Output of discrete-time LTI system guaranteed to be same form as input?

Question

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...

Answer 1

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.

Answer 2

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.

Answer 3

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 $a = e^{j \omega}$.