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February 25, 2013, 04:12 |
Transient analysis time step
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#1 |
Senior Member
Govindaraju
Join Date: Apr 2010
Posts: 209
Rep Power: 17 |
Dear all
I would like to clarify the time steps. I have a sinusoidal function and the frequency is 1 Hz. I did the transient simulation for one cycle one of my friends told me and insist me to do the multiple cycle simulations.Is it necessary to do the simulations for multiple cycles?. If I do the multiple cycles the solution time is too large. Is there any way to short cut. Thank you Regards |
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February 25, 2013, 05:15 |
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#2 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 17,872
Rep Power: 144 |
If you want the startup transient then you do not need to do multiple cycles. If you want the repeating cycle it depends how long it settles to a repeating pattern. It could be 1 cycle, it could be hundreds - it depends on how stable the system is and how long it takes to settle.
But you will need to run it long enough to show that you have run long enough - so the change in the last few cycles is small enough that you can call it converged. |
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February 25, 2013, 06:55 |
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#3 |
Senior Member
Lance
Join Date: Mar 2009
Posts: 669
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Glenn is (as always) correct, you will need to perform a sensitivity analysis on the number of cycles in order to avoid initial effects - otherwise the initial conditions will affect your solution.
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March 2, 2013, 21:01 |
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#4 | |
Senior Member
Govindaraju
Join Date: Apr 2010
Posts: 209
Rep Power: 17 |
Quote:
I have a set of data from 0 s to 2 s. I used one dimensional interpolation and activated extended minimum and maximum. My question is How to run this simulation for repeated cycles.( at least for 4 cycles) ? Thank you Regards Govind |
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March 3, 2013, 06:19 |
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#5 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 17,872
Rep Power: 144 |
The simple nasty approach is just to repeat the 1D function 4 times.
The better approach is to use the mod function, so the time input to the interpolation is t-mod(t,2) which should return a function which increase to 2, then goes to zero and increases to 2 again when t is 4, and repeats that forever. |
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March 3, 2013, 14:14 |
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#6 | |
Senior Member
Govindaraju
Join Date: Apr 2010
Posts: 209
Rep Power: 17 |
Quote:
Thank you . It works Govind |
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