Exercises — Heaviside step function and Dirac delta function

The symbol $u(t-a)$ means "0 while $t<a$, then 1 once $t>a$" — a light switch flipping on at $t=a$. (At exactly $t=a$ the value is left unspecified; it never affects an integral.)

• At $t=1$: since $1<3$, we are before the switch, so $u(1-3)=u(-2)=0$.
• At $t=3.5$: since $3.5>3$, we are after the switch, so $u(3.5-3)=u(0.5)=1$.
• The difference $u(t-3)-u(t-7)$ is a boxcar: it is $1-0=1$ for $3<t<7$ and $0$ everywhere else. Picture a flat-topped box sitting on the interval $(3,7)$ — the teal box in Figure s01, whose left edge rises at $t=3$ and right edge drops at $t=7$.

The delta's only job is to sift: $\int f(t)\delta (t-a)dt=f(a)$. It ignores the whole integrand except the single point $t=a=2$. Here $f(t)=t^2+1$, so the answer is $f(2)=2^2+1=5$. What it looks like: the plum spike in Figure s01 stands exactly at $t=2$; multiplying by $f$ then integrating just reads off the curve's height there and throws away everything else.

Both come straight from the defining integral $\mathcal{L}\{f\}={\int }_0^{\infty }{e}^{-st}f(t)dt$.

• Step: $\mathcal{L}\{u(t-4)\}={\int }_4^{\infty }{e}^{-st}dt=\frac{e^{-4s}}{s}$. (The lower limit is $4$ because $u(t-4)=0$ below it; the $1/s$ comes from doing the integral, the $e^{-4s}$ from starting at $4$.)
• Delta: $\mathcal{L}\{\delta (t-4)\}={\int }_0^{\infty }{e}^{-st}\delta (t-4)dt=e^{-4s}$. (Sift $e^{-st}$ at $t=4$; no $1/s$ because there is no interval to integrate — the delta does it for us.) The single difference — a $1/s$ or not — is the fingerprint of "step vs impulse."

This is already shifted: the thing being cubed is $(t-2)$, matching the argument of the switch. So we use the delay stamp directly.

• Identify $f(t-2)=(t-2{)}^3\Rightarrow f(t)=t^3$, with $a=2$.
• $F(s)=\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4}$ (from ${\int }_0^{\infty }{e}^{-st}{t}^{3}dt$, the standard $\mathcal{L}\{t^n\}=n!/s^{n+1}$).
• Stamp the delay: $\boxed{\mathcal{L}\{(t-2{)}^{3}u(t-2)\}=e^{-2s}\frac{6}{s^4}}$.

The switch is $u(t-2)$ but the multiplier is $t$, not $(t-2)$. The delay stamp demands the multiplier be written in terms of $(t-2)$. So rewrite: $t=(t-2)+2.$ Then $tu(t-2)=[(t-2)+2]u(t-2)=(t-2)u(t-2)+2u(t-2)$.

• First piece: $f(t)=t\Rightarrow F(s)=1/s^2$, stamped: $e^{-2s}/s^2$.
• Second piece: $2u(t-2)\Rightarrow 2\cdot e^{-2s}/s$.
• Total: $\boxed{e^{-2s}\left(\frac{1}{s^2}+\frac{2}{s}\right)}$.

Sifting picks $f(a)$ with $f=\cos$, $a=\pi /3$ (which is inside $(0,\infty )$, so it counts). $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}.$ Answer: $1/2$.

Build with boxcars (a boxcar $u(t-a)-u(t-b)$ is 1 only on $(a,b)$):

• $2$ on $(0,1)$: $2(u(t)-u(t-1))$.
• $0$ on $(1,3)$: contributes nothing.
• $t$ for $t>3$: $tu(t-3)$ (a half-line switched on at 3). So $f(t)=2-2u(t-1)+tu(t-3)$. (The values at $t=1,3$ are irrelevant to what follows — single points don't affect the integral.)

Transform. First recall the base fact $\mathcal{L}\{1\}={\int }_0^{\infty }{e}^{-st}dt=\frac{1}{s}$; by linearity $\mathcal{L}\{2\}=\frac{2}{s}$. For $u(t-1)$ the step formula gives $e^{-s}/s$. For $tu(t-3)$, split $t=(t-3)+3$ (so the multiplier becomes a function of $(t-3)$, as the delay stamp requires): $tu(t-3)=(t-3)u(t-3)+3u(t-3)\Rightarrow e^{-3s}\left(\frac{1}{s^2}+\frac{3}{s}\right).$ Total: $F(s)=\frac{2}{s}-\frac{2e^{-s}}{s}+e^{-3s}\left(\frac{1}{s^2}+\frac{3}{s}\right).$

The spike sits at $t=5$. A finite integral only "catches" it if $5$ lies inside the interval of integration.

• $I_1={\int }_0^2{e}^t\delta (t-5)dt$: the spike at $5$ is outside $[0,2]$, so the integral is $0$.
• $I_2={\int }_0^6{e}^t\delta (t-5)dt$: now $5\in [0,6]$, so sifting gives ${\mathbf{e}}^5$. Moral: the delta contributes only when its spike location is inside the limits — a case students forget when the limits are finite rather than $-\infty$ to $\infty$.

