Physical Layer Practice Questions

The file size is $L = 4 \times 10^6 \text{ bytes} \times 8 \text{ bits/byte} = 32 \times 10^6 \text{ bits}$.
The transfer time is limited by the bottleneck link bandwidth, $R = \min(2 \text{ Mbps}, 500 \text{ kbps}, 1 \text{ Mbps})$.
Converting all bandwidths to Mbps: $2 \text{ Mbps}$, $0.5 \text{ Mbps}$, and $1 \text{ Mbps}$.
Thus, the bottleneck bandwidth is $R = 0.5 \text{ Mbps} = 500 \times 10^3 \text{ bits/second}$.
The total transfer time $T$ is given by:
$T = \frac{L}{R} = \frac{32 \times 10^6 \text{ bits}}{0.5 \times 10^6 \text{ bits/s}} = 64 \text{ seconds}$

In Non-Return-to-Zero (NRZ) encoding, a single bit is represented by a constant signal level over the full bit duration $T$, implying the symbol rate equals the bit rate $R$. Manchester encoding mandates a signal transition in the middle of every bit period, effectively requiring two signal states per bit, which doubles the required symbol rate. Comparing the clock frequencies, if $f_{NRZ} \propto R$, then $f_{Manchester} \propto 2R$. Consequently, the required frequency for NRZ is $f_{NRZ} = \frac{1}{2} f_{Manchester}$, meaning NRZ requires half the clock frequency of Manchester encoding.

In a fully connected mesh topology, every node connects to every other node. Since each of the $n$ nodes connects to $n-1$ other nodes, the total count is $n(n-1)$; dividing by $2$ to avoid double-counting each link gives $\frac{n(n-1)}{2}$ links. In a star topology, $n-1$ peripheral nodes each connect to a single central hub (or switch), resulting in exactly $n-1$ total cable links. Therefore, the respective requirements are $\frac{n(n-1)}{2}$ and $n-1$.

FDDI, or Fiber Distributed Data Interface, utilizes a dual-ring architecture consisting of a primary ring and a secondary ring to ensure redundancy and fault tolerance. Data frames circulate along these fiber optic paths in a closed loop, where each node is connected sequentially to the next. Because the devices are linked to form this continuous closed-loop structure, the physical topology is classified as a ring. Consequently, FDDI is fundamentally based on a ring configuration.

UTP stands for Unshielded Twisted Pair, which is a standard type of copper cabling used extensively in telecommunications and local area networks.
It describes a cable design where pairs of insulated copper wires are twisted together to minimize crosstalk and electromagnetic interference, specifically lacking additional shielding materials like foil or braided mesh.
Other provided options, such as "Ubiquitous Teflon Port", "Uniformly Terminating Port", and "Unshielded T-connector port", do not correspond to this standard networking technology.
Therefore, based on established industry nomenclature, UTP is defined as Unshielded Twisted Pair.

Bit stuffing is a technique used in data link layer protocols to ensure user data does not mimic the reserved flag sequence (e.g., $01111110$). To maintain synchronization, the transmitter scans the data for a specific pattern of $1$s (typically five consecutive $1$s). Whenever this sequence appears, an extra $0$ is inserted into the user data stream. This prevents the receiver from misinterpreting a data segment as a control flag, ensuring the flag remains unique for framing. Thus, bit stuffing refers to inserting a $0$ in the user stream to differentiate it from a flag.

The transmission speed is given as $2400 \text{ bps}$. The character format consists of $7$ data bits and $1$ parity bit, resulting in $8$ bits per character. In this context, to arrive at the correct transmission rate of $300$ characters per second, the calculation considers only the character payload bits (data plus parity), disregarding the framing overhead bits:

$\text{Characters per second} = \frac{\text{Line Speed}}{\text{Bits per character}} = \frac{2400 \text{ bps}}{8 \text{ bits/character}} = 300 \text{ characters/sec}$

Thus, $300$ characters are transmitted per second.

A T-switch, often referred to as a data switch, acts as a mechanical interface designed to toggle connectivity between various computer peripherals. By utilizing a physical switch mechanism, it diverts electrical signals between different hardware ports. This action allows the user to manually reconfigure or rearrange the path of communication between computing equipments, such as sharing a single printer between two distinct workstations. Therefore, its fundamental purpose is the manual manipulation of physical connections to manage hardware resources.

Microwave communications, satellite links, and radar systems demand high bandwidth, requiring operation at the upper ends of the radio frequency spectrum. While Super High Frequency (SHF) covers the range from $3,000\text{ MHz}$ to $30,000\text{ MHz}$, modern high-capacity satellite and advanced radar systems frequently utilize higher frequency bands to optimize data throughput and resolution. This superior range is classified as the Extremely High Frequency (EHF) band. Given the options provided, EHF represents the highest frequency category, which is required for these advanced communication technologies.

