case that happens. As I am driving this part, I see the amplitude is around 1 volt pk-pk at 10Hz and 10mV at 50 MHz. Is anything I am missing in these calculations?

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Best Answer

In addition to the normal attenuation at increasing frequencies (as @ChrisStratton and OP pointed out), there is the potential to unwittingly add a low-pass filter to the output by decoupling it. This is based on the fact that the trace (or wire) between the DDS output and the decoupling capacitor will have some resistance, which combines with the capacitor to form an RC filter. The cutoff frequency is defined as: $$ f_{c}=\frac{1}{2 \pi RC} $$ Taking an arbitrary trace resistance of \$60 m\Omega\$ (consistant with a ½ inch, 8 mil, 0.5 oz trace) with your 100nF capacitor, we get: $$ f_{c}=\frac{1}{2 \pi \cdot 0.06 \cdot 100n} \approx 26.5MHz $$ This will change based on the actual resistance, as well as the capacitance of the scope probe (as discussed by @ChrisStratton), but it shows that the capacitance could be having a very real effect on the output amplitude at higher frequencies.