Add cheatsheet for midexam

.vscode/settings.json

@@ -1,9 +1,12 @@
 {
   "markdown-pdf.format": "Letter",
   "markdown-pdf.styles": [
-    "~/Documents/GitHub/markdown-pdf-styles/styles/font-sans.css",
-    "~/Documents/GitHub/markdown-pdf-styles/styles/text-small.css",
+    "~/Documents/GitHub/markdown-pdf-styles/cheatsheet.css",
+    "~/Documents/GitHub/markdown-pdf-styles/styles/column-three.css",
   ],
-  "markdown-pdf.headerTemplate": "<div style='font-size: 7pt; margin-left: 1cm;'> <span>CS455 - Hendra Wijaya (A20529195)</span> </div> <div style='margin-left: auto; margin-right: 1cm;'> <img src='data:image/png;base64,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'> </div>",
-  "markdown-pdf.footerTemplate": "<div style='font-size: 7pt; margin-left: 1cm;'> <span>Homework 1</span> </div> <div style='font-size: 7pt; margin-left: auto; margin-right: 1cm;'> <span class='pageNumber'></span> / <span class='totalPages'></span> </div>",
+  "markdown-pdf.margin.top": "0.25in",
+  "markdown-pdf.margin.bottom": "0.25in",
+  "markdown-pdf.margin.right": "0.25in",
+  "markdown-pdf.margin.left": "0.25in",
+  "markdown-pdf.displayHeaderFooter": false,
 }

assignments/exam1_cheatsheet.md

@@ -0,0 +1,14 @@
+<!-- hotfix: KaTeX -->
+<!-- https://github.com/yzane/vscode-markdown-pdf/issues/21/ -->
+<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML"></script>
+<script type="text/x-mathjax-config">MathJax.Hub.Config({ tex2jax: { inlineMath: [['$', '$']] }, messageStyle: 'none' });</script>
+
+:[Internet](../internet.md)
+
+:[Networks](../networks.md)
+
+:[Signals](../signals.md)
+
+:[Transmission](../transmission.md)
+
+:[Homework 1](hw1.md)

networks.md

@@ -36,7 +36,7 @@ relationship of the links and devices to one another.
 
 $$
 \begin{array}{rcl}
-  w & = & \displaystyle \frac{n.(n-1)}{2} \\
+  w & = & \displaystyle \frac{n.(n-1)}{2} \\\\
   \sf WAN \ links & = & \sf \frac{sites.(sites-1)}{2}
 \end{array}
 $$

signals.md

@@ -8,14 +8,14 @@ Superposition of two sine waves:
 
 $$
 \begin{array}{rcl}
-  S & = & \displaystyle A . \sin(\frac{2 \pi}P + \alpha) \\
+  S & = & \displaystyle A . \sin(\frac{2 \pi}P + \alpha) \\\\
   \sf position & = & \sf amplitude \times \sin(\frac{2 \pi}{period} + phase)
 \end{array}
 $$
 
 $$
 \begin{array}{rcl}
-  F & = & 2S \\
+  F & = & 2S \\\\
   \sf superposition & = & \sf 2 \times position
 \end{array}
 $$
@@ -26,7 +26,7 @@ The wavelength is a distance a signal can travel in one period.
 
 $$
 \begin{array}{rcl}
-  \lambda & = & \displaystyle \frac{v}{f} \\
+  \lambda & = & \displaystyle \frac{v}{f} \\\\
   \sf wavelength & = & \sf \frac{propagation \ speed}{frequency}
 \end{array}
 $$
@@ -37,8 +37,8 @@ $$
 >
 > $$
   \begin{array}{rcl}
-    \lambda & = & \frac{v}{f} \\
-    & = & \frac{3.10^8}{4.10^{14}} \\
+    \lambda & = & \frac{v}{f} \\\\
+    & = & \frac{3.10^8}{4.10^{14}} \\\\
     & = & \bf 0,75 \sf \ mm
   \end{array}
   $$
@@ -62,7 +62,7 @@ If a signal has $L$ levels, each level needs $\log_2{L}$ bits.
 
 $$
 \begin{array}{rcl}
-  L & = & B^2 \\
+  L & = & B^2 \\\\
   \sf signal \ length & = & \sf bits^2
 \end{array}
 $$
@@ -74,7 +74,7 @@ The bit rate is the number of bits sent in 1 s expressed in bits per second
 
 $$
 \begin{array}{rcl}
-  B & = & F . R . P \\
+  B & = & F . R . P \\\\
   \sf bits & = & \sf frames \times resolution \times bits \ per \ pixel
 \end{array}
 $$
@@ -89,7 +89,7 @@ $$
 >
 > $$
   \begin{array}{rcl}
-    100 . 24 . 80 . 8 & = & \frac{1,536,000}{60} \\
+    100 . 24 . 80 . 8 & = & \frac{1,536,000}{60} \\\\
     & = & \bf 25,600 \sf \ bps
   \end{array}
   $$
@@ -123,7 +123,7 @@ bandwidth that does not start from $0$.
 
