Multiplication term $ \frac{ 1}{T_s} $ in sampling theorem

Question

\begin{equation} X(\Omega) = \frac{ 1}{T_s} \sum ^{\infty}_{k=-\infty} X_a\left \lbrace \frac{\Omega /( 2 \pi) - k}{T_s}\right \rbrace \end{equation}

What is the purpose of multiplying sampled spectrum by sampling frequency ($ \frac{ 1}{T_s} $). Is it for making the equation dimensionally correct Or the purpose is scaling of the analog spectrum? Or is it averaging the effect of summation when CTFT spectrum get converted into DTFT spectrum?

When $ x_a(t) $ get sampled to $ x[k] $ dimensionally there is both entities are same. Accordingly when $ x_a (t) $ gets converted to $ X_a(\omega) $ and $ x[k] $ get converted to $X(\Omega)$ both spectrum values should agree the dimensional correctness.

Then why is there a $ \frac{ 1}{T_s} $ term ?

Edit(20220401):
To bring clarity and avoid confusion down bellow:

$X(\Omega)$ is spectrum of sampled signal (DTFT)
$X_a(\omega)$ is spectrum of original continuous time signal
$k$ is integer
$T_s$ is sampling time in seconds( 1/$F_s$)
$\Omega$ is normalized angular frequency (radians/sample) not angular frequency(radians/second).
Whatever inside the curly brackets ( argument of $X_a(.)$) is in dimension of frequency.

Answer 1

Later I figured out the necessity of the sampling term $T_s$ before the summation.
Lets assume unit of $x_a(t)$ is of dimension $[G]$ ($G$ may be in the units of mass, length, voltage or anything you can imagine) when you sample it uniformly using sampling time $T_s$ then the resulting digital signal $x[m]$ will also have the same dimension $[G]$.
When you calculate DTFT of $x[m]$ then also there is no Dimension change since exponential term is having no dimension and the dimension of $X(\Omega)$ will be remain $[G]$.

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

But the dimension changes will happen when you convert $x_a(t)$ into CTFT. Since the integration involves the term $dt$ of dimesion time $[T]$. So the dimension of $X_a(F)$ will be $[G][T]$

$$ X_a(F) = \int_{ - \infty}^{ \infty } x_a(t) e^{j 2 \pi F t } dt $$

So when you equate terms involving $X_a(F)$ and $X(\Omega)$ for dimensional equality the term $\frac{1}{T_s}$ have to be there.

$\underline{\text{Proof :}} $

Taken from Jeffrey A. Fessler ,EECS 451 Lecture Notes, Chapter 04, Section Proofs of Spectral Replication

$$ x[m] = x_a(m{T_s}) $$

sampling

$$ = \int_{- \infty}^{\infty} X_a(F) e^{j 2 \pi F (m T_s) } dF $$

inverse CTFT

$$ = \sum_{k=-\infty}^{\infty} \int_{2 \pi (k-1/2)/ T_s}^{2 \pi (k+1/2)/ T_s} X_a(F) e^{j 2 \pi F (m T_s) } dF $$

changing into summation of interval $2 \pi $

$$ = \sum_{k=-\infty}^{\infty} \int_{ - \pi}^{ \pi } X_a(\frac{\Omega /( 2 \pi) + k}{T_s}) e^{j 2 \pi (\frac{\Omega /( 2 \pi) + k}{T_s}) (m T_s) } \frac{d\Omega}{2 \pi T_s } $$

change in variable using $ \Omega = 2 \pi F {T_s} - 2 \pi {k} $ and then the F will be $ F = \frac{\Omega / (2 \pi) + k }{T_s} $

$$ = \frac{1}{2 \pi}\int_{ - \pi}^{ \pi } \bigg[ \frac{1}{T_s} \sum_{k=-\infty}^{\infty} X_a(\frac{\Omega /( 2 \pi) + k}{T_s}) \bigg] e^{j \Omega m } {d\Omega} $$

