# Difference between revisions of "Problem 4 - How do we decompose H(w)?"

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'''Q:''' How is it possible to decompose <math>S(\omega)</math>? | '''Q:''' How is it possible to decompose <math>S(\omega)</math>? | ||

'''A:''' There is a general observation to make. Given a transfer function <math>H(s)=\frac{1}{s+a}</math>, its power spectral density will be <math>S(\omega)=\frac{1}{ | '''A:''' There is a general observation to make. Given a transfer function <math>H(s)=\frac{1}{s+a}</math>, its power spectral density will be <math>S(\omega)=\frac{1}{\omega^2+a^2}</math>. If we define <math>\lambda:=\omega^2</math>, then we see that we have <math>S(\lambda)=\frac{1}{\lambda+a^2}</math>. Qualitatively, we can argue that poles of <math>H(s)</math> in <math>-a</math> are mapped to poles of <math>S(\lambda)</math> in <math>-a^2</math>. | ||

Same for transfer functions having more than one pole and zeros. | Same for transfer functions having more than one pole and zeros. | ||

In the exercise you should therefore substitute <math> | In the exercise you should therefore substitute <math>\omega^2</math> with <math>\lambda</math>, find its poles and zeros, and then map back to a guess for <math>H(s)</math>. Such guess will not be unique in general, but it is if one assumes certain properties regarding the phase! | ||

## Latest revision as of 18:51, 17 January 2007

**Q:** How is it possible to decompose ?

**A:** There is a general observation to make. Given a transfer function , its power spectral density will be . If we define , then we see that we have . Qualitatively, we can argue that poles of in are mapped to poles of in .
Same for transfer functions having more than one pole and zeros.

In the exercise you should therefore substitute with , find its poles and zeros, and then map back to a guess for . Such guess will not be unique in general, but it is if one assumes certain properties regarding the phase!

--Elisa