Impulses, Transfer Functions, and Poles

Definition Fix a real number $a$. The Dirac delta at $a$, written $\delta(t - a)$, is the distribution that assigns to each test function $\varphi$, meaning a smooth function that vanishes outside some bounded interval, the number

It is not a function. No function $g$ that is integrable on bounded intervals satisfies $\int g(t)\varphi(t)\,dt = \varphi(a)$ for every test function $\varphi$, so it makes no sense to ask for the value of $\delta(t - a)$ at $t = a$. The familiar notation $\int \delta(t - a)\varphi(t)\,dt = \varphi(a)$ is shorthand for this rule, and it is also used for functions $\varphi$ that are merely continuous at $a$. In this shorthand $\int_\alpha^\beta \delta(t - a)\,dt$ equals $1$ when $\alpha < a < \beta$ and $0$ when $a < \alpha$ or $a > \beta$. An endpoint $a = \alpha$ or $a = \beta$ needs a stated convention.

Definition An impulse of strength $J$ at time $a > 0$ is the input $J\,\delta(t - a)$, and a unit impulse is the case $J = 1$. It is the limit of short pulses of area $J$. For $\varepsilon > 0$ let $d_\varepsilon$ equal $J/\varepsilon$ on $[a, a + \varepsilon)$ and $0$ elsewhere. For every $\varphi$ continuous at $a$, as $\varepsilon \to 0^+$,

since the middle line is $J$ times the average of $\varphi$ over a shrinking interval. The height $J/\varepsilon$ grows without bound while the area stays $J$, so the limit is described by its area and not by a height. For $a > 0$, applying the rule to $e^{-st}$ gives the Laplace transform

for every complex $s$, and the same value is the limit of the Laplace transforms of the pulses with $J = 1$. An impulse at $a = 0$ sits at the endpoint of the one-sided integral and needs a convention. The usual one starts the integral just before $0$ and gives $\mathcal{L}\{\delta(t)\} = 1$.

Definition Let $m > 0$, $c$ and $k$ be constants, let $a > 0$, and let $y$ solve

understood as the limit of the responses to the pulses $d_\varepsilon$ of area $J$. Then $y$ is continuous at $t = a$, its velocity jumps by

and on each side of $a$ the solution satisfies the homogeneous equation. To find the size of the jump, integrate the equation over $[a - \varepsilon, a + \varepsilon]$,

and let $\varepsilon \to 0^+$. Since $y$ is continuous and bounded near $a$, the second and third terms tend to $0$. The position itself cannot jump, because a jump in $y$ would put a multiple of $\delta$ into $y'$ and the derivative of $\delta$ into $y''$, and the right side contains no such term. For an equation of order $n$ the same argument makes the derivative of order $n - 1$ jump by $J$ divided by the leading coefficient, while all lower derivatives stay continuous.

Definition Consider again a linear equation with constant coefficients,

with input $f$ and output $y$. Zero initial conditions mean $y(0) = y'(0) = \cdots = y^{(n-1)}(0) = 0$, so the system is at rest when the input starts. Under them, the Laplace transform of the equation is $p(s)Y(s) = F(s)$ with $p(s) = a_n s^n + \cdots + a_1 s + a_0$, so

The function $H$ is the transfer function. It depends only on the equation, it is the Laplace transform of the impulse-response kernel $h$, and in time the relation reads $y = h * f$. With nonzero initial values the Laplace transform of $y$ gains a further term, a polynomial built from the initial values divided by $p(s)$, and the transfer function does not describe that part. When the right side of a model also contains derivatives of the input, the same construction, with zero initial values for the input as well, gives a ratio of two polynomials $H(s) = q(s)/p(s)$.

Definition Write a rational transfer function in lowest terms, $H(s) = q(s)/p(s)$ with polynomials $q$ and $p$ that have no common root. Its poles are the roots of $p$, real or complex, the points near which $\lvert H(s) \rvert$ grows without bound. If $\deg q < \deg p$, partial fractions show that the impulse-response kernel $h$ is a combination of the terms $t^j e^{\lambda t}$, one for each pole $\lambda$ and each $j$ smaller than its multiplicity. A conjugate pair $\alpha \pm i\beta$ gives the real terms $t^j e^{\alpha t}\cos\beta t$ and $t^j e^{\alpha t}\sin\beta t$. So the real part of each pole sets the growth or decay of a mode of $h$. A root of $p$ that cancels against a root of $q$ is not a pole, and its mode does not appear in $h$. When $H = 1/p$, as for the equations above, the poles are exactly the roots of the characteristic polynomial.

Definition A system whose zero-state response is $y = h * f$ is bounded-input bounded-output stable, or BIBO stable, if every input $f$ that is bounded on $[0, \infty)$ produces a response $y$ that is bounded on $[0, \infty)$. For a piecewise continuous kernel $h$ this holds exactly when

and then $\lvert f \rvert \le B$ gives $\lvert y(t) \rvert \le B\int_0^\infty \lvert h \rvert\,dt$ for all $t \ge 0$. For a rational transfer function in lowest terms with $\deg q \le \deg p$, BIBO stability holds exactly when every pole has negative real part. The statement concerns only the zero state and the chosen input and output. A mode that cancels from $H$, or that the input cannot excite or the output cannot see, may still grow from a nonzero initial state, so BIBO stability does not show that every internal state stays bounded. Poles on the imaginary axis fail the test, even though the free motion they describe stays bounded when they are simple.