Difference between revisions of "FAQ: In the displayed equation after equation (8.3), how can you set x(0), a constant, to a function of s?"

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{{FAQ page
 
{{FAQ page
| chapter = Chapter Name
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| chapter = Transfer Functions
 
| contributor = Richard Murray
 
| contributor = Richard Murray
| date = 17 May 2008
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| date = 10 September 2012
 
| answer =  
 
| answer =  
This is a template page for creating a FAQ. You should replace the information above with the appropriate information for your FAQ and replace this text with the answer to your FAQ.  If your FAQ page is complicated, you may want to substitute this page (use <nowiki>{{subst:FAQ page ...}}</nowiki> and then customize it.
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In the derivation of the exponential response of a linear system, equation (8.3) gives the following expression for <amsmath>y(t)</amsmath>:
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<center><amsmath>
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  y(t) = Ce^{At} \Bigl( x(0)-(sI-A)^{-1} B \Bigr) + \Bigl( C (sI-A)^{-1} B + D \Bigr) e^{st}.
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</amsmath></center>
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To obtain the pure exponential response (the right hand term), the text states that we should choose the initial state as
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<center><amsmath>
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  x(0) = (s I - A)^{-1} B,
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</amsmath></center>
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Since we are used to thinking of <amsmath>s</amsmath> as the argument for a transfer function, it may seem confusing to set <amsmath>x(0)</amsmath> to a non-constant value.
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However, in this derivation we are considering the response of a linear system to a fixed input of the form <amsmath>u = e^{st}</amsmath> where <amsmath>s</amsmath> is a ''constant'' of the form <amsmath>s = \sigma + i \omega</amsmath>. Setting <amsmath>x(0)</amsmath> to the value given above thus result in the first term in the exponential response being zero. Note that this value only works for this ''specific'' choice of <amsmath>s</amsmath>; a different value must be chosen for different exponential inputs.  
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Also note that <amsmath>x(0)</amsmath> is complex-valued in this expression. This is OK because in practice we will always add together combinations of signals that result in real-valued expressions. See [[FAQ: Why is the exponential response complex valued?  Shouldn't the inputs and outputs of a linear system always be real-valued?]]
 
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Latest revision as of 18:17, 10 September 2012

(Contributed by Richard Murray, 10 September 2012)

In the derivation of the exponential response of a linear system, equation (8.3) gives the following expression for math:

math

To obtain the pure exponential response (the right hand term), the text states that we should choose the initial state as

math

Since we are used to thinking of math as the argument for a transfer function, it may seem confusing to set math to a non-constant value.

However, in this derivation we are considering the response of a linear system to a fixed input of the form math where math is a constant of the form math. Setting math to the value given above thus result in the first term in the exponential response being zero. Note that this value only works for this specific choice of math; a different value must be chosen for different exponential inputs.

Also note that math is complex-valued in this expression. This is OK because in practice we will always add together combinations of signals that result in real-valued expressions. See FAQ: Why is the exponential response complex valued? Shouldn't the inputs and outputs of a linear system always be real-valued?