2.4 KiB
#start of lec 21
From ma=F
m\frac{dv}{dt}=f(t)
integrate both sides:
m\int_{t_{0}} ^{t_{1}} \frac{dv}{dt}dt =\int _{{t_{0}}} ^{t_{1}}f(t) \, dt
mv(t_{1})-mv(t_{0})=\int _{t_{0}}^{t_{1}}f(t) \, dt
that is, change in momentum on the LHS equates to an impulse on the RHS.
(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.)
The definition of the Dirak delta function:
\delta(t-a)=\begin{cases}0, & t\ne a \\''\infty'', & t=a\end{cases}
however, a more useful definition is:
\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)
properties:
\int_{{-\infty}}^{\infty} \delta(t-a)\, dt=1
!Drawing 2023-10-25 13.16.20.excalidraw
\int _{\infty} ^t \delta(x-a)\, dx=\begin{cases}0, & t<a \\ 1, & a\leq t\end{cases}=u(t-a)
u'(t-a)=\delta(t-a)
What is \mathcal{L}\{\delta(t-a)\}?
\mathcal{L}\{\delta(t-a)\}=\int _{0} ^\infty \delta(t-a)e^{-st} \, dt for a>0
=\int _{-\infty} ^{\infty} e^{-st} \delta(t-a) \, dt=e^{-as} using the definition earlier
#ex
w''+6w'+5w=e^t\delta(t-1) \qquad w(0)=0 \qquad w'(0)=4solving this using something like #voparam would be very difficult, using LT should be very easy!
s^2W-sw(0)-w'(0)+6sW+5W=\int _{0} ^\infty e^{-st}e^t\delta(t-1)\, dt
=s^2-4+6sW+5W=\int _{-\infty} ^\infty e^{-st}e^t\delta(t-1)\, dt
(we can extend the range of the integral as the integrand is 0 for t<1)
this allows us to use the earlier definition of the delta function ie:
=e^{-(-s-1)} using definition of delta function earlier.
W(s^2+6s+5)=4+ee^{-s}
W(s)=\frac{4}{(s+1)(s+5)}+\frac{ee^{-s}}{(s+1)(s+5)}
w(t)=\frac{1}{4}\mathcal{L}^{-1}\left\{ \frac{1}{s+1} - \frac{1}{s+5} \right\}+e\mathcal{L}^{-1}\{e^{-s}\left( \frac{1}{s+1} - \frac{1}{s+5} \right)\}
=\underbrace{ \frac{1}{4}(e^{-t}-e^{ -5t }) }_{ \text{this came from initial conditions} }+\dots
\mathcal{L}^{-1}\{e^{as}F(s)\}=f(t-a)u(t-a)
y(t)= \frac{1}{4}(e^{-t}-e^{ -5t }) +\frac{e}{4}u(t-1)(e^{ -(t-1) }-e^{ -5(t-1) })notice that this RHS came from the impulse delta and the effect it has on the system. side note: delta functions are useful for quantum physics.
Lets start modelling some electric circuits again:
!Drawing 2023-10-25 13.43.26.excalidraw
0.2I_{1}+0.1I_{3}'+2I_{1}=g(t)
-I_{2}+0.1I_{3}'=0
I_{1}=I_{2}+I_{3}
voila, a system of three equations.
#end of lec 21