Concept:
In the M-Array modulation scheme, the minimum bandwidth required for ideal transmission is given by:
\({\left( {BW} \right)_{min}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz\)
Where,
Rb = bit rate in bps
N = number of levels in M-Array scheme
Calculation:
Given that,
Bit rate = M kbps
Number of levels = N = 16
\(\therefore {\left( {BW} \right)_{{\rm{min}}}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz = \frac{2M}{{{{\log }_2}16}}kHz\)
\( = \frac{2M}{{{{\log }_2}{2^4}}}kHz\)
\( = \frac{2M}{{4\; \times\; {{\log }_2}2}}kHz\)
\( (BW)_{min}= \frac{M}{{2}}kHz\)
So. The minimum bandwidth will be M/2 kHz, for ideal transmission.
For Baseband |
For Passband |
Binary: 1) B.W. = Rb |
Binary: 1) BW = 2 Rb |
Raised cosine (α) : 2) \(BW = \frac{{{R_b}}}{2}\left( {1 + \alpha } \right)\) |
Raised cosine (α) : \(2)\;BW = \frac{{2{R_b}}}{2}\left( {1 + \alpha } \right)\) = Rb (1 + α) |
M-ary: 1) \(B.W. = \frac{{{R_b}}}{{{{\log }_2}M}}\) |
M-ary: 1) \(B.W = \frac{{2{R_b}}}{{{{\log }_2}M}}\) |
Raised cosine (α): 2) \(B.W. = \frac{{{R_b}\left( {1 + \alpha } \right)}}{{2{{\log }_2}M}}\) |
Raised cosine (α) : 2) \(B.W = \frac{{{R_b}\left( {1 + \alpha } \right)}}{{{{\log }_2}M}}\) |
________ is mostly preferred for telegraphy.
Digital to Analog Modulation technique is as shown.
Important Points:
1) As shown, QAM is a mixture of both ASK and PSK.
2) QAM uses two carrier signals with the same frequency but with which are in quadrature. Quadrature here means out phase by 90°.
3) Since the two signals have the same frequency, they are detected using synchronous detection.
4) Hence, amplitude and the phase of the carrier frequency both vary with the message signal.
A constellation diagram of a QAM signal with 2 different amplitude levels and 8 different phases is shown:
Following constellation diagram represents:
Explanation:
B.W. of PSK = 2R_{b}
Whereas for M-ary PSK,
\(B.W. =\frac{2R_b}{log_2m}\)
Hence M-ary is preferred due to less bandwidth.
16 APSK 16 QAM 16 QPSK
Hence the given figure is 16 QAM.
So the solution is option (2).
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:
A single bit, equally likely to be 0 and 1, is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density N0/2. Binary signaling, with 0 ↦ p(t) and 1 ↦ q(t), is used for the transmission, along with an optimal receiver that minimizes the bit-error probability. Let φ_{1} (t), φ_{2}(t) form an orthonormal signal set. If we choose p(t) = φ_{1} (t) and q(t) = −φ_{1} (t), we would obtain a certain bit-error probability P_{b}. If we keep p(t) = φ_{1}(t), but take \(q\left( t \right) = \sqrt E\) φ_{2} (t) for what value of E would we obtain the same bit-error probability P_{b}?
Concept: In constellation diagram d_{min} is the minimum distance between two adjacent points and Probability of error is given by \({P_e} = Q\left( {\sqrt {\frac{{d_{min}^2}}{{2No}}} } \right)\)
Analysis: \(P\left( 0 \right) = P\left( 1 \right) = \frac{1}{2}\)
Case 1:
p(t) = Φ_{1}(t)
q(t) = -Φ_{1}(t)
The system is binary PSK
Probability of error is given by:
\({P_e} = Q\left( {\sqrt {\frac{{d_{min}^2}}{{2No}}} } \right)\)
d_{min} = 1 – (-1) = 2
\({P_e} = Q\left( {\sqrt {\frac{4}{{2No}}} \;} \right) = Q\left( {\sqrt {\frac{2}{{No}}} \;} \right) \ldots \left( 1 \right)\)
Case 2: When p(t) = Φ_{1}(t)
\(q\left( t \right) = \sqrt E {\phi _2}\left( t \right)\)
Since Φ_{1}(t) and Φ_{2}(t) are orthonormal, the signalling scheme is BFSK.
