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How to find the final time value of a signal?

swdesigner

( I did search the forum and support.cadence.. no luck)

Say my simulation terminates earlier than the input file specifies, I want to analyze the last 10 us of a waveform. 

How can I get tmax, which is the final timepoint?

xmax gives the time value corresponding to the max value that the signal attained.

Thanks

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  • Abhishek D said:
    Hi Andrew, I was also searching for the same answer and came across this thread. I would really appreciate if you can explain why that code is a better option for getting the time rather than the small command that can easily be done through Cadence Calc. Please pardon my ignorance in skill language. I am asking this question because I have 100s of GB data for waveform and I have to use the time as a variable in my ocean script. Best Regards Abhishek

     

    If you use the code that I provided, it will register some new functions in the calculator, so they are just as easy to use as the built-in functions. The ymax() approach is less efficient, because whilst it doesn't have to really generate a new waveform (the xval function simply takes the x-vector of the existing waveform and  creates a new waveform the x and y vectors the same as the original waveform), it still has to search along the entire waveform to find the greatest y value (which is in fact the greatest x value). My code simply looks at the last value, so is more efficient.

    Regards,

    Andrew.

Reply

  • Abhishek D said:
    Hi Andrew, I was also searching for the same answer and came across this thread. I would really appreciate if you can explain why that code is a better option for getting the time rather than the small command that can easily be done through Cadence Calc. Please pardon my ignorance in skill language. I am asking this question because I have 100s of GB data for waveform and I have to use the time as a variable in my ocean script. Best Regards Abhishek

     

    If you use the code that I provided, it will register some new functions in the calculator, so they are just as easy to use as the built-in functions. The ymax() approach is less efficient, because whilst it doesn't have to really generate a new waveform (the xval function simply takes the x-vector of the existing waveform and  creates a new waveform the x and y vectors the same as the original waveform), it still has to search along the entire waveform to find the greatest y value (which is in fact the greatest x value). My code simply looks at the last value, so is more efficient.

    Regards,

    Andrew.

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