From 11300cacd8bd0653b28fbc1f2f5b7e8351d44503 Mon Sep 17 00:00:00 2001 From: Hendra Anggrian Date: Thu, 22 Jun 2023 16:22:32 -0500 Subject: [PATCH] Submit HW2 --- assignments/exam1_cheatsheet.md | 10 ++--- assignments/exam2_cheatsheet.md | 10 ++--- assignments/hw1.md | 14 +++--- assignments/hw2.md | 77 ++++++++++++++++++++++++++++++++- bandwidth_utilization.md | 19 +++++--- digital_transmission.md | 15 +++++-- signals.md | 6 +-- switching.md | 2 +- telephone_and_cable.md | 24 +++++----- 9 files changed, 135 insertions(+), 42 deletions(-) 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 | $\underbrace{31 \to 255}_{225 \sf \ channels}$ |