A Very Special Function for LTIC Systems: The Everlasting Exponential
Sources:
- B. P. Lathi & Roger Green. (2021). Chapter 2: Time-Domain Analysis of Continuous-Time Systems. Signal Processing and Linear Systems (2nd ed., pp. 193-195). Oxford University Press.
In this section I will illustrate that, for a system specified by the differential equation \[ \begin{equation} \label{eq_2_2} Q(D) y(t)=P(D) x(t) , \end{equation} \]
its transfer function is \(H(s)\), the bilateral Laplace transform of \(h(t)\), which is the unit impulse response of the system, and that \(H(s)\) also satisfies: \[ \color{teal} {H(s)=\frac{P(s)}{Q(s)}} . \]
The response to the everlasting exponential
There is a very special connection of LTIC systems with the everlasting exponential function \(e^{s t}\), where \(s\) is a complex variable, in general.
We now show that the LTIC system's (zero-state) response to everlasting exponential input \(e^{s t}\) is also the same everlasting exponential (within a multiplicative constant). Moreover, no other function can make the same claim (I think it is not proved here).
Proof:
Note that we are talking here of an everlasting exponential (or sinusoid), which starts at \(\ell=-\infty\).
If \(h(t)\) is the system's unit impulse response, then system response \(y(t)\) to an everlasting exponential \(e^{s t}\) is given by \[ y(t)=h(t) * e^{s t}=\int_{-\infty}^{\infty} h(\tau) e^{s(t-\tau)} d \tau=e^{s t} \int_{-\infty}^{\infty} h(\tau) e^{-s \tau} d \tau \]
The integral on the right-most side can be denoted as \(H(s)\), which is also complex, in general.
Thus, \[ \begin{equation} \label{eq_2_38} \textcolor{blue} { y(t)=H(s) e^{s t} } \end{equation} \] where \[ \begin{equation} \label{eq_2_38b} \textcolor{green} { H(s)=\int_{-\infty}^{\infty} h(\tau) e^{-s \tau} d \tau } . \end{equation} \]
Later, we will elaborate that, by the definition of the bilateral Laplace transform, \(\eqref{eq_2_38b}\) is the bilateral Laplace transform of \(h(t)\). Also, note that equation \(\eqref{eq_2_38}\) is valid iff \(\eqref{eq_2_38b}\) converges, the domain of \(H(s)\) is also called the region of convergence (ROC).
For a given \(s\), note that \(H(s)\) is a constant. Thus, the input and the output are the same (within a multiplicative constant) for the everlasting exponential signal. Q.E.D.
The transfer function
The transfer function of the system is a function of complex variable \(s\) that is defined as \[ \text{Transfer function}=\left.\frac{\text { output signal }}{\text { input signal }}\right|_{\text {input} = e^{s t}} \]
The transfer function is defined for, and is meaningful to, LTIC systems only. It does not exist for nonlinear or time-varying systems, in general.
We repeat again that this discussion is about the everlasting exponential, which starts at \(t=-\infty\), not the causal exponential \(e^{s t} u(t)\), which starts at \(t=0\).
# TODO this only holds for x(t) = e^{st}, not general x(t).
Now we prove that, for a system specified by \[ \begin{equation} \label{eq_2_2} Q(D) y(t)=P(D) x(t) , \end{equation} \]
its transfer function is \(H(s)\) as in \(\eqref{eq_2_38b}\), i.e., the bilateral Laplace transform of \(h(t)\), and that \(H(s)\) also satisfies: \[ \color{teal} {H(s)=\frac{P(s)}{Q(s)}} . \]
This follows readily by considering an everlasting input \(\textcolor{purple} {x(t)=e^{s t}}\). According to \(\eqref{eq_2_38}\), the output is \[ \textcolor{pink} { y(t)=H(s) e^{s t}} . \]
Substitution of this \(x(t)\) and \(y(t)\) in \(\eqref{eq_2_2}\) yields \[ \textcolor{pink} {H(s) }\left[Q(D) \textcolor{pink} {e^{s t} }\right]=P(D) \textcolor{purple} { e^{s t} } . \]
Moreover, \[ D^r e^{s t}=\frac{d^r e^{s t}}{d t^r}=s^r e^{s t} \]
Hence, \[ P(D) e^{s t}=P(s) e^{s t} \quad \text { and } \quad Q(D) e^{s t}=Q(s) e^{s t} \]
Consequently, \[ H(s)=\frac{P(s)}{Q(s)} . \]