[SOLVED] Direct Digital Synthesizer

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ali ghafoor

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i want some help about SFDR of DDS. how does phase noise of reference CLOCK affect the SFDR of DDS? would SFDR be same or different in following two cases:
1- if i use a clock source of 100MHz and use internal PLL of DDS to make the system reference clock equal to 1GHz(multiplier of 10). clock source phase noise is -140dBc/Hz @ 100kHz offset(after multiplication it will become -140+20log(10)=-120dBc/Hz)
2- if i use a clock source of 1GHz and bypass internal PLL of DDS. clock source phase noise is -120dBc/Hz @ 100kHz offset

would bypassing internal PLL of DDS improve SFDR of DDS (even if the ultimate phase noise of reference clock is same)??
 

The DDS doesn't know if its clock is coming from a PLL or not, so if the output of the PLL is identical to the output of some other clock source, why would it make a difference? 120dBc/Hz is 120dBc/Hz.
 
Well, if you had a choice of using an extremely low noise crystal or saw oscillator at 100 MHz, or a crappy integrated VCO with a pll trying to clean it up inside of the DDS, which one would u choose?

The phase noise will be AT LEAST 20 Log (Neffective) worse. But there are additional terms...quantization noise, small discrete spurs, digital jitter noise, aliasing noise/spurs, etc etc, so it will be worse than that in real life.
 

dear barry, i agree with you..but i found a paper contradicting this argument..i have attached the paper inhere..please refer to its section 5..and provide me with your answer
 

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  • TECHNICAL_TUTORIAL_dds.pdf
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I'm not sure where you are seeing a contradiction. It states that using the internal multiplier (PLL) will increase phase noise (as would be expected), and thus your performance would degrade. You were proposing using a source clock versus a PLL-derived clock where both had the SAME phase noise. Basically, you can get a reference oscillator with much better phase noise than the output of a PLL will give you.
 
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