Binary Asymmetric Channel MCQ Quiz - Objective Question with Answer for Binary Asymmetric Channel - Download Free PDF

Explanation:

Average Probability of Error in a Binary Asymmetric Channel:

To solve the given problem, we need to calculate the average probability of error when transmitting two symbols, 0 and 1, over a binary asymmetric channel. The channel is characterized by different probabilities of error when transmitting 0 or 1. Let’s begin by understanding the problem and calculating the required values step-by-step.

Definitions and Notations:

• Let P(0) and P(1) be the source probabilities of symbols 0 and 1, respectively.
• Let P(error | 0) and P(error | 1) be the conditional probabilities of error when transmitting 0 and 1, respectively.
• The average probability of error, denoted as P(error), can be calculated using the formula:
  
  P(error) = P(0) × P(error | 0) + P(1) × P(error | 1)

Given Data:

• Source probabilities: P(0) = 0.2, P(1) = 0.8
• Conditional probabilities of error: P(error | 0) = 0.1, P(error | 1) = 0.4

Step-by-Step Calculation:

P(error) = P(0) × P(error | 0) + P(1) × P(error | 1)

= (0.2 × 0.1) + (0.8 × 0.4)

(0.2 × 0.1) = 0.02

(0.8 × 0.4) = 0.32

P(error) = 0.02 + 0.32 = 0.38

1. Substitute the given values into the formula for P(error):
2. Perform the calculations:
3. Add the results to find the average probability of error:

Final Answer: The average probability of error is 0.38.

Y₀ is detected to X₀ if:

\(P\left( {\frac{{{Y_0}}}{{{X_0}}}} \right)P\left( {{X_0}} \right) > P\left( {\frac{{{Y_0}}}{{{X_1}}}} \right)P\left( {{X_1}} \right)\)

\(\frac{1}{5} \times \frac{4}{{10}} > \frac{3}{5} \times \frac{6}{{10}}\) (Not satisfied)

Y₀ is detected as X₁ If:

\(P\left( {\frac{{{Y_0}}}{{{X_1}}}} \right)P\left( {{X_1}} \right) > P\left( {\frac{{{Y_0}}}{{{X_0}}}} \right)P\left( {{X_0}} \right)\)

\(\frac{3}{5} \times \frac{6}{{10}} > \frac{1}{5} \times \frac{4}{{10}}\) (Satisfied)

∴ Y0 is detected as X1

Y₁ is detected as X₀  If:

\(P\left( {\frac{{{Y_1}}}{{{X_0}}}} \right)P\left( {{X_0}} \right) > P\left( {\frac{{{Y_1}}}{{{X_1}}}} \right)P\left( {{X_1}} \right)\)

\(\frac{4}{5} \times \frac{4}{{10}} > \frac{2}{5} \times \frac{6}{{10}}\) (Satisfied)

∴ Y 1 is detected as X0.

Y₁ is detected as X₁ If:

\(P\left( {\frac{{{Y_1}}}{{{X_1}}}} \right)P\left( {{X_1}} \right) > P\left( {\frac{{{Y_1}}}{{{X_0}}}} \right)P\left( {{X_0}} \right)\)

\(\frac{2}{5} \times \frac{6}{{10}} > \frac{4}{5} \times \frac{4}{{10}}\) (Not satisfied)

∴ Y₀ is detected as X₁ and Y₁ is detected as X₀.

Concept:

Information in event x may be defined as:

\(I\left( x \right) \buildrel \Delta \over = {\log _2}\frac{1}{{p\left( x \right)}}\)

Where p(x) is the probability of event x.

For more than one event we use average information H[x] which may be defined as:

\(H\left[ x \right] \buildrel \Delta \over = \left( {{x_i}} \right){\log _2}\frac{1}{{p\left( {{x_i}} \right)}}\)

Application:

Given that x₀ is transmitted.