The bracketed expression must be $X (\Omega)$ since the integral is the inverse DTFT formula:

$$ x[m] = \frac{1}{2 \pi}\int_{ - \pi}^{ \pi } X(\Omega) e^{j \Omega m } {d\Omega} $$

Thus, when $x[m] = x_a (m T_s )$ (regardless of whether $x_a (t)$ is bandlimited) we have:

$$ X(\Omega) = \frac{ 1}{T_s} \sum_{k=-\infty}^{\infty} X_a\{\frac{\Omega /( 2 \pi) - k}{T_s}\} $$

units of bracketed term $X_a(.)$ is in units of frequency (Hz). Proof in terms of angular frequency $\Omega$ is also similar, there also $\frac{1}{T_s}$ term will come before the summation.

Answer 2

This is the equation I get:

\begin{equation} X(\omega) = \sum ^{\infty}_{k=-\infty} X_a\left \lbrace \frac{\omega - 2\pi k}{T_s}\right \rbrace \end{equation}

Here I used the convention of lower case to represent normalized frequencies, determined by dividing the frequency in cycles/sec or radians/sec by the sampling rate to get cycles/sample or radians/sample.

The above relationship assumes using $X_a(\Omega)$ with $\Omega$ as the angular frequency in radians/sec. It appears the equation posted by the OP is using $X_a(F)$ with $F$ in cycles/sec or Hz, and scaling the output by the sampling rate. If that is what was intended, then the equation appears correct for that purpose- but that should be specified.

My assumption is the equation is describing the result of sampling in time (at sampling rate $F_s = 1/T_s$) creating periodicity in frequency, here over index $k$.

Similarly, as a function of $\Omega$ in radians/sample we would get:

$$X(\Omega) = \sum ^{\infty}_{k=-\infty} X_a\left \lbrace {\Omega - \frac{2\pi k}{T_s}\right \rbrace $$

The difference between this and the OP’s answer below is I am assuming “sampling” is done as a product with an impulse train while the alternate approach uses discrete time approximation of integration in that each “sample” is the area under the curve over interval $T_s$ and dividing by $T_s$ provides the average amplitude over that time interval.

After sampling, we expect the analog spectrum from $F=0$ (in Hz) to $F=F_s$ (in Hz) to periodically repeat as $f$ extends to $\pm \infty$.

$\omega = 2\pi f$ is the continuous frequency domain in units of radians/sample and here extending periodically to $\pm \infty$. $f= F/F_s$ and is the normalized frequency domain in units of cycles/sample.

$X_a(\Omega)$ is the analog spectrum, with $\omega= 2\pi F$ in units of radians/sec. (And bandlimited to zero outside the range from $-2\pi F_s/2$ to $+2\pi F_s/2$ if we wish to avoid aliasing)

$X(\omega)$ is the resulting continuous and periodic spectrum of the discrete time signal that is periodic over any interval from $(k-1) 2\pi$ to $k 2 \pi$ for any $k$. (Integer $k$ used in the original formula), with $\omega$ in units of radians/sample.

With that, we are mapping the normalized frequency units of $\omega$ as radians/sample used for the discrete time spectrum $X(\omega)$ to the frequency units of $\Omega$ as radians/second and capturing the additive effects of the aliasing and periodicity over the $kf_s$ intervals in $\Omega$.

This is given by

$$\omega \rightarrow \Omega$$

$$\omega \rightarrow \frac{\omega-2\pi}{T_s}$$

With that we get the expected periodic spectrum for $X(\omega)$ as we index through each $k$.

This would result in the resulting spectrum having the same scale as the original spectrum at all the original frequency locations. If we were to multiply this result by $1/Ts$ then that would scale the result by the sampling rate, which alone does not make sense to me as an artifact of the sampling process specifically: Sampling in time is equivalent to multiplying the time domain waveform with a stream of impulses spaced by the sampling rate, and each with an area of 1. This product in time is a convolution in frequency resulting in a copy of the original spectrum with the same scale centered around the frequency of each of those impulses (which are spaced by the sampling rate), with no further scaling.