revised periodic functions

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Sasserisop 2023-11-09 15:08:28 -07:00
parent 71ba65ecdd
commit 3663c2779a
2 changed files with 21 additions and 15 deletions

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@ -48,21 +48,21 @@ $s^2Y+3sY+2Y=\mathcal{L}\{f\}= \frac{1}{s(1+e^{-1s})}$
$(s+1)(s+2)Y= \frac{1}{s(1+e^{-1s})}$
$Y(s)=\frac{1}{s(s+1)(s+2)} \frac{1}{1+e^{-s}}$
What property can we use to find the inverse of that?
> psst. We can use $\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\} \frac{1}{1-e^{-Ts}}$ ;)
> psst. We can use $\mathcal{L}\{y\}=\mathcal{L}\{y_{T}\} \frac{1}{1-e^{-Ts}}$ ;)
$Y(s)=F(s) \frac{1}{1+e^{-s}}$
But that second term has a $1+e^{-Ts}$ term in the denominator, it doesn't match up in the formula. There is a fix, peep this:
$Y(s)=F(s) \frac{1-e^{-s}}{(1+e^{-s})(1-e^{-s})}$
$Y(s)=F(s) \frac{1-e^{-s}}{1-e^{-2s}}$ <- We are rather lucky, the 2 in the denominator matches $T$, the period of our function.
We can tuck away the numerator into $F(s)$ :
$Y(s)=\underbrace{ \frac{1-e^{-s}}{s(s+1)(s+2)} }_{ F(s) } \frac{1}{1-e^{-2s}}$
Our equation is now in the correct form. We can now calculate the inverse of $F(s)$
Split up $F(s)$ :
$Y(s)=Y_{T}(s) \frac{1}{1+e^{-s}}$ <- But this is not true! because:
That second term has a $1+e^{-Ts}$ term in the denominator, it doesn't match up in the formula. There is a fix, peep this:
$Y(s)=Y_{T}(s) \frac{1-e^{-s}}{(1+e^{-s})(1-e^{-s})}$
$Y(s)=Y_{T}(s) \frac{1-e^{-s}}{1-e^{-2s}}$ <- We are rather lucky, the 2 in the denominator matches $T$, the period of our function.
We can tuck away the numerator into $Y_{T}(s)$ (this does redefine $Y_{T}$ to the correct expression, and the equation below is now true.)
$Y(s)=\underbrace{ \frac{1-e^{-s}}{s(s+1)(s+2)} }_{ Y_{T}(s) } \frac{1}{1-e^{-2s}}$
Our equation is now in the correct form. We can now calculate the inverse of $Y_{T}(s)$
Split up $Y_{T}(s)$ :
$Y(s)= \underbrace{ (\frac{1}{s(s+1)(s+2)} }_{ F_{1}(s) }-\underbrace{ \frac{e^{-s}}{s(s+1)(s+2)} )}_{F_{2}(s) } \frac{1}{1-e^{-2s}}$
We can use partial fractions for the first term and
$\mathcal{L}\{u(t-a)f_{1}(t-a)\}=e^{-as}F_{1}(s)$ for the second term. (Where $a=1$)
using partial fractions:
Using partial fractions:
$\frac{1}{s(s+1)(s+2)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+2}$
$A(s+1)(s+2)+Bs(s+2)+Cs(s+1)=(A+B+C)s^2+(3A+2B+C)s+2A=1$
$2A=1\implies A=\frac{1}{2}$
@ -72,12 +72,18 @@ subtract the two equations.
$1+B=0\implies B=-1$
$\implies C=\frac{1}{2}$
$\mathcal{L}^{-1}\{F_{1}\}=\mathcal{L}^{-1}\{\frac{1}{2} \frac{1}{s}-\frac{1}{s+1}+\frac{1}{2} \frac{1}{s+2}\}$
$\mathcal{L}^{-1}\{F_{1}\}=f_{1}=\frac{1}{2}-e^{-t}+\frac{1}{2}e^{-2t}$
$\mathcal{L}^{-1}\{F_{1}\}=f_{1}(t)=\frac{1}{2}-e^{-t}+\frac{1}{2}e^{-2t}$
Second term ($F_{2}(s)$): use $\mathcal{L}\{u(t-a)f_{1}(t-a)\}=e^{-as}F_{1}(s)=F_{2}(s)$
Second term, $F_{2}(s)$, use: $\mathcal{L}\{u(t-a)f_{1}(t-a)\}=e^{-as}F_{1}(s)=F_{2}(s)$
$f_{2}(t)=u(t-1)(\frac{1}{2}-e^{-(t-1)}+\frac{1}{2}e^{-2(t-1)})$
$y(t)=p(t)=f_{2a}(t)$ meaning $y(t)$ is a periodic function, with period of $T=2$
$$f_{2a}(t)=\mathcal{L}^{-1}\{F_{1}(s)-F_{2}(s)\}=\frac{1}{2}+\frac{1}{2}e^{-2t}-e^{-t}-u(t-1)(\frac{1}{2}-e^{-(t-1)}+\frac{1}{2}e^{-2(t-1)})$$
recombine the two parts:
$Y_{T}=F_{1}-F_{2}$
$\mathcal{L}^{-1}\{Y_{T}\}=\mathcal{L}^{-1}\{F_{1}\}-\mathcal{L}^{-1}\{F_{2}\}$
$y_{T}(t)=f_{1}(t)-f_{2}(t)$
$y(t)$ is a periodic function, with period of $T=2$
It's windowed form, $y_{T}$, is:
$$y_{T}(t)=\frac{1}{2}+\frac{1}{2}e^{-2t}-e^{-t}-u(t-1)(\frac{1}{2}-e^{-(t-1)}+\frac{1}{2}e^{-2(t-1)})$$
Peep da plot!
![Plot](drawings/PeriodicGraph.png)

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