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In http://www.eng.ucy.ac.cy/cpitris/courses/ECE623/presentations/DSP-LECT-10-11-12.pdf, it says that when sinusoidal input $X(z)$ starts at n=0 (with n<0 having zero input) and the input passes through digitl filter $H(z)$, resulting in output z-transform $Y(z) = H(z)X(z)$, transient response is associated with poles of the filter while steady-state response is associated with the poles of the input. Is this true? If so, how does one prove this?

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Yes, that's true if the $\mathcal{Z}$-transform of the input signal has poles on the unit circle (i.e., the input is a constant or a sinusoid, starting at $n=0$), and if the filter is stable, i.e. all its poles are inside the unit circle. If you split up $Y(z)$ by a partial fraction expansion, you get a contribution for each pole. The contributions from the poles of $H(z)$ all decay to zero because we've assumed that the filter is stable, i.e. all its pole magnitudes are smaller than $1$, and consequently, these terms decay exponentially. On the other hand, the contributions from the poles of $X(z)$ do not decay because the corresponding poles lie on the unit circle. This means that after a while, all contributions associated with the poles of $H(z)$ will be very small (practically zero; transients), whereas the contributions from the poles of $X(z)$ remain at a constant amplitude (steady-state response).

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  • $\begingroup$ unit signal ? => unit "circle" you mean ( in the first line) $\endgroup$ – Fat32 Jun 4 '15 at 17:59

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