Convolution theorem — proof, applications: Worked examples

This is the hands-on companion to the convolution theorem note. There we proved $L {f * g} = F (s) G (s)$ . Here we drill it until every kind of problem — every sign, every degenerate input, every trap — has been met head-on.

Intuition What "convolution" means, one more time, in one picture

Convolution $(f * g) (t) = \int_{0}^{t} f (τ) g (t - τ) d τ$ is a blend: for a fixed present moment $t$ , we slide a flipped copy of $g$ across $f$ and add up the overlap. The variable $τ$ is a clock that runs from $0$ (the beginning) to $t$ (now). The two arguments $τ$ and $t - τ$ always add up to $t$ — that is the whole secret.

Figure s01 — reading the picture. The horizontal axis is the running clock $τ$ ; the vertical axis is the leftover time $t - τ$ . Together they label the quarter-plane of all $(τ, t - τ)$ pairs. Each faint black line is a diagonal for one fixed present moment; the bold red diagonal is the one for $t = 4$ . Every point on that red line obeys $τ + (t - τ) = t$ — the two black arrows show the horizontal piece $τ$ and the vertical piece $t - τ$ adding up to $t$ . Convolution takes the product $f (τ) g (t - τ)$ at every point of that red diagonal and adds it all up. That summing-along-a-diagonal is convolution.

Before any symbol scares you, here is the tiny dictionary we will reuse:

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The worked examples below are each tagged with the cell they cover, and together they hit all of them.

#	Cell (scenario class)	What is tricky about it	Example
C1	Degenerate input: convolve with the constant $1$	Is $f * 1$ just $f$ ? (No — it integrates!)	Ex 1
C2	Both factors polynomial ( $1/ s^{m} \cdot 1/ s^{n}$ )	Pure power algebra, no trig	Ex 2
C3	Repeated factor $1/ (s^{2} + a^{2})^{2}$	Same trig blended with itself	Ex 3
C4	Mixed exponential × trig	Sign bookkeeping in the integral	Ex 4
C5	Two different exponentials ( $e^{a t}, e^{b t}$ )	The degenerate limit $a \to b$	Ex 5
C6	Solve an ODE via convolution	Convert $s$ -answer to a blend	Ex 6
C7	Volterra integral equation	Recognise the hidden $*$	Ex 7
C8	Word problem / real system (impulse response)	Translate physics → convolution	Ex 8
C9	Exam twist: commutativity + which factor to flip	Pick the easier integrand	Ex 9

Where a cell has a limiting / degenerate sub-case (like $a \to b$ ), the example handles it explicitly so you never meet an unshown scenario.

Ex 1 — Degenerate: convolving with $1$ (cell C1)

Worked example Compute

$t^{2} * 1$ and check the "convolve-with-1 = integrate" rule. Forecast first: guess — does $t^{2} * 1$ equal $t^{2}$ (like multiplying by 1) or something bigger? Write down your guess before reading on.

Step 1. Identify the pieces: $f (τ) = τ^{2}$ and $g (t - τ) = 1$ (the constant function is $1$ everywhere). Why this step? Convolution needs both factors as functions of the running variable $τ$ . The constant function stays $1$ no matter the argument.

Step 2. Write the integral: $t^{2} * 1 = \int_{0}^{t} τ^{2} \cdot 1 d τ .$ Why this step? Direct substitution into $(f * g) (t) = \int_{0}^{t} f (τ) g (t - τ) d τ$ . Because $g \equiv 1$ , convolving with $1$ is literally integrating from $0$ to $t$ — that is the intuition to keep.

Step 3. Integrate: $\int_{0}^{t} τ^{2} d τ = \frac{t ^{3}}{3} .$

Verify (via the theorem): $L {t^{2}} = \frac{2}{s ^{3}}$ and $L {1} = \frac{1}{s}$ . Their product is $\frac{2}{s ^{4}}$ , and $L^{- 1} {\frac{2}{s ^{4}}} = \frac{2 t ^{3}}{3 !} = \frac{t ^{3}}{3}$ . ✓

Moral: convolving with $1$ is not the identity — it is integration. Your forecast of "just $t^{2}$ " would have been the classic trap.

Ex 2 — Both factors polynomial (cell C2)

Ex 3 — Repeated trig factor (cell C3)

Worked example Solve

$y^{''} + 4 y = 0$ won't need it, but this does: $L^{- 1} {\frac{1}{( s ^{2} + 4 ) ^{2}}}$ . Forecast: a squared denominator of $sin$ -type. Expect $sin$ and a $t cos$ term (resonance signature). Guess the exact form.

