Simulink

2011 9 6
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Why the integration of a sin() function, with the integrator block of simulink, doesn't give a waveform of -cos()??

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Grzegorz Knor
Grzegorz Knor 2011-9-6

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What is your initial condition? Maybe 0?
In this case should be -1.
Integrator

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Grzegorz Knor
Grzegorz Knor 2011-9-6
hint: Initial condition you could set in Parameters and Dialog Box.
Paulo Silva
Paulo Silva 2011-9-6
Good answer, having initial condition -1 instead of 0 solves the problem, +1 vote

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Fabio
Fabio 2011-9-7

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Thanks Grzegorz,
I can see a cosine waveform but there is a peak upper than 1(almost 6), when the sine has an amplitude of 1. Do you know the reason of that behaviour?

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