Return to Answer

I don't understand how do you get your expression for the impulse response $w(t)$: however I'd solve the problem considering $w(t)$ as a functional $w(t)=\mathfrak{w}[p](t)$ of $p(t)$ and then calculating the functional derivative, i.e. $$ \frac{\partial w}{\partial p}=\frac{\delta w}{\delta p}=\frac{\delta\mathfrak{w}[p]}{\delta p}\triangleq\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0} $$ where $\delta p=\varepsilon v$ is a variation of the output singnal when the input $d(t)$ is kept constant (from the engineering point of view, possibly due to a parametric variation of the impulse response function). Precisely, by applying formally the Fourier transform to the Input/Output relation above we have we have $$ \hat{d}(\omega)\cdot\hat{w}(\omega)=\hat{p}(\omega)\iff w(t)=\mathscr{F}^{-1}\left[\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right](t)\label{1}\tag{1} $$ and thus $$ \begin{split} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}&=\mathscr{F}^{-1}\left[\frac{\hat{v}(\omega)}{\hat{d}(\omega)}\right](t)\\ &=\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast v(t) \end{split} $$ Finally (and formally) we can say that $$ \begin{split} \frac{\partial w}{\partial p}(\:\cdot\:)& \triangleq\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast (\:\cdot\:)\\ &=\int\limits_{-\infty}^{+\infty}\!\!\! \mathscr{F}^{\!-1}\left[\big({\hat{d}(\omega)\big)^{-1}}\right](t-s)(\:\cdot\:)\,\mathrm{d}s \end{split}\label{2}\tag{2} $$ i.e. the functional derivative $\frac{\delta w}{\delta p}$ is a convolution operator that maps the variations $v$ of the output signal $p$ to the linear space of (first order) variations of the impulse response $w$ allowing for, from the engineering point of view, an estimation of the magnitude of this quantity.

Notes

• All the above analysis is formal since without any assumption on $d(t), p(t)$ the Fourier transforms have no meaning. However, simply assuming $d\in\mathscr{E}^\prime$ (i.e., from the engineering point view, a finite duration of the input signal) and $w, p\in \mathscr{S}^\prime$ (a mild condition which allows to consider even non causal systems) the formal steps done above become perfectly rigorous as an application of the Hörmander/Łojasiewicz solution of the division problem (see this Q&A for more info on this point).
• Edit: the meaning of the variation $v$ of the output signal. As the development above shows and as it is explicitly said in the last sentence of the answer, $\frac{\delta w}{\delta p}$ is a convolution operator, not a function. Its argument $v(t)$ is therefore a datum: you cannot deduce it from the input $d$ and the output $p$ of the ideal system. Thus the methodology you may follow in order to use formula \eqref{2} is the following one
  1. Take a real system whose input signal is $d(t)$ and whose attended output signal is $p(t)$: the ideal impulse response of this system is $w(t)$ as expressed by \eqref{1}.
  2. Measure, or generally evaluate, the effective output signal $p_e(t)$.
  3. Put $v(t)=p_e(t)-p(t)$: now you can evaluate $\frac{\delta w}{\delta p}[v](t)$ by using a convolution calculation algorithm, and then you can estimate the effective impulse response $$ w_e(t)\simeq w(t)+\frac{\delta w}{\delta p}[v](t) $$
• From an engineering point of view, the $\varepsilon$ parameter could be though as the magnitude of the cause that makes $w$ vary, i.e. it could be the working temperature of the system ($\varepsilon=T$), the relative humidity ($\varepsilon=\theta\%$), the parameter variation ($\varepsilon=\text{tol.}\%$), etc. Thus functional derivative allows for an a posteriori (after a performing a set of measurements, for example) determination of the influence the change of those parameter has on the system.
• For more information on functional derivatives it is possible to have a look at this Q&A.