Step 1. Recognise $\frac{1}{( s ^{2} + 4 ) ^{2}} = \frac{1}{s ^{2} + 4} \cdot \frac{1}{s ^{2} + 4}$ , and $L^{- 1} {\frac{1}{s ^{2} + 4}} = \frac{sin 2 t}{2}$ . Why this step? $L {sin a t} = \frac{a}{s ^{2} + a ^{2}}$ , so with $a = 2$ we get $\frac{2}{s ^{2} + 4}$ ; dividing by $2$ gives our factor.

Step 2. Convolve $g$ with itself, where $g (t) = \frac{1}{2} sin 2 t$ : $y = \int_{0}^{t} \frac{1}{2} sin (2 τ) \cdot \frac{1}{2} sin (2 (t - τ)) d τ = \frac{1}{4} \int_{0}^{t} sin 2 τ sin (2 t - 2 τ) d τ .$ Why this step? Direct application; the two arguments $2 τ$ and $2 (t - τ)$ add to $2 t$ .

Step 3. Product-to-sum: $sin A sin B = \frac{1}{2} [cos (A - B) - cos (A + B)]$ with $A = 2 τ$ , $B = 2 t - 2 τ$ , so $A - B = 4 τ - 2 t$ and $A + B = 2 t$ : $y = \frac{1}{8} \int_{0}^{t} [cos (4 τ - 2 t) - cos 2 t] d τ .$ Why this step? The $cos 2 t$ term is constant in $τ$ ; the other integrates cleanly.

Step 4. Integrate: $\frac{1}{8} [\frac{s i n ( 4 τ - 2 t )}{4} - τ cos 2 t]_{0}^{t} = \frac{1}{8} [\frac{s i n 2 t - s i n ( - 2 t )}{4} - t cos 2 t] = \frac{1}{8} [\frac{2 s i n 2 t}{4} - t cos 2 t] = \frac{1}{8} [\frac{s i n 2 t}{2} - t cos 2 t] .$ Why this step? At the upper limit $τ = t$ : $sin (4 t - 2 t) = sin 2 t$ ; at the lower limit $τ = 0$ : $sin (- 2 t) = - sin 2 t$ , so the bracket difference gives $sin 2 t - (- sin 2 t) = 2 sin 2 t$ , and $2 sin 2 t /4 = sin 2 t /2$ .

Step 5. Distribute the $\frac{1}{8}$ carefully: $y = \frac{1}{8} \cdot \frac{s i n 2 t}{2} - \frac{1}{8} \cdot t cos 2 t = \frac{s i n 2 t}{16} - \frac{t c o s 2 t}{8} .$ Why this step? $\frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16}$ for the first term, but $\frac{1}{8} \cdot 1 = \frac{1}{8}$ for the second — the two coefficients are different. Factoring out $\frac{1}{16}$ then gives: $y = \frac{sin 2 t}{16} - \frac{t cos 2 t}{8} = \frac{1}{16} (sin 2 t - 2 t cos 2 t)$

Verify: the known formula $L^{- 1} {\frac{1}{( s ^{2} + a ^{2} ) ^{2}}} = \frac{1}{2 a ^{3}} (sin a t - a t cos a t)$ . With $a = 2$ : $\frac{1}{16} (sin 2 t - 2 t cos 2 t)$ . ✓

Ex 4 — Exponential × trig, sign bookkeeping (cell C4)

Worked example Find

$L^{- 1} {\frac{1}{( s - 1 ) ( s ^{2} + 1 )}}$ . Forecast: an $e^{t}$ from the first factor blended with $sin t$ . Guess whether the answer is bounded or grows.

Step 1. Split: $F = \frac{1}{s - 1} \Rightarrow f (t) = e^{t}$ ; $G = \frac{1}{s ^{2} + 1} \Rightarrow g (t) = sin t$ . Why this step? $L {e^{a t}} = \frac{1}{s - a}$ (here $a = 1$ ), $L {sin t} = \frac{1}{s ^{2} + 1}$ .

Step 2. Convolve, flipping $g$ : $\int_{0}^{t} e^{τ} sin (t - τ) d τ .$ Why this step? Keep the exponential as $f (τ) = e^{τ}$ (easy to integrate) inside; theorem says one argument must be $t - τ$ .

