Exercises — Convolution theorem — proof, applications

The test: in a convolution the two arguments must ==add up to $t$== ($\tau +(t-\tau )=t$), and the upper limit must be $t$.

• (a) Arguments are $\tau$ and $t-\tau$ — they add to $t$. Limits are $0$ to $t$. ✅ This is a convolution with $f(\tau )={\tau }^2$ (so $f(t)=t^2$) and $g(t-\tau )=t-\tau$ (so $g(t)=t$).
• (b) Both factors use $\tau$, $\tau$ — arguments are $\tau$ and $\tau$, they do not add to $t$. ❌ Just an ordinary integral, not a convolution.
• (c) Upper limit is $\infty$, not $t$. ❌ Not the causal convolution we defined (it's a different object). Answer: only (a); $f(t)=t^2,g(t)=t$.

Split the product. $\frac{1}{s^3(s-2)}=\underset{F(s)}{\underbrace{\frac{1}{s^3}}}\cdot \underset{G(s)}{\underbrace{\frac{1}{s-2}}}$. Recall the standard pairs (from Laplace Transform — definition and properties): ${\mathcal{L}}^{-1}\left\{\frac{1}{s^3}\right\}=\frac{t^2}{2}=f(t),{\mathcal{L}}^{-1}\left\{\frac{1}{s-2}\right\}=e^{2t}=g(t).$ Why $t^2/2$? Because $\mathcal{L}\{t^n\}=n!/s^{n+1}$, so $\mathcal{L}\{t^2\}=2/s^3$, and we need to divide by $2$. Apply the theorem: ${\mathcal{L}}^{-1}\left\{\frac{1}{s^3(s-2)}\right\}={\int }_0^t\frac{{\tau }^2}{2}{e}^{2(t-\tau )}d\tau .$ Done — this is the requested convolution form.

Direct. Here $f(t)=1,g(t)=1$: $(1\ast 1)(t)={\int }_0^{t}1\cdot 1d\tau =[\tau {]}_0^t=t.$ Theorem check. $\mathcal{L}\{1\}=\frac{1}{s}$, so the product in the $s$-world is $\frac{1}{s}\cdot \frac{1}{s}=\frac{1}{s^2}$, and ${\mathcal{L}}^{-1}\{1/s^2\}=t$. ✅ Same answer. This is the cleanest proof that convolution ≠ pointwise product: $1\cdot 1=1$ but $1\ast 1=t$.

$f(t)=t,g(t)=e^t$. By commutativity ($f\ast g=g\ast f$) put the easier-to-integrate factor with the flip. Let's keep $f(\tau )=\tau$: $(t\ast e^t)(t)={\int }_0^t\tau e^{t-\tau }d\tau =e^t{\int }_0^t\tau e^{-\tau }d\tau .$ Why pull out $e^t$? Because $e^{t-\tau }=e^t{e}^{-\tau }$ and $e^t$ is constant with respect to the running clock $\tau$. Integrate by parts ($u=\tau ,dv=e^{-\tau }d\tau ,v=-e^{-\tau }$): ${\int }_0^t\tau e^{-\tau }d\tau =[-\tau e^{-\tau }{]}_0^t+{\int }_0^t{e}^{-\tau }d\tau =-t{e}^{-t}+[-e^{-\tau }{]}_0^t=-t{e}^{-t}-e^{-t}+1.$ Multiply back by $e^t$: $(t\ast e^t)(t)=e^t(1-e^{-t}-t{e}^{-t})=\boxed{e^t-1-t}.$ Sanity via theorem: $\mathcal{L}\{t\}\mathcal{L}\{e^t\}=\frac{1}{s^2}\cdot \frac{1}{s-1}$, and $\mathcal{L}\{e^t-1-t\}=\frac{1}{s-1}-\frac{1}{s}-\frac{1}{s^2}$. Common denominator $s^2(s-1)$: numerator $s^2-s(s-1)-(s-1)=s^2-s^2+s-s+1=1$. So it equals $\frac{1}{s^2(s-1)}$. ✅

Split: $\frac{1}{s}\cdot \frac{1}{s^2+4}$. Inverses: ${\mathcal{L}}^{-1}\{1/s\}=1$ and ${\mathcal{L}}^{-1}\{1/(s^2+4)\}=\frac{1}{2}\sin 2t$ (since $\mathcal{L}\{\sin at\}=a/(s^2+a^2)$, here $a=2$, so we divide by $2$). ${\mathcal{L}}^{-1}={\int }_0^{t}1\cdot \frac{1}{2}\sin (2(t-\tau ))d\tau =\frac{1}{2}{\int }_0^t\sin (2t-2\tau )d\tau .$ Let $w=2t-2\tau ,dw=-2d\tau$; as $\tau :0\to t$, $w:2t\to 0$: $=\frac{1}{2}\cdot \frac{-1}{2}[-\cos w]...\Rightarrow \frac{1}{2}[\frac{\cos (2t-2\tau )}{2}{]}_0^t=\frac{1}{4}(\cos 0-\cos 2t)=\boxed{\frac{1}{4}(1-\cos 2t)}.$ Check: $\mathcal{L}\{\frac{1}{4}(1-\cos 2t)\}=\frac{1}{4}(\frac{1}{s}-\frac{s}{s^2+4})=\frac{1}{4}\cdot \frac{(s^2+4)-s^2}{s(s^2+4)}=\frac{1}{4}\cdot \frac{4}{s(s^2+4)}=\frac{1}{s(s^2+4)}$. ✅

