What is the expression for an oversampled signal?

Question

Problem: A simplified schematic of an oversampling circuit is shown below. A data sequence, \$x[n]\$, is transmitted by REG1 which is in clock domain \$C_{tx}\$. \$x[n]\$ is oversampled by REG2 in the clock domain \$C_{rx}\$, resulting in the sample sequence \$y[m]\$. The clock \$C_{tx}\$ has frequency \$f_{tx}\$ and the clock \$C_{rx}\$ has frequency \$f_{rx}\$ and the oversampling rate is defined as \$ \beta \in \mathbb{R} \$.

^{simulate this circuit – Schematic created using CircuitLab}

Definitions:

1. \$n = 0,1,2,\ldots \$
2. \$m = 0,1,2,\ldots \$
3. \$ \beta = f_{rx}/f_{tx} \$

Question:

What is \$y[m]\$, expressed in terms of \$x, n, m\$ and/or \$ \beta \$? Assume \$ \beta \geq 1\$. Also, REG1 and REG2 may be treated as zero-delay models: \$T_{propagation} = T_{setup} = T_{hold} = 0\$. For simplicity, you may assume \$\phi_{tx}(0) = \phi_{rx}(0) = 0\$, where \$\phi_{tx}(t)\$ is the phase of \$C_{tx}\$ at time \$t\$, and \$\phi_{rx}(t)\$ is the phase of \$C_{rx}\$ at time \$t\$.

All answers are welcome, but the accepted answer must provide analytical proof of the proposed expression. Intuitive explanations or a few evaluated cases are not analytical proof.

Answer 1

I go with the answer of Vasiliy and proofed it.

$$y\left[m\right] = x\left[{\Big\lfloor{\frac{m}{\beta}}\Big\rfloor}\right]$$

As I really can't think while writing TeX and you may need this before the end of the month, I wrote on a tablet.

The solution is to describe the input signal x in the time domain. Then the output signal y is derived from x using natural sampling. Based on the knowledge about the clock domains (

• synchronized at time zero and
• receiver clock equal or faster then transmitter clock
• both clocks are exactly periodically

), the expression can be simplified.

Edit: The clock relation expression $$n\left[m\right] = {\Big\lfloor{\frac{m}{\beta}}\Big\rfloor}$$ is not really obvious, but can be expressed very well using system theory. The idea is to describe the clock cycle count of the TX clock (TX clock value) as an integer value signal over time. This signal can be expressed using the floor function.

Then this signal is sampled using natural sampling at the positions of the RX clock events. In this way the counter value n of the slower TX clock can be expressed for all RX clock event positions mT₁. Because the positions only depend on m, n can be expressed as a function of m.

Answer 2

Maybe this:

$$y\left[m\right] = x\left[{\Big\lfloor{\frac{m}{\beta}}\Big\rfloor}\right]$$

Draw one or two exmples with different Betha to perform a sanity check.