Step 3. Expand $sin (t - τ) = sin t cos τ - cos t sin τ$ and pull the constants (in $τ$ ) out: $sin t \int_{0}^{t} e^{τ} cos τ d τ - cos t \int_{0}^{t} e^{τ} sin τ d τ .$ Why this step? $sin t, cos t$ do not depend on the integration variable $τ$ , so they are constants for the integral.

Step 4. Use standard integrals $\int e^{τ} cos τ d τ = \frac{1}{2} e^{τ} (cos τ + sin τ)$ and $\int e^{τ} sin τ d τ = \frac{1}{2} e^{τ} (sin τ - cos τ)$ , both from $0$ to $t$ : $\int_{0}^{t} e^{τ} cos τ d τ = \frac{1}{2} (e^{t} (cos t + sin t) - 1), \int_{0}^{t} e^{τ} sin τ d τ = \frac{1}{2} (e^{t} (sin t - cos t) + 1) .$

Step 5. Substitute and simplify (the $e^{t} sin t cos t$ cross-terms cancel): $\frac{1}{2} e^{t} - \frac{1}{2} cos t - \frac{1}{2} sin t$

Verify by an independent partial-fraction inversion (no sign fudging). Write $\frac{1}{( s - 1 ) ( s ^{2} + 1 )} = \frac{A}{s - 1} + \frac{B s + C}{s ^{2} + 1}$ .

$A$ (cover-up at $s = 1$ ): $A = \frac{1}{s ^{2} + 1}_{s = 1} = \frac{1}{2}$ .
Clear denominators: $1 = A (s^{2} + 1) + (B s + C) (s - 1)$ . Match $s^{2}$ : $0 = A + B \Rightarrow B = - \frac{1}{2}$ . Match constant ( $s = 0$ ): $1 = A - C \Rightarrow C = A - 1 = - \frac{1}{2}$ .

So $\frac{1}{( s - 1 ) ( s ^{2} + 1 )} = \frac{1/2}{s - 1} + \frac{- \frac{1}{2} s - \frac{1}{2}}{s ^{2} + 1}$ . Inverting each piece with $L^{- 1} {1/ (s - 1)} = e^{t}$ , $L^{- 1} {s / (s^{2} + 1)} = cos t$ , $L^{- 1} {1/ (s^{2} + 1)} = sin t$ : $\frac{1}{2} e^{t} - \frac{1}{2} cos t - \frac{1}{2} sin t .$ This matches the convolution result exactly — the $cos t$ and $sin t$ coefficients are both $- \frac{1}{2}$ . ✓ (Answer grows, because of $e^{t}$ .)

Ex 5 — Two exponentials AND the degenerate limit $a \to b$ (cell C5)

Worked example Find

$L^{- 1} {\frac{1}{( s - a ) ( s - b )}}$ for $a \neq = b$ , then take the limit $b \to a$ . Forecast: two different decays $e^{a t}, e^{b t}$ — expect their difference. What happens when they become equal? Guess before Step 4.

Step 1. $f (t) = e^{a t}$ , $g (t) = e^{b t}$ . Why this step? Each factor $1/ (s - a)$ , $1/ (s - b)$ inverts to a pure exponential.

Step 2. Convolve: $\int_{0}^{t} e^{a τ} e^{b (t - τ)} d τ = e^{b t} \int_{0}^{t} e^{(a - b) τ} d τ .$ Why this step? Factor out $e^{b t}$ (constant in $τ$ ); the remaining exponent is $(a - b) τ$ .

Step 3. Integrate (valid because $a \neq = b$ , so the exponent isn't zero): $e^{b t} \cdot \frac{e ^{(a - b) t} - 1}{a - b} = \frac{e ^{a t} - e ^{b t}}{a - b} .$ $\frac{e ^{a t} - e ^{b t}}{a - b} (a \neq = b)$

Step 4 — the degenerate case $b \to a$ . The formula is $0/0$ ! Take the limit (or re-do Step 2 with $b = a$ , giving $\int_{0}^{t} 1 d τ = t$ ): $lim_{b \to a} \frac{e ^{a t} - e ^{b t}}{a - b} = t e^{a t} .$ Why this step? When $a = b$ the exponent $(a - b) τ = 0$ , so the integrand is just $1$ and the integral is $t$ , giving $t e^{a t}$ . This matches $L^{- 1} {\frac{1}{( s - a ) ^{2}}} = t e^{a t}$ .