I don't understand how do you get your expression for the impulse response $w(t)$: however I'd solve the problem considering $w(t)$ as a functional $w(t)=\mathfrak{w}[p](t)$ of $p(t)$ and then calculating the functional derivative, i.e. $$ \frac{\partial w}{\partial p}=\frac{\delta w}{\delta p}=\frac{\delta\mathfrak{w}[p]}{\delta p}\triangleq\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0} $$ where $\delta p=\varepsilon v$ is a variation of the output singnal when the input $d(t)$ is kept constant (from the engineering point of view, possibly due to a parametric variation of the impulse response function). Precisely, by applying formally the Fourier transform to the Input/Output relation above we have we have $$ \hat{d}(\omega)\cdot\hat{w}(\omega)=\hat{p}(\omega)\iff w(t)=\mathscr{F}^{-1}\left[\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right](t)\label{1}\tag{1} $$ and thus $$ \begin{split} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}&=\mathscr{F}^{-1}\left[\frac{\hat{v}(\omega)}{\hat{d}(\omega)}\right](t)\\ &=\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast v(t) \end{split} $$ Finally (and formally) we can say that $$ \begin{split} \frac{\partial w}{\partial p}(\:\cdot\:)& \triangleq\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast (\:\cdot\:)\\ &=\int\limits_{-\infty}^{+\infty}\!\!\! \mathscr{F}^{\!-1}\left[\big({\hat{d}(\omega)\big)^{-1}}\right](t-s)(\:\cdot\:)\,\mathrm{d}s \end{split}\label{2}\tag{2} $$ i.e. the functional derivative $\frac{\delta w}{\delta p}$ is a convolution operator that maps the variations $v$ of the output signal $p$ to the linear space of (first order) variations of the impulse response $w$ allowing for, from the engineering point of view, an estimation of the magnitude of this quantity.

Notes

• All the above analysis is formal since without any assumption on $d(t), p(t)$ the Fourier transforms have no meaning. However, simply assuming $d\in\mathscr{E}^\prime$ (i.e., from the engineering point view, a finite duration of the input signal) and $w, p\in \mathscr{S}^\prime$ (a mild condition which allows to consider even non causal systems) the formal steps done above become perfectly rigorous as an application of the Hörmander/Łojasiewicz solution of the division problem (see this Q&A for more info on this point).
• Edit: the meaning of the variation $v$ of the output signal. As the development above shows and as it is explicitly said in the last sentence of the answer, $\frac{\delta w}{\delta p}$ is a convolution operator, not a function. Its argument $v(t)$ is therefore a datum: you cannot deduce it from the input $d$ and the output $p$ of the ideal system. Thus the methodology you may follow in order to use formula \eqref{2} is the following one
  1. Take a real system whose input signal is $d(t)$ and whose expected output signal is $p(t)$: the ideal impulse response of this system is $w(t)$ as expressed by \eqref{1}.
  2. Apply to the system the input signal $d(t)$ and measure, or more generally evaluate, the effective output signal $p_e(t)$.
  3. Put $v(t)=p_e(t)-p(t)$: now you can evaluate $\frac{\delta w}{\delta p}[v](t)$ by using a convolution calculation algorithm, and then you can estimate the effective impulse response $$ w_e(t)\simeq w(t)+\frac{\delta w}{\delta p}[v](t) $$
• From an engineering point of view, the $\varepsilon$ parameter could be though as the magnitude of the cause that makes $w$ vary, i.e. it could be the working temperature of the system ($\varepsilon=T$), the relative humidity ($\varepsilon=\theta\%$), the parameter variation ($\varepsilon=\text{tol.}\%$), etc. Thus functional derivative allows for an a posteriori (after a performing a set of measurements, for example) determination of the influence the change of those parameter has on the system.
• For more information on functional derivatives it is possible to have a look at this Q&A.

I don't understand how do you get your expression for the impulse response $w(t)$: however I'd solve the problem considering $w(t)$ as a functional $w(t)=\mathfrak{w}[p](t)$ of $p(t)$ and then calculating the functional derivative, i.e. $$ \frac{\partial w}{\partial p}=\frac{\delta w}{\delta p}=\frac{\delta\mathfrak{w}[p]}{\delta p}\triangleq\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0} $$ where $\delta p=\varepsilon v$ is a variation of the output singnal when the input $d(t)$ is kept constant (from the engineering point of view, possibly due to a parametric variation of the impulse response function). Precisely, by applying formally the Fourier transform to the Input/Output relation above we have we have $$ \hat{d}(\omega)\cdot\hat{w}(\omega)=\hat{p}(\omega)\iff w(t)=\mathscr{F}^{-1}\left[\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right](t)\label{1}\tag{1} $$ and thus $$ \begin{split} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}&=\mathscr{F}^{-1}\left[\frac{\hat{v}(\omega)}{\hat{d}(\omega)}\right](t)\\ &=\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast v(t) \end{split} $$ Finally (and formally) we can say that $$ \begin{split} \frac{\partial w}{\partial p}(\:\cdot\:)& \triangleq\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast (\:\cdot\:)\\ &=\int\limits_{-\infty}^{+\infty}\!\!\! \mathscr{F}^{\!-1}\left[\big({\hat{d}(\omega)\big)^{-1}}\right](t-s)(\:\cdot\:)\,\mathrm{d}s \end{split}\label{2}\tag{2} $$ i.e. the functional derivative $\frac{\delta w}{\delta p}$ is a convolution operator that maps the variations $v$ of the output signal $p$ to the linear space of (first order) variations of the impulse response $w$ allowing for, from the engineering point of view, an estimation of the magnitude of this quantity.

