Consider the LCCDE : \begin{equation} \sum_{k=0}^{N}\alpha_{k}\frac{d^{k}f(t)}{dt^{k}}=\sum_{m=0}^{M}\beta_{m}\frac{d^{k}g(t)}{dt^{k}} \end{equation} Taking the Fourier transform on both sides, we get : \begin{equation} \sum_{k=0}^{N}\alpha_{k}(2j\pi\lambda)^{k}\widehat{f}(\lambda)=\sum_{m=0}^{M}\beta_{m}(2j\pi\lambda)^{m}\widehat{g}(\lambda) \tag{1}\end{equation} Consider the two polynomials : $$ P(x):=\sum_{k=0}^{N}\alpha_{k}x^{k}\qquad\text{and}\qquad Q(x):=\sum_{m=0}^{M}\beta_{m}x^{m} $$ Then since $\displaystyle\frac{P(x)}{Q(x)}$ has no poles then $\displaystyle\frac{P(2j\pi\lambda)}{Q(2j\pi\lambda)}$ has no poles for real $\lambda$ and thus $(1)$ becomes : \begin{equation} \widehat{g}(\lambda)=\frac{P(2j\pi\lambda)}{Q(2j\pi\lambda)}\widehat{f}(\lambda) \end{equation}

The impulse response $h=\mathcal{F}^{-1}\{H\}$ is computed by decomposing $H$ into partial fractions. The poles $\displaystyle\frac{P}{Q}$ are assumed to lie off the imaginary axis. There are thus two cases to consider :

  1. ($\frac{P}{Q}$ has only simple poles) : In this case, $H$ can be decomposed in the form : \begin{equation} H(\lambda)=\sum_{m=0}^{M}\frac{\beta_{m}}{2j\pi\lambda-z_{m}} \end{equation} where $z_{1},z_{2},\cdots,z_{M}$ are the poles. Therefore : \begin{equation} h(t)=\left(\sum_{m\in \mathcal{M}_{-}}\beta_{m}e^{z_{m}t}\right)u(t)-\left(\sum_{m\in \mathcal{M}_{+}}\beta_{m}e^{z_{m}t}\right)u(-t) \end{equation} where we define : \begin{align*} &\mathcal{M}_{-}:=\{m\in\{1,2,\cdots,M\}\;|\;\Re(z_{m})<0\}\\ &\mathcal{M}_{+}:=\{m\in\{1,2,\cdots,M\}\;|\;\Re(z_{m})>0\} \end{align*}
  2. ($\frac{P}{Q}$ has multiple poles) : In this case, let $z_{1},z_{2},\cdots,z_{p}$ the poles and let $\gamma_{1},\gamma_{2},\cdots,\gamma_{p}$ be their multiplicites. Then we can write $H$ as : \begin{equation} H(\lambda)=\sum_{m=1}^{p}\sum_{\gamma=1}^{\gamma_{m}}\frac{\beta_{m,\gamma}}{(2j\pi\lambda-z_{p})^{\gamma}} \end{equation} We can see that : \begin{equation} h(t)=\left(\sum_{m\in \mathcal{M}_{-}}P_{m}e^{z_{m}t}\right)u(t)-\left(\sum_{m\in \mathcal{M}_{+}}P_{m}e^{z_{m}t}\right)u(-t) \end{equation} where : $$ P_{m}:=\sum_{\gamma=1}^{\gamma_{m}}\beta_{m,\gamma}\frac{t^{\gamma-1}}{(\gamma-1)!} $$

Remark : The case where we have pure imaginary poles, can not be solved using the two cases above, for a differential equation of the form : $$ g''+\omega^{2} g=f $$ where $\displaystyle\frac{P(x)}{Q(x)}=\frac{1}{x^{2}+\omega^{2}}$ has two poles are on the imaginary axis. In this case, $h$ is a sinusoid and the Fourier transform of $H$ (when $H$ is considered to be a function) is no longer defined

I didn't understand why we would be concerned in finding Fourier transform of $H$ as I didn't understand what the author meant by saying when $H$ is considered to be a function.

[Reference] : C. Gasquet, P. Witomski, Fourier Analysis and Applications: Filtering, Numerical Computations, Wavelets. Translated by R. Ryan

  • $\begingroup$ There may be more elsewhere in the book, otherwise it's not clear why the Fourier transform of a sinusoid, or the inverse of its equivalent Laplace transform, are not possible. $\endgroup$ – a concerned citizen Mar 22 at 22:04
  • $\begingroup$ What's remaining of the statement was "This problern will be resolved in the context of distributions" which the author wrote there an example of the resonator saying that we were not able to analyze it using the Fourier transform because the poles were on the imaginary axis @aconcernedcitizen $\endgroup$ – CynthiaZ1998 Mar 22 at 22:13
  • $\begingroup$ @aconcernedcitizen I was thinking what if he meant that $H$ would be an input in a cascade system and that it's Fourier transform makes a problem? $\endgroup$ – CynthiaZ1998 Mar 22 at 22:15
  • $\begingroup$ There must be a larger context that is not necessarily in that paragraph, or even in that chapter. I can't say anything else based on what I see, and quoting a whole book is not very practical. Maybe some clues are in the past, or yet to come. $\endgroup$ – a concerned citizen Mar 22 at 22:25
  • $\begingroup$ Alright, I shall update my question in a moment to include the details. $\endgroup$ – CynthiaZ1998 Mar 22 at 22:26

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