amplitude of upsampled and downsampled signal without filter

Question

Given: $$ DTFT\{x[n]\}=X(\omega)= \begin{cases} 1 & |\omega| \leq 2/\pi \\ 0 & 2/ \pi < |\omega| < \pi \end{cases}\ \ \ \ \ (periodic\ 2\pi) $$

If I downsample $X(\omega)$ by M. I get:

$$ \begin{aligned} x_d[n] &= x[nM] \\ \\ X_d(\omega) &= 1/M\ X(\omega/M) \end{aligned} $$

Now what happens if I up-sample $x[n]$ again by $L$?

why can't I just let $M = 1/L$ and substitute into down-sampling equation to get up-sampling equation?

$$ \begin{aligned} x_i[n] &= x[n/L] \\ \\ X_i(\omega) &= L\ X(L \omega) \end{aligned} $$

I'm a little bit confused because my book is telling me that upsampling equation doesn't scale the amplitude by L:

$$ X_i(\omega) = X(L\omega) $$

Just wondering what happened to the L term when upsampling?

(page 112, Schaum's Outline, Digital Signal Processing, Second Edition)

If I look up DTFT time scaling property, I find:

$$ DTFT\{\ x[an]\ \} = 1/a X(\omega / a) $$

Which seems to confirm that upsampler should have an L term ..

Answer 1

Close, but not the right formula for the up-sampler:

$$x_i[n] = x[n/L]$$

The "up sampler" is really an expander that inserts zeros between the samples of x[n], and it has the formula:

$$ x_u[n] =\begin{cases} x[n/L] & n=0, \pm L, \pm 2L, ... \\ \\ 0 & otherwise \end{cases} $$

The DTFT of the up-sampler "expander" can be derived as follows:

using $\delta(n)$ we can rewrite $x_u[n]$ as:

$$ x_u[n] = \sum_{k=-\infty}^{\infty}x[k]\delta[n-kL] $$

Applying definition of DTFT to $x_u[n]$:

$$ \begin{aligned} X_u(\omega) &= \sum_{n=-\infty}^{\infty} \left(x_u[n] e^{-j\omega n} \right) \\ \\ X_u(\omega) &= \sum_{n=-\infty}^{\infty} \left( \sum_{k=-\infty}^{\infty}x[k]\delta[n-kL] \right) e^{-j\omega n} \\ \end{aligned} $$

Applying sifting property of $\delta$:

$$ \begin{aligned} X_u(\omega) &= \sum_{n=-\infty}^{\infty} x[k] e^{-j\omega L k} \\ \\ X_u(\omega) &= X(\omega L) \end{aligned} $$

Thus, the gain of up-sampler "expander" in freq domain is unity.