# Relationship Between Sampled Continuous and Discrete Time Signals

Consider the sketched system below. $x_c(t)$ is an arbitrary, continuous-time signal at the input and $s(t)$ is an impulse train, defined as $s(t)=\sum_{n=-\infty}^{\infty} \delta(t-nT)$, where T is the sampling period. Hence:

$$x_s(t) = x_c(t)s(t)=\sum_{n=-\infty}^{\infty} x_c(nT)\delta(t-nT)$$

Now I calculate and compare the fourier transform of $x_s(t)$ and $x_c(nT)$.

$$X_s(j\Omega)=\int_{-\infty}^{\infty} x_s(t)e^{-j\Omega t}dt=\int_{-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x_c(nT)\delta(t-nT)e^{-j\Omega t}dt=\sum_{n=-\infty}^{\infty}x_c(nT)e^{-j\Omega nT}$$

$$X(e^{j\omega})=\sum_{n=-\infty}^{\infty} x[n]e^{-j\omega n}$$

The results look quite similar. In fact, they are equal if $\omega=\Omega T$.

The calculation is no problem, but the issue I have is understanding what this really means. So obviously it is a scaling. If I define the angular frequency as $\Omega_s=\frac{2\pi}{T}$ and plug it into the above formula for $\omega$, I get $\omega_s=2\pi$. So it seems that through the process of discretization the actual frequency information is lost and the fourier transform goes from 0 to $2\pi$. In order to retrieve the information, I need to multiply $\omega$ with the sampling frequency. Is this statement correct? Why does this actually happen? What meaning is behind this? Probably I am missing an obvious thing, I feel like missing the forest for the trees.

• sampling in time domain corresponds to periodizing in frequency domain – user1952009 Dec 27 '16 at 17:48
• Yep, but I am talking about the discretization step. $X_s(j\omega)$ is already periodic. I am wondering about the scaling $\omega=\Omega T$. – Daiz Dec 27 '16 at 17:53
• What you call $x_c(nT)$ and $x[n]$ is unclear. And everything is in the link. – user1952009 Dec 27 '16 at 18:23
• To address the question about losing the sampling frequency information, this should be no surprise as you lose the sampling interval information when forming $x[n] = x_c(nT)$. This is only a function of $n$ (sample "number") and the time "units" have been lost. – Bungo Dec 27 '16 at 18:37
• do you want to know why the Laplace transform of the sampled continuous-time signal is essentially the Z transform of the discrete-time sample? that's pretty straight forward. or do you want to know why sampling in the time domain causes the spectrum in the frequency domain to be periodically extended? (or sampling the spectrum in the frequency domain causes periodic extension to the time domain signal?) that comes right outa the sampling theorem or Poisson's summation formula. – robert bristow-johnson Dec 27 '16 at 23:58

$$\omega=\Omega T\tag{1}$$
where $\Omega$ is the angular frequency ($\Omega=2\pi f$), $T=1/f_s$ is the sampling period ($f_s$ is the sampling frequency), and $\omega$ is the normalized angular frequency, which is often used as the independent frequency variable when describing spectra of discrete-time signals. Note that you could also use $\Omega$ as the independent variable for discrete-time signals, but it always appears together with the sampling interval $T$. This is an important property of sampled or discrete-time signals, and that's also why the two variables $\Omega$ and $T$ are commonly combined into a single variable $\omega=\Omega T=2\pi f/f_s$.
Note that sampling results in a periodization of the spectrum, and this is why information can be lost. This is true for both spectra $X_s(j\Omega)$ and $X(e^{j\omega})$. The periodicity of the spectra implies that the fundamental frequency interval $f\in [-f_s/2,f_s/2]$, or, equivalently, $\Omega\in [-\pi/T,\pi/T]$ or $\omega\in [-\pi,\pi]$ contains all information of the sampled signal, and all frequencies outside that interval are redundant. This is an inherent property of sampled (or discrete-time) signals, and it is the cause of aliasing, i.e. the mapping of different continuous-time signals to the same sampled signal.