This Jupyter notebook is part of a collection of notebooks in the bachelors module Signals and Systems, Communications Engineering, Universität Rostock. Please direct questions and suggestions to Sascha.Spors@uni-rostock.de.
An eigenfunction of a system is defined as the input signal $x(t)$ which produces the output signal $y(t) = \mathcal{H}\{ x(t) \} = \lambda \cdot x(t)$ with $\lambda \in \mathbb{C}$. The weight $\lambda$ associated with $x(t)$ is known as scalar eigenvalue of the system. Hence, besides a weighting factor, an eigenfunction is not modified by passing through the system.
Complex exponential signals $e^{s t}$ with $s \in \mathbb{C}$ are eigenfunctions of linear time-invariant (LTI) systems. This can be proven by applying the properties of LTI systems. Lets assume a generic LTI system with input signal $x(t) = e^{s t}$ and output signal $y(t) = \mathcal{H}\{ x(t) \}$. The response of the LTI system to the shifted input signal $x(t-\tau) = e^{s (t-\tau)}$ reads
\begin{equation} y(t - \tau) = \mathcal{H}\{ x(t-\tau) \} = \mathcal{H}\{ e^{-s \tau} \cdot e^{s t} \} \end{equation}due to the implied shift-invariance. Now considering the implied linearity this can be reformulated as
\begin{equation} y(t - \tau) = e^{-s \tau} \cdot \mathcal{H}\{ e^{s t} \} = e^{-s \tau} \cdot y(t) \end{equation}It is straightforward to show that $y(t) = \lambda e^{st}$ fulfills above equation
\begin{equation} \lambda e^{s t} e^{-s \tau} = e^{-s \tau} \lambda e^{st} \end{equation}Example
An LTI system whose input/output relation is given by the following inhomogeneous linear ordinary differential equation (ODE) with constant coefficients is investigated
\begin{equation} a_0 y(t) + a_1 \frac{d y(t)}{dt} + a_2 \frac{d^2 y(t)}{dt^2} = x(t) \end{equation}with $a_i \in \mathbb{R} \quad \forall i$. In the remainder, the output signal $y(t)$ of the system is computed by explicit solution of the ODE for $x(t) = e^{s t}$ as input signal. Integration constants are discarded for ease of illustration.
import sympy as sym
%matplotlib inline
sym.init_printing()
t, s, a0, a1, a2 = sym.symbols('t s a:3')
x = sym.exp(s * t)
y = sym.Function('y')(t)
ode = sym.Eq(a0*y + a1*y.diff(t) + a2*y.diff(t, 2), x)
solution = sym.dsolve(ode)
solution.subs({'C1': 0, 'C2': 0})
Exercises
Example
The following inhomogeneous linear ODE with time-dependent coefficient is considered as an example for a time-variant but linear system
\begin{equation} t \cdot \frac{d y(t)}{dt} = x(t) \end{equation}The output signal $y(t)$ of the system for a complex exponential signal at the input $x(t) = e^{st}$ is computed by explicit solution of the ODE. Again integration constants are discarded.
ode = sym.Eq(t*y.diff(t), x)
solution = sym.dsolve(ode)
solution.subs('C1', 0)
Note, $\text{Ei}(\cdot)$ denotes the exponential integral. The response $y(t)$ of the time-variant system is not equal to a weighted complex exponential signal $\lambda \cdot e^{s t}$. It can be concluded that complex exponentials are no eigenfunctions of this particular time-variant system.
Example
A final example considers the following non-linear inhomogeneous ODE with constant coefficients
\begin{equation} \left( \frac{d y(t)}{dt} \right)^2 = x(t) \end{equation}as example for a non-linear but time-invariant system. Again, the output signal $y(t)$ of the system for a complex exponential signal at the input $x(t) = e^{st}$ is computed by explicit solution of the ODE. As before, integration constants are discarded.
ode = sym.Eq(y.diff(t)**2, x)
solution = sym.dsolve(ode)
[si.subs('C1', 0) for si in solution]
Obviously for this non-linear system complex exponential signals are no eigenfunctions.
The complex eigenvalue $\lambda$ constitutes the weight of a complex exponential signal $e^{st}$ (using complex frequency $s$) experiences when passing through an LTI system. It is commonly termed as transfer function and is denoted by $H(s)=\lambda(s)$. Using this definition, the output signal $y(t)$ of an LTI system for a complex exponential signal at the input reads
\begin{equation} y(t) = \mathcal{H} \{ e^{st} \} = H(s) \cdot e^{st} \end{equation}Note that the concept of the transfer function is directly linked to the linearity and time-invariance of a system. Only in this case, complex exponential signals are eigenfunctions of the system and $H(s)$ describes the properties of an LTI system with respect to these.
Above equation can be rewritten in terms of the magnitude $| H(s) |$ and phase $\varphi(s) = \arg \{ H(s) \}$ of the complex transfer function $H(s)$
\begin{equation} y(t) = | H(s) | \cdot e^{s t + j \varphi(s)} \end{equation}The magnitude $| H(s) |$ provides the frequency dependent attenuation/amplification of the eigenfunction $e^{st}$ by the system, while $\varphi(s)$ provides the introduced phase-shift.
In order to establish a link between the transfer function $H(s)$ and the impulse response $h(t)$, the output signal $y(t) = \mathcal{H} \{ x(t) \}$ of an LTI system with input signal $x(t)$ is considered. It is given by convolving the input signal with the impulse response
\begin{equation} y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(t-\tau) \cdot h(\tau) \; d\tau \end{equation}For a complex exponential signal as input $x(t) = e^{st}$, the output of an LTI system is given as $y(t) = \mathcal{H} \{ e^{st} \} = H(s) \cdot e^{st}$. Introducing both signals into the convolution integral yields
\begin{equation} H(s) \cdot e^{st} = \int_{-\infty}^{\infty} e^{st} e^{-s \tau} \cdot h(\tau) \; d\tau \end{equation}which after canceling $e^{s t}$ (the integral depends not on $t$) results in
\begin{equation} H(s) = \int_{-\infty}^{\infty} h(\tau) \cdot e^{-s \tau} \; d\tau \end{equation}under the assumption that the integral converges. The transfer function $H(s)$ can be computed from the impulse response $h(t)$ by integrating over the impulse response multiplied with the complex exponential function $e^{- s \tau}$. This constitutes an integral transformation, which is later introduced in more detail as Laplace transform. Usually the temporal variable $t$ is then used
\begin{equation} H(s) = \int_{-\infty}^{\infty} h(t) \cdot e^{-s t} \; d t \end{equation}rather than $\tau$ which remained from the convolution integral calculus.
Copyright
The notebooks are provided as Open Educational Resource. Feel free to use the notebooks for your own educational purposes. The text is licensed under Creative Commons Attribution 4.0, the code of the IPython examples under the MIT license. Please attribute the work as follows: Lecture Notes on Signals and Systems by Sascha Spors.