Heaviside step function and Dirac delta function: Worked examples

Definition The Laplace transform notation used on this page

The symbol $L {\cdot}$ is the Laplace transform — an operator that turns a function of time $t$ into a function of a new variable $s$ . In plain words, it is a machine: you feed in $f (t)$ , out comes $L {f (t)} = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t .$

What is $s$ ? A helper variable (think "frequency-like knob"). We never measure it in the real world; it exists only in the transformed picture, and it must be large enough ( $s > 0$ for the functions here) for the integral to converge. Full details in Laplace Transform — Definition and Existence.
What is $F (s)$ ? Just the name of the output; the capital letter matches the lower-case input ( $f \to F$ , $y \to Y$ ).
The inverse $L^{- 1} {\cdot}$ is the machine run backwards: given $F (s)$ it recovers the original $f (t)$ . So $L^{- 1} {F (s)} = f (t)$ undoes $L {f (t)} = F (s)$ , the way $arctan$ undoes $tan$ .
Why use it at all? It converts a differential equation (hard: involves derivatives) into an algebra problem in $s$ (easy: just divide), which we then invert back to time. See Solving IVPs with Laplace Transforms.

The scenario matrix

Every problem on this topic is one (or a mix) of the cells below. The final column names the worked example that covers it.

Cell	Scenario class	What makes it tricky	Example
A	Un-shifted $g (t) u (t - a)$	$g$ is not already written as $g (t - a)$ — must rewrite first	Ex 1
B	Multi-piece step assembly	several boxcars, several delay stamps	Ex 2
C	Delta inside an integral (sifting)	plain evaluation, no ODE	Ex 3
D	Degenerate $a = 0$	$u (t)$ and $δ (t)$ at the very start	Ex 4
E	Delta at a shifted time driving an ODE	gate + shift on the output	Ex 5
F	Step forcing an ODE, watch steady state	limiting value as $t \to \infty$	Ex 6
G	Both a step and a delta on the RHS	superpose two responses	Ex 7
H	Word problem (real impulse: a hammer/kick)	translate physics → $δ$	Ex 8
I	Exam twist: delta before $t = 0$ , or evaluate a delta outside its window	the sift lands where nothing is measured	Ex 9

We now sweep the matrix top to bottom.

Ex 1 — Cell A: un-shifted $g (t) u (t - a)$

Worked example Transform

$L {t u (t - 2)}$ Forecast: Guess first. The naive (wrong) instinct is $e^{- 2 s} \cdot \frac{1}{s ^{2}}$ . Is $g (t) = t$ already in the form $g (t - 2)$ ? No — the theorem needs the argument shifted. So we expect an extra term to appear once we fix that. Predict something like $e^{- 2 s} (\frac{1}{s ^{2}} + \frac{something}{s})$ .

Steps.

Recall the delay stamp: $L {f (t - a) u (t - a)} = e^{- a s} F (s)$ . Why this step? It is the only rule that handles a $u (t - a)$ factor — but it demands the multiplier be a function of $(t - a)$ , not of $t$ .
Rewrite $t$ using $(t - 2)$ : $t = (t - 2) + 2$ . Why this step? This forces the " $(t - 2)$ " the theorem requires; the leftover $+ 2$ is a constant we handle separately.
Split: $t u (t - 2) = (t - 2) u (t - 2) + 2 u (t - 2)$ . Why this step? Now each piece is genuinely a function of $(t - 2)$ times $u (t - 2)$ .
Transform each piece. For $(t - 2) u (t - 2)$ : here $f (t) = t$ , $F (s) = \frac{1}{s ^{2}}$ , giving $e^{- 2 s} \frac{1}{s ^{2}}$ . For $2 u (t - 2)$ : here $f (t) = 2$ , $F (s) = \frac{2}{s}$ , giving $e^{- 2 s} \frac{2}{s}$ . Why this step? Apply the stamp separately to each shifted piece.
Add: $L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) .$

Verify: The forecast's "extra term" is exactly $\frac{2}{s}$ , matching the $+ 2$ from the rewrite. Sanity: if we had forgotten step 2 we'd have dropped that $\frac{2}{s}$ — this is precisely the classic mistake steel-manned in the parent note. ✓

Takeaway for cell A: whenever the multiplier is a bare $g (t)$ , substitute $t = (t - a) + a$ and expand before touching the theorem.

