Difference between revisions of "FAQ: Why is the exponential response complex valued? Shouldn't the inputs and outputs of a linear system always be real-valued?"

From FBSwiki
Jump to: navigation, search
(Created page with "{{subst:FAQ template}}")
 
 
Line 1: Line 1:
 
{{FAQ page
 
{{FAQ page
| chapter = Chapter Name
+
| chapter = Transfer Functions
 
| contributor = Richard Murray
 
| contributor = Richard Murray
| date = 17 May 2008
+
| date = 10 September 2012
 
| answer =  
 
| answer =  
This is a template page for creating a FAQYou 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.
+
'''Q''': In the derivation of the exponential response, the output signal corresponding to input <amsmath>u(t) = e^{st}</amsmath> is given by
 +
<center><amsmath>
 +
  y(t) = Ce^{At} \Bigl( x(0)-(sI-A)^{-1} B \Bigr) + \Bigl( C (sI-A)^{-1} B + D \Bigr) e^{st}.
 +
</amsmath></center>
 +
If <amsmath>s</amsmath> is a complex number <amsmath>s = \sigma + i \omega</amsmath>, then the inputs and outputs are both complexShouldn't they be real?
 +
 
 +
'''A''': In practice, we are interested in the response of the system to real-value inputs.  For example, we might want to obtain the response of a linear system to a sinusoidal input <amsmath>u(t) = a \sin(\omega t)</amsmath>.  Note that we can break up this signal into the sum of two complex exponentials:
 +
<center><amsmath>
 +
u(t) = -i a e^{st} +i a e^{\bar s t}, \qquad\text{where $s = i \omega$, $\bar s = -i \omega$}
 +
</amsmath></center>
 +
Since our system is linear, we can compute the output by summing up the (scaled) responses to the individual inputs.  If we let <amsmath>y(t)</amsmath> represent the output to <amsmath>e^{s t}</amsmath> and <amsmath>\bar y(t)</amsmath> the output to <amsmath>e^{\bar s t}</amsmath>, then the response to the input <amsmath>u(t)</amsmath> will be given by
 +
<center><amsmath>
 +
u(t) = -i a y(t) +i a \bar y(t).
 +
</amsmath></center>
 +
Plugging in the expressions for the exponential response, it can be shown the this turns out to be a real-valued signal.
 +
 
 +
Similarly, the initial condition required to cancel out the transient response and obtain a pure exponential response is given by
 +
<center><amsmath>
 +
  x(0) = -i a (s I - A)^{-1} B + i a (\bar s I - A)^{-1} B,
 +
</amsmath></center>
 +
which can also be shown to be real-valued since it is the sum of a complex number and its complex conjugate.
 
}}
 
}}

Latest revision as of 18:33, 10 September 2012

(Contributed by Richard Murray, 10 September 2012)

Q: In the derivation of the exponential response, the output signal corresponding to input math is given by

math

If math is a complex number math, then the inputs and outputs are both complex. Shouldn't they be real?

A: In practice, we are interested in the response of the system to real-value inputs. For example, we might want to obtain the response of a linear system to a sinusoidal input math. Note that we can break up this signal into the sum of two complex exponentials:

math

Since our system is linear, we can compute the output by summing up the (scaled) responses to the individual inputs. If we let math represent the output to math and math the output to math, then the response to the input math will be given by

math

Plugging in the expressions for the exponential response, it can be shown the this turns out to be a real-valued signal.

Similarly, the initial condition required to cancel out the transient response and obtain a pure exponential response is given by

math

which can also be shown to be real-valued since it is the sum of a complex number and its complex conjugate.