Take Laplace, using $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)=s^{2}Y$ (both zero): $s^{2}Y+4Y=F(s)\Rightarrow Y=\underset{\text{transfer function }H(s)}{\underbrace{\frac{1}{s^2+4}}}F(s).$ $Y$ is a product of two $s$-functions, so its inverse is a convolution: $y(t)=h(t)\ast f(t),h(t)={\mathcal{L}}^{-1}\{H(s)\}=\frac{1}{2}\sin 2t.$ Why this is beautiful: $h(t)=\frac{1}{2}\sin 2t$ is the system's response to a unit impulse. Any input $f$ produces $\boxed{y(t)=\frac{1}{2}{\int }_0^t\sin (2(t-\tau ))f(\tau )d\tau }$ — the output is the input, smeared by the impulse response. That "smearing" is convolution.

Spot the convolution. ${\int }_0^t(t-\tau )y(\tau )d\tau =(t\ast y)(t)$ because arguments $t-\tau$ and $\tau$ add to $t$. (See Volterra Integral Equations.) Laplace both sides. $\mathcal{L}\{e^t\}=\frac{1}{s-1}$, $\mathcal{L}\{t\}=\frac{1}{s^2}$, and convolution → product: $Y=\frac{1}{s-1}+\frac{1}{s^2}Y.$ Solve algebraically (this is the whole point — the integral became algebra): $Y\left(1-\frac{1}{s^2}\right)=\frac{1}{s-1}\Rightarrow Y\cdot \frac{s^2-1}{s^2}=\frac{1}{s-1}\Rightarrow Y=\frac{s^2}{(s-1)(s^2-1)}.$ Note $s^2-1=(s-1)(s+1)$, so $Y=\frac{s^2}{(s-1{)}^2(s+1)}$. Partial fractions (see Inverse Laplace Transform — partial fractions): $\frac{s^2}{(s-1{)}^2(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1{)}^2}+\frac{C}{s+1}.$

• $s=1$: $B(1+1)=1\Rightarrow B=\frac{1}{2}$.
• $s=-1$: $C(-1-1{)}^2=1\Rightarrow C\cdot 4=1\Rightarrow C=\frac{1}{4}$.
• Compare $s^2$ coefficients: $A+C=1\Rightarrow A=1-\frac{1}{4}=\frac{3}{4}$. $Y=\frac{3/4}{s-1}+\frac{1/2}{(s-1{)}^2}+\frac{1/4}{s+1}.$ Invert term by term: $\boxed{y(t)=\frac{3}{4}{e}^t+\frac{1}{2}t{e}^t+\frac{1}{4}{e}^{-t}}.$ Spot check at $t=0$: $y(0)=\frac{3}{4}+0+\frac{1}{4}=1=e^0$. And the RHS at $t=0$ is $e^0+0=1$. ✅

Laplace with $y(0)=0,y^{\prime }(0)=1$: $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-s\cdot 0-1=s^{2}Y-1$. $s^{2}Y-1+Y=\frac{1}{s^2+1}\Rightarrow (s^2+1)Y=1+\frac{1}{s^2+1}.$ $Y=\underset{\text{from initial velocity}}{\underbrace{\frac{1}{s^2+1}}}+\underset{\text{driven part}}{\underbrace{\frac{1}{(s^2+1{)}^2}}}.$ First term inverts to $\sin t$. Second term is a product $\frac{1}{s^2+1}\cdot \frac{1}{s^2+1}$, so it's $\sin t\ast \sin t$. From the parent note (App 2): $\sin t\ast \sin t=\frac{1}{2}(\sin t-t\cos t).$ Add: $\boxed{y(t)=\sin t+\frac{1}{2}\sin t-\frac{1}{2}t\cos t=\frac{3}{2}\sin t-\frac{1}{2}t\cos t}.$ Check IVs: $y(0)=0$ ✅. $y^{\prime }(t)=\frac{3}{2}\cos t-\frac{1}{2}\cos t+\frac{1}{2}t\sin t=\cos t+\frac{1}{2}t\sin t$, so $y^{\prime }(0)=1$ ✅.

