# amplitude of upsampled and downsampled signal without filter

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 ..

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.