Submit HW2

assignments/exam1_cheatsheet.md

@@ -3,12 +3,12 @@
 <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>
 
-:[](../introduction.md)
+:[Introduction](../introduction.md)
 
-:[](../network_models.md)
+:[Network models](../network_models.md)
 
-:[](../signals.md)
+:[Data and signals](../signals.md)
 
-:[](../digital_transmission.md)
+:[Digital transmission](../digital_transmission.md)
 
-:[](hw1.md)
+:[Homework 1](hw1.md)

assignments/exam2_cheatsheet.md

@@ -3,12 +3,12 @@
 <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>
 
-:[](../analog_transmission.md)
+:[Analog transmission](../analog_transmission.md)
 
-:[](../bandwidth_utilization.md)
+:[Bandwidth utilization](../bandwidth_utilization.md)
 
-:[](../switching.md)
+:[Switching](../switching.md)
 
-:[](../telephone_and_cable.md)
+:[Telephone and cable](../telephone_and_cable.md)
 
-:[](hw2.md)
+:[Homework 2](hw2.md)

assignments/hw1.md

@@ -27,7 +27,7 @@ $$
 \begin{array}{rcl}
   L & = & B^2 \\\\
   & = & 4^2 \\\\
-  & = & \bf 16 \sf\ levels
+  & = & \bf 16 \sf \ levels
 \end{array}
 $$
 
@@ -52,7 +52,7 @@ $$
 \begin{array}{rcl}
   B & = & F . R . P \\\\
   & = & 1 \times 1,200 \times 1,000 \times \log_{2}{1,026} \\\\
-  & = & \bf 12,000,000,000 \sf\ bits
+  & = & \bf 12,000,000,000 \sf \ bits
 \end{array}
 $$
 
@@ -143,7 +143,7 @@ $$
 \begin{array}{rcl}
   P_2 & = & P_1 . 10^{G / 10} \\\\
   & = & 1 \times 10^{10 / 10} \\\\
-  & = & \bf 10 \sf\ mW
+  & = & \bf 10 \sf \ mW
 \end{array}
 $$
 
@@ -154,7 +154,7 @@ $$
   P_3 & = & P_2 . 10^{G / 10} \\\\
   & = & 10 \times 10^{-(0.5 \times 10) / 10} \\\\
   & = & 10 \times 10^{-5 / 10} \\\\
-  & = & \bf 3.16 \sf\ mW
+  & = & \bf 3.16 \sf \ mW
 \end{array}
 $$
 
@@ -164,7 +164,7 @@ $$
 \begin{array}{rcl}
   P_4 & = & P_3 . 10^{G / 10} \\\\
   & = & 3.16 \times 10^{15 / 10} \\\\
-  & = & \bf 99.92 \sf\ mW
+  & = & \bf 99.92 \sf \ mW
 \end{array}
 $$
 
@@ -214,7 +214,7 @@ $$
   PT & = & \displaystyle \frac{D}{PS} \\\\
   & = & \frac{1,000 \textsf{ km}}{(2*10^8) \textsf{ m/s}} \\\\
   & = & \frac{1,000,000,000 \textsf{ cm}}{200,000,000 \textsf{ m/s}} \\\\
-  & = & \bf 5 \sf\ ms
+  & = & \bf 5 \sf \ ms
 \end{array}
 $$
 
@@ -225,7 +225,7 @@ $$
   TT & = & \displaystyle \frac{MS}{B} \\\\
   & = & \frac{2.5 \textsf{ Mb} \times 8}{5 \textsf{ Mbps}} \\\\
   & = & \frac{20 \textsf{ Mb}}{5 \textsf{ Mbps}} \\\\
-  & = & \bf 4 \sf\ s
+  & = & \bf 4 \sf \ s
 \end{array}
 $$
 

assignments/hw2.md

@@ -9,11 +9,25 @@
 
 > What is the carrier signal? Explain its role in the analog transmission.
 
