Amplitude Shift Keying (ASK) MCQ Quiz - Objective Question with Answer for Amplitude Shift Keying (ASK) - Download Free PDF

Explanation:

Numerical Solution for Eb/N0 Calculation

Given Data:

• CNDR (Carrier-to-Noise Density Ratio): 100 dBHz
• Modulation Scheme: 256-QAM
• Symbol Rate: 12.5 M Symbols/s (or 12.5 × 10⁶ Symbols/s)

Step-by-Step Solution:

To calculate the energy-per-bit to noise density ratio (Eb/N0), we can use the relationship between CNDR, symbol rate, and the modulation scheme. The formula is as follows:

Formula:

Eb/N0 = CNDR - 10 × log₁₀(R × log₂(M))

Where:

• CNDR: Carrier-to-Noise Density Ratio (in dBHz)
• R: Symbol rate (in Symbols/s)
• M: Modulation order (number of distinct symbols in the modulation, for 256-QAM, M = 256)

Step 1: Convert CNDR to dB Scale

In this question, CNDR is already provided in dBHz (100 dBHz), so no additional conversion is necessary.

Step 2: Calculate the Logarithmic Term for M

The modulation order (M) for 256-QAM is 256. The logarithmic term log₂(M) is calculated as:

log₂(256) = log₂(2⁸) = 8

Step 3: Calculate the Logarithmic Term for Symbol Rate

The symbol rate (R) is given as 12.5 M Symbols/s, which is equivalent to:

R = 12.5 × 10⁶ Symbols/s

The logarithmic term 10 × log₁₀(R × log₂(M)) is calculated as:

10 × log₁₀(12.5 × 10⁶ × 8)

First, calculate the product:

12.5 × 10⁶ × 8 = 100 × 10⁶ = 10⁸

Now, calculate the logarithm:

log₁₀(10⁸) = 8

Finally, multiply by 10:

10 × log₁₀(R × log₂(M)) = 10 × 8 = 80 dB

Step 4: Subtract the Logarithmic Term from CNDR

Eb/N0 = CNDR - 10 × log₁₀(R × log₂(M))

Substitute the values:

Eb/N0 = 100 dB - 80 dB

Eb/N0 = 20 dB

The value of Eb/N0 is 20 dB.

ASK System:

1. For ASK Transmitter on-off keying is used.

2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.

1 : s1(t) = Ac cos 2πfct

0 : s2 (t) = 0

In ASK modulation each symbol is transmitted using a single bit

Hence, the bit rate = baud rate

Baud rate = 400 baud

If the question would have asked QPSK in case of ASK

Then QPSK transfers 2 bit for 1 symbol

Analysis:

The probability of error for ASK, PSK, and FSK is given as

\(P_{e_{ASK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}})} \space \)

\(P_{e_{PSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{N_0}}})} \space \)

\(P_{e_{FSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}})} \space \)

Q(x)  is a decreasing function therefore as x increases the value of Q(x) decreases

\( \sqrt {\dfrac {A^2_c T_b \space }{N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{2N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{4N_0}} \space \)

Therefore, 

\( Q (\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}) \space > \space Q(\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}) \space > \space Q( \sqrt {\dfrac {A^2_c T_b \space }{N_0}}) \space \)

Pe ASK  >  PeFSK   > Pe PSK

ASK modulation scheme gives Maximum probability of error.

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space.

In these schemes, bit-by-bit transmission through free space occurs.

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S1 (t) = Acos 2π fct

For binary ‘0’ → S2 (t) = 0

The Constellation Diagram Representation is as shown:

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space.

In these schemes, bit-by-bit transmission through free space occurs.

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S1 (t) = Acos 2π fct

For binary ‘0’ → S2 (t) = 0

The Constellation Diagram Representation is as shown:

ASK System:

1. For ASK Transmitter on-off keying is used.

2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.

1 : s1(t) = Ac cos 2πfct

0 : s2 (t) = 0

Analysis:

The probability of error for ASK, PSK, and FSK is given as

\(P_{e_{ASK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}})} \space \)

\(P_{e_{PSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{N_0}}})} \space \)

\(P_{e_{FSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}})} \space \)

Q(x)  is a decreasing function therefore as x increases the value of Q(x) decreases

\( \sqrt {\dfrac {A^2_c T_b \space }{N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{2N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{4N_0}} \space \)

Therefore, 

\( Q (\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}) \space > \space Q(\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}) \space > \space Q( \sqrt {\dfrac {A^2_c T_b \space }{N_0}}) \space \)

Pe ASK  >  PeFSK   > Pe PSK

ASK modulation scheme gives Maximum probability of error.

