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I need help solving the following Fourier transform question.

Given,

$$ X_s(f) = \frac{1}{\Delta T} \sum_{n = -\infty}^{\infty} X\left(f - \frac{n}{\Delta T} \right) $$

$$ H(f) = \begin{cases} 1 & -\frac{1}{2 \Delta T} \leq f \leq \frac{1}{2\Delta T} \\ 0 & \text{otherwise}\end{cases} $$

Show that

$$ x = \mathcal F^{-1} \left\{ X_s H \right\} = \sum_{n = -\infty}^{\infty} x(n \Delta T)\text{Sinc}\left(\frac{t - n\Delta T}{ \Delta T}\right) $$

Attempt: Well I know from Fourier tables that

$$ \mathcal F \left( \sum_{n = -\infty}^{\infty} \delta (t - n\Delta T)\right) = \frac{1}{\Delta T} \sum_{n = -\infty}^{\infty} \delta \left( f - \frac{n}{\Delta T} \right) $$

$$ \mathcal F ( \text{Sinc}(at) ) = \frac{1}{|a|} \text{rect}\left(\frac{f}{a}\right) $$

where

$$ \text{rect}(f) = \begin{cases} 1 & -0.5 \leq f \leq 0.5 \\ 0 & otherwise \end{cases} $$

so I can set up the equation to be solved as

$$ x = \mathcal F^{-1} \left\{ X_sH\right\} = \mathcal F^{-1} \sum_{n = -\infty}^{\infty} X(f - \frac{n}{\Delta T} \text{rect}(\Delta f) $$

but I do not how I can use this information to get x.Any help will be appreciated.

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1 Answer 1

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Note that if $X(f)$ is the Fourier transform of $x(t)$, then $X_s(f)$ is the Fourier transform of the sampled version of $x(t)$:

$$X_s(f)=\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}x(n\Delta T)\delta(t-n\Delta T)\right\}=\mathcal{F}\left\{x_s(t)\right\}$$

$H(f)$ is an ideal low-pass filter with impulse response

$$h(t)=\frac{1}{\Delta T}\textrm{sinc}(t/\Delta T)$$

Since $\mathcal{F}^{-1}\{X_s(f)H(f)\}=(x_s*h)(t)$ we get

$$\mathcal{F}^{-1}\{X_s(f)H(f)\}=\frac{1}{\Delta T}\sum_{n=-\infty}^{\infty}x(n\Delta T)\textrm{sinc}\left(\frac{t-n\Delta T}{\Delta T}\right)$$

Note that in the result you gave in the question there is a factor $1/\Delta T$ missing.

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