Step Functions, Time Shifts, and Periodic Inputs

Definition For a real number $a$, the Heaviside step function at $a$ is

Its value at $t = a$ is a matter of convention, and $0$, $1$ and $\tfrac{1}{2}$ are all in use. That value changes no integral, so it changes no Laplace transform. When the behavior of a solution at the switch matters, the differential equation decides it, not the value assigned to the step. Multiplying a function $g$ by $u(t - a)$ gives a function equal to $0$ for $t < a$ and to $g(t)$ for $t > a$. For $a \le 0$ the step equals $1$ at every $t > 0$, so only $a > 0$ produces a real switch in a one-sided signal. For $a \ge 0$ and $\operatorname{Re} s > 0$,

Definition Let $f$ be defined for $t \ge 0$ and let $a \ge 0$. The signal $f$ delayed by $a$ is

which is $0$ for $0 \le t < a$ and $f(t - a)$ for $t > a$. Its graph is the graph of $f$ moved $a$ units to the right. It reads $f$ at the elapsed time $\tau = t - a$ since the switch, so after the switch it repeats the history of $f$ from its own time $0$. The step factor is part of the shift. Without it, the formula $f(t - a)$ on $0 \le t < a$ would need values of $f$ at negative times, which a one-sided signal does not have. A delay is different from switching on the original signal, $u(t - a)\,f(t)$, which reads $f$ on the original clock. The two coincide only if $f(t - a) = f(t)$ for every $t > a$, as for a constant $f$, so in general they are different inputs.

Definition Let $0 < a_1 < a_2 < \cdots < a_n$ be switching times. Suppose an input $f$ is given by a formula $g_0(t)$ for $0 \le t < a_1$, by $g_k(t)$ for $a_k < t < a_{k+1}$ when $1 \le k < n$, and by $g_n(t)$ for $t > a_n$. Then at every $t \ge 0$ other than the switching times

where $c_k(t) = g_k(t) - g_{k-1}(t)$ is the correction at the $k$th switch, the new formula minus the old one. For $a_j < t < a_{j+1}$, or for $t > a_n$ when $j = n$, exactly the first $j$ corrections are on, and their sum telescopes to $g_j(t) - g_0(t)$, which is why the formula is right. Before taking the Laplace transform, rewrite each correction in its own elapsed time, as $c_k(t) = d_k(t - a_k)$ with $d_k(\tau) = c_k(\tau + a_k)$. Then every term has the form of a delayed signal.

Definition Let $f$ be defined for $t \ge 0$ and integrable on every bounded interval, let $F = \mathcal{L}\{f\}$, and let $a \ge 0$. At every $s$ where $F(s)$ exists,

The integrand vanishes for $t < a$. On $[a, a + b]$ substitute $\tau = t - a$, which splits the weight as $e^{-st} = e^{-as}e^{-s\tau}$,

and let $b \to \infty$. Read from right to left, the theorem inverts delays. If $f = \mathcal{L}^{-1}\{F\}$, then $\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t - a)\,f(t - a)$, so invert $F$ first and then delay the whole function. For a signal that is switched on rather than delayed, the same substitution gives $\mathcal{L}\{u(t - a)\,g(t)\} = e^{-as}\,\mathcal{L}\{g(t + a)\}$, which is the rewriting step done in one line.

Definition Let $f$ be piecewise continuous on $[0, \infty)$ and periodic with period $T > 0$, so $f(t + T) = f(t)$ for all $t \ge 0$. Then for $\operatorname{Re} s > 0$

The function $f$ is bounded, because it repeats its values on $[0, T]$, where it is bounded. So it is of exponential order $0$, and $F(s)$ exists for $\operatorname{Re} s > 0$. Split $[0, \infty)$ into the blocks $[nT, (n + 1)T]$ and substitute $t = \tau + nT$ in the $n$th block. That block then contributes $e^{-nsT}F_T(s)$, and the blocks form a geometric series with ratio $e^{-sT}$, whose modulus $e^{-T\operatorname{Re} s}$ is less than $1$. Summing the series gives the formula. Only one period enters the numerator, and the denominator accounts for all the repeats.