Binary Signalling Schemes MCQ Quiz in தமிழ் - Objective Question with Answer for Binary Signalling Schemes - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation:

Among all ASK has the least noise immunity due to the following:

1. FSK is less susceptible to errors than ASK – the receiver looks for specific frequency changes over a number of intervals, so voltage (noise) spikes can be ignored.

2. PSK allows information to be transmitted in the radio communication in a way more efficiently as compared to that of FSK and it is also less prone to error when we compare to ASK modulation.

3. ASK technique is not suitable for high bit rate data transmission and has poor bandwidth efficiency.

Hence it is highly susceptible to noise and other external factors.

Hence option (2) is the correct answer.

Important Points

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.

FSK (Frequency Shift Keying):

In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.

For binary ‘1’ → S1 (A) = Acos 2π fHt

For binary ‘0’ → S2 (t) = A cos 2π fLt . The constellation diagram is as shown:

ASK(Amplitude Shift Keying):

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:

where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.

PSK(Phase Shift Keying):

In PSK (phase shift keying) binary 1 is represented with a carrier signal and binary O is represented with 180° phase shift of a carrier

For binary ‘1’ → S1 (A) = Acos 2π fct

For binary ‘0’ → S2 (t) = A cos (2πfct + 180°) = - A cos 2π fct

The Constellation Diagram Representation is as shown:

Amplitude Shift Keying: In this modulation scheme, the amplitude of the carrier changes according to the bit amplitude.

The probability of error of ASK is given as:

$P_e=\frac{1}{2}erfc\left(\sqrt{\frac{E_b}{4N_0}}\right)$

It has the maximum probability of error when compared to ASK and FSK.

ASK is known as ON-OFF keying ((b) is incorrect)

Frequency Shift Keying: It stands for Frequency Shift Keying. The frequency of the carrier is changed between two levels depending upon the incoming bit.

DPSK:

• It stands for Differential phase-shift keying.
• It does not need a synchronous (coherent) carrier at the demodulation ((c) is correct)
• The input sequence of the binary bit is modified such that the next bit depends upon the previous bit. Therefore in the receiver previous received bits are used to detect the present bit.

Given, the bandwidth of the Gaussian noise channel:

B_T = 100 kHz

When $\frac{sinx}{x}$ pulse waveform is used, the minimum transmission bandwidth for 4 phase PSK system is defined as:

$B_T=\frac{R_b}{{\log }_{2}M}$

Since, for the 4-phase PSK, we have M = 4. Therefore, the maximum data rate for the system is given by:

R_b = (log₂ 4)(100 kHz)

R_b = 200 kHz

Again, we have the channel bandwidth (modulation bandwidth) as:

B_T = 100 kHz

For binary FSK, we have:

M = 2

Therefore, the maximum data rate that can be transmitted through the channel is given by:

$B_T=\frac{R_b}{{\log }_{2}M}$

$R_b=100\times lo{g}_2(2)$

R_b = 100 kHz

We have the channel bandwidth (modulation bandwidth)

B_T = 100 kHz

For M-frequency orthogonal signals, the relation between modulation bandwidth and data rate is given by:

$R_b=\frac{2{\log }_{2}M}{M}{B}_T$

Since, for 4-frequency orthogonal FSK, we have:

M = 4

Thus, the maximum data rate is obtained as:

$R_b=\frac{2{\log }_{2}4}{4}\left(100\right)$

R_b = 100 kHz

The probability of error of $\mathrm{A}\mathrm{S}\mathrm{K}$, ${\mathrm{P}}_{\mathrm{e}\mathrm{A}\mathrm{S}\mathrm{K}}=\mathrm{Q}\left(\sqrt{\frac{{\mathrm{A}}^2{\mathrm{T}}_{\mathrm{b}}}{4\eta }}\right)$

For $\mathrm{P}\mathrm{S}\mathrm{K}$, ${\mathrm{P}}_{\mathrm{e}\mathrm{P}\mathrm{S}\mathrm{K}}=\mathrm{Q}\left(\sqrt{\frac{{\mathrm{A}}^2{\mathrm{T}}_{\mathrm{b}}}{\eta }}\right)$

For $\mathrm{F}\mathrm{S}\mathrm{K}$, ${\mathrm{P}}_{\mathrm{e}\mathrm{F}\mathrm{S}\mathrm{K}}=\mathrm{Q}\left(\sqrt{\frac{{\mathrm{A}}^2{\mathrm{T}}_{\mathrm{b}}}{2\eta }}\right)$

Thus, we have ${\mathrm{P}}_{\mathrm{e}\mathrm{A}\mathrm{S}\mathrm{K}}>{\mathrm{P}}_{\mathrm{e}\mathrm{F}\mathrm{S}\mathrm{K}}>{\mathrm{P}}_{\mathrm{e}\mathrm{P}\mathrm{S}\mathrm{K}}$ as $\mathrm{Q}\left(\mathrm{x}\right)$ is a decreasing function.