Strip the delay: let $F(s)=\frac{1}{s^2+4}$. Since $\mathcal{L}\{\sin (2t)\}=\frac{2}{s^2+4}$, we have $F(s)=\frac{1}{2}\cdot \frac{2}{s^2+4}$, so $f(t)=\frac{1}{2}\sin (2t)$. The factor $e^{-\pi s}$ means "shift $f$ right by $\pi$ AND gate it with $u(t-\pi )$": ${\mathcal{L}}^{-1}\left\{\frac{e^{-\pi s}}{s^2+4}\right\}=\frac{1}{2}\sin (2(t-\pi ))u(t-\pi ).$ Now simplify the phase using the periodicity of sine (period $2\pi$): $\sin (2t-2\pi )=\sin (2t)$ because subtracting a full period $2\pi$ from the angle leaves sine unchanged. So this equals $\frac{1}{2}\sin (2t)u(t-\pi )$. Written by case (boundary value at $t=\pi$ read as a limit, hence $0$): $y(t)=\{\begin{array}{ll}0& 0\le t<\pi \\ \frac{1}{2}\sin (2t)& t>\pi .\end{array}$

Transform both sides. Using $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)$ and $\mathcal{L}\{\delta (t-\pi )\}=e^{-\pi s}$: $s^{2}Y+4Y=e^{-\pi s}\Rightarrow (s^2+4)Y=e^{-\pi s}\Rightarrow Y=\frac{e^{-\pi s}}{s^2+4}.$ Invert. This is exactly Exercise 3.3's expression: $y(t)=\frac{1}{2}\sin (2(t-\pi ))u(t-\pi )=\frac{1}{2}\sin (2t)u(t-\pi ).$ By cases (the boundary value at $t=\pi$ is a limit, so $y(\pi )=0$): $y(t)=\{\begin{array}{ll}0& 0\le t<\pi \\ \frac{1}{2}\sin (2t)& t>\pi .\end{array}$ Motion: the mass sits perfectly still until $t=\pi$; the hammer-kick at $t=\pi$ launches a steady oscillation of amplitude $1/2$ and angular frequency $2$. See the plum kick and orange oscillation in Figure s02.

Transform. $\mathcal{L}\{y^{\prime }\}=sY-y(0)=sY$, and $\mathcal{L}\{u(t-2)\}=e^{-2s}/s$: $sY+3Y=\frac{e^{-2s}}{s}\Rightarrow Y=\frac{e^{-2s}}{s(s+3)}.$ Partial fractions on the un-delayed core: $\frac{1}{s(s+3)}=\frac{1}{3}\left(\frac{1}{s}-\frac{1}{s+3}\right)$, whose inverse is $\frac{1}{3}(1-e^{-3t})$. Apply the delay stamp (shift by 2 + gate): $y(t)=\frac{1}{3}(1-e^{-3(t-2)})u(t-2).$ By cases (boundary at $t=2$ read as a limit, giving $y(2)=0$): $y(t)=\{\begin{array}{ll}0& 0\le t<2\\ \frac{1}{3}(1-e^{-3(t-2)})& t>2.\end{array}$ As $t\to \infty$, $e^{-3(t-2)}\to 0$, so $y\to 1/3$ — the steady state of $y^{\prime }+3y=1$.

Transform each piece.

• $\mathcal{L}\{u(t)\}=1/s$, $\mathcal{L}\{u(t-2\pi )\}=e^{-2\pi s}/s$, $\mathcal{L}\{\delta (t-\pi )\}=e^{-\pi s}$.
• LHS with zero ICs: $(s^2+1)Y$. So $(s^2+1)Y=\frac{1}{s}-\frac{e^{-2\pi s}}{s}+e^{-\pi s}.$ $Y=\frac{1}{s(s^2+1)}-\frac{e^{-2\pi s}}{s(s^2+1)}+\frac{e^{-\pi s}}{s^2+1}.$

Invert each term. First, $\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}$, whose inverse is $1-\cos t$. Call this $g(t)=1-\cos t$. Also ${\mathcal{L}}^{-1}\{1/(s^2+1)\}=\sin t$.

• Term 1: $g(t)=1-\cos t$ (no delay).
• Term 2: $-g(t-2\pi )u(t-2\pi )=-(1-\cos (t-2\pi ))u(t-2\pi )$. Now $\cos (t-2\pi )=\cos t$ because cosine has period $2\pi$ (subtracting a full turn leaves it unchanged), so this is $-(1-\cos t)u(t-2\pi )$.
• Term 3: $\sin (t-\pi )u(t-\pi )$. Now $\sin (t-\pi )=-\sin t$ — shifting sine by half a period $\pi$ flips its sign (an odd-symmetry / half-period reflection). So this term is $-\sin tu(t-\pi )$.

Assemble: $y(t)=(1-\cos t)-(1-\cos t)u(t-2\pi )-\sin tu(t-\pi ).$

By cases (evaluate the switches on each interval; boundary values at $t=\pi ,2\pi$ are limits, and we verify continuity below):

Step transform carries a 1/s; delta transform is a bare exponential. Delay stamp requires the multiplier written as a function of (t − a). Inverting $e^{-as}F(s)$ always keeps the gate u(t − a). A finite integral catches $\delta (t-a)$ only if $a$ lies inside the limits. A delta forcing a 2nd-order ODE jumps the velocity, not the position.