Constellation Diagram:
(d_{min})^{2} = (√E)^{2} + (1)^{2}
d_{min}^{2} = E + 1
\({P_e} = Q\left( {\sqrt {\frac{{d_{min}^2}}{{2No}}} } \right)\)
\({P_e} = Q\sqrt {\frac{{E + 1}}{{2No}}} \ldots \left( 2 \right)\)
Comparing (1) and (2)
\(\frac{2}{{No}} = \frac{{E + 1}}{{2No}}\)
4 = E + 1
E = 3Coherent detection: In it, the receiver exploits the exact knowledge of the phase of the carrier to detect the signal better.
Non-coherent detection: It involves making some approximations to the phase information that results in a loss in performance. However, it simplifies the circuitry.
There are different methods for demodulating an FSK wave. The main methods of demodulating FSK signals are:
1) Asynchronous detector
2) Synchronous (Coherent) detector.
The block diagram of a coherent detector is as shown:
Given:
R_{b} = 4 kbps
In FSK, the best possible interval between the carriers is given as:
\(ΔT = \frac{2}{R_b}\)
\(ΔT = \frac{2}{4 \ kbps}\)
ΔT = 0.5 msec
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:
PSK(Phase Shift Keying):
In PSK (phase shift keying), binary 1 is represented with a carrier signal and binary 0 is represented with 180° phase shift of a carrier, i.e. the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.
For binary ‘1’ → S1 (A) = Acos 2π f_{C}t
For binary ‘0’ → S2 (t) = A cos (2πf_{C}t + 180°) = - A cos 2π f_{C}t
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.
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.
The transmission bandwidth for a m-ary PSK is given by:
\(BW = \frac{{2{R_b}}}{N}\)
Rb = Bit rate = 1/Tb
Tb = bit duration
N = log_{2}m
For Binary transmission:
\(BW = \frac{2R_b}{log_22}= \frac{2R_b}{1}\)
∴ We can see that the m-ary transmission bandwidth is smaller than binary by: log2m
Analogue modulation techniques:
Digital Modulation Techniques:
Notes:
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:
Important Points
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:
In amplitude shift keying, transmission bandwidth is equal to Base bandwidth.
Additional Information
The bandwith of differnet passband modulation scheme is as shown:
Modulation Scheme |
Bandwidth |
ASK and PSK |
2Rb (Same as baseband BW) |
FSK |
fH - fL + 2Rb where fH is the higher cut-off frequency, fL is the lower cut-off frequency. |
DSB-FC & DSB-SC |
2fm Where fm is the frequency of the message signal. |
SSB-SC |
fm |
VSB-SC |
fm + fv where fv is the vestige frequency |
NBFM (β < 1) |
2fm |
WBFM (β < 1) |
2fm(1 + βf) Where βf is the modulation index of WBFM, given by βf = Δf/fm, where Δf is the frequency deviation. |
PM |
2fm(1 + βp) Where βp is the modulation index of pm, given by βp = Kp × Am, where Kp is amplitude sensitivity of PM signal, Am is the amplitude of message signal. |
M - array PSK |
2Rb/n Where n is the number of bits. |
M – array QAM |
For rectangular pulse, 2Rb/n For raised cosine signal, Rb(1 + α)/n, where α is the roll-off. |
A 70 MHz carrier is QPSK modulated by a 1.544 Mbps T1 data stream. The transmitter employs a raised-cosine filter with α = 0.2. What is the transmitted bandwidth of the signal?
