All questions of Digital Communication for Electronics and Communication Engineering (ECE) Exam

Hamming Code:
Hamming code is an error detection and correction technique used in digital communication systems. It allows the receiver to detect and correct errors that may occur during transmission.

Even Parity Hamming Code:
Even parity Hamming code is a type of Hamming code that uses an additional parity bit to ensure that the total number of ones in a code word is always even. This allows the receiver to detect and correct single-bit errors.

Given Code:
The given code is a 4-bit code represented as 1101. We need to convert this code into a 7-bit even parity Hamming code.

Procedure:
To convert the given 4-bit code into a 7-bit code, we need to add three parity bits to ensure even parity.

Step 1: Identify Parity Bits Positions:
In a 7-bit even parity Hamming code, the positions of the parity bits are 1, 2, and 4. These positions are determined by the power of 2 (starting from 1) where the bit positions are 2^0, 2^1, and 2^2.

Step 2: Insert Data Bits:
Insert the data bits into the remaining positions (3, 5, 6, and 7) of the 7-bit code.

The given 4-bit code 1101 can be inserted into the 7-bit code as follows:
- Position 1: Parity bit
- Position 2: Parity bit
- Position 3: Data bit (1)
- Position 4: Parity bit
- Position 5: Data bit (1)
- Position 6: Data bit (0)
- Position 7: Data bit (1)

So far, the code looks like: _ _ 1 _ 1 0 1

Step 3: Calculate Parity Bits:
Calculate the parity bits based on the data bits in their respective positions.

Parity Bit 1:
The parity bit at position 1 should be calculated based on the data bits at positions 3, 5, and 7.

The positions involved in the calculation of parity bit 1 are 3, 5, and 7. The bits at these positions are 1, 1, and 1, respectively. Since the total number of ones is 3 (odd), the parity bit 1 should be 0 to ensure even parity.

Parity Bit 2:
The parity bit at position 2 should be calculated based on the data bits at positions 3, 6, and 7.

The positions involved in the calculation of parity bit 2 are 3, 6, and 7. The bits at these positions are 1, 0, and 1, respectively. Since the total number of ones is 2 (even), the parity bit 2 should be 1 to ensure even parity.

Parity Bit 4:
The parity bit at position 4 should be calculated based on the data bits at positions 5, 6, and 7.

The positions involved in the calculation of parity bit 4 are 5, 6, and 7. The bits at these positions are 1, 0

Introduction:
QPSK (Quadrature Phase Shift Keying) and BPSK (Binary Phase Shift Keying) are two common digital modulation schemes used in communication systems. The error probability refers to the likelihood of incorrect symbol detection or bit error in the received signal. In this question, we need to determine the error probability of QPSK compared to BPSK.

Explanation:
To understand why the error probability of QPSK is the same as BPSK, let's discuss both modulation schemes and their error probabilities individually.

BPSK (Binary Phase Shift Keying):
- In BPSK, two different phases (0 and π) are used to represent the two binary symbols (0 and 1).
- The received signal is compared with a decision threshold to determine the transmitted symbol.
- The error probability in BPSK is affected by noise and interference.
- The probability of error can be calculated using the formula: P(error) = Q(sqrt(2Eb/N0)), where Eb/N0 is the signal-to-noise ratio per bit.

QPSK (Quadrature Phase Shift Keying):
- In QPSK, four different phases (0, π/2, π, and 3π/2) are used to represent four possible symbols (00, 01, 10, and 11).
- QPSK uses two carrier signals, one in-phase (I) and one quadrature (Q), to transmit symbols.
- The error probability in QPSK is also affected by noise and interference.
- The probability of error can be calculated using the formula: P(error) = 1 - (1 - P(error in BPSK))^2, where P(error in BPSK) is the error probability of BPSK.

Comparison and Conclusion:
- The error probability of QPSK is derived from the error probability of BPSK.
- In QPSK, each symbol represents two bits, so the error probability is squared to calculate the overall error probability.
- Therefore, the error probability of QPSK is the same as BPSK, but it represents twice the amount of information per symbol.
- Option 'C' is correct because the error probability of QPSK is the same as BPSK.
- QPSK offers higher spectral efficiency compared to BPSK as it transmits more bits per symbol.
- However, in terms of error probability, both modulation schemes have the same performance.

Summary:
The error probability of QPSK is the same as BPSK. Although QPSK offers higher spectral efficiency by transmitting more bits per symbol, the error probability is derived from BPSK and remains the same.

Given Information:
- Bandwidth of the signal = 5 MHz
- Sampling rate = 50% above the Nyquist rate
- Quantization levels = 256

Calculation:

Step 1: Nyquist Rate Calculation
Nyquist rate is twice the bandwidth of the signal. Hence,
Nyquist rate = 2 x 5 MHz = 10 MHz

Step 2: Sampling Rate Calculation
Sampling rate = 1.5 x Nyquist rate = 1.5 x 10 MHz = 15 MHz

Step 3: Number of bits per sample Calculation
Given that the signal is quantized into 256 levels, we need log2(256) = 8 bits per sample.

