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Coherent Signals

In the attached tutorial, I see the quote below:

In fact, the only time the DDS output is an exact integer division of its system clocking frequency is when the

division factor is a power of two, 2N. Only then will the input clock and the output frequency be “phase-locked” or synchronized.”


Which made me wonder…what’s the difference between phase locked and coherent? 

I want to convince myself that if I put the reference clock on an Oscope and the output clock on an Oscope, the two may have different frequencies, but will not be drifting with respect to each other regardless of the frequency the DDS is synthesizing.  They will both appear stationary on the Oscope.  Is that true??? 

  • Hi lockeba,

    In my understanding, whenever the 2^N condition is satisfied and assuming that the accumulator starts at zero, the signal output exactly hits the zero crossing at each end and start of the cycle. If the condition is not satisfied then it might not hit the zero crossing at exactly one cycle. I think the phase coherent term is usually used in modulations. Let's say you have an FSK and you're alternating frequencies. Every time you start at a new frequency, the signal always starts with a fixed initial phase. For example, if you have a zero offset sine, no matter how fast you alternate between two frequencies, the signal always starts with zero offset.



  • The short answer is that the system clock and DDS output are always synchronized regardless of the tuning word, but not necessarily at each zero-crossing of the output sinusoid. That degree of synchronization only occurs for tuning words that are an exact power of 2, which is the point of the statement in the "tutorial".

    Zero-crossings of the "Nyquist filtered" DDS output sinusoid coincide with a system clock edge at intervals of the grand repitition rate (GRR). The GRR is tuning word dependent and is given by:

    GRR = 2^L    (L is number of the right-most non-zero bit of the tuning word)

    That is, consider the following 32-bit tuning word:

    FTW = 0001 1001 0001 1001 1111 0000 1010 0000

    To find L, count FTW bits from left to right, until you encounter the last "1" in the tuning word. In the above example, L = 27. Therefore, the GRR for the above tuning word is 2^27. That is, it takes 134,217,728 system clock cycles between zero-crossings of the DDS output sinusoid before one of them exactly coincides with a system clock edge.

    Proving this with an oscilloscope is challenging. The general setup would be to connect Channel A of the 'scope to the Nyquist filtered DDS output and trigger Channel A at the zero-crossing point of the sinusoid. Connect Channel B to the system clock. With both traces displayed, you should see a system clock edge on the far left of the B-trace (coinciding with the trigger caused by the zero-crossing). Note, there may be a slight, but constant, time offset due to the precision of the 'scope's triggering circuitry. If you can capture multiple sinusoidal cycles on a single sweep, you will note that zero-crossings of the DDS output and system clock edges do not necessarily align (expect for the special case when the FTW is an exact power of 2).

    This is not an easy experiment to set up. Especially for the case of L=27 above, because it's not possible to display 2^27 (134,217,728) system clock cycles on the screen. Picking a more suitable tuning word helps.

    For example, the tuning word below will yield DDS output zero-crossings that line up with every 16-th system clock edge.

    FTW = 0011 0000 0000 0000 0000 0000 0000 0000

    Whereas you can get alignment on every zero-crossing with a slight change:

    FTW = 0001 0000 0000 0000 0000 0000 0000 0000   <== an exact power of 2

    Coherency is another topic altogether.

    Hope this helps.