Why are those two systems equivalent?

  • System 1:$$\boxed{x[n]}\longrightarrow\boxed{\uparrow 3}\longrightarrow\boxed{\downarrow 4} \longrightarrow\boxed{y[n]}$$

  • System 2:$$\boxed{x[n]}\longrightarrow\boxed{\downarrow 4}\longrightarrow\boxed{\uparrow 3} \longrightarrow\boxed{y[n]}$$

By drawing in time the first system's output my result for the first $4$ samples: $$y[0]=x[0], y[1]=0,y[2]=0,y[3]=x[12]$$

by drawing in time the second system's output my result for the first $4$ samples: $$y[0]=x[0], y[1]=0,y[2]=0,y[3]=x[4]$$

Why the statement is correct?

  • 1
    $\begingroup$ your output sample for the first system y[3]=x[12] is wrong, it should be y[3]=x[4]. $\endgroup$ – Fat32 Jun 20 '16 at 15:04

Upsampling (by $M$) and downsampling (by $N$) are interchangeable if and only if $M$ and $N$ are co-prime (i.e., they have no common factors other than $1$). This is the case in your example because $2$ and $3$ are co-prime.

Think about your specific example in the time domain. Introduce a signal $u[n]$ between the upsampler and the downsampler. In the first case you have

$$u[n]=\begin{cases}x[n/3],& n=3k,\quad k\in\mathbb{Z}\\0,&\text{otherwise}\end{cases}$$


$$y[n]=u[4n]=\begin{cases}x[4n/3],& 4n=3k,\quad k\in\mathbb{Z}\\0,&\text{otherwise}\end{cases}\tag{1}$$

In the second case you have



$$y[n]=\begin{cases}u[n/3],& n=3k,\quad k\in\mathbb{Z}\\0,&\text{otherwise}\end{cases}\tag{3}$$

With $(2)$ you get

$$y[n]=\begin{cases}x[4n/3],& 4n=3k,\quad k\in\mathbb{Z}\\0,&\text{otherwise}\end{cases}\tag{4}$$

which is identical with $(1)$.


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