Concept:
The bandwidth for a raised cosine pulse with a roll-ff factor of α, is given by:
Bandwidth = \(\frac{{{R_b}}}{2}\left( {1\; + \;α } \right)\) ----(1)
Rb = Bita/Data rate
Calculation:
Given:
α = 0.2, f_{c} = 70 MHz, R_{b} = 1.544 Mbps
From equation (1);
\(BW=\frac{{{1.544 \ Mbps}}}{2}\left( {1\; + 0.2 } \right)\)
BW = 0.9264 Mbps
BW = 926.4 kHz
Calculation:
Observe first 5 consecutive 0s
Summation rule: the first 5 consecutive 0’s could start at position 1, 2, 3, 4, 5, or 6
Start at the first position
Other 5 bits can be anything: 2^{5} = 32
Start at the second position
The first bit must be a 1
There are various possibilities that can be included in it.
Remaining bits can be anything: 2^{4} = 16
Start at third position
The second bit must be a 1 due to above-mentioned reason.
First bit and last 3 bits can be anything: 2^{4} = 16
Starting at 4,5 and 6 positions
Same as starting at positions 2 or 3: 16 each
Total = 32 + 16 + 16 + 16 + 16 + 16 = 112
The five consecutive 1’s ensue the same pattern and have different like 112 possibilities
There would be two cases counted twice (that we thus need to exclude): 0000011111 and 1111100000
Total = 112 + 112 - 2 = 222
Arrange in increasing order of Null-to-Null bandwidth (Hz) of the following binary bandpass signals.
(i) QPSK
(ii) MSK
(iii) FSK
(iv) ASK
For ASK: BWASK = 2Rb
For M-PSK BW is calculated as:
\(Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}\)
For QPSK M = 4 bandwidth = Rb
For FSK: \(B{W_{FSK}} = 2{R_b} + \left| {{f_1} - {f_2}} \right|\)
f1 and f2 are the two lower and upper frequencies.
for MSK bandwidth B = 1.5 Rb
f1 and f2 are the two lower and upper frequencies.
Observation:
BW_{QPSK} < BWMSK < BW_{ASK}< BWFSK
Concept:
In Binary Phase-Shift Keying (BPSK) system, the waveforms for binary 1 and 0 are represented by:
S_{1}(t) = A_{c} cos (2πf_{c}t)
S_{2}(t) = -A_{c} cos (2πf_{c}t)
When the receiver detects waveform with a matched filter, the average bit-error probability (P_{e}) is given by:
\({P_e} = Q\left[ {\sqrt {\frac{{A_c^2{T_b}}}{{{N_0}}}} } \right]\)
Where,
A_{c} = Amplitude of signal at the receiver.
T_{b} = Bit duration of the signal.
\({T_b} = \frac{1}{{{R_b}}}\)
R_{b} = Bit rate of the signal.
N_{0} = one-sided noise power spectral density
Q(x) = Q-function
Calculation:
Given that,
R_{b} = 1 Mbps = 1 × 10^{6} bps
\(\therefore {T_b} = \frac{1}{{{R_b}}} = {10^{ - 6}}\;sec\)
A = 1 mv = 1 × 10^{-3} V
N_{0} = 10^{-11} W/Hz
So,
\({P_e} = Q\left[ {\sqrt {\frac{{{{\left( {1 \times {{10}^{ - 3}}} \right)}^2} \times {{10}^{ - 6}}}}{{{{10}^{ - 11}}}}} } \right]\)
\( = Q\left[ {\sqrt {\frac{{{{10}^{ - 12}}}}{{{{10}^{ - 11}}}}} } \right]\)
\( = Q\left[ {\sqrt {{{10}^{ - 1}}} } \right]\)
\(\therefore {P_e} = Q\left[ {\sqrt {0.1} } \right]\)
Important Points:
Scheme |
P_{e} |
BASK |
\({P_e} = Q\left[ {\sqrt {\frac{{A_c^2{T_b}}}{{4{N_0}}}} } \right]\) |
BFSK |
\({P_e} = Q\left[ {\sqrt {\frac{{A_c^2{T_b}}}{{2{N_0}}}} } \right]\) |
BPSK |
\({P_e} = Q\left[ {\sqrt {\frac{{A_c^2{T_b}}}{{{N_0}}}} } \right]\) |
From the above table, we say that BPSK gives the lowest P_{e} and BASK gives the highest P_{e}.
So BPSK is mostly used for digital modulation.