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#start of lec 19 This lecture we will learn about periodic functions, specifically, non-sinusoidal periodic functions.
Periodic function
Definition:
f is periodic with period T \quad (T>0) if:
f(t)=f(t+T), \quad \forall\ t\in \mathbb{R}!Drawing 2023-10-20 13.06.35.excalidraw We will now compute laplace transforms of these periodic functions. Computing DE's containing these periodic functions using something like #voparam would not be easy. If we take the windowed version of the function (one period, where everywhere else is 0, ie:)
f_{T}(t)=\begin{cases}f(t)\ ,\ & 0\leq t\leq T \\0\ ,\ & \text{otherwise}\end{cases}
we can "glue together" many of these windows together to rebuild our f(t), like this:
f(t)=f_{T}(t)+f_{T}(t-T)u(t-T)+f_{T}(t-2T)u(t-2T)+\dots
\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}+\mathcal{L}\{f(t-T)u(t-T)\}+\dots
recall the formula from last lec: \mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)
then:
\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}(1+e^{-TS}+e^{-2TS}+e^{-3TS}+\dots)
\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}(1+e^{-TS}+(e^{-TS})^{2}+(e^{-TS})^{3}+\dots)
This is a geometric series! 1+r+r^2+\dots
Geometric series are convergent when |r|<1
and equal to: \frac{1}{1-r}
in this case, r=e^{-Ts}
so:
\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\} \frac{1}{1-e^{-Ts}}handy formula! ^ will be used again.
#ex
imagine another function: (image is of a square wave with a period of 2a, oscillates between 1 and 0, starts at 1 when t=0.)
!Drawing 2023-10-20 13.27.58.excalidraw
\mathcal{L}\{f\}=\mathcal{L}\{f_{2a}\} \frac{1}{1-e^{-2as}}
f_{2a}=u(t)-u(t-a) (this is the first period piece)
\implies \mathcal{L}\{f_{2a}\}=\mathcal{L}\{u(t)\}-\mathcal{L}\{u(t-a)\}=\frac{1}{s}- \frac{e^{-as}}{s}
plug back in:
\mathcal{L}\{f\}=\mathcal{L}\{f_{2a}\} \frac{1}{1-e^{-2as}}=\frac{1}{s}\cancel{ (1-e^{-as}) } \frac{1}{\cancel{ (1-e^{-as}) }(1+e^{-as})}
\mathcal{L}\{f\}=\frac{1}{s(1+e^{-as})}#ex
y''+3y'+2y=f(t) where f(t) is from the previous example
y(0)=y'(0)=0,\ a=1 (a is width of 1/2 period in the function f(t))
s^2Y+3sY+2Y=\mathcal{L}\{f\}= \frac{1}{s(1+e^{-s})}
Y(s)=\frac{1}{s(s+1)(s+2)} \frac{1}{1+e^{-s}}
=F(s) \frac{1}{1+e^{-s}}
\mathcal{L}^{-1}\{F\}=\mathcal{L}^{-1}\{\frac{1}{2} \frac{1}{s}+\frac{1}{2} \frac{1}{s+2}-\frac{1}{s+1}\}
=\frac{1}{2}+\frac{1}{2}e^{-2t}-e^{-t}
y(t)= \dots
we want to use formula from earlier
so we need to change F(s) \frac{1}{1+e^{-s}}
to: F(s) \frac{1e^{-s}}{1-e^{-2s}}
y(t)=p(t), periodic with period of 2 (T=2)
f_{2a}(t)=\mathcal{L}^{-1}\{F(s)-F(s)e^{-s}\}=\frac{1}{2}+\frac{1}{2}e^{-2t}-e^{-t}-\left( \frac{1}{2}+\frac{1}{2} e^{-2(t-1)}-e^{-(t-1)})u(t-1 \right)