Physical Layer Practice Questions

Manchester encoding represents each bit with a transition in the middle of the bit period. To achieve this, each bit requires two signal changes: one to define the bit's logic (e.g., high-to-low for '1', low-to-high for '0') and another to ensure a transition for synchronization.

Thus, for every 1 bit transmitted, 2 signal changes (symbols) occur.
Let $R_b$ be the bit rate and $R_s$ be the baud rate (symbol rate).
The relationship is:
$R_s = 2 \times R_b$
Therefore, the bit rate is half the baud rate:
$R_b = \frac{R_s}{2}$

An 8-bit optical encoder utilizes $n = 8$ bits to represent positions within one full rotation, resulting in $2^n$ distinct angular positions.
The number of distinct positions is calculated as:
$2^8 = 256$
The resolution is obtained by dividing the total degrees in a circle ($360^\circ$) by the number of positions:
$\text{Resolution} = \frac{360^\circ}{256} = 1.40625^\circ$
Rounding this value gives approximately $1.4^\circ$.

Manchester encoding is a digital line coding method where binary data is represented by voltage transitions rather than static signal levels. In this scheme, a binary bit is encoded by a signal transition occurring exactly at the midpoint of the bit interval. Specifically, a $0$ is typically defined by a high-to-low transition, and a $1$ by a low-to-high transition. Since a phase transition is required within each bit duration to define the logical value, it enables self-clocking. Unlike NRZ encoding, which maintains static levels, Manchester encoding fundamentally relies on these per-bit phase transitions to represent data.

The bandwidth ($BW$) of a signal is defined as the difference between its highest frequency ($f_{high}$) and its lowest frequency ($f_{low}$), expressed as:
$BW = f_{high} - f_{low}$
Substituting the given values, where $BW = 5 \text{ kHz}$ and $f_{low} = 52 \text{ kHz}$:
$5 \text{ kHz} = f_{high} - 52 \text{ kHz}$
Solving for the highest frequency ($f_{high}$):
$f_{high} = 5 \text{ kHz} + 52 \text{ kHz} = 57 \text{ kHz}$

The output capacity $C$ is $\frac{5000}{1000} = 5$ packets per time unit. Let $N_t$ be the number of packets arriving at time $t$, and $B_t$ be the backlogged packets at the end of time unit $t$, defined as $B_t = \max(0, B_{t-1} + N_t - 5)$ with $B_0 = 0$.
Calculating $B_t$ for $t=1$ to $20$: $1, 5, 3, 5, 2, 0, 0, 0, 0, 1, 0, 5, 7, 7, 10, 8, 9, 6, 10, 10$.
The sum of backlogged packets is $\sum_{t=1}^{20} B_t = 1+5+3+5+2+0+0+0+0+1+0+5+7+7+10+8+9+6+10+10 = 89$.
The average number of backlogged packets per time unit is $\frac{89}{20} = 4.45$.

In Manchester encoding (waveform a), a bit is determined by the direction of the transition in the middle of the interval. Using the convention where a High-to-Low transition represents $0$ and a Low-to-High transition represents $1$, decoding waveform (a) yields the bit stream $0111101000$. Waveform (b) represents the same bit stream using Differential Manchester encoding, characterized by a transition at the start of the bit interval for a $0$ and no transition for a $1$. Comparing these results with the given options confirms the correct bit stream and coding scheme.

In serial transmission, a frame typically includes $1$ start bit, $8$ data bits, and $2$ stop bits, resulting in a total frame length of:
$1 (\text{start}) + 8 (\text{data}) + 2 (\text{stop}) = 11 \text{ bits/character}$
Given the baud rate of $2400$ bits per second, the number of characters transmitted per second is calculated by dividing the baud rate by the bits per character:
$\text{Characters per second} = \left\lfloor \frac{2400}{11} \right\rfloor$
$\text{Characters per second} = \lfloor 218.18 \rfloor = 218$
Thus, $218$ characters can be transmitted per second.

In asynchronous serial communication, the total number of bits per frame is calculated by summing the start, data, stop, and parity bits: $1 \text{ (start)} + 8 \text{ (data)} + 2 \text{ (stop)} + 1 \text{ (parity)} = 12 \text{ bits per character}$. Given a baud rate of $9600$ bits per second, we determine the number of characters transmitted per second by dividing the baud rate by the bits per character:

$\text{Characters per second} = \frac{9600 \text{ bits/s}}{12 \text{ bits/character}} = 800 \text{ characters/s}$

Thus, $800$ characters can be transmitted per second.

For T1, each character requires $8 \text{ (data)} + 2 \text{ (start)} + 1 \text{ (stop)} + 1 \text{ (parity)} = 12$ bits. The transfer rate is $\frac{1200}{12} = 100 \text{ characters/sec}$.
For T2, a transmission block consists of $3 \text{ (sync)} + 30 \text{ (info)} = 33$ characters, totaling $33 \times 8 = 264$ bits.
The time required to transmit this block is $\frac{264}{1200} = 0.22 \text{ seconds}$.
The effective information transfer rate is $\frac{30 \text{ info characters}}{0.22 \text{ seconds}} \approx 136.36 \text{ characters/sec}$, which rounds to 136 characters/sec.
Thus, the transfer rates are 100 characters/sec for T1 and 136 characters/sec for T2.

In asynchronous serial transmission, the communication line is maintained at a logical '1' (idle state) when inactive. The '0' start bit forces the signal to low, notifying the receiver that a byte is arriving and allowing it to synchronize its sampling timing with the sender. Subsequently, one or more '1' stop bits ensure the signal returns to the idle state, signifying the completion of the byte and readying the system for the next frame. This framing mechanism is essential for the receiver to identify the boundary of each byte, effectively synchronizing it for data reception. Thus, the padded bits serve the purpose of timing alignment rather than error recovery or parity computation.

