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I know that in the continuous-time context, if I supply a complex exponential input to a Linear Time Invariant system, the output will be of the same form as the input - for example, if the input is $x(t)=Ae^{j \omega_0 t+\phi}$, the output will be of the form $y(t)=|H(j\omega_o)|\centerdot A \centerdot e^{j(H(s)\omega_0 t+\phi +\theta_H)}$ where $H(s)|_{s=j\omega_0} = |H(j\omega_0)|e^{j\theta_H}$ is the transfer function of the system evaluated at the frequency $j\omega_0$. My point is that the system does not modify the input frequency.

Does the same hold true in the discrete-time context? If I supply an input $x[n]=a^n$ to an LTI system, will the output also contain $a^n$? Or is it possible to supply $a^n$ and get the output $b^n +c^n$ where $a\ne b\ne c$ (for example, $x[n]=(1/6)^n$, $y[n]=B(1/2)^n + C(1/3)^n$)?

Edit: I've posted a related question that contains the issue that actually prompted this question...

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Despite the two other answers, I'll add one more, because I have the feeling that your final paragraph has not yet been answered satisfactorily.

A discrete-time LTI system with impulse response $h[n]$ has a transfer function

$$H(z)=\sum_{n=-\infty}^{\infty}h[n]z^{-n}\tag{1}$$

Its output $y[n]$ for a given input sequence $x[n]$ is given by the convolution sum

$$y[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\tag{2}$$

For $x[n]=a^n$ we get from $(2)$

$$y[n]=\sum_{k=-\infty}^{\infty}h[k]a^na^{-k}=a^n\sum_{k=-\infty}^{\infty}h[k]a^{-k}=a^nH(a)\tag{3}$$

where Eq. $(1)$ was used in the last equality. Eq. $(3)$ shows that for $x[n]=a^n$, the output of a discrete-time LTI system is just a weighted version of the input signal. Hence, the sequence $a^n$ is an eigenfunction of a discrete-time LTI system.

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  • $\begingroup$ I've posted a related question that I probably should have asked in the first place - would appreciate your take on that one as well... $\endgroup$ – Westerley Aug 6 '15 at 16:21
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The point is that a complex exponential function is an eigenfunction of an LTI system link. Therefore the frequency is not changing, i.e., the LTI system translates to a (complex-valued) gain $H(j\omega_0)$.

For discrete systems, the eigenfunctions are sampled exponential functions link, i.e., $x[n] = A\cdot e^{j\omega_n n}$, where $\omega_0=2\pi f/f_s$ is a normalized frequency. The output of a discrete LTI system is then $y[n]=H(e^{j\omega_n})\cdot A \cdot e^{j\omega_nn}$.

EDIT: A discrete input signal x[n] can be written as a weighted sum of complex exponentials, i.e., $X(e^{j\omega}) = \sum_{n}x[n]e^{j\omega}$ link. This transformation is in fact a change of basis. Driving such a signal thru an LTI system will weight each exponential function with the complex-valued gain $H(e^{j\omega})$. Thus, no additional frequencies will be introduced.

Note (a bit out of scope, but good to realize): When dealing with sampling a contuous-time signal and dealing with finite length signals, then you may observe additional frequencies due to aliasing and windowing.

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  • $\begingroup$ So then, is it possible for a discrete-time LTI system to produce a different form output than input (as per the last paragraph of my question)? $\endgroup$ – Westerley Aug 6 '15 at 1:39
  • $\begingroup$ If you can decompose your input signal into its projection onto some orthogonal set of eigenfunctions (e.g. an orthogonal basis made up of complex exponential functions), then the system's output will contain scaled versions of each of those basis functions. This is the property that you're alluding to, and is valid for both continuous and discrete-time systems. This model of an LTI system as a linear operator is analogous to how one might interpret multiplication by a matrix via its spectral decomposition. $\endgroup$ – Jason R Aug 6 '15 at 2:11
  • $\begingroup$ Ahso! So, as per Matt L. answer above, since $a^n$ is by itself an eigenfunction of a discrete-time LTI system, there is no way/need for an output of $b^n + c^n$ to created with an input of $a^n$. $\endgroup$ – Westerley Aug 6 '15 at 15:11
  • $\begingroup$ Indeed. You can also have a look at the following link on the z-transform, where also Region of Convergence for your sequence is discussed. $\endgroup$ – Brian Aug 6 '15 at 15:45
  • $\begingroup$ I've posted a related question that I probably should have asked in the first place - would appreciate your take on that one... $\endgroup$ – Westerley Aug 6 '15 at 16:21
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it turns out that, solely because it's Linear and Time-Invariant, an LTI discrete-time system has output $y[n]$ mapped from input $x[n]$ using convolution:

$$ y[n] = \sum\limits_{i=-\infty}^{+\infty} x[i] \, h[n-i] $$

where $h[n]$ is the impulse response of the due to an input that is the Kronecker delta

$$ \delta[n] = \begin{cases} 1, & \text{if }n = 0 \\ 0, & \text{if }n \ne 0 \end{cases} $$

so, if $x[n]=\delta[n]$, then $y[n]=h[n]$.

now, if $x[n]=e^{j \omega n}$, it is not terribly hard ('specially for discrete summation rather than the convolution integral for continuous-time systems) to show that if

$$x[n]=e^{j \omega n}$$

then

$$ y[n] = H(e^{j \omega}) e^{j \omega n} = H(e^{j \omega}) x[n]$$

where

$$ H(z) = \sum\limits_{n=-\infty}^{+\infty} h[n] \, z^{-n} $$

because i'm lazy (and need to get going), i'm happy for anyone to edit this answer to include and explicit derivation. but it's easy. then substitute $a = e^{j \omega}$.

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  • $\begingroup$ This I'm aware of, however what is not clear to me is if all that means a discrete-time LTI system can (or can not) produce an output that is of different form than the input, as indicated in the last paragraph of my question... $\endgroup$ – Westerley Aug 6 '15 at 1:42
  • $\begingroup$ Having understood what the other posters responded, now I see what you were getting at - thanks! $\endgroup$ – Westerley Aug 6 '15 at 15:20
  • $\begingroup$ I've posted a related question that I probably should have asked in the first place - would appreciate your take on that one... $\endgroup$ – Westerley Aug 6 '15 at 16:22

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