AD9910 Nonlinearity

Hi there,

1. Digital aliasing is the effect when new frequencies appearing in the sampled signal after reconstruction which were not present in the original signal. It can be also caused by either: too low sample rate for sampling a particular signal or too high frequencies present in the signal for a particular sample rate. You should be seeing mirrored signal at the unfiltered output.

2. Upon checking on the values that you have given, R_int = 100Ω while R_L= 50Ω and is using 1.5: 1 transformer. Based on these given values, the impedance didn’t match. You must change the values for this. If we remain R_L= 50Ω then we must use R_int = 22.2Ω. However, if we use R_int = 100Ω we need to use R_L= 225Ω.

This is the equation for impedance matching:

k = √(R_L / R_s)

Any linear-time invariant system does not cause any intermodulation. There will be just difference in the phase and amplitude given that the input frequency is still the same at the output. Only difference in phase and amplitude will happen given that scenario.

3. Possible reasons for nonlinearity include:

• The use of passive filter. No amplifying element was used in the passive filters thus it offers low signal gain. Thus, comparatively lower signal than its input is expected at the output
• Nonlinear systems do not depend on its input. Its output is not directly proportional to its input.

4. A faster clock requires more power. Voltage required is affected significantly more than clock speed. Thus, in general, the higher the speed the more voltage it requires. But clock power does not affect the output in some ways. So, to answer your question, it would be NO. Clock power does not affect the behavior of the DDS output.

Thank you.

I have some more questions about transformer impedance. I'm considering changing the transformer to ADT-1T-1P+(1:2 ratio), and have the secondary coil on the differential side. If I want the output load to be 50ohm, am I correct to have each resistor on the differential side to have 100ohm to have 200ohm in total?
The evaluation board for the transistor states I have two 50ohm resistors to have 100ohm in total, and I don't understand why if impedance depends on turn number squared.https://www.minicircuits.com/pcb/WTB-430_P02.pdf is the evaluation board pdf link.

The ADT-1T-1P+ has an impedance ratio (Zr=Zs/Zp) of 2. That is, Zs:Zp = 2:1. The impedance ratio and turns ratio relate as: Nr=sqrt(Zr)=sqrt(Zs)/sqrt(Zp). Written in ratio form: sqrt(Zs):sqrt(Zp), or sqrt(2):1, in this case.

However, the transformer PRI and SEC are used in reverse in this application. In keeping with the application, Ns and Np, as follows, denote the "new" secondary and primary, per the application. Thus, the turns ratio (Ns:Np) is 1:sqrt(2). By convention, Np=1, so Ns:Np becomes, (1/sqrt(2)):1.

This means that if you want the impedance reflected on the filter side (the new secondary) to be 50Ω, then each of the resistors on the IOUT_P/N side (the new primary) must be 50Ω.

You might find this Application Note helpful.

Technically, regarding the AD9910 Evaluation Board, the 50Ω resistors on the DDS side the of transformer should have been 25Ω each for a proper impedance match to the 50Ω on the filter side. However, there appears to be little to no degradation in spectral characteristics of the output signal even with the impedance mismatch.

The ADT-1T-1P+ has an impedance ratio (Zr=Zs/Zp) of 2. That is, Zs:Zp = 2:1. The impedance ratio and turns ratio relate as: Nr=sqrt(Zr)=sqrt(Zs)/sqrt(Zp). Written in ratio form: sqrt(Zs):sqrt(Zp), or sqrt(2):1, in this case.

However, the transformer PRI and SEC are used in reverse in this application. In keeping with the application, Ns and Np, as follows, denote the "new" secondary and primary, per the application. Thus, the turns ratio (Ns:Np) is 1:sqrt(2). By convention, Np=1, so Ns:Np becomes, (1/sqrt(2)):1.

This means that if you want the impedance reflected on the filter side (the new secondary) to be 50Ω, then each of the resistors on the IOUT_P/N side (the new primary) must be 50Ω.

You might find this Application Note helpful.

Technically, regarding the AD9910 Evaluation Board, the 50Ω resistors on the DDS side the of transformer should have been 25Ω each for a proper impedance match to the 50Ω on the filter side. However, there appears to be little to no degradation in spectral characteristics of the output signal even with the impedance mismatch.