Notes

• All the above analysis is formal since without any assumption on $d(t), p(t)$ the Fourier transforms have no meaning. However, simply assuming $d\in\mathscr{E}^\prime$ (i.e., from the engineering point view, a finite duration of the input signal) and $w, p\in \mathscr{S}^\prime$ (a mild condition which allows to consider even non causal systems) the formal steps done above become perfectly rigorous as an application of the Hörmander/Łojasiewicz solution of the division problem (see this Q&A for more info on this point).
• Edit: the meaning of the variation $v$ of the output signal. As the development above shows and as it is explicitly said in the last sentence of the answer, $\frac{\delta w}{\delta p}$ is a convolution operator, not a function. Its argument $v(t)$ is therefore a datum: you cannot deduce it from the input $d$ and the output $v$ of the ideal system. Thus the methodology you may follow in order to use formula \eqref{2} is the following one
  1. Take a real system whose input signal is $d(t)$ and whose attended output signal is $p(t)$: the ideal impulse response of this system is $w(t)$ as expressed by \eqref{1}.
  2. Measure, or generally evaluate, the effective output signal $p_e(t)$.
  3. Put $v(t)=p_e(t)-p(t)$: now you can evaluate $\frac{\delta w}{\delta p}[v](t)$ by using a convolution calculation algorithm, and then you can estimate the effective impulse response $$ w_e(t)\simeq w(t)+\frac{\delta w}{\delta p}[v](t) $$
• From an engineering point of view, the $\varepsilon$ parameter could be though as the magnitude of the cause that makes $w$ vary, i.e. it could be the working temperature of the system ($\varepsilon=T$), the relative humidity ($\varepsilon=\theta\%$), the parameter variation ($\varepsilon=\text{tol.}\%$), etc. Thus functional derivative allows for an a posteriori (after a performing a set of measurements, for example) determination of the influence the change of those parameter has on the system.
• For more information on functional derivatives it is possible to have a look at this Q&A.

I don't understand how do you get your expression for the impulse response $w(t)$: however I'd solve the problem considering $w(t)$ as a functional $w(t)=\mathfrak{w}[p](t)$ of $p(t)$ and then calculating the functional derivative, i.e. $$ \frac{\partial w}{\partial p}=\frac{\delta w}{\delta p}=\frac{\delta\mathfrak{w}[p]}{\delta p}\triangleq\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0} $$ where $\delta p=\varepsilon v$ is a variation of the output singnal when the input $d(t)$ is kept constant (from the engineering point of view, possibly due to a parametric variation of the impulse response function). Precisely, by applying formally the Fourier transform to the Input/Output relation above we have we have $$ \hat{d}(\omega)\cdot\hat{w}(\omega)=\hat{p}(\omega)\iff w(t)=\mathscr{F}^{-1}\left[\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right](t)\label{1}\tag{1} $$ and thus $$ \begin{split} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}&=\mathscr{F}^{-1}\left[\frac{\hat{v}(\omega)}{\hat{d}(\omega)}\right](t)\\ &=\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast v(t) \end{split} $$ Finally (and formally) we can say that $$ \begin{split} \frac{\partial w}{\partial p}(\:\cdot\:)& \triangleq\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast (\:\cdot\:)\\ &=\int\limits_{-\infty}^{+\infty}\!\!\! \mathscr{F}^{\!-1}\left[\big({\hat{d}(\omega)\big)^{-1}}\right](t-s)(\:\cdot\:)\,\mathrm{d}s \end{split}\label{2}\tag{2} $$ i.e. the functional derivative $\frac{\delta w}{\delta p}$ is a convolution operator that maps the variations $v$ of the output signal $p$ to the linear space of (first order) variations of the impulse response $w$ allowing for, from the engineering point of view, an estimation of the magnitude of this quantity.

