# Is the sum of discrete sinusoids an eigenfunction of an LTI system?

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

Given an LTI system with impulse response $h(n)$, a complex sinusoid of the form $e^{j\omega_0n}$ is always an eigenfunction:
$$\mathcal{T}\{ e^{j\omega_0 n} \} = \sum_{k=-\infty}^{\infty} h(k) e^{j \omega_0 (n-k)} = e^{j \omega_0 n} \left( \sum_{k=-\infty}^{\infty} h(k) e^{-j \omega_0 k}\right) = H\left(e^{j\omega_0}\right) \cdot e^{j \omega_0 n}$$
$$\mathcal{T}\{ e^{j\omega_0 n} + e^{j2\omega_0 n} \} = H\left(e^{j\omega_0}\right) \cdot e^{j \omega_0 n} + H\left(e^{j2\omega_0}\right) \cdot e^{j 2\omega_0 n}$$
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.
• 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. – robert bristow-johnson Dec 23 '17 at 4:20
• @robertbristow-johnson Thanks for your observation, I replaced it with the $\mathcal{T}$ for Transfer. – Tendero Dec 23 '17 at 4:35