realistic sampling - where am I wrong?

Question

I’m given a signal $x(t)$, it's convolved with $h(t)$ and sampled at rate T=1. The result is called $\tilde{x}[n]$.

For $$h(t) = \begin{cases} 1 & -0.5<t\le 0.5 \\ 0 & \text{else} \end{cases},$$ find $\tilde{X}^f(\theta)$ as an expression of $X^f(\theta)$.

My attempt

I noticed that I can write for a sampled signal $$\tilde{X}^f(\theta) = \frac{1}{T} \sum_{k=-\infty}^{\infty} \tilde{X}^F\left(\omega = \frac{(\theta - 2\pi k)}{T}\right).$$

All that I had left to do was simplify using the convolution to multiplication relation and this identity:

$$\begin{cases} k & |t|<t_0 \\ 0 & \text{else} \end{cases} \to 2kt_0 \text{sinc}(\omega t_0),$$ I substituted $\tilde{X}^F(\omega)$ into the expression.

Results

$$\sum_{k=-\infty}^{\infty} X^F(\theta - 2\pi k)\operatorname*{sinc}(0.5(\theta-2\pi k))$$

whereas what they got: $$\sum_{k=-\infty}^{\infty} X^F(\theta - 2\pi k)\operatorname*{sinc}\left(\frac{\theta-2\pi k}{2\pi}\right)$$

I'm losing my mind over the $\pi$ in the denominator, why is it there?

Answer 1

It is hard to follow what you did as your nomenclature is inconsistent and you didn't explicitly define any of it.

I believe the bit you got wrong was the Fourier transform of $h\left[t\right]$, with the correct expression for the more general case of a rectangular pulse of height $k$ centred on $t=0$ with width $t_0$ being: $$\frac{k}{t_0}{\rm sinc}\left[\frac{\omega}{2\pi t_0}\right]$$ Where $\omega$ is the frequency (radians per unit time).

Here is a full solution assuming that $X^f$ is the Fourier transform of the signal prior to convolution, $\bar X^f$ is the final discrete time Fourier transform, and $\theta$ is the frequency (radians per unit time):

Firstly, the Fourier transform of $h(t)$ is: $$H\left[f\right]={\rm sinc}\left[f\right]$$ Where $f$ is the frequency (cycles per unit time).

Secondly, using the Poisson summation formula and that convolution in the time domain is multiplication in the transform domain: $$\bar X^f\left[2\pi f\right]=\sum^\infty_{k=-\infty}X^f\left[2\pi\left(f-\frac{k}{T}\right)\right]{\rm sinc}\left[f-\frac{k}{T}\right]\\ =\sum^\infty_{k=-\infty}X^f\left[2\pi\left(f-k\right)\right]{\rm sinc}\left[f-k\right]$$ Where I have used that the sampling interval, $T$, is equal to $1$, and $\theta=2\pi f$.

Finally, rewriting in terms of $\theta$: $$\bar X^f\left[\theta\right]=\sum^\infty_{k=-\infty}X^f\left[\theta-2\pi k\right]{\rm sinc}\left[\frac{\theta}{2\pi}-k\right]\\ =\sum^\infty_{k=-\infty}X^f\left[\theta-2\pi k\right]{\rm sinc}\left[\frac{\theta-2\pi k}{2\pi}\right]$$

This matches the correct expression that you were given.