# 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 \(S(\omega)\)?

**A:** There is a general observation to make. Given a transfer function \(H(s)=\frac{1}{s+a}\), its power spectral density will be \(S(\omega)=\frac{1}{\omega^2+a^2}\). If we define \(\lambda:=\omega^2\), then we see that we have \(S(\lambda)=\frac{1}{\lambda+a^2}\). Qualitatively, we can argue that poles of \(H(s)\) in \(-a\) are mapped to poles of \(S(\lambda)\) in \(-a^2\).
Same for transfer functions having more than one pole and zeros.

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

--Elisa