A Very Special Function for LTIC Systems: The Everlasting Exponential

Sources:

1. 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 $$\begin{array}{}\text{(1)}& Q(D)y(t)=P(D)x(t),\end{array}$$

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: $$H(s)=\frac{P(s)}{Q(s)}.$$

The response to the everlasting exponential

There is a very special connection of LTIC systems with the everlasting exponential function $e^{st}$, where $s$ is a complex variable, in general.

We now show that the LTIC system's (zero-state) response to everlasting exponential input $e^{st}$ 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 $\ell =-\mathrm{\infty }$.

If $h(t)$ is the system's unit impulse response, then system response $y(t)$ to an everlasting exponential $e^{st}$ is given by $$y(t)=h(t)\ast e^{st}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h(\tau )e^{s(t-\tau )}d\tau =e^{st}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h(\tau )e^{-s\tau }d\tau$$

The integral on the right-most side can be denoted as $H(s)$, which is also complex, in general.

Thus, $$\begin{array}{}\text{(2)}& y(t)=H(s)e^{st}\end{array}$$ where $$\begin{array}{}\text{(3)}& H(s)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h(\tau )e^{-s\tau }d\tau .\end{array}$$

Later, we will elaborate that, by the definition of the bilateral Laplace transform, $\text{(3)}$ is the bilateral Laplace transform of $h(t)$. Also, note that equation $\text{(2)}$ is valid iff $\text{(3)}$ 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 $$\text{Transfer function}={\frac{\text{ output signal }}{\text{ input signal }}|}_{\text{input}=e^{st}}$$

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 $t=-\mathrm{\infty }$, not the causal exponential $e^{st}u(t)$, 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 $$\text{Label 'eq\_2\_2' multiply defined}$$

its transfer function is $H(s)$ as in $\text{(3)}$, i.e., the bilateral Laplace transform of $h(t)$, and that $H(s)$ also satisfies: $$H(s)=\frac{P(s)}{Q(s)}.$$

This follows readily by considering an everlasting input $x(t)=e^{st}$. According to $\text{(2)}$, the output is $$y(t)=H(s)e^{st}.$$

Substitution of this $x(t)$ and $y(t)$ in $\text{(1)}$ yields $$H(s)[Q(D)e^{st}]=P(D)e^{st}.$$

Moreover, $$D^r{e}^{st}=\frac{d^r{e}^{st}}{d{t}^r}=s^r{e}^{st}$$

Hence, $$P(D)e^{st}=P(s)e^{st}\text{ and }Q(D)e^{st}=Q(s)e^{st}$$

Consequently, $$H(s)=\frac{P(s)}{Q(s)}.$$