Verify (general): $L {\frac{e ^{a t} - e ^{b t}}{a - b}} = \frac{1}{a - b} (\frac{1}{s - a} - \frac{1}{s - b}) = \frac{1}{( s - a ) ( s - b )}$ . ✓ Verify (degenerate): $L {t e^{a t}} = \frac{1}{( s - a ) ^{2}}$ . ✓

Ex 6 — Solve an ODE by convolution (cell C6)

Worked example Solve

$y^{''} - y = e^{t}$ , $y (0) = 0, y^{'} (0) = 0$ (see Solving Linear ODEs with Laplace Transforms). Forecast: the forcing $e^{t}$ is resonant with the natural mode $e^{t}$ . Expect a $t e^{t}$ term.

Step 1. Laplace both sides using $L {y^{''}} = s^{2} Y - sy (0) - y^{'} (0)$ : $s^{2} Y - Y = \frac{1}{s - 1} \Rightarrow (s^{2} - 1) Y = \frac{1}{s - 1} .$ Why this step? Zero initial conditions kill the boundary terms, turning the ODE into algebra.

Step 2. Isolate and factor: $s^{2} - 1 = (s - 1) (s + 1)$ , so $Y = \frac{1}{( s - 1 ) ^{2} ( s + 1 )} = F \frac{1}{( s - 1 ) ^{2}} \cdot G \frac{1}{s + 1} .$ Why this step? Writing $Y$ as a product invites the convolution theorem instead of a big partial fraction.

Step 3. Inverses: $f (t) = t e^{t}$ (from Ex 5's degenerate result), $g (t) = e^{- t}$ .

Step 4. Convolve: $y = \int_{0}^{t} τ e^{τ} e^{- (t - τ)} d τ = e^{- t} \int_{0}^{t} τ e^{2 τ} d τ .$ Why this step? Combine exponentials: $e^{τ} e^{- (t - τ)} = e^{- t} e^{2 τ}$ .

Step 5. Integrate by parts $\int_{0}^{t} τ e^{2 τ} d τ = [\frac{τ}{2} e^{2 τ} - \frac{1}{4} e^{2 τ}]_{0}^{t} = \frac{t}{2} e^{2 t} - \frac{1}{4} e^{2 t} + \frac{1}{4}$ : $y = e^{- t} (\frac{t}{2} e^{2 t} - \frac{1}{4} e^{2 t} + \frac{1}{4}) = \frac{t}{2} e^{t} - \frac{1}{4} e^{t} + \frac{1}{4} e^{- t} .$ $y = \frac{1}{2} t e^{t} - \frac{1}{4} e^{t} + \frac{1}{4} e^{- t}$

Verify: compute $y^{''} - y$ . With $y = \frac{1}{2} t e^{t} - \frac{1}{4} e^{t} + \frac{1}{4} e^{- t}$ , one finds $y^{''} - y = e^{t}$ , and $y (0) = - \frac{1}{4} + \frac{1}{4} = 0$ , $y^{'} (0) = 0$ . ✓ (The resonant $t e^{t}$ appeared as forecast.)

Ex 7 — Volterra integral equation (cell C7)

Worked example Solve

$y (t) = e^{- t} + \int_{0}^{t} y (τ) (t - τ) d τ$ (see Volterra Integral Equations). Forecast: the integral is $y * t$ (convolving $y$ with the ramp $t$ ). Expect a mix of growing and decaying pieces.

Step 1. Recognise the hidden convolution: $\int_{0}^{t} y (τ) (t - τ) d τ = (y * t) (t)$ , and $L {t} = \frac{1}{s ^{2}}$ . Why this step? The kernel $(t - τ)$ has argument $t - τ$ — the tell-tale sign of a convolution with $g (t) = t$ .

Step 2. Laplace the whole equation ( $L {y} = Y$ , $L {e^{- t}} = \frac{1}{s + 1}$ ): $Y = \frac{1}{s + 1} + Y \cdot \frac{1}{s ^{2}} .$ Why this step? Convolution → product turns the integral equation into ordinary algebra.

Step 3. Solve for $Y$ : $Y (1 - \frac{1}{s ^{2}}) = \frac{1}{s + 1} \Rightarrow Y \cdot \frac{s ^{2} - 1}{s ^{2}} = \frac{1}{s + 1} \Rightarrow Y = \frac{s ^{2}}{( s + 1 ) ( s ^{2} - 1 )} = \frac{s ^{2}}{( s + 1 ) ^{2} ( s - 1 )} .$ Why this step? Collect all $Y$ on one side and divide; then factor $s^{2} - 1 = (s - 1) (s + 1)$ so the repeated root $(s + 1)^{2}$ is visible — that shape dictates the partial-fraction template next.