Notes

• All the above analysis is formal since without any assumption on $d(t), p(t)$ the Fourier transforms have no meaning. However, simply assuming $d\in\mathscr{E}^\prime$ (i.e., from the engineering point view, a finite duration of the input signal) and $w, p\in \mathscr{S}^\prime$ (a mild condition which allows to consider even non causal systems) the formal steps done above become perfectly rigorous as an application of the Hörmander/Łojasiewicz solution of the division problem (see this Q&A for more info on this point).
• Edit: the meaning of the variation $v$ of the output signal. As the development above shows and as it is explicitly said in the last sentence of the answer, $\frac{\delta w}{\delta p}$ is a convolution operator, not a function. Its argument $v(t)$ is therefore a datum: you cannot deduce it from the input $d$ and the output $p$ of the ideal system. Thus the methodology you may follow in order to use formula \eqref{2} is the following one
  1. Take a real system whose input signal is $d(t)$ and whose attended output signal is $p(t)$: the ideal impulse response of this system is $w(t)$ as expressed by \eqref{1}.
  2. Measure, or generally evaluate, the effective output signal $p_e(t)$.
  3. Put $v(t)=p_e(t)-p(t)$: now you can evaluate $\frac{\delta w}{\delta p}[v](t)$ by using a convolution calculation algorithm, and then you can estimate the effective impulse response $$ w_e(t)\simeq w(t)+\frac{\delta w}{\delta p}[v](t) $$
• From an engineering point of view, the $\varepsilon$ parameter could be though as the magnitude of the cause that makes $w$ vary, i.e. it could be the working temperature of the system ($\varepsilon=T$), the relative humidity ($\varepsilon=\theta\%$), the parameter variation ($\varepsilon=\text{tol.}\%$), etc. Thus functional derivative allows for an a posteriori (after a performing a set of measurements, for example) determination of the influence the change of those parameter has on the system.
• For more information on functional derivatives it is possible to have a look at this Q&A.

I don't understand how do you get your expression for the impulse response $w(t)$: however I'd solve the problem considering $w(t)$ as a functional $w(t)=\mathfrak{w}[p](t)$ of $p(t)$ and then calculating the functional derivative, i.e. $$ \frac{\partial w}{\partial p}=\frac{\delta w}{\delta p}=\frac{\delta\mathfrak{w}[p]}{\delta p}\triangleq\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0} $$ where $\delta p=\varepsilon v$ is a variation of the output singnal when the input $d(t)$ is kept constant (from the engineering point of view, possibly due to a parametric variation of the impulse response function). Precisely, by applying formally the Fourier transform to the Input/Output relation above we have we have $$ \hat{d}(\omega)\cdot\hat{w}(\omega)=\hat{p}(\omega)\iff w(t)=\mathscr{F}^{-1}\left[\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right](t) $$ and thus $$ \begin{split} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}&=\mathscr{F}^{-1}\left[\frac{\hat{v}(\omega)}{\hat{d}(\omega)}\right](t)\\ &=\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast v(t) \end{split} $$ Finally (and formally) we can say that $$ \begin{split} \frac{\partial w}{\partial p}(\:\cdot\:)& \triangleq\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast (\:\cdot\:)\\ &=\int\limits_{-\infty}^{+\infty}\!\!\! \mathscr{F}^{\!-1}\left[\big({\hat{d}(\omega)\big)^{-1}}\right](t-s)(\:\cdot\:)\,\mathrm{d}s \end{split} $$ i.e. the functional derivative $\frac{\delta w}{\delta p}$ is a convolution operator that maps the variations $v$ of the output signal $p$ to the linear space of (first order) variations of the impulse response $w$ allowing for, from the engineering point of view, an estimation of the magnitude of this quantity.