Ex 2 — Cell B: assembling a multi-piece input

Worked example Transform the piecewise

$f (t) = ⎩ ⎨ ⎧ 3 t 0 0 < t < 2 2 < t < 5 t > 5$ Forecast: We will build $f$ from boxcars and predict a plain $\frac{3}{s}$ plus several $e^{- 2 s} (\dots)$ and $e^{- 5 s} (\dots)$ stamps.

Steps.

Build the boxcar form from scratch. A boxcar $u (t - a) - u (t - b)$ equals $1$ on $(a, b)$ and $0$ elsewhere (switch on at $a$ , off at $b$ ). Isolate each piece:
- The constant $3$ lives on $(0, 2)$ : write $3 (u (t) - u (t - 2))$ .
- The ramp $t$ lives on $(2, 5)$ : write $t (u (t - 2) - u (t - 5))$ .
- After $t = 5$ the function is $0$ , so no term is needed. Adding and using $u (t) = 1$ for $t > 0$ : $f = 3 - 3 u (t - 2) + t u (t - 2) - t u (t - 5)$ . Why? Each boxcar reveals exactly one interval and hides the rest, so their sum reproduces the piecewise graph.
$L {3} = \frac{3}{s}$ ; $L {- 3 u (t - 2)} = - 3 \cdot \frac{e ^{- 2 s}}{s}$ . Why? Direct step transform $L {u (t - a)} = \frac{e ^{- a s}}{s}$ .
$t u (t - 2)$ : reuse Ex 1's trick with $a = 2$ : $= e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s})$ . Why? Un-shifted multiplier.
$- t u (t - 5)$ : rewrite $t = (t - 5) + 5$ , giving $- e^{- 5 s} (\frac{1}{s ^{2}} + \frac{5}{s})$ . Why? Same fix, now $a = 5$ .
Collect: $F (s) = \frac{3}{s} - \frac{3 e ^{- 2 s}}{s} + e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) - e^{- 5 s} (\frac{1}{s ^{2}} + \frac{5}{s}) .$ Simplify the $e^{- 2 s}$ group: $- \frac{3}{s} + \frac{2}{s} = - \frac{1}{s}$ , so $F (s) = \frac{3}{s} + e^{- 2 s} (\frac{1}{s ^{2}} - \frac{1}{s}) - e^{- 5 s} (\frac{1}{s ^{2}} + \frac{5}{s}) .$

Verify (value continuity check): At $t \to 2^{-}$ , $f = 3$ ; at $t \to 2^{+}$ , $f = t = 2$ . The figure below shows the input; the jump down from $3$ to $2$ at $t = 2$ is genuine (the function is discontinuous there), and our formula reproduces it since the $3 u$ term switches off while the $t u$ term switches on. ✓

Figure — Ex 2 input graph. The blue segment is the constant $3$ on $(0, 2)$ ; the yellow line is the ramp $t$ on $(2, 5)$ ; the green segment is $0$ after $t = 5$ . The open circle at $(2, 2)$ and filled circle at $(2, 3)$ mark the genuine downward jump (red arrow); the second red arrow marks the switch-off at $t = 5$ . This is exactly the graph our boxcar sum reproduces.

Figure — Heaviside step function and Dirac delta function

Ex 3 — Cell C: pure sifting, no ODE

Worked example Evaluate

$\int_{- \infty}^{\infty} cos (t) δ (t - \frac{π}{3}) d t$ and $\int_{0}^{\infty} e^{- 2 t} δ (t - 1) d t$ . Forecast: The sifting rule says a delta at $t = a$ just reads off the value of the rest of the integrand at $t = a$ . So predict $cos (π /3) = \frac{1}{2}$ and $e^{- 2}$ .

Steps.