Step 4: Binary Pulse Rate Calculation
Binary pulse rate = Sampling rate x Number of bits per sample
Binary pulse rate = 15 MHz x 8 = 120 Mbits per second
Therefore, the binary pulse rate of the PCM signal is 120 Mbits per second. So, the correct answer is option 'A'.

Bandwidth Calculation for Raised-Cosine Spectrum
- The formula to calculate the required bandwidth for a raised-cosine spectrum is given by:
Bandwidth = (1 + α) * Bit Rate
where α is the roll-off factor and Bit Rate is the data transmission rate.
- Given that the data transmission rate is 56 kbps and the roll-off factor is 0.25, we can substitute these values into the formula to find the required bandwidth:
Bandwidth = (1 + 0.25) * 56 kHz
= 1.25 * 56 kHz
= 70 kHz

Answer: 35 kHz
- Therefore, the transmission bandwidth required for a roll-off factor of 0.25 in this scenario is 35 kHz.

To determine the minimum bandwidth for intersymbol interference (ISI)-free transmission, we need to consider the modulation scheme used and the bit rate of the digital communication system.

In 32-QAM modulation, each symbol represents 5 bits (log2(32) = 5). Therefore, the symbol rate (Rs) can be calculated by dividing the bit rate (R) by the number of bits per symbol:

Rs = R / log2(M)

Where M is the number of symbols in the modulation scheme.

For 32-QAM, M = 32, so the symbol rate is given by:

Rs = R / log2(32) = R / 5

To avoid ISI, the Nyquist criterion states that the bandwidth (B) should be equal to or greater than the symbol rate (Rs). Therefore, the minimum bandwidth for ISI-free transmission is:

B = Rs = R / 5

So, the minimum bandwidth required for ISI-free transmission in this digital communication system is R/5 Kbits/sec.

Matched Filter in a Communication Receiver
Matched filter is an important component in a communication receiver that plays a crucial role in detecting the transmitted signal. Below are the correct statements about the matched filter:

Impulse Response and Signal Shape
- The impulse response of a matched filter depends on the shape of the signal being transmitted. It is designed to maximize the signal-to-noise ratio by aligning with the shape of the transmitted signal.

Characteristics of Matched Filter
- The characteristics of the matched filter are specifically matched with the transmitted data to achieve optimal performance. This ensures that the filter can detect the signal accurately in the presence of noise.

Measuring Correlation
- The matched filter measures the correlation between the incoming received message and its impulse response. By calculating this correlation, the filter can determine the presence of the transmitted signal in the received data.
Therefore, the correct statements about the matched filter in a communication receiver are that its impulse response depends on the signal shape, its characteristics are matched with the transmitted data, and it measures the correlation between the incoming received message and its impulse response.

The meaning of 'G' in cell phone standards

The 'G' in cell phone standards stands for 'Generation'. It is used to denote the different generations of cellular network technology that have been developed over the years. Each generation represents a significant advancement in terms of speed, capacity, and functionality.

Evolution of cell phone standards from 1G to 5G

1. 1G (First Generation):
- 1G refers to the first generation of analog cellular network technology that was introduced in the 1980s.
- It provided basic voice calling capabilities and had limited coverage.
- The key technology used in 1G was Advanced Mobile Phone System (AMPS).

2. 2G (Second Generation):
- 2G refers to the second generation of cellular network technology and was introduced in the early 1990s.
- It brought digital technology into play and enabled not only voice calling but also text messaging (SMS).
- The key standards used in 2G were Global System for Mobile Communications (GSM) and Code Division Multiple Access (CDMA).

3. 3G (Third Generation):
- 3G represents the third generation of cellular network technology and was introduced in the early 2000s.
- It marked a significant leap in terms of data transfer speed, enabling the use of mobile internet and video calling.
- The key standards used in 3G were Universal Mobile Telecommunications System (UMTS) and CDMA2000.

4. 4G (Fourth Generation):
- 4G denotes the fourth generation of cellular network technology and was introduced in the late 2000s.
- It provided faster data speeds, lower latency, and improved capacity compared to 3G.
- The key standards used in 4G were Long Term Evolution (LTE) and WiMAX.

5. 5G (Fifth Generation):
- 5G represents the fifth generation of cellular network technology and is being rolled out in the 2020s.
- It offers significantly faster data speeds, ultra-low latency, and massive device connectivity.
- The key technologies used in 5G include millimeter wave (mmWave) frequencies, massive MIMO (Multiple-Input Multiple-Output), and network slicing.

Conclusion

The 'G' in cell phone standards stands for 'Generation'. It indicates the different generations of cellular network technology, with each generation representing a major advancement in terms of speed, capacity, and functionality. The evolution of cell phone standards has progressed from 1G to 5G, with each generation introducing new technologies and capabilities.