Notes

• All the above analysis is formal since without any assumption on $d(t), p(t)$ the Fourier transforms have no meaning. However, simply assuming $d\in\mathscr{E}^\prime$ (i.e., from the engineering point view, a finite duration of the input signal) and $w, p\in \mathscr{S}^\prime$ (a mild condition which allows to consider even non causal systems) the formal steps done above become perfectly rigorous as an application of the Hörmander/Łojasiewicz solution of the division problem (see this Q&A for more info on this point).
• From an engineering point of view, the $\varepsilon$ parameter could be though as the magnitude of the cause that makes $w$ vary, i.e. it could be the working temperature of the system ($\varepsilon=T$), the relative humidity ($\varepsilon=\theta\%$), the parameter variation ($\varepsilon=\text{tol.}\%$), etc. Thus functional derivative allows for an a posteriori (after a performing a set of measurements, for example) determination of the influence the change of those parameter has on the system.
• For more information on functional derivatives it is possible to have a look at this Q&A.

I don't understand how do you get your expression for the impulse response $w(t)$: however I'd solve the problem considering $w(t)$ as a functional $w(t)=\mathfrak{w}[p](t)$ of $p(t)$ and then calculating the functional derivative, i.e. $$ \frac{\partial w}{\partial p}=\frac{\delta w}{\delta p}=\frac{\delta\mathfrak{w}[p]}{\delta p}\triangleq\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0} $$ where $\delta p=\varepsilon v$ is a variation of the output singnal when the input $d(t)$ is kept constant (from the engineering point of view, possibly due to a parametric variation of the impulse response function). Precisely, by applying formally the Fourier transform to the Input/Output relation above we have we have $$ \hat{d}(\omega)\cdot\hat{w}(\omega)=\hat{p}(\omega)\iff w(t)=\mathscr{F}^{-1}\left[\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right](t)\label{1}\tag{1} $$ and thus $$ \begin{split} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}&=\mathscr{F}^{-1}\left[\frac{\hat{v}(\omega)}{\hat{d}(\omega)}\right](t)\\ &=\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast v(t) \end{split} $$ Finally (and formally) we can say that $$ \begin{split} \frac{\partial w}{\partial p}(\:\cdot\:)& \triangleq\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast (\:\cdot\:)\\ &=\int\limits_{-\infty}^{+\infty}\!\!\! \mathscr{F}^{\!-1}\left[\big({\hat{d}(\omega)\big)^{-1}}\right](t-s)(\:\cdot\:)\,\mathrm{d}s \end{split}\label{2}\tag{2} $$ i.e. the functional derivative $\frac{\delta w}{\delta p}$ is a convolution operator that maps the variations $v$ of the output signal $p$ to the linear space of (first order) variations of the impulse response $w$ allowing for, from the engineering point of view, an estimation of the magnitude of this quantity.

Notes

• All the above analysis is formal since without any assumption on $d(t), p(t)$ the Fourier transforms have no meaning. However, simply assuming $d\in\mathscr{E}^\prime$ (i.e., from the engineering point view, a finite duration of the input signal) and $w, p\in \mathscr{S}^\prime$ (a mild condition which allows to consider even non causal systems) the formal steps done above become perfectly rigorous as an application of the Hörmander/Łojasiewicz solution of the division problem (see this Q&A for more info on this point).
• Edit: the meaning of the variation $v$ of the output signal. As the development above shows and as it is explicitly said in the last sentence of the answer, $\frac{\delta w}{\delta p}$ is a convolution operator, not a function. Its argument $v(t)$ is therefore a datum: you cannot deduce it from the input $d$ and the output $v$ of the ideal system. Thus the methodology you may follow in order to use formula \eqref{2} is the following one
  1. Take a real system whose input signal is $d(t)$ and whose attended output signal is $p(t)$: the ideal impulse response of this system is $w(t)$ as expressed by \eqref{1}.
  2. Measure, or generally evaluate, the effective output signal $p_e(t)$.
  3. Put $v(t)=p_e(t)-p(t)$: now you can evaluate $\frac{\delta w}{\delta p}[v](t)$ by using a convolution calculation algorithm, and then you can estimate the effective impulse response $$ w_e(t)\simeq w(t)+\frac{\delta w}{\delta p}[v](t) $$
• From an engineering point of view, the $\varepsilon$ parameter could be though as the magnitude of the cause that makes $w$ vary, i.e. it could be the working temperature of the system ($\varepsilon=T$), the relative humidity ($\varepsilon=\theta\%$), the parameter variation ($\varepsilon=\text{tol.}\%$), etc. Thus functional derivative allows for an a posteriori (after a performing a set of measurements, for example) determination of the influence the change of those parameter has on the system.
• For more information on functional derivatives it is possible to have a look at this Q&A.