+A carrier signal is a high-frequency signal produced by the sending device. The
+signal would then go through modulation, that is, modifying amplitude,
+frequency, or phase.
+
 ## Problem 2
 
 > Consider a low-pass channel with the carrier frequency of $15$ kHz. What is
   the BASK bit rate if $d=0.5$?
 
+**Using [Bandwidth for ASK](https://github.com/hendraanggrian/IIT-CS455/blob/main/analog_transmission.md#bandwidth-for-ask)**:
+
+$$
+\begin{array}{rcl}
+  B & = & (1+d).S \\\\
+  & = & (1+0.5).15 \sf \ KHz \\\\
+  & = & \bf 22.5 \sf \ KHz
+\end{array}
+$$
+
 ## Problem 3
 
 > Calculate the baud rate for the given bit rate and type of modulation:
@@ -23,40 +37,87 @@
 > - $60$ kbps, 16-QAM
 > - $9000$ bps, QPSK
 
+| Bit rate | Baud rate |
+| --- | ---: |
+| $3000$ bps, BFSK | $3,000 \textsf{ bps}\ /\ 1 = \bf 3,000 \sf \ bps$ |
+| $50$ kbps, BASK | $50 \textsf{ kbps}\ /\ 1 = \bf 50 \sf \ kbps$ |
+| $60$ kbps, 16-QAM | $60 \textsf{ kbps}\ /\ 4 = \bf 15 \sf \ kbps$ |
+| $9000$ bps, QPSK | $9,000 \textsf{ bps}\ /\ 2 = \bf 4,500 \sf \ bps$ |
+
 ## Problem 4
 
 > The constellation diagram has two points whose coordinates are $(2,0)$
   and $(8,0)$. What modulation does this diagram present? What is the
   interpretation of these coordinates?
 
+In constellation diagram, $X$-axis is related to in-phase career while $Y$-axis
+is quadrature career. Because $X$ axis is moving $(2 \to 8)$ and $Y$ stays the
+same $(0)$, it suggests that **BFSK modulation** is in use.
+
 ## Problem 5
 
 > How is a channel related to a link?
 
+In the context of multiplexing, many channels share one link. Channel refers to
+the portion of a link that carries a transmission between a given
+sender-receiver pair.
+
 ## Problem 6
 
-> How is a channel related to a link?
+> Why is a filter needed in the FDM demultiplexing process?
+
+Filters are needed in demultiplexing FDM to decompose the multiplexed signals
+into their constituent component signals, separating the input signals from
+their carriers and passing them to the output lines.
 
 ## Problem 7
 
 > We would like to digitize a certain analog signal using $5$ bits per sample.
   The lowest frequency of this signal is $100$ kHz and its bandwidth is $200$
   kHz. What is the bit rate?
 
+**Using [Nyquist theorem](https://github.com/hendraanggrian/IIT-CS455/blob/main/digital_transmission.md#nyquist-theorem)**:
+
+$$
+\begin{array}{rcl}
+  C & = & 2B . S \\\\
+  & = & 2.(100+200) \textsf{ KHz} / 5 \\\\
+  & = & 600 \textsf{ KHz} / 5 \\\\
+  & = & \bf 3,000 \sf \ kbps
+\end{array}
+$$
+
 ## Problem 8
 
 > Can packets in a datagram network arrive at their destination out of order? If
   so, why?
 
+A packet in a datagram network is created independently even if a packet is part
+of multipacket transmission. Therefore yes, they are unordered.
+
 ## Problem 9
 
 > What are the basic differences between circuit switching and datagram
   switching?
 
+| | Circuit switching | Datagram switching |
+| --- | --- | --- |
+| Type | Connection-oriented | Connection-less |
+| Flow | Connection setup → Data transfer → Connection teardown | Carry header → Routing table → Forward packet |
+| Efficiency | **Bad** — Resources are allocated for the entire duration of the connection. | **Better** — Resources are allocated only when there are packets to be transferred. |
+| Delay | **Minimal** — Due to the time needed to create the connection, transfer data and disconnect the circuit. | **Significant** — Each packet may wait at a switch before it is forwarded. |
+
 ## Problem 10
 
 > Explain the function of a dial-up modem.
 
