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Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density SN(f) = N0/2 = 10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Yt, represent the random variable y(t1)
Yt = Ns if transmitted bit bk = 0
YK = a + Nk if transmitted bit bk = 1
where Nk represents the noise sample value. The noise sample has a probability density function, PN (n) = 0.5αe (This has mean zero and variance 2/α2), Assume transmitted bits lo be equiprobable and threshold z is set to a/2 = 10-6 V

‘The probability of bit error is
  • a)
    0.5 × e-3.5
  • b)
    0.5 × e-5
  • c)
    0.5 × e-7
  • d)
    0.5 × e-10
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Consider a baseband binary PAM receiver shown below. The additive chan...
Concept:
Consider the block diagram as shown below to represent the noise which is treated as error:
The received signal is:
r(t) = Si(t) + n(t)
The output of the filter is:
z(t) = ai(t) + n0(t)
ai(t): Signal component
n0(t): Noise component
The probability of error is represented in terms of the Q function:

Pe = Q[x]
When the pdf's are not having an equal width then the probability of error will not be the same for both symbols. In this case, the Bit error rate is calculated.
BER = P(1) × Pe1 + P(0) × Pe0 
P(1): Probability when 1 is transmitted
P(0): Probability when 0 transmitted
Pe0: Probability of error when 0 transmitted
Pe1: Probability of error when 1 transmitted.
Calculation:
The system is defined as shown:

The given mean is 0 and variance is 2/α2 
V[X] = E[X2] - (E[X])2 
2/α2 = E[X2]
The area under the curve is:

E[X2] = 10-20 × 2 × 106 
2/α2 = 10-14 × 2
α2 = 1014 
α = 107 
Given Yt, represent the random variable y(t1)
Yt = Ns if transmitted bit bk = 0
YK = a + Nk if transmitted bit bk = 1 and threshold z is set to a/2 = 10-6 V
P(0) = P(1) = 1/2
Pe = P(1) × Pe1 + P(0) × Pe0 
Pe1 calculation
We know that when the output signal is less than the threshold it is considered as 0 and there is an error when 1 is transmitted.
Yk < z ⇒ error
a + Nk < 10-6 
Nk < -10-6 


Pe0 calculation
We know that when the output signal is more than the threshold it is considered as 1 and there is an error when 0 is transmitted.
Yk > z ⇒ error
Ns > 10-6 
Ns > -10-6 

The probability of error is:
Free Test
Community Answer
Consider a baseband binary PAM receiver shown below. The additive chan...
Concept:
Consider the block diagram as shown below to represent the noise which is treated as error:
The received signal is:
r(t) = Si(t) + n(t)
The output of the filter is:
z(t) = ai(t) + n0(t)
ai(t): Signal component
n0(t): Noise component
The probability of error is represented in terms of the Q function:

Pe = Q[x]
When the pdf's are not having an equal width then the probability of error will not be the same for both symbols. In this case, the Bit error rate is calculated.
BER = P(1) × Pe1 + P(0) × Pe0 
P(1): Probability when 1 is transmitted
P(0): Probability when 0 transmitted
Pe0: Probability of error when 0 transmitted
Pe1: Probability of error when 1 transmitted.
Calculation:
The system is defined as shown:

The given mean is 0 and variance is 2/α2 
V[X] = E[X2] - (E[X])2 
2/α2 = E[X2]
The area under the curve is:

E[X2] = 10-20 × 2 × 106 
2/α2 = 10-14 × 2
α2 = 1014 
α = 107 
Given Yt, represent the random variable y(t1)
Yt = Ns if transmitted bit bk = 0
YK = a + Nk if transmitted bit bk = 1 and threshold z is set to a/2 = 10-6 V
P(0) = P(1) = 1/2
Pe = P(1) × Pe1 + P(0) × Pe0 
Pe1 calculation
We know that when the output signal is less than the threshold it is considered as 0 and there is an error when 1 is transmitted.
Yk < z ⇒ error
a + Nk < 10-6 
Nk < -10-6 


Pe0 calculation
We know that when the output signal is more than the threshold it is considered as 1 and there is an error when 0 is transmitted.
Yk > z ⇒ error
Ns > 10-6 
Ns > -10-6 

The probability of error is:
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Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density SN(f) = N0/2 = 10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Yt, represent the random variable y(t1)Yt = Ns if transmitted bit bk = 0YK = a + Nk if transmitted bit bk = 1where Nk represents the noise sample value. The noise sample has a probability density function, PN (n) = 0.5αe (This has mean zero and variance 2/α2), Assume transmitted bits lo be equiprobable and threshold z is set to a/2 = 10-6 V‘The probability of bit error isa)0.5 × e-3.5b)0.5 × e-5c)0.5 × e-7d)0.5 × e-10Correct answer is option 'D'. Can you explain this answer?
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Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density SN(f) = N0/2 = 10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Yt, represent the random variable y(t1)Yt = Ns if transmitted bit bk = 0YK = a + Nk if transmitted bit bk = 1where Nk represents the noise sample value. The noise sample has a probability density function, PN (n) = 0.5αe (This has mean zero and variance 2/α2), Assume transmitted bits lo be equiprobable and threshold z is set to a/2 = 10-6 V‘The probability of bit error isa)0.5 × e-3.5b)0.5 × e-5c)0.5 × e-7d)0.5 × e-10Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density SN(f) = N0/2 = 10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Yt, represent the random variable y(t1)Yt = Ns if transmitted bit bk = 0YK = a + Nk if transmitted bit bk = 1where Nk represents the noise sample value. The noise sample has a probability density function, PN (n) = 0.5αe (This has mean zero and variance 2/α2), Assume transmitted bits lo be equiprobable and threshold z is set to a/2 = 10-6 V‘The probability of bit error isa)0.5 × e-3.5b)0.5 × e-5c)0.5 × e-7d)0.5 × e-10Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density SN(f) = N0/2 = 10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Yt, represent the random variable y(t1)Yt = Ns if transmitted bit bk = 0YK = a + Nk if transmitted bit bk = 1where Nk represents the noise sample value. The noise sample has a probability density function, PN (n) = 0.5αe (This has mean zero and variance 2/α2), Assume transmitted bits lo be equiprobable and threshold z is set to a/2 = 10-6 V‘The probability of bit error isa)0.5 × e-3.5b)0.5 × e-5c)0.5 × e-7d)0.5 × e-10Correct answer is option 'D'. Can you explain this answer?.
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