forked from Sasserisop/MATH201
50 lines
2.7 KiB
Markdown
50 lines
2.7 KiB
Markdown
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#start of lec 21
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From newton's second law: $ma=F$
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$m\frac{dv}{dt}=f(t)$
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integrate both sides:
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$m\int_{t_{0}} ^{t_{1}} \frac{dv}{dt}dt =\int _{{t_{0}}} ^{t_{1}}f(t) \, dt$
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$mv(t_{1})-mv(t_{0})=\int _{t_{0}}^{t_{1}}f(t) \, dt$
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that is, change in momentum on the LHS equates to an impulse on the RHS.
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(picture shown, you can have the same impulse, the same area under the graph if you squish down $f(t)$ to be narrower, as long as you make it taller. If we take it to the extreme we get the Dirak delta function.)
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# Dirak delta function
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#dirak_delta
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The definition of the Dirak delta function is:
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$\delta(t-a)=\begin{cases}0, & t\ne a \\''\infty'', & t=a\end{cases}$
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however, a more useful definition is:
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$\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$
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properties:
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$\int_{{-\infty}}^{\infty} \delta(t-a)\, dt=1$
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$\int _{-\infty} ^t \delta(t-a)\, dt=\begin{cases}0, & t<a \\ 1, & t\geq a\end{cases}=u(t-a)$
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$u'(t-a)=\delta(t-a)$
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What is $\mathcal{L}\{\delta(t-a)\}$?
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$\mathcal{L}\{\delta(t-a)\}=\int _{0} ^\infty \delta(t-a)e^{-st} \, dt$ for $a>0$
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Using the definition of Dirak $\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$:
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$$\mathcal{L}\{\delta(t-a)\}=\int _{-\infty} ^{\infty} e^{-st} \delta(t-a) \, dt=e^{-as}$$
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## Examples of DE's with Dirak
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#ex #second_order_nonhomogenous #dirak_delta #IVP
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Solve for $w(t)$:
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$$w''+6w'+5w=e^t\delta(t-1) \qquad w(0)=0 \qquad w'(0)=4$$
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Solving this using something like #voparam would be very difficult, using LT should be very easy!
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Hit it with the LT!
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$s^2W-\cancel{ sw(0) }-\cancelto{ 4 }{ w'(0) }+6sW-\cancel{ 6w(0) }+5W=\int _{0} ^\infty e^{-st}e^t\delta(t-1)\, dt$
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Now we try to isolate for $W$
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$=s^2W-4+6sW+5W=\int _{-\infty} ^\infty e^{-st}e^t\delta(t-1)\, dt$
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(we can extend the range of the integral as the integrand is 0 for all t<1)
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this allows us to use the definition of Dirak on the RHS, namely: $\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$
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> Alternatively, you could have taken the laplace of the RHS by using properties 2 and 14 from the [big LT table](drawings/bigLTtable.png).
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$W(s^2+6s+5)=4+ee^{-s}$
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$W(s)=\frac{4}{(s+1)(s+5)}+\frac{ee^{-s}}{(s+1)(s+5)}$
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Using partial fractions:
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$w(t)=\mathcal{L}^{-1}\left\{ \frac{1}{s+1} - \frac{1}{s+5} \right\}+\frac{e}{4}\mathcal{L}^{-1}\left\{ e^{-s}\left( \frac{1}{s+1} - \frac{1}{s+5} \right) \right\}$
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$=\underbrace{(e^{-t}-e^{ -5t }) }_{ \text{this came from initial conditions} }+\frac{e}{4}\mathcal{L}^{-1}\{\dots\}$
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for second term use property:
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$\mathcal{L}^{-1}\{e^{as}F(s)\}=f(t-a)u(t-a)$
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$$y(t)= (e^{-t}-e^{ -5t }) +\frac{e}{4}u(t-1)(e^{ -(t-1) }-e^{ -5(t-1) })$$
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notice that the right most term came from the impulse and the effect it had on the system.
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side note: delta functions are useful for quantum physics. |