+The function of a dial-up modem is to modulate and demodulate.
+
+| Function | Conversion |
+| --- | --- |
+| Modulate | Binary data → Bandpass analog signal |
+| Demodulate | Modulated signal → Recovered binary data |
+
 ## Problem 11
 
 > The FHSS technique with $16$ different carrier frequencies is used to create a
@@ -65,6 +126,20 @@
   sender-receiver pairs from $1$ to $8$. What is the resulting expanded
   bandwidth $B2$?
 
+**Using [FHSS](https://github.com/hendraanggrian/IIT-CS455/blob/main/bandwidth_utilization.md#fhss)**:
+
+$$
+\begin{array}{rcl}
+  B_2 & = & B_1 . N \\\\
+  & = & 16 \textsf{ carriers} . 8 \\\\
+  & = & \bf 128 \sf \ carriers
+\end{array}
+$$
+
 ## Problem 12
 
 > Explain why ADSL is asymmetric.
+
+- Faster and inexpensive — Compared to a dial-up modem.
+- Focus on what's important — Most users require download instead of
+  upload speed.

bandwidth_utilization.md

@@ -41,7 +41,7 @@ lines onto higher-bandwidth lines.
 ### TDM
 
 **Time division multiplexing (TDM)** is a digital process that allows several
-connections to share the high bandwidth of a link. INstead of sharing a portion
+connections to share the high bandwidth of a link. Instead of sharing a portion
 of the bandwidth as in FDM, **time is shared**. Each connection occupies a
 portion of time in the link.
 
@@ -55,13 +55,13 @@ The data rate of the output link must be in $n$ times higher than the data rate
 of a single input link to guarantee the flow of data.
 
 TDM can be visualized as two fast-rotating switches, one on the multiplexing
-side and the other on the de-multiplexing side. And the switch is
+side and the other on the demultiplexing side. And the switch is
 **unsynchronized** and rotate at the same speed but in positive direction.
 
 | Side | Description |
 | --- | --- |
 | Multiplexing | As the switch open in front of a connection, this connection can send a unit of data onto the link. This process is called interleaving. |
-| De-multiplexing | As the switch open, the connection can receive a unit of data from the link. |
+| Demultiplexing | As the switch open, the connection can receive a unit of data from the link. |
 
 If a source does not have data to send, the corresponding slot in the output
 frame is empty.
@@ -93,7 +93,7 @@ digital signal (DS) service of digital hierarchy.
 
 | Service | Description | Equation |
 | --- | --- | --- |
-| DS-0 | Single digital channel of $64$ Kbps. | $(2.400) . 8 \textsf{ bit} = 64,000 \sf\ bps$ |
+| DS-0 | Single digital channel of $64$ Kbps. | $(2.400) . 8 \textsf{ bit} = 64,000 \sf \ bps$ |
 | DS-1 | $1,544$ Mbps source. | $$1,544 \textsf{ Mbps} = \underbrace{24 . 64 \textsf{ Kbps}}_\textsf{phone channels} + \underbrace{8 \textsf{ Kbps}}_\textsf{overhead}$$ |
 | DS-2 | $6,312$ Mbps source. | $6,312 \textsf{ Mbps} = 96 . 64 \textsf{ Kbps} + 168 \textsf{ Kbps}$ |
 | DS-3 | $44,376$ Mbps source. | $44,376 \textsf{ Mbps} = 672 . 64 \textsf{ Kbps} + 1,368 \textsf{ Kbps}$ |
@@ -105,8 +105,8 @@ The telephone companies use T-lines to implement the DS services.
 
