diff --git a/assignments/exam1_cheatsheet.md b/assignments/exam1_cheatsheet.md
index 5ca7492..5ab603a 100644
--- a/assignments/exam1_cheatsheet.md
+++ b/assignments/exam1_cheatsheet.md
@@ -3,12 +3,12 @@
-:[](../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)
diff --git a/assignments/exam2_cheatsheet.md b/assignments/exam2_cheatsheet.md
index 7bb6ecd..fc84df8 100644
--- a/assignments/exam2_cheatsheet.md
+++ b/assignments/exam2_cheatsheet.md
@@ -3,12 +3,12 @@
-:[](../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)
diff --git a/assignments/hw1.md b/assignments/hw1.md
index 9113d9c..90d3060 100644
--- a/assignments/hw1.md
+++ b/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 @@ $$
TT & = & \displaystyle \frac{MS}{B} \\\\
& = & \frac{2.5 \textsf{ Mb} \times 8}{5 \textsf{ Mbps}} \\\\
& = & \frac{20 \textsf{ Mb}}{5 \textsf{ Mbps}} \\\\
- & = & \bf 4 \sf\ s
+ & = & \bf 4 \sf \ s
\end{array}
$$
diff --git a/assignments/hw2.md b/assignments/hw2.md
index 3345570..54e449a 100644
--- a/assignments/hw2.md
+++ b/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:
@@ -23,19 +37,38 @@
> - $60$ 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
@@ -43,20 +76,48 @@
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.
diff --git a/bandwidth_utilization.md b/bandwidth_utilization.md
index 6ef9552..f6aa39a 100644
--- a/bandwidth_utilization.md
+++ b/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.
+
+$$
+\begin{array}{rcl}
+ B_2 & = & B_1 . N \\\\
+ \sf bandwidth\ \#2 & = & \sf bandwidth\ \#1 \times num\ of\ pairs
+\end{array}
+$$
diff --git a/digital_transmission.md b/digital_transmission.md
index 5877f4e..1a3e0fe 100644
--- a/digital_transmission.md
+++ b/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.
$$
\begin{array}{rcl}
- SNR_{dB} & = & 6.02 \times B + 1.76 \sf\ dB \\\\
- \sf depends\ on\ the\ num\ of\ quantization\ levels\ L & = & 6.02 \times \textsf{num of bits per sample} + 1.76 \sf\ dB
+ SNR_{dB} & = & 6.02 \times B + 1.76 \sf \ dB \\\\
+ \sf depends\ on\ the\ num\ of\ quantization\ levels\ L & = & 6.02 \times \textsf{num of bits per sample} + 1.76 \sf \ dB
\end{array}
$$
diff --git a/signals.md b/signals.md
index 330081f..759bedb 100644
--- a/signals.md
+++ b/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}
$$
@@ -166,7 +166,7 @@ respectively.
& = & 10 (\log\frac{P_3}{P_2} + \log\frac{P_2}{P_1}) \\\\
& = & 10 \log\frac{P_3}{P_2} + 10 \log\frac{P_2}{P_1} \\\\
& = & 10 - 3 \\\\
- & = & \bf 7 \sf\ dB
+ & = & \bf 7 \sf \ dB
\end{array}
$$
diff --git a/switching.md b/switching.md
index da35e29..bb8c12c 100644
--- a/switching.md
+++ b/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
diff --git a/telephone_and_cable.md b/telephone_and_cable.md
index 6e3091a..095c283 100644
--- a/telephone_and_cable.md
+++ b/telephone_and_cable.md
@@ -18,30 +18,34 @@ is $2,400$ Hz, covering the range from $600$ Hz to $3,000$ Hz.
## Dial-up modems
The term **modem** is a word that refers to the two entities that make up the
-devices: a signal modulator and a signal de-modulator. A **modulator** creates a
-bandpass analog signal from binary data. A **de-modulator** recovers the binary
-data from the modulated signal.
+devices: a signal modulator and a signal de-modulator.
+
+| Entity | Role |
+| --- | --- |
+| Modulator | Creates a bandpass analog signal from binary data. |
+| Demodulator | Recovers the binary data from the modulated signal. |
### V.32
32-QAM with a baud rate of $2,400$. Because only $4$ bits of each $5$-bit
-pattern represent data, the resulting data rate is $4.2,400 = 9,600 \sf\ Kbps$.
+pattern represent data, the resulting data rate is $4.2,400 = 9,600 \sf \ Kbps$.
### V.32BIS
128-QAM ($7$ bits).
-$$2,400.6 = 14,400 \sf\ bps$$
+$$2,400.6 = 14,400 \sf \ bps$$
## Digital subscriber line
-The modulation technique that has become standard for ADSL is called the
-**discrete multitone technique (DMT)**. It combines QAM with FDM. An available
-bandwidth of $1,1$ MHz is divided into $256$ channels.
+The modulation technique that has become standard for *Asymmetric Digital
+Subscriber Line (ADSL)* is called the **discrete multitone technique (DMT)**. It
+combines QAM with FDM. An available bandwidth of $1,1$ MHz is divided into $256$
+channels.
| Type | Channel |
| --- | --- |
| Voice | $0$ |
| Idle | $1 \to 5$ |
-| Upstream data | $\underbrace{6 \to 30}_{25 \sf\ channels}$ |
-| Downstream data | $\underbrace{31 \to 255}_{225 \sf\ channels}$ |
+| Upstream data | $\underbrace{6 \to 30}_{25 \sf \ channels}$ |
+| Downstream data | $\underbrace{31 \to 255}_{225 \sf \ channels}$ |