What is the Fourier transform of the impulse response of the following relation between y and x?
y(t)= $\sum_{m=0}^{\infty}[u(t-mT_s)-u(t-mT_s-T_h)][x(t)-x(mT_s)]$.
$Y(f)=H_0(f)X(f)+ \underbrace{\sum_{n=-\infty}^{\infty}X_n(f-n/T_s)H_n(f)}_{n\neq0}$
Derive expressions for $H_0(f)$ and $H_n(f)$
I have tried a lot to solve it, but getting nowhere. Please help.
The answer is given in the IEEE paper Circuit Techniques for Reducing the Effects of Op-Amp Imperfections: Autozeroing,Correlated Double Sampling and Chopper Stabilization. by Christian Enz and Gabor Temes. I need the method please.
\begin{equation} y(t)=\sum_{m=-\infty}^{\infty} [u(t-mT_s)-u(t-mT_S-T_H)][x(t)-x(mT_S)] \label{AZt} \end{equation} Let us define \begin{equation} g(t)=[u(t)-u(t-T_H)] \label{gt} \end{equation} Then \begin{eqnarray} G(f) &=& \int_{-\infty}^{\infty}g(t)e^{j2\pi ft}dt \\ & = & \int_{0}^{T_H}g(t)e^{j2\pi ft}dt \end{eqnarray}
Then \begin{equation} y(t)=\sum_{m=-\infty}^{\infty} g(t-mT_S)[x(t)-x(mT_S)] \end{equation} Taking fourier transform on both sides, \begin{equation} Y(f)=X(f)*\sum_{m=-\infty}^{\infty} g(t-mT_S) - \sum_{m=-\infty}^{\infty}x(mT_S)G(f)e^{-j2\pi fmT_s} \end{equation} Since $\sum_{m=-\infty}^{\infty} g(t-mT_S)$ is periodic, its fourier transform is the sequence of dirac delta functions, weighted by its fourier series coefficients. Thus, \begin{equation} \mathcal{F}\left(\sum_{m=-\infty}^{\infty} g(t-mT_S)\right)=\sum_{n=-\infty}^{\infty}a_n\delta(f-\frac{n}{T_S}) \end{equation} where \begin{eqnarray} a_n &=& \frac{1}{T_S}\int_{0}^{T_S}g(t)e^{-jn\frac{2\pi}{T_S}t}dt\\ & =& \frac{1}{T_S}\int_{0}^{T_H}g(t)e^{-jn\frac{2\pi}{T_S}t}dt \\ & =& \frac{1}{T_S}G(\frac{n}{T_S}) \end{eqnarray} for $n\neq 0$. For n=0,
\begin{equation} a_0=\frac{1}{T_S}\int_{0}^{T_H}g(t)dt = \frac{T_H}{T_S} \end{equation} Thus, the first term of the equation is \begin{eqnarray} X(f)*\sum_{m=-\infty}^{\infty} g(t-mT_S) & = & X(f)* \sum_{n=-\infty}^{\infty}\frac{1}{T_S}G(\frac{n}{T_S})\delta(f-\frac{n}{T_S}) \\ & = & \frac{1}{T_S}G(\frac{n}{T_S})\sum_{n=-\infty}^{\infty}X(f-\frac{n}{T_S}) \label{term1} \end{eqnarray}
The second term of the equation is \begin{eqnarray} \sum_{m=-\infty}^{\infty}x(mT_S)G(f)e^{-j2\pi fmT_s} & = & G(f)\sum_{m=-\infty}^{\infty}\left(\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}\delta(t-mT_S)dt \right)\\ & = & G(f) \mathcal{F}\left(x(t)\sum_{m=-\infty}^{\infty}\delta(t-mT_S)\right) \\ & = & G(f)\left[X(f)*\sum_{n=-\infty}^{\infty}(\frac{n}{T_S})\delta(f-\frac{n}{T_S})\right] \\ & = & G(f)\sum_{n=-\infty}^{\infty}(\frac{1}{T_S}) X(f-\frac{n}{T_S}) \label{term2} \end{eqnarray} Now, to find G(f) is pretty straightforward. \begin{eqnarray} G(f) & = &\frac{1}{j2\pi f}(1-e^{-j2\pi fT_H}) \\ & = & \frac{2sin(\pi fT_H)e^{-j\pi fT_H}}{2\pi f} \\ & = & T_Hsinc(\pi fT_H)e^{-j\pi fT_H} \end{eqnarray} Thus, the overall transform is written as \begin{equation} Y(f) = d \sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty} ((sinc(\pi nd)e^{-j\pi nd} - sinc(\pi fT_H)e^{-j\pi fT_H})X(f-\frac{n}{T_S}))+d (1 - sinc(\pi fT_H)e^{-j\pi fT_H})X(f) \end{equation}
where $d=T_H/T_S$