Sifting: $\int_{- \infty}^{\infty} f (t) δ (t - a) d t = f (a)$ . Why? This is the entire definition of $δ$ — it picks out one value.
First integral: $f (t) = cos t$ , $a = \frac{π}{3}$ $\Rightarrow$ answer $= cos \frac{π}{3} = \frac{1}{2}$ . Why the answer is a plain number? The delta collapses the whole integral to one evaluation.
Second integral: $f (t) = e^{- 2 t}$ , $a = 1$ . Check the spike $t = 1$ lies inside the range $[0, \infty)$ — it does. $\Rightarrow$ answer $= e^{- 2}$ . Why check the range? If $a$ fell outside the limits the integral would be $0$ (Ex 9 covers that).

Verify: $cos (π /3) = 0.5$ exactly; $e^{- 2} \approx 0.1353$ . Both are checked numerically below. ✓

Ex 4 — Cell D: the degenerate $a = 0$

Worked example Find

$L {u (t)}$ and $L {δ (t)}$ , and solve $y^{'} + 3 y = δ (t)$ , $y (0^{-}) = 0$ . Forecast: With $a = 0$ the delay stamp $e^{- a s}$ becomes $e^{0} = 1$ . So expect the plainest possible transforms: $\frac{1}{s}$ for the step, $1$ for the delta. The ODE should then just be a decaying exponential started by the kick at $t = 0$ .

Steps.

$L {u (t)} = \frac{e ^{- 0 \cdot s}}{s} = \frac{1}{s}$ — identical to $L {1}$ for $t > 0$ . Why? $u (t) = 1$ for all $t > 0$ ; the point $t = 0$ has zero width and cannot change an integral.
$L {δ (t)} = e^{- 0 \cdot s} = 1$ . Why beautiful? The delta at the origin is the input whose transform is a flat $1$ — "all frequencies equally," so the output is the system's raw impulse response.
ODE: $s Y - y (0^{-}) + 3 Y = 1 \Rightarrow (s + 3) Y = 1 \Rightarrow Y = \frac{1}{s + 3}$ . Why $y (0^{-}) = 0$ ? We take the initial condition just before the kick, so the delta's punch shows up in the response, not the IC.
Invert: $y (t) = L^{- 1} {\frac{1}{s + 3}} = e^{- 3 t} u (t)$ . Why the $u (t)$ ? Nothing exists before $t = 0$ ; the kick creates a sudden jump of the response to $1$ at $t = 0^{+}$ .

Verify: Immediately after the kick, $y (0^{+}) = 1$ — a unit jump caused by a unit impulse, exactly what an impulse of strength $1$ should do to a first-order system. As $t \to \infty$ , $y \to 0$ : the system forgets the kick. ✓

Ex 5 — Cell E: a delta at a shifted time drives an ODE

Worked example Solve

$y^{''} + 4 y = 3 δ (t - 2)$ , $y (0) = 0, y^{'} (0) = 0$ . Forecast: A still oscillator (natural frequency $ω = 2$ , since $y^{''} + 4 y = 0$ oscillates as $cos 2 t, sin 2 t$ ) gets a kick of strength $3$ at $t = 2$ . Predict: dead flat until $t = 2$ , then a sine of angular frequency $2$ , scaled by the kick, and gated by $u (t - 2)$ .

Steps.

Transform: $s^{2} Y + 4 Y = 3 e^{- 2 s}$ . Why $e^{- 2 s}$ ? $L {δ (t - 2)} = e^{- 2 s}$ ; zero ICs kill the $sy (0), y^{'} (0)$ terms.
Solve algebraically: $Y = \frac{3 e ^{- 2 s}}{s ^{2} + 4}$ . Why? Just divide.
Recognise $L^{- 1} {\frac{1}{s ^{2} + 4}} = \frac{1}{2} sin 2 t$ . Why the $\frac{1}{2}$ ? Because $L {sin ω t} = \frac{ω}{s ^{2} + ω ^{2}}$ with $ω = 2$ , so we divide by $ω = 2$ to isolate $\frac{1}{s ^{2} + 4}$ .
Apply the delay stamp (shift and gate): $y (t) = 3 \cdot \frac{1}{2} sin (2 (t - 2)) u (t - 2) = \frac{3}{2} sin (2 (t - 2)) u (t - 2) .$ Why this step? Inverting the factor $e^{- 2 s}$ (see the delay stamp in Ex 1) must produce a shifted-and-gated version $f (t - 2) u (t - 2)$ ; forgetting the gate would wrongly make $y$ nonzero before the kick.

