# Fourier transform of sampling function

Calculate the Fourier transform of $f_{ZOH}$ (the zero-order hold reconstruction of a sampled signal).

Where $f_{ZOH} (t)= f(kT), \ \ kT \leq t < (k+1)T,$ and the sampled signal is

$$f_s = f(t) \sum^\infty_{-\infty} \delta (t - kT).$$

Attempt

First writing the $f_{ZOH} (t)$ as a convolution between impulse response and sampled signal:

$$f_{ZOH} (t) = \left( \sum^{\infty}_{k=-\infty} f(kT) \delta(t-kT) \right) * h(t) = \sum^{\infty}_{k=-\infty} f(kT)h(t-kT).$$

But the shape function for the impulse response is the rectangular pulse function:

$$f_{ZOH} (t) = \sum^{\infty}_{k=-\infty} f(kT) \Pi (t-kT).$$

Now we want to calculate the Fourier transform of $f_{ZOH}$:

$$FT \Big[ f_{ZOH} \Big] = \frac{1}{T} F(\nu ) \sum^{\infty}_{k=-\infty} \Pi \left( \nu - \frac{k}{T} \right) = \frac{1}{T} \sum^{\infty}_{k=-\infty} F\left(\frac{k}{T}\right) FT \left[ \Pi \left(\nu - \frac{k}{T}\right) \right]$$

$$= \frac{1}{T} \sum^{\infty}_{k=-\infty} F\left(\frac{k}{T}\right) \textrm{sinc} \left(\nu - \frac{k}{T}\right).$$

But I was told that this is not the correct answer. What is wrong here?

Any explanation would be greatly appreciated.

The Fourier transform of the sampled signal $$f_s = f(t) \sum^\infty_{-\infty} \delta (t - kT)$$ is \begin{align} F_s(\nu)&=F(\nu)*\mathcal F\left\{\sum^\infty_{k=-\infty} \delta (t - kT)\right\}\\ &=F(\nu)*\sum^\infty_{k=-\infty} \frac{1}{T}\delta \left(\nu - \frac{k}{T}\right)\\ &=\frac{1}{T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right) \end{align} so the Fourier transform $F_{\mathrm{ZOH}}(\nu)$ of $f_{\mathrm{ZOH}}(t)=f_s(t)*h_{\mathrm{ZOH}}(t)$ is $$F_{\mathrm{ZOH}}(\nu)=\mathcal F\left\{(f_s*h_{\mathrm{ZOH}})(t)\right\}=F_s(\nu)H_{\mathrm{ZOH}}(\nu)=H_{\mathrm{ZOH}}(\nu)\frac{1}{T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right)$$ The Fourier transform $\mathcal F\left\{f_{\mathrm{ZOH}}(t)\right\}$ can be also evaluated as \begin{align} F_{\mathrm{ZOH}}(\nu)&=F(\nu)*\mathcal F\left\{\sum^\infty_{k=-\infty} h_{\mathrm{ZOH}}(t - kT)\right\}\\ &=F(\nu)*\sum^\infty_{k=-\infty} \frac{1}{T} H_{\mathrm{ZOH}}\left(\frac{k}{T}\right)\delta\left(\nu - \frac{k}{T}\right)\\ &=\sum^\infty_{k=-\infty} \frac{1}{T} H_{\mathrm{ZOH}}\left(\frac{k}{T}\right)F\left(\nu - \frac{k}{T}\right) \end{align} so we have $$F_{\mathrm{ZOH}}(\nu)=H_{\mathrm{ZOH}}(\nu)\frac{1}{T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right)=\sum^\infty_{k=-\infty} \frac{1}{T} H_{\mathrm{ZOH}}\left(\frac{k}{T}\right)F\left(\nu - \frac{k}{T}\right)$$ For the ZOH we use $$h_{\mathrm{ZOH}}(t)=\begin{cases}1 & 0\le t\le T\\ 0 & \text{otherwise}\end{cases}$$ with Fourier tansform $$H_{\mathrm{ZOH}}(\nu)=\frac{\sin(\pi\nu T)}{\pi\nu}\mathrm{e}^{-\mathrm j \pi\nu T}=T\mathrm{sinc(\pi\nu T)}\mathrm{e}^{-\mathrm j \pi\nu T}$$ and then $$F_{\mathrm{ZOH}}(\nu)=\mathrm{sinc(\pi\nu T)}\mathrm{e}^{-\mathrm j \pi\nu T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right)$$