 $$
 \begin{array}{rcl}
-  24.8 + 1 & = & 193 \sf\ bits \\\\
-  1,544 \sf\ Mbps & = & 24 . 64 \textsf{ Kbps} + 8 \textsf{ Kbps}
+  24.8 + 1 & = & 193 \sf \ bits \\\\
+  1,544 \sf \ Mbps & = & 24 . 64 \textsf{ Kbps} + 8 \textsf{ Kbps}
 \end{array}
 $$
 
@@ -149,3 +149,10 @@ If the number of hopping frequencies is $m$, we can multiplex $m$ channels by
 using the same $B_{SS}$ bandwidth. This is possitive because a station uses just
 one frequency in each hopping period. $m-1$ other frequencies can be used by
 other $m-1$ stations.
+
+$$
+\begin{array}{rcl}
+  B_2 & = & B_1 . N \\\\
+  \sf bandwidth\ \#2 & = & \sf bandwidth\ \#1 \times num\ of\ pairs
+\end{array}
+$$

digital_transmission.md

@@ -186,15 +186,22 @@ least twice the highest frequency in the original signal.
 | Low-pass | The same as the highest frequency. |
 | Bandpass | Lower than the value of maximum frequency. |
 
+$$
+\begin{array}{rcl}
+  C & = & 2B / S \\\\
+  \sf \ bit\ rate & = & 2 \textsf{ bandwidth} / \sf bits\ per\ sample
+\end{array}
+$$
+
 > #### Example: Telephone
 >
 > $$
   \begin{array}{rcl}
-    f_\textsf{max} & = & 4 \sf\ KHz \\\\
+    f_\textsf{max} & = & 4 \sf \ KHz \\\\
     f_S & = & 4 . 4000 \\\\
     & = & 8000 . \frac{\textsf{samples}}{S} \\\\
     & = & 8000 . 8 \\\\
-    & = & \bf 64 \sf\ Kbps
+    & = & \bf 64 \sf \ Kbps
   \end{array}
   $$
 
@@ -227,8 +234,8 @@ The quantization error affects the signal-to-noise ratio of the signal.
 
 $$
 \begin{array}{rcl}
-  SNR_{dB} & = & 6.02 \times B + 1.76 \sf\ dB \\\\
-  \sf depends\ on\ the\ num\ of\ quantization\ levels\ L & = & 6.02 \times \textsf{num of bits per sample} + 1.76 \sf\ dB
+  SNR_{dB} & = & 6.02 \times B + 1.76 \sf \ dB \\\\
+  \sf depends\ on\ the\ num\ of\ quantization\ levels\ L & = & 6.02 \times \textsf{num of bits per sample} + 1.76 \sf \ dB
 \end{array}
 $$
 

signals.md

@@ -39,7 +39,7 @@ $$
   \begin{array}{rcl}
     \lambda & = & \frac{v}{f} \\\\
     & = & \frac{3.10^8}{4.10^{14}} \\\\
-    & = & \bf 0,75 \sf\ mm
+    & = & \bf 0,75 \sf \ mm
   \end{array}
   $$
 
@@ -90,7 +90,7 @@ $$
 > $$
   \begin{array}{rcl}
     100 . 24 . 80 . 8 & = & \frac{1,536,000}{60} \\\\
-    & = & \bf 25,600 \sf\ bps
+    & = & \bf 25,600 \sf \ bps
   \end{array}
   $$
 
@@ -166,7 +166,7 @@ respectively.
     & = & 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
+    & = & \bf 7 \sf \ dB
   \end{array}
   $$
 

switching.md

@@ -12,7 +12,7 @@ more devices lined to the switch.
   dedicated during the entire duration of a phone call until the teardown phase.
 3. There is a continuous flow of data sent by the source station and received by
   the destination station.
-4. The switches route the data (call) based on their occupied band (FDM) or time
+4. The switches route the data (call) based on their occupied band (TDM) or time
   slot (FDM).
 
 ### Three phases