diff --git a/content/Periodic functions (lec 19).md b/content/Periodic functions (lec 19).md
index 77520a2..b18c2c8 100644
--- a/content/Periodic functions (lec 19).md	
+++ b/content/Periodic functions (lec 19).md	
@@ -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)
 
diff --git a/content/drawings/PeriodicGraph.png b/content/drawings/PeriodicGraph.png
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