Binary Phase Shift Keying (BPSK) MCQ Quiz - Objective Question with Answer for Binary Phase Shift Keying (BPSK) - Download Free PDF

For coherent BFSK, the condition for orthogonality is,

Given that, f₀ = 50 kHz and f₁ = 60 kHz. So,

f₁ - f₀ = 10 kHz

For option (a) ⇒  kbps ⇒ 10 ≠ n(8)

For option (b) ⇒  kbps ⇒ 10 = n(5)

For option (c) ⇒  kbps ⇒ 10 = 5(2) = n(2)

For option (d) ⇒  kbps ⇒ 10 ≠ n(6)

So, only option (b), (c) satisfies the required condition.

Concept:

The spectrum of BPSK is represented as:

Max upper side frequency 

Minimum lower side freq. 

Where,

f_c = carrier frequency

R_b = Input but rate

Calculation:

Given:

R_b = 10 Mbps, f_c = 70 MHz

Max upper side frequency will be:

Min lower side frequency will be:

Hence, the BPSK spectrum  will be:

Concept: For a minimum probability of error value of α(Threshold detection value) lies in the common region of conditional pdfs.

Calculation: Given, X = {-0.5, +0.5}

Y = X + Z

The probability density function of f(z) is defined as:

Since the area under pdf is always 1, the amplitude of the above PDF is 1/2.

Now,the noisy observation when x = -0.5 is transmitted, i.e.

Y = -0.5 + Z

The PDFof Y knowing that X= -0.5 is as shown:

Similarly, the noisy observation when x = +0.5 is transmitted is;

Y = 0.5 + Z

The PDF of Y knowing that X= 0.5 is as shown:

For a minimum probability of error value of α lies in the common region.

i.e. -0.5 ≤ α ≤ 0.5

probability of error:

For minimum prob of error: α should be minimum

When, 

P_e is minimum and the minimum probability of error is:

Since

i.e. area under noise PSD = 1

When 1 is transmitted received signal

When (-1) is transmitted received signal

Probability of error

Concept: 1. If modulated signals are added then only signal trength increases but the noise level remains the same. 

                 2. γ is nothing but the ratio of Signal energy to power spectral density.

Application: Bit error rate (BER) of BPSK system with AWGN channel  additive white Gaussian Noise (AWGN) with power spectral density N₀/2 γ parameter is the function of bit energy and Noise power spectral density.

The demodulator receives the output of both channels.

So, 

By comparing we find

So, b = √2 = 1.41 

Concept: For a minimum probability of error value of α(Threshold detection value) lies in the common region of conditional pdfs.

Calculation: Given, X = {-0.5, +0.5}

Y = X + Z

The probability density function of f(z) is defined as:

Since the area under pdf is always 1, the amplitude of the above PDF is 1/2.

Now,the noisy observation when x = -0.5 is transmitted, i.e.

Y = -0.5 + Z

The PDFof Y knowing that X= -0.5 is as shown:

Similarly, the noisy observation when x = +0.5 is transmitted is;

Y = 0.5 + Z

The PDF of Y knowing that X= 0.5 is as shown:

For a minimum probability of error value of α lies in the common region.

i.e. -0.5 ≤ α ≤ 0.5

probability of error:

For minimum prob of error: α should be minimum

When, 

P_e is minimum and the minimum probability of error is:

Concept:

The spectrum of BPSK is represented as:

Max upper side frequency 

Minimum lower side freq. 

Where,

f_c = carrier frequency

R_b = Input but rate

Calculation:

Given:

R_b = 10 Mbps, f_c = 70 MHz

Max upper side frequency will be:

Min lower side frequency will be:

Hence, the BPSK spectrum  will be:

Concept: 1. If modulated signals are added then only signal trength increases but the noise level remains the same. 

                 2. γ is nothing but the ratio of Signal energy to power spectral density.

Application: Bit error rate (BER) of BPSK system with AWGN channel  additive white Gaussian Noise (AWGN) with power spectral density N₀/2 γ parameter is the function of bit energy and Noise power spectral density.

The demodulator receives the output of both channels.

So, 

By comparing we find

So, b = √2 = 1.41 

Concept: For a minimum probability of error value of α(Threshold detection value) lies in the common region of conditional pdfs.

Calculation: Given, X = {-0.5, +0.5}

Y = X + Z

The probability density function of f(z) is defined as:

Since the area under pdf is always 1, the amplitude of the above PDF is 1/2.

Now,the noisy observation when x = -0.5 is transmitted, i.e.

Y = -0.5 + Z

The PDFof Y knowing that X= -0.5 is as shown:

Similarly, the noisy observation when x = +0.5 is transmitted is;

Y = 0.5 + Z

The PDF of Y knowing that X= 0.5 is as shown:

For a minimum probability of error value of α lies in the common region.

i.e. -0.5 ≤ α ≤ 0.5

probability of error:

For minimum prob of error: α should be minimum

When, 

P_e is minimum and the minimum probability of error is:

For coherent BFSK, the condition for orthogonality is,

Given that, f₀ = 50 kHz and f₁ = 60 kHz. So,

f₁ - f₀ = 10 kHz

For option (a) ⇒  kbps ⇒ 10 ≠ n(8)

For option (b) ⇒  kbps ⇒ 10 = n(5)

For option (c) ⇒  kbps ⇒ 10 = 5(2) = n(2)

For option (d) ⇒  kbps ⇒ 10 ≠ n(6)

So, only option (b), (c) satisfies the required condition.

Since

i.e. area under noise PSD = 1

When 1 is transmitted received signal

When (-1) is transmitted received signal

Probability of error

Concept:

The spectrum of BPSK is represented as:

Max upper side frequency 

Minimum lower side freq. 

Where,

f_c = carrier frequency

R_b = Input but rate

Calculation:

Given:

R_b = 10 Mbps, f_c = 70 MHz

Max upper side frequency will be:

Min lower side frequency will be:

Hence, the BPSK spectrum  will be:

R_b = 5000 bits/sec

The amplitude of input waveform is A = 1 mV = 10^-3 V.

Single-sided noise power spectral density N₀ = 10^-11 W/Hz

Since the signal power is normalized relative to 1 Ω resistive load, we have the average signal power as:

Therefore, the bit energy is given by:

So, the bit error probability for BPSK receiver is obtained as:

= 4.05 × 10^-16

Now, the average number of errors in one day is given by:

Errors/day = P_eR_b × (86400 sec/day)

= (4.05 × 10^-6) × 5000 × (86400)

= 1750 bits in error