Therefore average information in received symbols will be:

\(= \mathop \sum \limits_{i = 0}^1 p\left[ {\frac{{{y_i}}}{{{x_0}}}} \right]\log \frac{1}{{p\left[ {\frac{{{y_i}}}{{{x_0}}}} \right]}}\)

∴ Information will be:

\(= p\left[ {\frac{{{y_0}}}{{{x_0}}}} \right]{\log _2}\frac{1}{{p\left[ {\frac{{{y_0}}}{{{x_0}}}} \right]}} + p\left[ {\frac{{{y_1}}}{{{x_0}}}} \right]{\log _2}\frac{1}{{p\left[ {\frac{{{y_1}}}{{{x_0}}}} \right]}}\)

\( \Rightarrow \frac{3}{4}\log \frac{4}{3} + \frac{1}{2}\log 2\)

⇒ 0.311 + 0.50

⇒ 0.811 bits

Concept:

Information in event x may be defined as:

\(I\left( x \right) \buildrel \Delta \over = {\log _2}\frac{1}{{p\left( x \right)}}\)

Where p(x) is the probability of event x.

For more than one event we use average information H[x] which may be defined as:

\(H\left[ x \right] \buildrel \Delta \over = \left( {{x_i}} \right){\log _2}\frac{1}{{p\left( {{x_i}} \right)}}\)

Application:

Given that x₀ is transmitted.

Therefore average information in received symbols will be:

\(= \mathop \sum \limits_{i = 0}^1 p\left[ {\frac{{{y_i}}}{{{x_0}}}} \right]\log \frac{1}{{p\left[ {\frac{{{y_i}}}{{{x_0}}}} \right]}}\)

∴ Information will be:

\(= p\left[ {\frac{{{y_0}}}{{{x_0}}}} \right]{\log _2}\frac{1}{{p\left[ {\frac{{{y_0}}}{{{x_0}}}} \right]}} + p\left[ {\frac{{{y_1}}}{{{x_0}}}} \right]{\log _2}\frac{1}{{p\left[ {\frac{{{y_1}}}{{{x_0}}}} \right]}}\)

\( \Rightarrow \frac{3}{4}\log \frac{4}{3} + \frac{1}{2}\log 2\)

⇒ 0.311 + 0.50

⇒ 0.811 bits

Explanation:

Average Probability of Error in a Binary Asymmetric Channel:

To solve the given problem, we need to calculate the average probability of error when transmitting two symbols, 0 and 1, over a binary asymmetric channel. The channel is characterized by different probabilities of error when transmitting 0 or 1. Let’s begin by understanding the problem and calculating the required values step-by-step.

Definitions and Notations:

• Let P(0) and P(1) be the source probabilities of symbols 0 and 1, respectively.
• Let P(error | 0) and P(error | 1) be the conditional probabilities of error when transmitting 0 and 1, respectively.
• The average probability of error, denoted as P(error), can be calculated using the formula:
  
  P(error) = P(0) × P(error | 0) + P(1) × P(error | 1)

Given Data:

• Source probabilities: P(0) = 0.2, P(1) = 0.8
• Conditional probabilities of error: P(error | 0) = 0.1, P(error | 1) = 0.4

Step-by-Step Calculation:

P(error) = P(0) × P(error | 0) + P(1) × P(error | 1)

= (0.2 × 0.1) + (0.8 × 0.4)

(0.2 × 0.1) = 0.02

(0.8 × 0.4) = 0.32

P(error) = 0.02 + 0.32 = 0.38

1. Substitute the given values into the formula for P(error):
2. Perform the calculations:
3. Add the results to find the average probability of error:

Final Answer: The average probability of error is 0.38.

Concept:

Information in event x may be defined as:

\(I\left( x \right) \buildrel \Delta \over = {\log _2}\frac{1}{{p\left( x \right)}}\)

Where p(x) is the probability of event x.

For more than one event we use average information H[x] which may be defined as:

\(H\left[ x \right] \buildrel \Delta \over = \left( {{x_i}} \right){\log _2}\frac{1}{{p\left( {{x_i}} \right)}}\)

Application:

Given that x₀ is transmitted.