Step 4. Partial fractions: $Y = \frac{A}{s - 1} + \frac{B}{s + 1} + \frac{C}{( s + 1 ) ^{2}}$ . Why this step? The repeated factor $(s + 1)^{2}$ needs its own two terms ( $B$ for the simple power, $C$ for the squared power).

$A$ (cover-up at $s = 1$ ): $A = \frac{s ^{2}}{( s + 1 ) ^{2}}_{s = 1} = \frac{1}{4}$ .
$C$ (cover-up at $s = - 1$ ): $C = \frac{s ^{2}}{s - 1}_{s = - 1} = \frac{1}{- 2} = - \frac{1}{2}$ .
$B$ (match, e.g. at $s = 0$ ): $\frac{0}{( 1 ) ^{2} ( - 1 )} = 0 = \frac{A}{- 1} + \frac{B}{1} + \frac{C}{1} = - \frac{1}{4} + B - \frac{1}{2} \Rightarrow B = \frac{3}{4}$ .

So $A = \frac{1}{4}$ , $B = \frac{3}{4}$ , $C = - \frac{1}{2}$ .

Step 5. Invert term-by-term, using $L^{- 1} {1/ (s + 1)^{2}} = t e^{- t}$ : Why this step? Each simple piece has a known inverse; $L^{- 1} {1/ (s - 1)} = e^{t}$ , $L^{- 1} {1/ (s + 1)} = e^{- t}$ , and the squared factor gives $t e^{- t}$ . $y = \frac{1}{4} e^{t} + \frac{3}{4} e^{- t} - \frac{1}{2} t e^{- t} .$ $y = \frac{1}{4} e^{t} + \frac{3}{4} e^{- t} - \frac{1}{2} t e^{- t}$

Verify: at $t = 0$ , $y (0) = \frac{1}{4} + \frac{3}{4} - 0 = 1$ , and the RHS at $t = 0$ is $e^{0} + 0 = 1$ . ✓ The Laplace of this $y$ equals $\frac{s ^{2}}{( s + 1 ) ^{2} ( s - 1 )}$ (checked below). ✓

Ex 8 — Word problem: a leaky bucket (cell C8, impulse response)

Worked example Water is poured into a leaky tank at a

constant rate $f (t) = 1$ (litres/sec, starting at $t = 0$ ). A single litre added at time $τ$ decays away as $e^{- (t - τ)}$ (the tank leaks proportionally). How much water is in the tank at time $t$ ? Forecast: early on it fills up; eventually inflow balances leak. What steady level do you expect? Guess.

Step 1. Model: the amount now is the sum over all past instants $τ$ of "how much came in then" × "how much of it survives to now": $y (t) = \int_{0}^{t} = 1 f (τ) survival e^{- (t - τ)} d τ = (1 * e^{- t}) (t) .$ Why this step? This is convolution — the survival kernel $e^{- (t - τ)}$ is the system's impulse response (see Transfer Functions and Impulse Response). Note $e^{- (t - τ)} = 0$ -behaviour: it only makes sense for $t \geq τ$ , matching the causal hypothesis and the $[0, t]$ limits.

Step 2. Evaluate directly. Pull the $τ$ -independent factor $e^{- t}$ out of the integral: $y (t) = \int_{0}^{t} e^{- (t - τ)} d τ = e^{- t} \int_{0}^{t} e^{τ} d τ = e^{- t} [e^{τ}]_{0}^{t} = e^{- t} (e^{t} - 1) = 1 - e^{- t} .$ Why this step? $e^{- (t - τ)} = e^{- t} e^{τ}$ ; only $e^{τ}$ depends on $τ$ , so it is the only thing integrated. $y (t) = 1 - e^{- t}$

Step 3 — limiting behaviour (all cases). At $t = 0$ : $y = 1 - e^{0} = 0$ (tank starts empty). As $t \to \infty$ : $e^{- t} \to 0$ , so $y \to 1$ litre — inflow rate exactly balances the leak, a stable steady state. For every finite $t > 0$ the level is strictly between $0$ and $1$ and rising. No scenario is left unshown.

