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  3. PPV Plot for flat squarish waveform

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PPV Plot for flat squarish waveform

Srayan
Srayan over 2 years ago

Hello,

I am trying to plot the ISF for a squarish waveform generated by an oscillator. I am aware that the simulator plots PPV not ISF. However, the shape of PPV and ISF will be more or less the same (please correct me if I'm wrong). The red waveform is the transient waveform for an output node for my oscillator. The blue waveform is the PPV plot for the same node. It has been discussed in literature that for the flat part of a squarish waveform the ISF will be 0. I have highlighted the flat-ish part of the transient output but the PPV plot is not 0 (neither constant) for that region. 

Why is it so?

Please note I have made 2 assumptions:

1. PPV and ISF look the same (they have different definitions)

2. The red waveform is treated as a squarish waveform by the simulator.

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  • ShawnLogan
    ShawnLogan over 2 years ago

    Dear Srayan,

    Srayan said:
    However, the shape of PPV and ISF will be more or less the same (please correct me if I'm wrong).

    Correct.

    Srayan said:

    It has been discussed in literature that for the flat part of a squarish waveform the ISF will be 0. I have highlighted the flat-ish part of the transient output but the PPV plot is not 0 (neither constant) for that region. 

    Why is it so?

    Your comment regarding the slope of the PPV approaching zero when the output waveform shows a zero slope is accurate. An example for a ring based VCO whose output is a square wave from reference [1] is shown as Figure 1.

    Figure 1 from Figure 7 of reference [1]

    However, what is a little puzzling to me concerning your overlaid plot is the units of your PPV. The PPV should should have amplitudes that extend above and below zero and it appears your function does not - its units are all positive. This does not seem consistent with its definition. I annotated your figure and made included a possible location for the zero level of your PPV and included it as Figure 2 . This is consistent with the location of the relatively low slope region of your oscillator output waveform.

    Shawn

    Figure 2

    Reference [1]

    S. Galeone and M. P. Kennedy, "A comparison of simulation strategies for estimating phase noise in oscillators," 2017 13th Conference on Ph.D. Research in Microelectronics and Electronics (PRIME), Giardini Naxos - Taormina, Italy, 2017, pp. 213-216,

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  • Srayan
    Srayan over 2 years ago in reply to ShawnLogan

    Hello,

    I apologize for the misleading units for the PPV plot (blue waveform), it does oscillate around the 0 crossing. The units on the left is the voltage amplitude for the o/p wave (red waveform).

    So in conclusion you are saying the simulator shows the correct PPV plot? I

    Why isn't the PPV flat when the o/p waveform is constant (almost)?

    Regards,

    Srayan Sankar Chatterjee

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  • ShawnLogan
    ShawnLogan over 2 years ago in reply to Srayan

    Dear Srayan,

    Srayan said:
    The units on the left is the voltage amplitude for the o/p wave (red waveform).

    Ahah! Thank you for clarifying the y-axis units of the PPV curve. I examined the screen shot you included in your initial Forum post and did not see any other y axis other than the y axis on the left, so perhaps the second y axis is not visible in your screen capture. Nevertheless, I was happy to read that it is centered about 0 as expected.

    Srayan said:
    So in conclusion you are saying the simulator shows the correct PPV plot?

    Without knowing the details of your oscillator other than its output waveform, the shape of your PPV waveform does look reasonable. 

    Srayan said:
    Why isn't the PPV flat when the o/p waveform is constant (almost)?

    Looking at your output waveform, I see two features that suggest the PPV should not be flat when the oscillator is above or below its logic threshold. I've try to better illustrate the impact of feedthrough evident on your output waveform and its impact on the location of the instant the transition times cross the logic threshold in Figure 1. I hope this makes some sense to you and apologize if it is not clear!

    Shawn

    Figure 1

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