Therefore average information in received symbols will be:

\(= \mathop \sum \limits_{i = 0}^1 p\left[ {\frac{{{y_i}}}{{{x_0}}}} \right]\log \frac{1}{{p\left[ {\frac{{{y_i}}}{{{x_0}}}} \right]}}\)

∴ Information will be:

\(= p\left[ {\frac{{{y_0}}}{{{x_0}}}} \right]{\log _2}\frac{1}{{p\left[ {\frac{{{y_0}}}{{{x_0}}}} \right]}} + p\left[ {\frac{{{y_1}}}{{{x_0}}}} \right]{\log _2}\frac{1}{{p\left[ {\frac{{{y_1}}}{{{x_0}}}} \right]}}\)

\( \Rightarrow \frac{3}{4}\log \frac{4}{3} + \frac{1}{2}\log 2\)

⇒ 0.311 + 0.50

⇒ 0.811 bits

Given:

P(x₀) = 0.5, P(x₁) = 0.5, P(y₀|x₀) = 0.75, P(y₁|x₁) = 0.5, P(y₁|x₀) = 0.25, P(y₁|x₁) = 0.5

For the uncertainty with regards to the received symbol, given that "one" has been transmitted, we find

\(H = - P \left(\dfrac{y_0}{x_1} \right) \log_2 P \left( \dfrac{y_0}{x_1} \right) - P \left(\dfrac{y_1}{x_1} \right) \log_2 P \left( \dfrac{y_1}{x_1} \right)\)

= \( - \dfrac{1}{2} \log_2 \dfrac{1}{2} - \dfrac{1}{2} \log_2 \dfrac{1}{2}\)

= 1 bit.

Y₀ is detected to X₀ if:

\(P\left( {\frac{{{Y_0}}}{{{X_0}}}} \right)P\left( {{X_0}} \right) > P\left( {\frac{{{Y_0}}}{{{X_1}}}} \right)P\left( {{X_1}} \right)\)

\(\frac{1}{5} \times \frac{4}{{10}} > \frac{3}{5} \times \frac{6}{{10}}\) (Not satisfied)

Y₀ is detected as X₁ If:

\(P\left( {\frac{{{Y_0}}}{{{X_1}}}} \right)P\left( {{X_1}} \right) > P\left( {\frac{{{Y_0}}}{{{X_0}}}} \right)P\left( {{X_0}} \right)\)

\(\frac{3}{5} \times \frac{6}{{10}} > \frac{1}{5} \times \frac{4}{{10}}\) (Satisfied)

∴ Y0 is detected as X1

Y₁ is detected as X₀  If:

\(P\left( {\frac{{{Y_1}}}{{{X_0}}}} \right)P\left( {{X_0}} \right) > P\left( {\frac{{{Y_1}}}{{{X_1}}}} \right)P\left( {{X_1}} \right)\)

\(\frac{4}{5} \times \frac{4}{{10}} > \frac{2}{5} \times \frac{6}{{10}}\) (Satisfied)

∴ Y 1 is detected as X0.

Y₁ is detected as X₁ If:

\(P\left( {\frac{{{Y_1}}}{{{X_1}}}} \right)P\left( {{X_1}} \right) > P\left( {\frac{{{Y_1}}}{{{X_0}}}} \right)P\left( {{X_0}} \right)\)

\(\frac{2}{5} \times \frac{6}{{10}} > \frac{4}{5} \times \frac{4}{{10}}\) (Not satisfied)

∴ Y₀ is detected as X₁ and Y₁ is detected as X₀.

For r₀

(0.7) (0.9) > (0.3) (0.4)

r₀ decoded as m₀

for r₁

(0.3) (0.6) > (0.7) (0.1)

r₁ decoded as m₁

probability of correct

(0.7) (0.9) + (0.3) (0.6)

= 0.81

P_error = 1 – 0.81

= 0.19

Explanation:

Average Probability of Error in a Binary Asymmetric Channel:

To solve the given problem, we need to calculate the average probability of error when transmitting two symbols, 0 and 1, over a binary asymmetric channel. The channel is characterized by different probabilities of error when transmitting 0 or 1. Let’s begin by understanding the problem and calculating the required values step-by-step.

