diff --git a/Chapter1.pdf b/Chapter1.pdf
index 8dcb690..13f8b38 100644
Binary files a/Chapter1.pdf and b/Chapter1.pdf differ
diff --git a/Chapter1.tex b/Chapter1.tex
index 41679a0..d8097f0 100644
--- a/Chapter1.tex
+++ b/Chapter1.tex
@@ -256,5 +256,11 @@
     \[C \ge \frac{10\m\A}{2\times 120\Hz \times 0.1\V} = \mans{417 \u\F}\]
     Now we need to find the AC input voltage. The peak voltage after rectification must be 10V (per the requirements). Since each phase of the AC signal must pass through 2 diode drops, we have to add this to find out what our AC peak-to-peak voltage must be. Thus we have
     \[V_{\in, \pp} = 10\V + 2(0.6\V) = \mans{11.2\V}\]
+
+    \ex{1.30}
+    $V_\out$ is simply the voltage at the output of an impedance voltage divider. We know that $Z_R = R$ and $Z_C = \frac{1}{j\omega C}$. Thus we have 
+    \[V_\out = \frac{Z_C}{Z_R + Z_C} V_\in = \frac{\frac{1}{j \omega C}}{R + \frac{1}{j \omega C}} V_\in = \frac{1}{1 + j \omega R C} V_\in\]
+    The magnitude of this expression can be found by multiplying by the complex conjugate and taking the square root.
+    \[sqrt{V_\out V_\out^*} = \frac{1}{\sqrt{1 + \omega^2R^2C^2}}V_\in\]
     
 \end{document}
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diff --git a/taoesolutions.sty b/taoesolutions.sty
index 949d69d..ff15a2b 100644
--- a/taoesolutions.sty
+++ b/taoesolutions.sty
@@ -12,6 +12,7 @@
 
 \newcommand{\todo}[1]{\textcolor{red}{\textbf{TODO: #1}}}
 \renewcommand{\title}[1]{\begin{center}\begin{Huge}#1\end{Huge}\end{center}}
+\newcommand{\subtitle}[1]{\begin{center}\begin{Large}#1\end{Large}\end{center}}
 \newcommand{\ex}[1]{\section*{Exercise #1}}
 \newcommand{\mans}[1]{\boxed{\mathbf{#1}}}
 \newcommand{\tans}[1]{\framebox{\textbf{#1}}}