Explanation:

For the signals to be orthogonal, the cross correlation between ${\mathrm{f}}_1\left(\mathrm{t}\right)$ and ${\mathrm{f}}_2\left(\mathrm{t}\right)$ should be zero, i.e,

 $\underset{0}{\overset{\mathrm{T}}{\int }}{\mathrm{f}}_1\left(\mathrm{t}\right){\mathrm{f}}_2\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}=0$

$$\begin{gathered} \Rightarrow \underset{0}{\overset{\mathrm{T}}{\int }}{\mathrm{A}}^2\mathrm{c}\mathrm{o}\mathrm{s}\left(2\pi {\mathrm{f}}_1\mathrm{t}+{\varphi }_1\right)\mathrm{c}\mathrm{o}\mathrm{s}\left(2\pi {\mathrm{f}}_2\mathrm{t}+{\varphi }_2\right)\mathrm{d}\mathrm{t} \\ \Rightarrow \frac{{\mathrm{A}}^2}{2}\left[\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{c}\mathrm{o}\mathrm{s}\left[2\pi ({\mathrm{f}}_1+{\mathrm{f}}_2)\mathrm{t}+{\varphi }_1+{\varphi }_2\right]\mathrm{d}\mathrm{t} \\ +\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{c}\mathrm{o}\mathrm{s}\left[2\pi ({\mathrm{f}}_1-{\mathrm{f}}_2)\mathrm{t}+{\varphi }_1-{\varphi }_2\right]\mathrm{d}\mathrm{t}\right] \\ \Rightarrow \frac{{\mathrm{A}}^2/2}{2\pi ({\mathrm{f}}_1+{\mathrm{f}}_2)}\left[\mathrm{s}\mathrm{i}\mathrm{n}\left[2\pi ({\mathrm{f}}_1+{\mathrm{f}}_2)\mathrm{T}+{\varphi }_1+{\varphi }_2\right]-\mathrm{s}\mathrm{i}\mathrm{n}\left[{\varphi }_1+{\varphi }_2\right]\right] \\ +\frac{{\mathrm{A}}^2/2}{2\pi ({\mathrm{f}}_1-{\mathrm{f}}_2)}\left[\mathrm{s}\mathrm{i}\mathrm{n}\left[2\pi ({\mathrm{f}}_1-{\mathrm{f}}_2)\mathrm{T}+{\varphi }_1-{\varphi }_2\right]-\mathrm{s}\mathrm{i}\mathrm{n}\left[{\varphi }_1-{\varphi }_2\right]\right] \end{gathered}$$

Now, the first term is zero as ${\mathrm{f}}_1$ and ${\mathrm{f}}_2$ are integral multiples of $\frac{1}{\mathrm{T}}$ So, ${\mathrm{f}}_1+{\mathrm{f}}_2=\frac{\mathrm{r}}{\mathrm{T}}\left(\mathrm{l}\mathrm{e}\mathrm{t}\right)$ where $\mathrm{r}$ is an integer.

$\Rightarrow \mathrm{s}\mathrm{i}\mathrm{n}\left[2\pi ({\mathrm{f}}_1+{\mathrm{f}}_2)\mathrm{T}+{\varphi }_1+{\varphi }_2\right]-\mathrm{s}\mathrm{i}\mathrm{n}\left[{\varphi }_1+{\varphi }_2\right]=0$

Now, for the second term to be zero, $2\pi ({\mathrm{f}}_1-{\mathrm{f}}_2)\mathrm{T}+{\varphi }_1-{\varphi }_2={\varphi }_1-{\varphi }_2+2\pi \mathrm{n}$

$\Rightarrow {\mathrm{f}}_1-{\mathrm{f}}_2=\frac{\mathrm{K}}{\mathrm{T}}$

$\Rightarrow {\left|{\mathrm{f}}_1-{\mathrm{f}}_2\right|}_{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{m}\mathrm{u}\mathrm{m}}=\frac{1}{\mathrm{T}}$

When the orthogonal signals of an $\mathrm{M}-\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{F}\mathrm{S}\mathrm{K}$ signal are detected coherently, the adjacent signals need only be separated from each other by a frequency difference $1/2\mathrm{T}$ so as to maintain orthogonality. This is the minimum separation and hence the minimum bandwidth possible for $\mathrm{M}-\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{F}\mathrm{S}\mathrm{K}$. Now, bandwidth efficiency $\rho =\frac{{\mathrm{R}}_{\mathrm{b}}}{\mathrm{B}\mathrm{W}}$

For $\mathrm{M}-\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{F}\mathrm{S}\mathrm{K}$ $\rho =\frac{\frac{{\log }_2\mathrm{M}}{\mathrm{T}}}{\frac{\mathrm{M}}{2\mathrm{T}}}=\frac{2{\log }_2\mathrm{M}}{\mathrm{M}}$

Thus, $\rho =\frac{2{\log }_{2}8}{8}=\frac{2\times 3}{8}=0.75$.

$\begin{array}{l}B{W}_{FSK}=f_{max}-f_{min}+\frac{2}{T_b}\\ =10-6+\frac{2}{0.001}=6\mathrm{k}\mathrm{H}\mathrm{z}\end{array}$