Does $$x[n]=e^{j\omega n}+e^{2j\omega n}$$ represent an eigenfunction of an LTI system?


1 Answer 1


Given an LTI system with impulse response $h(n)$, a complex sinusoid of the form $e^{j\omega_0n}$ is always an eigenfunction:

$$\mathcal{T}\{ e^{j\omega_0 n} \} = \sum_{k=-\infty}^{\infty} h(k) e^{j \omega_0 (n-k)} = e^{j \omega_0 n} \left( \sum_{k=-\infty}^{\infty} h(k) e^{-j \omega_0 k}\right) = H\left(e^{j\omega_0}\right) \cdot e^{j \omega_0 n}$$

Note, however, that the sum of eigenfunctions corresponding to different eigenvalues is not an eigenfunction. You can see this question for a demonstration in a generic vector space. In this case it can be easily seen that, because the system is LTI (in particular, linear), the sum of the respective eigenfunctions will appear at the output:

$$\mathcal{T}\{ e^{j\omega_0 n} + e^{j2\omega_0 n} \} = H\left(e^{j\omega_0}\right) \cdot e^{j \omega_0 n} + H\left(e^{j2\omega_0}\right) \cdot e^{j 2\omega_0 n} $$

As you can see, there is no way to express the output as $K\cdot\left(e^{j\omega_0 n} + e^{j2\omega_0 n}\right)$, except in the case where the two eigenvalues (as stated above, $H\left(e^{j\omega_0}\right)$ and $H\left(e^{j2\omega_0}\right)$ in this case) are the same.

  • $\begingroup$ Tendero, i approved your edit to the question, but it says it needs the approval of one more to take. in your answer, i am not too keen on your use of script $\mathcal{H}$ as it can be confused with a specific operator, the Hilbert Transform. $\endgroup$ Dec 23, 2017 at 4:20
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    $\begingroup$ @robertbristow-johnson Thanks for your observation, I replaced it with the $\mathcal{T}$ for Transfer. $\endgroup$
    – Tendero
    Dec 23, 2017 at 4:35

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