# realistic sampling - where am I wrong?

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 $$h(t) = \begin{cases} 1 & -0.5 find $$\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:

$$\begin{cases} k & |t| I substituted $$\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?

• I have no idea what this notation means. What is "$X^f(\theta)$"? What is $\theta$? Is $X(\cdot)$ being raised to the $f$ power? Why not just use standard electrical engineering notation? Commented Apr 20 at 21:35
• this is the notation I was taught for DTFT, $$X^f(\theta) = \sum_{-\infty}^{\infty}x[n] e^{-j\theta n}$$ Commented Apr 21 at 11:51

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.