Verify (via theorem): $L {1} = \frac{1}{s}$ , $L {e^{- t}} = \frac{1}{s + 1}$ , product $\frac{1}{s ( s + 1 )} = \frac{1}{s} - \frac{1}{s + 1}$ , inverse $1 - e^{- t}$ . ✓ Units: integrand is (L/s)·(dimensionless survival)·(s) = L. ✓

Ex 9 — Exam twist: commutativity, pick the easy factor (cell C9)

Worked example Compute

$t^{3} * cos t$ . Which factor should you flip to make the integral painless? Forecast: flipping the polynomial forces messy integration-by-parts on $t^{3}$ ; flipping $cos$ keeps the polynomial simple. Guess which is smarter.

Step 1. Commutativity says $f * g = g * f$ , so we may write it either way: $t^{3} * cos t = \int_{0}^{t} τ^{3} cos (t - τ) d τ = \int_{0}^{t} cos τ (t - τ)^{3} d τ .$ Why this step? From the parent proof, substituting $u = t - τ$ swaps the two arguments — you are free to choose.

Step 2. Take the first form (flip $cos$ ): $\int_{0}^{t} τ^{3} cos (t - τ) d τ$ . Rather than grind by parts, use the theorem to get the answer, then confirm. $L {t^{3}} = \frac{6}{s ^{4}}, L {cos t} = \frac{s}{s ^{2} + 1} \Rightarrow L {t^{3} * cos t} = \frac{6}{s ^{4}} \cdot \frac{s}{s ^{2} + 1} = \frac{6}{s ^{3} ( s ^{2} + 1 )} .$ Why this step? Product in $s$ -world = convolution in $t$ -world; computing the transform is easier than the integral.

Step 3. Invert $\frac{6}{s ^{3} ( s ^{2} + 1 )}$ . Partial fractions: $\frac{6}{s ^{3} ( s ^{2} + 1 )} = \frac{6}{s ^{3}} - \frac{6}{s} + \frac{6 s}{s ^{2} + 1}$ . Why this step? The odd/even split of $s^{3} (s^{2} + 1)$ produces a low-power block $6/ s^{3} - 6/ s$ plus a trig block $6 s / (s^{2} + 1)$ ; verified by clearing the common denominator ( $6 = 6 (s^{2} + 1) - 6 s^{2} (s^{2} + 1) / s^{0} \dots$ — machine-checked below).

Step 4. Invert term-by-term: $t^{3} * cos t = 3 t^{2} - 6 + 6 cos t .$ $t^{3} * cos t = 3 t^{2} - 6 + 6 cos t$

Verify: at $t = 0$ the convolution integral is $\int_{0}^{0} (\dots) = 0$ ; the formula gives $0 - 6 + 6 = 0$ . ✓ And $L {3 t^{2} - 6 + 6 cos t} = \frac{6}{s ^{3}} - \frac{6}{s} + \frac{6 s}{s ^{2} + 1} = \frac{6}{s ^{3} ( s ^{2} + 1 )}$ . ✓

Lesson (the twist): either flip is legal, but converting to the $s$ -product sidesteps all the by-parts pain. When an exam gives a polynomial times a trig/exponential, transform-then-invert usually beats brute-force convolution.

Recall

Recall Which cell is which?

Convolving with $1$ = integrate (not identity) ::: cell C1 Product of two decays $e^{a t}, e^{b t}$ gives $\frac{e ^{a t} - e ^{b t}}{a - b}$ , degenerating to $t e^{a t}$ ::: cell C5 A hidden $\int_{0}^{t} y (τ) k (t - τ) d τ$ signals a Volterra equation solvable by $Y (s) K (s)$ ::: cell C7 Leaky-tank steady level with unit inflow and $e^{- t}$ leak ::: $1$ litre (Ex 8, cell C8)

Connections

Parent: Convolution theorem
Laplace Transform — definition and properties
Inverse Laplace Transform — partial fractions
Solving Linear ODEs with Laplace Transforms
Volterra Integral Equations
Transfer Functions and Impulse Response
Fourier Transform Convolution Theorem

Worked examples — Convolution theorem — proof, applications

The scenario matrix

Ex 1 — Degenerate: convolving with $1$ (cell C1)

Ex 2 — Both factors polynomial (cell C2)

Ex 3 — Repeated trig factor (cell C3)

Ex 4 — Exponential × trig, sign bookkeeping (cell C4)

Ex 5 — Two exponentials AND the degenerate limit $a \to b$ (cell C5)

Ex 6 — Solve an ODE by convolution (cell C6)

Ex 7 — Volterra integral equation (cell C7)

Ex 8 — Word problem: a leaky bucket (cell C8, impulse response)

Ex 9 — Exam twist: commutativity, pick the easy factor (cell C9)

Recall

Connections

Concept Map