To transmit one character in asynchronous serial communication, a start bit is required in addition to the specified framing bits. The total number of bits per character is 1 start bit + 8 data bits + 1 parity bit + 2 stop bits, resulting in $1 + 8 + 1 + 2 = 12$ bits per character.
Given a transfer rate of 300 characters per second, the required transmission throughput is $300 \text{ characters/s} \times 12 \text{ bits/character} = 3600 \text{ bits/s (bps)}$.
Since standard baud rates are fixed, we must choose the lowest standard rate that is equal to or greater than the calculated requirement of 3600 bps.
Comparing the options, 2400 baud is insufficient, while 4800 baud is the next standard rate that can sustain the required throughput.

RS-232 operates as an asynchronous protocol where the transmitter and receiver do not share a common clock signal. Because there is no external synchronization, each data frame must be self-clocking. The start bit, typically represented as a logical $0$, transitions the signal from the idle state (logical $1$) to alert the receiver that a new data frame is beginning. This specific transition forces the receiver's internal sampling clock to align with the transmitter's bit timing for that frame. Consequently, the start bit is essential to synchronize the receiver for the successful processing of every individual byte.

Nyquist's maximum data rate = 2B × log₂M. Here B = 6 MHz = 6×10⁶ Hz, M = 64 signal levels. log₂(64) = log₂(2⁶) = 6 bits/symbol. Maximum data rate = 2 × 6×10⁶ × 6 = 72×10⁶ bps = 72 Mbps. This is the theoretical maximum for a noiseless 6 MHz channel with 64-QAM. In practice, the actual throughput is limited by the SNR (Shannon's limit). This calculation is used in cable TV (DOCSIS) where downstream channels use 64-QAM or 256-QAM over 6 MHz channels.

QPSK (Quadrature Phase Shift Keying): uses 4 distinct phase shifts (0°, 90°, 180°, 270°) in the carrier signal. Each phase value represents a 2-bit symbol (dibit): 00→0°, 01→90°, 10→180°, 11→270°. Bits per symbol = log₂(4) = 2. Baud rate (symbol rate) = bit rate / 2. QPSK achieves twice the spectral efficiency of BPSK (Binary PSK, 1 bit/symbol). Higher-order modulations: 8-PSK (3 bits/symbol), 16-QAM (4 bits/symbol), 64-QAM (6 bits/symbol), 256-QAM (8 bits/symbol). Trade-off: higher-order modulation carries more bits/symbol but is more susceptible to noise (requires higher SNR).

Coaxial cable structure and properties: A: TRUE - coaxial cable's cylindrical metallic shield provides excellent EMI shielding, supporting higher frequencies (up to GHz range) and longer distances than unshielded twisted pair (UTP). B: TRUE - coaxial structure: center conductor (copper), dielectric insulator (polyethylene), metallic braid/foil shield, outer plastic jacket. The concentric design provides inherent shielding - the outer conductor acts as a Faraday cage around the inner conductor. C: TRUE - coaxial cable is stiffer, heavier, and more expensive than UTP. Modern LANs use UTP (Cat5e/Cat6) because structured cabling with UTP is cheaper and more flexible. Coaxial remains in use for: cable TV (CATV), cable internet (DOCSIS), RF connections, antenna feeds. D: FALSE - coaxial cable carries both analog (cable TV) and digital signals (cable internet).

Transmission delay = file size / link bandwidth. File size = 1 MB = 1 × 10⁶ × 8 bits = 8 × 10⁶ bits = 8 Mbits. Link bandwidth = 100 Mbps = 100 × 10⁶ bps. Transmission delay = 8 × 10⁶ / (100 × 10⁶) = 0.08 seconds = 80 ms. Propagation delay = distance / propagation speed = 10,000 m / (2 × 10⁸ m/s) = 5 × 10⁻⁵ s = 0.05 ms. Total delay = 80 + 0.05 ≈ 80.05 ms. Here transmission delay dominates because the file is large and the link is relatively slow.

Shannon theorem gives maximum achievable data rate.

IEEE 802.11b (Wi-Fi 1, 11 Mbps) and 802.11g (Wi-Fi 3, 54 Mbps) both use the 2.4 GHz ISM (Industrial, Scientific, Medical) band. Characteristics: unlicensed band (no license required), 2.400-2.4835 GHz (83.5 MHz total), 14 channels (22 MHz wide, 5 MHz spacing), 3 non-overlapping channels (1, 6, 11 in North America/Asia, or 1, 7, 13 in Europe). Subject to interference from: microwave ovens (2.45 GHz), Bluetooth (also 2.4 GHz ISM), baby monitors, cordless phones. Longer range than 5 GHz due to lower frequency (better wall penetration). 802.11a/n/ac/ax use 5 GHz (option B, used by 802.11a). 802.11ad/ay use 60 GHz mmWave.

Baud rate (symbol rate): the number of signal changes (symbols) per second, measured in baud (Bd). Each symbol change represents one signaling event. Bit rate: the number of bits transmitted per second (bps). Relationship: bit rate = baud rate × log₂M, where M is the number of distinct signal levels. Example: if a modem uses 4 voltage levels (M=4) at 2400 baud: bit rate = 2400 × log₂4 = 2400 × 2 = 4800 bps. If M = 2 (binary): baud rate = bit rate. If M > 2: bit rate > baud rate. Early telephone modems increased bit rate by using more signal levels while keeping the baud rate within the phone line's bandwidth limitation.