How to reduce the spur of the DDS?

I uesed AD9956 to make a circuit, and the spur of the signal outputing from AD9956 is bad. Could anyone can help me reduce the spur? Please tell me what I can do for the better spur and phase noise. Thanks!

In this document they explain all sources of spurs in DDS and how you can minimize them:
http://www.edn.com/contents/images/415103.pdf

In a recently design I use the AD9954 and I archieve the wanted more then 70dB SFDR. I think the most problems are in the output circiut design. I use an center tapped transformer and an elliptic lowpass to get this good result over a wide output frequency range.
Maybe you are not working in the right voltage compiliance range.

Take a look in AD's DDS-Tutorial its a good source to understand this topic.
Or use the Design-Tools at Analog-Website.

Thank you for your reply.
Can you tell me more clearly what the "voltage compiliance" mean? What voltage it refer? The filter just can fliter the signal of the stopband, And the spur in the passband how to reduce it by the output circuit?
The a good filter is too difficuit to design by lump resistor, capacitor and inductor. Do you have any good advice?

I don't know the range of your output frequency, but its clearly that there are spurs. But the SFDR should be same or better than in the datasheet.

The voltage compiliance of the output mean, that a the IOUT and the /IOUT Pin of the device is able to swing around +/- 0,5V from AVVD.

So if you connect two resistors from AVVD to this pins, and you have an reference current of 10mA these resistors are limited to a value of R<=50Ohm.

If you use an center tapped transformer with ratio 1:1 you have to terminate the secondary side with equal or less then 100Ohm.

To calculate the best filter you can use the tool Filter Solution. It works fine. I use it to start my simulations of the passive output section. So you are able to find the best filter.

I know its the hardest part to get the best results. I spent most of my time with spectral measurements.

My output frequency are 102.4±0.512MHz. I connect a 1:1 transformer the to IOUT and /IOUT pins, and the secondary port of the transformer is the utput.

Should I need a 100ohm resistor between the output pin and GND, then connect to the filter?
Thank you!

hi, see this paper http://www.intersil.com/data/tb/tb318.pdf

Sorry I was long away from this board.

Yes connect a resistor 100Ohms or lower between the transformer and the filter.
Should work good.

You could also connect only a resistor after the filter. But this is in cnnection to your filter design the current source sees the imput impedance of your filter. So you should pay attention for the calculated imput impedance!

I noticed the most noise comes from the Sync_clk so you should programm this clock outside of the passband of your filter.

Hope this helps you!