Why are complex exponentials e^{j ω0 t} eigenfunctions of LTI systems, and what is the eigenvalue?

Why are complex exponentials e^{j ω0 t} eigenfunctions of LTI systems, and what is the eigenvalue?

• A. Because they are orthogonal; eigenvalue is zero.
• B. Because convolution with h(t) yields y(t) = (H(jω0)) e^{jω0 t}; eigenvalue is H(jω0).
• C. Because they are energy signals; eigenvalue is energy.
• D. Because they are time-limited; eigenvalue is impulse.

Answer: (B)

Complex exponentials are eigenfunctions of LTI systems because the system treats each frequency component independently and simply scales it. If the input is x(t) = e^{j ω0 t}, the output is the convolution with the impulse response h(t): y(t) = h * x = ∫ h(τ) e^{j ω0 (t−τ)} dτ = e^{j ω0 t} ∫ h(τ) e^{−j ω0 τ} dτ = H(jω0) e^{j ω0 t}, where H(jω) = ∫ h(τ) e^{−j ω τ} dτ is the Fourier transform of the impulse response, i.e., the system’s frequency response at ω0. Thus e^{j ω0 t} comes out scaled by H(jω0); that scaling factor is the eigenvalue corresponding to this eigenfunction. In other words, each frequency component is amplified or attenuated by the frequency response, with no change in its exponential form.

Complex exponentials are eigenfunctions of LTI systems because the system treats each frequency component independently and simply scales it. If the input is x(t) = e^{j ω0 t}, the output is the convolution with the impulse response h(t): y(t) = h * x = ∫ h(τ) e^{j ω0 (t−τ)} dτ = e^{j ω0 t} ∫ h(τ) e^{−j ω0 τ} dτ = H(jω0) e^{j ω0 t}, where H(jω) = ∫ h(τ) e^{−j ω τ} dτ is the Fourier transform of the impulse response, i.e., the system’s frequency response at ω0. Thus e^{j ω0 t} comes out scaled by H(jω0); that scaling factor is the eigenvalue corresponding to this eigenfunction. In other words, each frequency component is amplified or attenuated by the frequency response, with no change in its exponential form.