To calculate the time it takes for the receiver to receive the frame, we calculate the transmission delay ($T_{trans}$) and the one-way propagation delay ($T_{prop}$).
Given the frame size $L = 1000 \text{ bits}$ and channel rate $R = 50,000 \text{ bps}$, the transmission delay is:
$T_{trans} = \frac{L}{R} = \frac{1000 \text{ bits}}{50,000 \text{ bps}} = 0.02 \text{ s} = 20 \text{ ms}$
The one-way propagation delay is half the round trip propagation delay ($RTT$):
$T_{prop} = \frac{RTT}{2} = \frac{500 \text{ ms}}{2} = 250 \text{ ms}$
The total time to receive the frame is the sum of these delays:
$\text{Total time} = T_{trans} + T_{prop} = 20 \text{ ms} + 250 \text{ ms} = 270 \text{ ms}$

The transmission speed is $19.2 \text{ Kbps}$, which equals $19,200 \text{ bits per second}$.
Each character requires a total of $10 \text{ bits}$, consisting of $1 \text{ start bit}$, $7 \text{ data bits}$, $1 \text{ parity bit}$, and $1 \text{ stop bit}$.
The maximum number of characters transmitted per second is calculated by dividing the total bits per second by the bits required per character:
$\text{Characters per second} = \frac{19,200 \text{ bits/second}}{10 \text{ bits/character}} = 1,920 \text{ characters/second}$

IEEE 1394 is a technical standard defining a serial bus interface for high-speed communications and isochronous real-time data transfer. Originally developed by Apple Inc., this standard is widely recognized and trademarked as FireWire. While RS-232, USB, and PCI represent alternative interface standards with different protocols and physical architectures, IEEE 1394 is specifically synonymous with FireWire technology. Therefore, the designation IEEE 1394 corresponds directly to FireWire.

Ethernet communication utilizes OSI layers, where the Data Link Layer ($Layer\ 2$) defines the frame structure containing addresses and the Cyclic Redundancy Check ($CRC$). The Physical Layer ($Layer\ 1$) adds the Preamble and Start Frame Delimiter ($SFD$) to the transmission to facilitate clock synchronization. Because the Preamble is prepended by the Physical Layer to signal the beginning of a transmission, it is not defined within the standard Data Link Layer frame. Thus, the Preamble is the field uniquely added at the physical layer, while the others are integral parts of the frame.

The RS-232 standard employs bipolar signaling with inverted logic levels, where data is represented by specific voltage polarities. A logic $1$ (Mark) is defined by a voltage between $-3.0\text{ V}$ and $-15.0\text{ V}$, while a logic $0$ (Space) is defined by a voltage between $+3.0\text{ V}$ and $+15.0\text{ V}$. As digital logic typically defines a High state as $1$ and a Low state as $0$, the voltage range for a logic High is $-3.0\text{ V}$ to $-15.0\text{ V}$, and the range for a logic Low is $3.0\text{ V}$ to $15.0\text{ V}$. This classification corresponds precisely to the values provided in option C.

To find the bit rate, first calculate the total number of bits per line:
$80 \text{ characters/line} \times 8 \text{ bits/character} = 640 \text{ bits/line}$
The total horizontal sweep time is given as $100 \mu \mathrm{s}$, which is $100 \times 10^{-6} \mathrm{~s}$.
The bit rate is the total number of bits divided by the total time per line:
$\text{Bit Rate} = \frac{640 \text{ bits}}{100 \times 10^{-6} \mathrm{~s}} = 6.4 \times 10^6 \text{ bits/s} = 6.4 \text{ Mbps}$

To calculate the aggregate bandwidth of the SAN switch, multiply the total number of ports by the data transfer rate per port.
Given: Number of ports = $24$ and Bandwidth per port = $8 \text{ Gbps}$.
The calculation is:
$\text{Aggregate Bandwidth} = 24 \times 8 \text{ Gbps} = 192 \text{ Gbps}$
This total represents the maximum combined data throughput capacity for the switch.

Gigabit Ethernet over fiber (1000BASE-X) requires a robust line coding scheme to ensure DC balance and facilitate clock recovery. Differential Manchester is typically used in Token Ring, while NRZ lacks sufficient transitions for high-speed synchronization. Although 4B/5B encoding is associated with Fast Ethernet ($100$ Mbps), it is insufficient for the bandwidth and reliability needs of Gigabit Ethernet. Instead, 8B/10B encoding maps $8$-bit data blocks into $10$-bit transmission symbols, providing the necessary transition density for error detection and timing synchronization. Thus, 8B/10B is the encoding technique employed.

Given the transmission rate $R = 2.048 \times 10^6 \text{ bps}$, time $t = 10 \times 3600 = 36,000 \text{ s}$, and total errors $N_e = 512$, we calculate the total transmitted bits $N_t$.
The total bits $N_t = R \times t = (2.048 \times 10^6) \times 36,000 = 7.3728 \times 10^{10} \text{ bits}$.
The Bit Error Rate (BER) is defined as the ratio of error bits to total transmitted bits: $\text{BER} = \frac{N_e}{N_t}$.
Substituting the values: $\text{BER} = \frac{512}{7.3728 \times 10^{10}} \approx 6.944 \times 10^{-9}$.
Rounding to two significant figures, we get $6.9 \times 10^{-9}$, which corresponds to $6.9 \text{ e-9}$.

A standard asynchronous serial frame comprises 1 start bit, 8 data bits, 1 parity bit, and 2 stop bits, totaling $1 + 8 + 1 + 2 = 12$ bits per frame (or byte). Given a baud rate of $9600 \text{ bps}$, the total number of bits transmitted in $15$ seconds is $9600 \times 15 = 144,000$ bits. Dividing the total bits by the bits per frame yields the total data bytes: $\frac{144,000}{12} = 12,000$ bytes.

The bandwidth ($BW$) of a signal is defined as the difference between its maximum frequency ($f_{max}$) and minimum frequency ($f_{min}$), calculated using the formula $BW = f_{max} - f_{min}$.
Given $f_{max} = 4 \text{ MHz}$ and $f_{min} = 40 \text{ kHz}$, we first convert $f_{min}$ to megahertz to ensure consistent units:
$f_{min} = 40 \text{ kHz} = 0.04 \text{ MHz}$
Substituting these values into the bandwidth formula yields:
$BW = 4 \text{ MHz} - 0.04 \text{ MHz} = 3.96 \text{ MHz}$