Explanation:

Numerical Solution for Eb/N0 Calculation

Given Data:

• CNDR (Carrier-to-Noise Density Ratio): 100 dBHz
• Modulation Scheme: 256-QAM
• Symbol Rate: 12.5 M Symbols/s (or 12.5 × 10⁶ Symbols/s)

Step-by-Step Solution:

To calculate the energy-per-bit to noise density ratio (Eb/N0), we can use the relationship between CNDR, symbol rate, and the modulation scheme. The formula is as follows:

Formula:

Eb/N0 = CNDR - 10 × log₁₀(R × log₂(M))

Where:

• CNDR: Carrier-to-Noise Density Ratio (in dBHz)
• R: Symbol rate (in Symbols/s)
• M: Modulation order (number of distinct symbols in the modulation, for 256-QAM, M = 256)

Step 1: Convert CNDR to dB Scale

In this question, CNDR is already provided in dBHz (100 dBHz), so no additional conversion is necessary.

Step 2: Calculate the Logarithmic Term for M

The modulation order (M) for 256-QAM is 256. The logarithmic term log₂(M) is calculated as:

log₂(256) = log₂(2⁸) = 8

Step 3: Calculate the Logarithmic Term for Symbol Rate

The symbol rate (R) is given as 12.5 M Symbols/s, which is equivalent to:

R = 12.5 × 10⁶ Symbols/s

The logarithmic term 10 × log₁₀(R × log₂(M)) is calculated as:

10 × log₁₀(12.5 × 10⁶ × 8)

First, calculate the product:

12.5 × 10⁶ × 8 = 100 × 10⁶ = 10⁸

Now, calculate the logarithm:

log₁₀(10⁸) = 8

Finally, multiply by 10:

10 × log₁₀(R × log₂(M)) = 10 × 8 = 80 dB

Step 4: Subtract the Logarithmic Term from CNDR

Eb/N0 = CNDR - 10 × log₁₀(R × log₂(M))

Substitute the values:

Eb/N0 = 100 dB - 80 dB

Eb/N0 = 20 dB

The value of Eb/N0 is 20 dB.

ASK System:

1. For ASK Transmitter on-off keying is used.

2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.

1 : s1(t) = Ac cos 2πfct

0 : s2 (t) = 0

Notes:

FSK: 1 is represented by the high-frequency carrier and 0 is represented by low-frequency carrier

PSK: Symbols for 0 and 1 have the same frequency but there is a 180° phase difference between the two.

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space.

In these schemes, bit-by-bit transmission through free space occurs.

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S1 (t) = Acos 2π fct

For binary ‘0’ → S2 (t) = 0

The Constellation Diagram Representation is as shown:

Concept:

The transmission bandwidth for ASK scheme is given by:

Bandwidth = 2R_b (for Rectangular pulses)

R_b = Bit rate given by:

R_b = nf_s

n = number of bits used in PCM encoding.

f_s = Sampling frequency

Calculation:

Given n = 10

f_m = Message signal frequency \(= \frac{{2\pi \times {{10}^4}}}{{2\pi }} = 10kHz\) 

f_s = Nyquist Rate = 2 f_m = 20 kHz.

R_b = nf_s = 10 × 20 kbps

= 200 kbps.

So the Transmission Bandwidth = 200 k × 2

Concept:

The bandwidth of M-ary PSK and ASK is given by:

\(W = \frac{{{2R_b}}}{{{{\log }_2}M\;}}\left( {1 + α } \right)\)

R_b = Bitrate

α = Roll-off factor

For minimum transmission bandwidth α = 0

Also, the bandwidth of M-ary FSK is always greater than M-ary PSK and FSK.

Application:

The minimum bandwidth of 2-ary ASK and PSK will be:

\(W = \frac{{{2R_b}}}{{{{\log }_2}M\;}}=2R_b\)

The bandwidth of 2-ary FSK will always be greater than it.

The bandwidth of 4-ary ASK will be:

\(W = \frac{{{2R_b}}}{{{{\log }_2}4\;}}=R_b\)

So, the bandwidth of 4-ary ASK is minimum.

In ASK modulation each symbol is transmitted using a single bit

Hence, the bit rate = baud rate

Baud rate = 400 baud

If the question would have asked QPSK in case of ASK

Then QPSK transfers 2 bit for 1 symbol

Concept:

\(\rm B.W_{ASK}= 2{R_b} = \frac{2}{{{T_b}}}\)

Calculation:

Now,

\(\rm {R_b} = n{f_s} = 12 \times 5000 = 60000\ bps\)

So,

\(\rm BW = 120\ kHz\)