I am trying to get equivalent of this system using the Noble Identities:

$$x[n] \rightarrow \boxed{H(z^L)} \rightarrow \boxed{\uparrow L} \rightarrow y[n]$$

my answer is

$$x[n] \rightarrow \boxed{\uparrow L} \rightarrow \boxed{H(z^{L\cdot L})} \rightarrow y[n]$$

Is this answer correct?

  • 1
    $\begingroup$ Yes, this is a correct answer. $\endgroup$
    – jojeck
    Commented Mar 2, 2015 at 23:07


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