 $$
 \begin{array}{rcl}
-  P_2 & = & P_1 . 10^{G / 10} \\
+  P_2 & = & P_1 . 10^{G / 10} \\\\
   P_2 & = & P_1 . \sf 10^{gain / 10}
 \end{array}
 $$
@@ -145,7 +145,7 @@ signal at two different points.
 
 $$
 \begin{array}{rcl}
-  dB & = & L . D \\
+  dB & = & L . D \\\\
   \sf total \ loss & = & \sf loss \ per \ kilometer \times distance
 \end{array}
 $$
@@ -161,11 +161,11 @@ respectively.
 >
 > $$
   \begin{array}{rcl}
-    \log{a}.b & = & \log{a} + \log{b} \\
-    10 \log_{10}\frac{P_3}{P_1} & = & 10 \log_{10}\frac{P_3}{P_2} . \frac{P_2}{P_1} \\
-    & = & 10 (\log\frac{P_3}{P_2} + \log\frac{P_2}{P_1}) \\
-    & = & 10 \log\frac{P_3}{P_2} + 10 \log\frac{P_2}{P_1} \\
-    & = & 10 - 3 \\
+    \log{a}.b & = & \log{a} + \log{b} \\\\
+    10 \log_{10}\frac{P_3}{P_1} & = & 10 \log_{10}\frac{P_3}{P_2} . \frac{P_2}{P_1} \\\\
+    & = & 10 (\log\frac{P_3}{P_2} + \log\frac{P_2}{P_1}) \\\\
+    & = & 10 \log\frac{P_3}{P_2} + 10 \log\frac{P_2}{P_1} \\\\
+    & = & 10 - 3 \\\\
     & = & \bf 7 \sf \ dB
   \end{array}
   $$
@@ -190,7 +190,7 @@ The *Signal-to-Noise Ratio (SNR)* is actually the ratio of what is wanted
 
 $$
 \begin{array}{rcl}
-  SNR & = & \displaystyle \frac{ASP}{ANP} \\
+  SNR & = & \displaystyle \frac{ASP}{ANP} \\\\
   \sf signal-to-noise \ ratio & = & \sf \frac{average \ signal \ power}{average \ noise \ power}
 \end{array}
 $$
@@ -199,7 +199,7 @@ $SNR_{dB}$ is defined:
 
 $$
 \begin{array}{rcl}
-  SNR_{dB} & = & 10 \log_{10} SNR \\
+  SNR_{dB} & = & 10 \log_{10} SNR \\\\
   \sf (signal-to-noise \ ratio)_{dB} & = & 10 \log_{10} signal-to-noise \ ratio
 \end{array}
 $$
@@ -213,7 +213,7 @@ For a noiseless channel, the nyquist bit rate formula defines the
 
 $$
 \begin{array}{rcl}
-  R & = & 2 . B . \log_2 L \\
+  R & = & 2 . B . \log_2 L \\\\
   \sf bit \ rate (in \ bps) & = & \sf 2 \times bandwidth \times \log_2 signal \ levels \ used
 \end{array}
 $$
@@ -225,7 +225,7 @@ determine the **theoretical** highest data rate for a **noisy** channel:
 
 $$
 \begin{array}{rcl}
-  C & = & B . \log_2{(1 + SNR)} \\
+  C & = & B . \log_2{(1 + SNR)} \\\\
   \sf capacity (in \ bps) & = & \sf bandwidth \times \log_2{(1 + signal-to-noise \ ratio)}
 \end{array}
 $$
@@ -252,7 +252,7 @@ the source.
 
 $$
 \begin{array}{rcl}
-  L & = & PT + TT + QT + PD \\
+  L & = & PT + TT + QT + PD \\\\
   \sf latency & = & \sf propagation \ time + transmission \ time + queueing \ time + processing \ delay
 \end{array}
 $$
@@ -264,7 +264,7 @@ destination.
 
 $$
 \begin{array}{rcl}
-  PT & = & \displaystyle \frac{D}{PS} \\
+  PT & = & \displaystyle \frac{D}{PS} \\\\
   \sf propagation \ time & = & \sf \frac{distance}{propagation \ speed}
 \end{array}
 $$
@@ -279,7 +279,7 @@ message and the bandwidth of the channel.
 
 $$
 \begin{array}{rcl}
-  TT & = & \displaystyle \frac{MS}{B} \\
+  TT & = & \displaystyle \frac{MS}{B} \\\\
   \sf transmission \ time & = & \sf \frac{message \ size}{bandwidth}
 \end{array}
 $$

transmission.md

@@ -32,7 +32,7 @@ The relationship between data rate and signal rate:
 
 $$
 \begin{array}{rcl}
-  S & = & \frac{CN}{T} \\
+  S & = & \frac{CN}{T} \\\\
   \sf baud \ rate & = & \sf \frac{factor \times data \ rate (bps)}{data \ elements \ each \ signal}
 \end{array}
 $$