Way 1 — convolution. Split $\frac{s}{(s^2+1{)}^2}=\frac{s}{s^2+1}\cdot \frac{1}{s^2+1}$. Inverses: ${\mathcal{L}}^{-1}\{s/(s^2+1)\}=\cos t$, ${\mathcal{L}}^{-1}\{1/(s^2+1)\}=\sin t$. So we need $\cos t\ast \sin t$: $\cos t\ast \sin t={\int }_0^t\cos \tau \sin (t-\tau )d\tau .$ Product-to-sum: $\cos \tau \sin (t-\tau )=\frac{1}{2}[\sin t-\sin (2\tau -t)]$ (using $\sin X\cos Y=\frac{1}{2}[\sin (X+Y)+\sin (X-Y)]$ with $X=t-\tau ,Y=\tau$). $=\frac{1}{2}{\int }_0^t\sin td\tau -\frac{1}{2}{\int }_0^t\sin (2\tau -t)d\tau =\frac{1}{2}t\sin t-\frac{1}{2}[\frac{-\cos (2\tau -t)}{2}{]}_0^t.$ The second bracket: $-\frac{1}{2}(-\frac{1}{2})[\cos t-\cos (-t)]=\frac{1}{4}[\cos t-\cos t]=0$. $\Rightarrow \cos t\ast \sin t=\frac{1}{2}t\sin t.$ Way 2 — known transform. $\mathcal{L}\{t\sin t\}=\frac{2s}{(s^2+1{)}^2}$ (differentiate $\mathcal{L}\{\cos t\}$ trick / table). Hence ${\mathcal{L}}^{-1}\{s/(s^2+1{)}^2\}=\frac{1}{2}t\sin t$. Agree: both give $\boxed{\frac{1}{2}t\sin t}$. ✅

General $\omega >0$. Laplace with zero IVs: $(s^2+{\omega }^2)Y=F(s)$, so $Y=\frac{1}{s^2+{\omega }^2}F(s).$ Now ${\mathcal{L}}^{-1}\{1/(s^2+{\omega }^2)\}=\frac{1}{\omega }\sin \omega t$ (because $\mathcal{L}\{\sin \omega t\}=\omega /(s^2+{\omega }^2)$, so we divide by $\omega$). This is the impulse response $h(t)$. The product $H(s)F(s)$ inverts to the convolution: $y(t)=h\ast f=\frac{1}{\omega }{\int }_0^t\sin (\omega (t-\tau ))f(\tau )d\tau .■$ Degenerate $\omega \to 0$. The equation becomes $y^{\prime \prime }=f(t)$. Take the limit inside the kernel: as $\omega \to 0$, $\frac{\sin (\omega (t-\tau ))}{\omega }\to (t-\tau )$ (since $\sin x\approx x$ for small $x$). So $y(t)\to {\int }_0^t(t-\tau )f(\tau )d\tau .$ Interpretation / check: $y^{\prime \prime }=f$ with zero IVs means integrate $f$ twice; and indeed ${\int }_0^t(t-\tau )f(\tau )d\tau$ is exactly the double antiderivative (differentiate once ⇒ ${\int }_0^{t}f(\tau )d\tau$, twice ⇒ $f(t)$). ✅ The kernel degrades gracefully — no blow-up.

$y(t)={\int }_0^t{e}^{-(t-\tau )}\cdot 1d\tau =e^{-t}{\int }_0^t{e}^{\tau }d\tau =e^{-t}(e^t-1)=\boxed{1-e^{-t}}.$ Near $t=0$: $1-e^{-t}\approx t$ (rises linearly from $0$ — the system hasn't "filled up" yet). $y(0)=0$: no output before any input has accumulated. As $t\to \infty$: $e^{-t}\to 0$, so $y\to 1$. The output saturates at the input level — a classic first-order "charging" curve. This is the frequency-domain cousin's step response. Cross-check via theorem: $H(s)=\frac{1}{s+1}$, $F(s)=\frac{1}{s}$, product $\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}$, inverse $1-e^{-t}$. ✅

For $t>ϵ$: $(f\ast g_{ϵ})(t)={\int }_0^{t}f(t-\tau )g_{ϵ}(\tau )d\tau =\frac{1}{ϵ}{\int }_0^{ϵ}f(t-\tau )d\tau .$ Why only $[0,ϵ]$? Because $g_{ϵ}(\tau )=0$ for $\tau >ϵ$ — the kernel is a thin, tall pulse of area $1$. That last integral is the average of $f$ over a window of width $ϵ$ ending at $t$. As $ϵ\to 0$, the average of a continuous function over a shrinking window converges to the point value: ${\lim }_{ϵ\to 0}(f\ast g_{ϵ})(t)=f(t).$ Interpretation: the unit-area pulse acts as the identity of convolution — convolving with it returns $f$ unchanged. That's exactly the role of the Dirac delta: $f\ast \delta =f$. Every impulse-response derivation in L3/L5.1 secretly relies on this.