Definitions and Notations:

• Let P(0) and P(1) be the source probabilities of symbols 0 and 1, respectively.
• Let P(error | 0) and P(error | 1) be the conditional probabilities of error when transmitting 0 and 1, respectively.
• The average probability of error, denoted as P(error), can be calculated using the formula:
  
  P(error) = P(0) × P(error | 0) + P(1) × P(error | 1)

Given Data:

• Source probabilities: P(0) = 0.2, P(1) = 0.8
• Conditional probabilities of error: P(error | 0) = 0.1, P(error | 1) = 0.4

Step-by-Step Calculation:

P(error) = P(0) × P(error | 0) + P(1) × P(error | 1)

= (0.2 × 0.1) + (0.8 × 0.4)

(0.2 × 0.1) = 0.02

(0.8 × 0.4) = 0.32

P(error) = 0.02 + 0.32 = 0.38

1. Substitute the given values into the formula for P(error):
2. Perform the calculations:
3. Add the results to find the average probability of error:

Final Answer: The average probability of error is 0.38.

Concept:

MAP and ML Technique are used to decrease the bandwidth of transmission

MAP or ML decoding:

Consider a binary channel as shown

P(m₀), P(m₁) – Probability of transmission of symbol m₀, m₁

P(r|m) – Probability of receiving in r, when m symbol was Transmitted.

Case (1):

When Bit is r₀

If,

\(P\left( {{m_0}} \right)P\left( {\frac{{{r_0}}}{{{m_0}}}} \right) > P\left( {{m_1}} \right)P\left( {\frac{{{r_0}}}{{{m_1}}}} \right)\)

then it is decoded as m₀

\(P\left( {{m_0}} \right)P\left( {\frac{{{r_0}}}{{{m_0}}}} \right) < P\left( {{m_1}} \right)P\left( {\frac{{{r_0}}}{{{m_1}}}} \right)\)

decoded as m₁

Case (2):

When Bit is r₁

If,

\(P\left( {{m_1}} \right)P\left( {\frac{{{r_1}}}{{{m_1}}}} \right) > P\left( {{m_0}} \right)P\left( {\frac{{{r_1}}}{{{m_0}}}} \right)\)

decoded as m₁

\(P\left( {{m_0}} \right)P\left( {\frac{{{r_1}}}{{{m_0}}}} \right) > P\left( {{m_1}} \right)P\left( {\frac{{{r_1}}}{{{m_1}}}} \right)\)

decoded as m₀

Calculation:

Using the MAP technique

Case (1)

When the received bit is (r₀)

(0.7)(0.9) > (0.3)(0.4)

(m₀) is detected.

Case (2)

When received bit is (r₁)

(0.3)(0.6) > (0.7)(0.1)

(m₁) is detected

Probability of correct decision is given by

P_c = (0.7)(0.9) + (0.3)(0.6)

= 0.81

Now,

ML technique P(m₀) = P(m₁) = 0.5

Case (1)

When received bit (m₀)

(0.5)(0.9) > (0.5)(0.4)

(m₀) is detected

Case (2)

When received bit (r₁)

(0.5)(0.6) > (0.5)(0.1)

(m₁) is detected

Probability of correct decision is given by

P_c = (0.5)(0.6) + (0.5)(0.9)

P_c = 0.75

Difference between MAP (P_c) and ML (P_c) is:

= |0.81 – 0.75|

= |0.06|

0.06

For the decision to assign r₁ to input message compare probability

P (r₁/m₀) P (m₀) and P (r₁/m₁) P (m₁)

(0.25) (0.6) > (0.2) (0.4)

r₁ is assigned to m₀

The probability of receiving the right message is

\(P\left( c \right)\; = \;P\left( {\frac{{{r_0}}}{{{m_0}}}} \right)P\left( {{m_0}} \right) + P\left( {\frac{{{r_2}}}{{{m_1}}}} \right)P\left( {{m_1}} \right) + P\left( {\frac{{{r_1}}}{{{m_0}}}} \right)P\left( {{m_0}} \right)\)

= (0.5) (0.6) + (0.6) (0.4) + (0.25) (0.6)

= 0.3 + 0.24 + 0.15

= 0.69

Probability of error P(e) = 1 – 0.69

= 0.31