Eigenfunction Property of LTI Systems

Why Complex Exponentials Are Eigenfunctions of Every LTI System

Eigenfunction of an LTI System

For any LTI system, complex exponentials are eigenfunctions.

If the input is:

$$ x(t) = e^{j\omega_0 t} $$

then the output is:

$$ y(t) = H(j\omega_0)\, e^{j\omega_0 t} $$

where \(H(j\omega_0)\) is the system’s frequency response.

Why is it called an eigenfunction?

Because it satisfies the eigenvalue equation:

$$ T\{x(t)\} = \lambda x(t) $$

• Eigenfunction → \( e^{j\omega_0 t} \)
• Eigenvalue → \( H(j\omega_0) \)

So the eigenvalue is not \( e^{j\omega_0 t} \). The eigenvalue is the scalar \( H(j\omega_0) \).

What about \( e^{j\omega_0 t} u(t) \)?

The \(u(t)\) (unit step) is added to make the signal causal:

$$ e^{j\omega_0 t} u(t) $$

This is often used in Laplace transform analysis:

$$ \mathcal{L}\{e^{j\omega_0 t} u(t)\} = \frac{1}{s - j\omega_0} $$

Big Picture

There are two different meanings of eigenvalues in LTI systems:

State-space eigenvalues

• Eigenvalues of matrix \(A\)
• Determine stability and natural modes

System (operator) eigenvalues

• For input \( e^{j\omega t} \)
• Output is \( H(j\omega) e^{j\omega t} \)
• Eigenvalue = \( H(j\omega) \)

Step 1: What is an eigenfunction?

For a system \(T\{\cdot\}\), if

$$ T\{x(t)\} = \lambda x(t) $$

• \(x(t)\) = eigenfunction
• \(\lambda\) = eigenvalue

The system only scales the signal — it does not change its shape.

Step 2: Start from LTI system definition

Any continuous-time LTI system is defined by convolution:

$$ y(t) = x(t) * h(t) $$

$$ y(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau $$

Step 3: Apply complex exponential input

Let:

$$ x(t) = e^{j\omega_0 t} $$

Substitute into convolution:

$$ y(t) = \int_{-\infty}^{\infty} e^{j\omega_0 \tau} h(t-\tau) d\tau $$

Rewrite:

$$ e^{j\omega_0 \tau} = e^{j\omega_0 t} e^{-j\omega_0 (t-\tau)} $$

Factor out \( e^{j\omega_0 t} \):

$$ y(t) = e^{j\omega_0 t} \int_{-\infty}^{\infty} h(t-\tau)e^{-j\omega_0 (t-\tau)} d\tau $$

Step 4: Important Observation

The integral does not depend on \(t\).

It becomes:

$$ H(j\omega_0) = \int_{-\infty}^{\infty} h(\lambda) e^{-j\omega_0 \lambda} d\lambda $$

This is the Fourier Transform of \(h(t)\).

Summary

$$ y(t) = H(j\omega_0) e^{j\omega_0 t} $$

• Input = \( e^{j\omega_0 t} \)
• Output = scaled version of same signal
• Scaling factor = \( H(j\omega_0) \)

Physical Meaning

If you input a pure sinusoid:

$$ \cos(\omega_0 t) $$

• Amplitude changes
• Phase changes
• No new frequency is created

This is the foundation of:

• Fourier Transform
• Frequency response
• Bode plots
• Filter design
• Communication systems

Discrete-Time Version

Input:

$$ x[n] = e^{j\omega_0 n} $$

Output:

$$ y[n] = H(e^{j\omega_0}) e^{j\omega_0 n} $$

Same principle applies.