Graphing decaying sinusoids by hand?

Discussion in 'Math' started by dewasiuk, Feb 26, 2011.

  1. dewasiuk

    Thread Starter New Member

    Feb 14, 2011
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    How could this be roughly accomplished without making an extensive table of values?

    For example if I use the laplace method of analyzing an RLC circuit with a pulse excitation, I know that I will arrive at a decaying sinusoidal current represented by the following equation: [​IMG]

    I would like to be able to quickly graph the equation incase I didn't have access to software tools at that moment.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    For your given example

    1. Draw the exponential envelope bounded by the curves

    f1(t)=e^{-\alpha t}

    and

    f2(t)=-e^{-\alpha t}

    2. Draw a sinusoid constrained in amplitude by the bounded region with damped period

    T=\frac{2\pi}{\omega}

    Since it is a sine function in this particular case the value at t=0 will be zero. If it were a simple cosine function the initial value would be 1. A phase displaced sine or cosine function would have an initial value determined by the value of the sine or cosine term at t=0 sec.
     
    Last edited: Feb 27, 2011
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Here's an example sketch of a phase shifted exponentially decaying cosine function.
     
  4. dewasiuk

    Thread Starter New Member

    Feb 14, 2011
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    Oh I already know how to roughly sketch the actual shapes, but it's mostly about determining the amplitude of the first peak(for a sine wave) so I can draw it more accurately. I should have included that in my first post sorry.
     
  5. Georacer

    Moderator

    Nov 25, 2009
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    Peaks and zero-crossings still occur in the multiples of pi/2, so these are points in time you look for.

    For example, your first peak will be at the intersection of the envelope with the line x=pi/2 for a sinusoidal curve.
    That is, of course, for zero time-shift. Shift your times accordingly for θ<>0.
     
  6. dewasiuk

    Thread Starter New Member

    Feb 14, 2011
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    Damn I feel silly for not thinking about this first haha. Thanks :)
     
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