I am using an AD9833 DDS as a sine signal source for my circuit. The bandwidth will be 3.5MHz. I would like to calculate the RMS noise of the DDS for my bandwidth. The datasheet has a SNR specification for a test frequency.

AD9833 SNR

The AD9833 can create sine signals up to 12.5MHz with a clock signal of 25MHz. What bandwidth does the datasheet refer to? Is it 12.5 MHz?

The amplitude of the output signal is about 0.3V. I thought I could calculate the RMS noise like this:

$$ V_{RMS\,12.5MHz}= \cfrac{0.3\,V/\sqrt{2}}{10^{\frac{60dB}{20}}}= 212\mu V $$

$$ e_{AD9833}=\cfrac{212\mu V}{\sqrt{12.5MHz}}=60\frac{nV}{\sqrt{Hz}} $$

$$ V_{RMS\,3.5MHz} = 60\frac{nV}{\sqrt{Hz}}\cdot \sqrt{3.5\,MHz}=112\mu V $$

Is this calculation any good?

  • \$\begingroup\$ Just use the noise spectral graphs. Figures 14-20. \$\endgroup\$
    – DKNguyen
    Commented Jul 19, 2022 at 20:19

1 Answer 1


EE&O ...

How to calculate RMS noise from datasheet?

S/N ratio is just defined for 1 frequency \$Fmclk=25 MHz\$ and \$Fout=Fmclk/4096\$.
But I would not consider it is a "normal" s/n ratio.

Here, I should only use the definition of SFDR ...

The wideband SFDR gives the magnitude of the largest spur or harmonic relative to the magnitude of the fundamental frequency in the zero to Nyquist bandwidth, 0 to Fmclk/2.
The narrow-band SFDR gives the attenuation of the largest spur or harmonic in a bandwidth of ±200 kHz about the fundamental frequency.

And SNR ...

SNR is the ratio of the RMS value of the measured output signal to the RMS sum of all other spectral components below the Nyquist frequency

So, if I wanted a very "good" signal & s/n ratio, I should use a "tracking" band-pass filter centered on the output frequency.

  • \$\begingroup\$ I can't put a bandpass filter on the output because I will also need low frequencies. I did not consider the SDFR though. \$\endgroup\$ Commented Jul 20, 2022 at 6:26
  • \$\begingroup\$ Ok. for the band-pass, if you use "random" frequencies very different. \$\endgroup\$
    – Antonio51
    Commented Jul 20, 2022 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.