Verify (continuity of position): At $t = 2^{+}$ , $sin (2 \cdot 0) = 0$ , so $y (2^{+}) = 0$ — position is continuous, as it must be for a second-order kick (a delta in $y^{''}$ jumps the velocity, not the position). The velocity jump: $y^{'} (2^{+}) = \frac{3}{2} \cdot 2 cos 0 = 3$ , matching the impulse strength $3$ . ✓

Figure — Ex 5 kicked oscillator. The green segment shows $y = 0$ before the kick; the red arrow at $t = 2$ is the impulse $3 δ (t - 2)$ ; the blue curve is the resulting sine $\frac{3}{2} sin (2 (t - 2))$ that switches on only for $t > 2$ . Note the curve starts at height $0$ (position continuous) but with a nonzero slope (velocity kicked).

Ex 6 — Cell F: step forcing and the steady-state limit

Figure — Ex 6 step-forced rise. The green segment is $y = 0$ before the switch at $t = 1$ (red arrow); the yellow curve is the exponential rise; the dashed blue line marks the steady value $2$ that the curve approaches but never overshoots.

Ex 7 — Cell G: a step AND a delta together

Worked example Solve

$y^{'} + y = u (t - 1) + δ (t - 2)$ , $y (0) = 0$ . Forecast: Two events. First a switch turns on at $t = 1$ (drives $y$ toward $1$ ), then a kick at $t = 2$ adds a sudden jump. By linearity the total answer is the sum of the two individual responses. Predict a rising curve from $t = 1$ , then a discontinuous upward jump at $t = 2$ .

Steps.

Transform: $s Y + Y = \frac{e ^{- s}}{s} + e^{- 2 s} \Rightarrow Y = \frac{e ^{- s}}{s ( s + 1 )} + \frac{e ^{- 2 s}}{s + 1}$ . Why add them? The RHS is a sum, and Laplace is linear.
Step part: $\frac{1}{s ( s + 1 )} = \frac{1}{s} - \frac{1}{s + 1} \to 1 - e^{- t}$ ; delay $\Rightarrow (1 - e^{- (t - 1)}) u (t - 1)$ . Why? Standard partial fractions + delay stamp.
Delta part: $\frac{1}{s + 1} \to e^{- t}$ ; delay $\Rightarrow e^{- (t - 2)} u (t - 2)$ . Why no fraction to split? The delta's transform is just $e^{- 2 s}$ , leaving a clean $\frac{1}{s + 1}$ .
Sum: $y (t) = (1 - e^{- (t - 1)}) u (t - 1) + e^{- (t - 2)} u (t - 2) .$ Why simply add? Because the equation is linear: the response to a sum of inputs is the sum of the individual responses, so we superpose the step-response and the impulse-response.

Verify (jump at the kick): Just before $t = 2$ : $y (2^{-}) = 1 - e^{- 1} \approx 0.632$ . Just after: the delta adds $e^{- (2 - 2)} = e^{0} = 1$ , so $y (2^{+}) \approx 1.632$ . A first-order system jumps by exactly the impulse strength ( $= 1$ ) — confirmed. ✓ (See Impulse Response and Convolution for why each kick just shifts a copy of $e^{- t}$ .)

Ex 8 — Cell H: a real-world hammer blow

Worked example A mass–spring at rest satisfies

$y^{''} + 9 y = 0$ (position $y$ , natural angular frequency $3$ ). At $t = \frac{π}{6}$ a hammer delivers an impulse of strength $6$ . Find the motion. Forecast: The model is $y^{''} + 9 y = 6 δ (t - \frac{π}{6})$ , $y (0) = y^{'} (0) = 0$ . Nothing moves until the strike; then a sine of angular frequency $3$ . Amplitude should scale with the kick.