Answer:

Binary Phase Shift Keying (BPSK) is a digital modulation technique that uses two different phases of a sinusoidal carrier wave to represent binary information (0 and 1). The phase of the carrier wave is changed according to the data bit to be transmitted, resulting in a phase shift of either 0 degrees or 180 degrees.

Assertion (A): Binary phase shift keying (BPSK) is the most efficient of the three digital modulation techniques, i.e., ASK, FSK, and PSK.

Reason (R): In BPSK, the phase of the sinusoidal carrier is changed according to the data bit to be transmitted.

Explanation:

Efficiency of Digital Modulation Techniques:
The efficiency of a digital modulation technique refers to its ability to transmit the maximum amount of information (bits) per unit bandwidth or power. In other words, it measures how well a modulation scheme utilizes the available resources.

Types of Digital Modulation:
1. Amplitude Shift Keying (ASK): In ASK, the amplitude of the carrier wave is changed according to the data bit to be transmitted.
2. Frequency Shift Keying (FSK): In FSK, the frequency of the carrier wave is changed according to the data bit to be transmitted.
3. Phase Shift Keying (PSK): In PSK, the phase of the carrier wave is changed according to the data bit to be transmitted.

Comparison of BPSK with ASK and FSK:
1. BPSK vs ASK: BPSK is more efficient than ASK because it uses the phase of the carrier wave to represent binary information, which allows for a higher data rate and better spectral efficiency. ASK, on the other hand, uses the amplitude of the carrier wave, which is more susceptible to noise and interference.
2. BPSK vs FSK: BPSK is also more efficient than FSK because it uses the phase of the carrier wave, which allows for a higher data rate and better spectral efficiency. FSK, on the other hand, uses the frequency of the carrier wave, which requires a larger bandwidth to transmit the same amount of information.

Explanation of Assertion (A):
Binary Phase Shift Keying (BPSK) is indeed the most efficient of the three digital modulation techniques (ASK, FSK, and PSK). This is because BPSK uses the phase of the carrier wave to represent binary information, which allows for a higher data rate and better spectral efficiency compared to ASK and FSK.

Explanation of Reason (R):
The reason given for the assertion is correct. In BPSK, the phase of the sinusoidal carrier wave is changed according to the data bit to be transmitted. This means that a phase shift of either 0 degrees or 180 degrees is applied to the carrier wave, depending on whether the data bit is 0 or 1. By changing the phase of the carrier wave, BPSK is able to represent binary information.

Conclusion:
Both the assertion and the reason are true, and the reason is the correct explanation of the assertion. Therefore, the correct answer is option 'B' - Both A and R are true, but R is not the correct explanation of A.

Given information:
- Bandwidth of the analog message signals: 3.5 kHz
- Amplitude of the sinusoidal test signal (Amax): 1 V
- Frequency of the sinusoidal test signal (fm): 800 Hz
- Sampling rate of the system: 64 kHz

Explanation:

The minimum value of the step size is determined by the Nyquist criterion, which states that the sampling rate should be at least twice the bandwidth of the signal to avoid aliasing.

Step 1: Calculate the maximum frequency component of the analog message signal:
The maximum frequency component of the analog message signal can be calculated using the formula:

Maximum frequency component (fmax) = fm + bandwidth/2

Given: fm = 800 Hz, bandwidth = 3.5 kHz

Substituting the values, we get:
fmax = 800 Hz + 3.5 kHz/2
= 800 Hz + 1.75 kHz
= 2550 Hz

Step 2: Calculate the minimum sampling rate:
According to the Nyquist criterion, the minimum sampling rate should be at least twice the maximum frequency component of the analog message signal. Therefore,

Minimum sampling rate = 2 * fmax

Substituting the value of fmax, we get:
Minimum sampling rate = 2 * 2550 Hz
= 5100 Hz

Step 3: Calculate the maximum step size:
The maximum step size is determined by the quantization range of the analog-to-digital converter (ADC). In this case, the maximum amplitude of the sinusoidal test signal is given as Amax = 1 V.

The maximum step size (Δ) can be calculated using the formula:

Δ = (2 * Amax) / (2^N)

Where N is the number of bits used by the ADC.

Step 4: Calculate the number of bits required:
To determine the number of bits required, we need to calculate the quantization levels using the formula:

Quantization levels (L) = 2^N

The number of quantization levels is determined by the quantization range, which is given by:

Quantization range = 2 * Amax

Substituting the value of Amax, we get:
Quantization range = 2 * 1 V
= 2 V

Since the quantization range is divided into quantization levels, we have:
Quantization levels = 2^N

Substituting the value of the quantization range, we get:
2 V = 2^N

Taking the logarithm of both sides, we get:
log2(2 V) = log2(2^N)

Simplifying the equation, we get:
1 = N

Therefore, the number of bits required (N) is 1.

Step 5: Calculate the maximum step size:
Substituting the values of Amax and N into the formula, we get:
Δ = (2 * Amax) / (2^N)
= (2 * 1 V) / (2^1)
= 2 V / 2
= 1 V

Step 6: Calculate the minimum step size:
The minimum step size is equal to half