ece205: add convolution, impulse

docs/2a/ece205.md

@@ -162,6 +162,57 @@ $$
 \mathcal L\{u_c(t)f(t-c)\}=e^{-sc}\mathcal L\{f\}
 $$
 
+### Convolution
+
+Convolution is a weird thingy that does weird things.
+
+$$(f*g)(t)=\int^t_0f(\tau)g(t-\tau)d\tau$$
+
+It is commutative ($f*g=g*f$) and is useful in transforms:
+
+$$\mathcal L\{f*g\}=\mathcal L\{f\}\mathcal L\{g\}$$
+
+!!! example
+    To solve $4y''+y=g(t),y(0)=3, y'(0)=-7$:
+    
+    \begin{align*}
+    4\mathcal L\{y''\}+\mathcal L\{y\}&=\mathcal L\{g(t)\} \\
+    4(s^2\mathcal L\{y\}-s\cdot y(0) - y'(0))+\mathcal L\{y\} &=\mathcal L\{g(t)\} \\
+    \mathcal L\{y\}(4s^2+1)-12s+28&=\mathcal L\{g(t)\} \\
+    \mathcal L\{y\}&=\frac{\mathcal L\{g(t)\}}{4s^2+1} + \frac{12s}{4s^2+1} - \frac{28}{4s^2+1} \\
+    y&=\mathcal L^{-1}\left\{\frac{1}{4s^2+1}\mathcal L\{g(t)\}\right\} + \mathcal L^{-1}\left\{3\frac{s}{s^2+\frac 1 4}\right\}-\mathcal L^{-1}\left\{7\frac{1}{s^2+\frac 1 4}\right\} \\
+    &= \mathcal L^{-1}\left\{\frac 1 2\mathcal L\left\{\sin\left(\tfrac 1 2 t\right)\right\}\mathcal L\{g(t)\} \right\}+3\cos\left(\tfrac 1 2 t\right)-14\sin\left(\tfrac 1 2t\right) \\
+    &=\frac 1 2\left(\sin\left(\tfrac 1 2 t\right)*g(t)\right)+3\cos\left(\tfrac 1 2 t\right)-14\sin\left(\tfrac 1 2t\right) \\
+    &=\frac 1 2\int^t_0\sin(\tfrac 1 2\tau)g(t-\tau)d\tau + 3\cos(\tfrac 1 2 t)-14\sin(\tfrac 1 2 t)
+    \end{align*}
+
+### Impulse
+
+The **impulse for duration $\epsilon$** is defined by the **dirac delta function**:
+
+$$
+\delta_\epsilon(t)=\begin{cases}
+\frac 1\epsilon & \text{if }0\leq t\leq\epsilon \\
+0 & \text{else}
+\end{cases}
+$$
+
+As $\epsilon\to 0, \delta_\epsilon(t)\to\infty$. Thus:
+$$
+\delta(t-a)=\begin{cases}
+\infty & \text{if }t=a \\
+0 & \text{else}
+\end{cases} \\
+\boxed{\int^\infty_0\delta(t-a)dt=1}
+$$
+
+If a function is continuous, multiplying it by the impulse function is equivalent to turning it on at that particular point. For $a\geq 0$:
+
+$$\boxed{\int^\infty_0\delta(t-a)dt=g(a)}$$
+
+Thus we also have:
+
+$$\mathcal L\{\delta (t-a)\}=e^{-as}\implies\mathcal L^{-1}\{1\}=\delta(t)$$
 
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