Steps.

Model the hammer as a delta of strength $6$ at $t = \frac{π}{6}$ . Why a delta? A hammer acts over a near-zero time but delivers a fixed total impulse — exactly the parent note's "shrink the tap to zero" idea.
Transform: $s^{2} Y + 9 Y = 6 e^{- π s /6} \Rightarrow Y = \frac{6 e ^{- π s /6}}{s ^{2} + 9}$ . Why? Zero ICs, $L {δ} = e^{- a s}$ .
$L^{- 1} {\frac{1}{s ^{2} + 9}} = \frac{1}{3} sin 3 t$ . Why the $\frac{1}{3}$ ? $L {sin 3 t} = \frac{3}{s ^{2} + 9}$ , divide by $ω = 3$ .
Shift & gate: $y (t) = 6 \cdot \frac{1}{3} sin (3 (t - \frac{π}{6})) u (t - \frac{π}{6}) = 2 sin (3 t - \frac{π}{2}) u (t - \frac{π}{6}) = - 2 cos (3 t) u (t - \frac{π}{6}) .$ Why this step? Inverting $e^{- π s /6}$ forces a shift by $\frac{π}{6}$ plus the gate; then the identity $sin (3 t - \frac{π}{2}) = - cos 3 t$ rewrites it in a tidier form.

Verify (units & velocity jump): Position continuous: at $t = \frac{π}{6}^{+}$ , $y = - 2 cos \frac{π}{2} = 0$ . Velocity jump $y^{'} (\frac{π}{6}^{+}) = 2 \cdot 3 sin \frac{π}{2} = 6$ — equals the impulse strength divided by mass ( $= 1$ ). Amplitude $2 = impulse / ω = 6/3$ , as expected. ✓

Ex 9 — Cell I: exam twist, spike outside the window

Worked example Evaluate

$\int_{0}^{\infty} t^{2} δ (t + 1) d t$ and $\int_{0}^{5} e^{t} δ (t - 7) d t$ . Forecast: In both, the spike sits outside the integration range. A delta contributes only if its spike lies inside the limits; otherwise the integrand is zero everywhere that matters. Predict both integrals $= 0$ .

Steps.

First: $δ (t + 1) = δ (t - (- 1))$ spikes at $t = - 1$ . Why does this matter? The range starts at $t = 0$ ; the spike at $- 1$ is never included, so $\int_{0}^{\infty} (\dots) = 0$ . (Contrast: over $\int_{- \infty}^{\infty}$ it would give $(- 1)^{2} = 1$ .)
Second: spike at $t = 7$ , but the upper limit is $5$ . Since $7 > 5$ , the spike is excluded, so the integral $= 0$ . Why check both limits? The sift only fires when $a$ lies in $[lower, upper]$ ; here we must confirm the spike is neither below the lower limit nor above the upper one before concluding.

Verify: Both integrals return $0$ because $a \in /$ the integration interval. This is the trap most exams set — always locate the spike before applying sifting. ✓

Recall

Recall Which cell, which fix? (hide and answer)

For $g (t) u (t - a)$ with $g$ not shifted, first rewrite $g (t)$ as ::: $g ((t - a) + a)$ , then expand. A delta of strength $k$ at $t = a$ on a first-order $y^{'} + cy$ makes $y$ jump at $t = a$ by ::: exactly $k$ . A delta on the RHS of a second-order system jumps the ::: velocity (not the position — position stays continuous). $\int_{0}^{\infty} f (t) δ (t - a) d t$ equals $f (a)$ only when ::: $a \geq 0$ (the spike is inside $[0, \infty)$ ); otherwise it is $0$ . The delay stamp $e^{- a s} F (s)$ inverts to ::: $f (t - a) u (t - a)$ — never forget the gate $u (t - a)$ . $L^{- 1} {F (s)}$ means ::: run the Laplace machine backwards — recover the time function $f (t)$ from its transform $F (s)$ .