QPSK Transmitter MCQ Quiz - Objective Question with Answer for QPSK Transmitter - Download Free PDF

The correct answer is 1) QPSK (Quadrature Phase Shift Keying).

Explanation

• QPSK (Quadrature Phase Shift Keying):
  • QPSK is a multilevel modulation technique that uses four distinct phase shifts to represent four different symbols.
  • These four phases are typically separated by 90 degrees (e.g., 45°, 135°, 225°, and 315°).
  • Each symbol represents two bits of information, allowing for twice the data rate compared to BPSK.
• Other options:
  • BFSK (Binary Frequency Shift Keying): Uses two different frequencies.
  • BPSK (Binary Phase Shift Keying): Uses two phase shifts.
  • 8-PSK: Uses eight phase shifts.

Concept:

Probability of error (Pe) is given as;

\(P_e=Q\sqrt{\frac{E_b}{N_0}}\)

Pe is minimum when energy (Eb) is maximum. 

Now,

Eb ∝ d2 (distance between the symbol)

i.e., Pe is minimum when the distance is maximum. 

Calculation:

For M-ary PSK, Energy is calculated as:

E_s = E_blog₂M

\(d=2\sqrt{{E_s}\frac{sin\pi}{M}}\)

Where E_s gives the distance between adjacent signaling points.

1. For 8-PSK

Es = Eblog28

E_s = 3E_b

\(d=2√{3E_b}\frac{sin\pi}{8}\)

\(d=0.67√{E_b}\)  ---(1)

2.  For QPSK

M = 4

Es = Eblog24

Es = 2Eb

\(d=2√ {2E_b}\frac{sin\pi}{4}\)

d = 2√E_b   ---(2) 

For QAM:

\(d=\sqrt {0.2E_s}\)

Where E_s = 4E_b 

\(d=2\sqrt{0.2E_b}\)  ----(3)

So from all equations, we get that:

d is maximum for QPSK

then Pe is minimum when the distance is maximum. 

Hence Minimum power is required to transmit QPSK

Hence option (1) is the correct answer.

Important Points

1. QPSK is a modulation scheme that allows one symbol to transfer two bits of data. There are four possible two-bit numbers (00, 01, 10, 11).

2. \(Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}\)

QPSK modulation represents symbols by a constellation of four-phase angles of the carrier signal, orthogonal to each other.

There are two bits per symbol. So for example 00 = 0°, 01 = 90° , 10 = 180°m and 11= 270°. 

QPSK transmits twice the data rate in a given bandwidth compared to BPSK - at the same BER. 

The constellation diagram for 4 different symbols is as shown:

Quadrature Amplitude Modulation:

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:

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:

Given 64 kbps PCM coding is used to code each channel, hence bandwidth of QPSK modulation is

BW_QPSK=R_S(1+0.4)=1.4R_s

_{\({R_s} = \frac{{{R_B}}}{{{{\log }_2}M}} = \frac{{{R_b}}}{{{{\log }_2}4}} = \frac{{{R_b}}}{2} = 32kbps\)}

BW _QPSK = 1.4 X 32 = 44.8 kHz

Number of channels supported by 36 MHz bandwidth is \( = \frac{{36 \times {{10}^6}}}{{44.8\times {{10}^3}}}\;\)

= 803 channels

Concept:

Probability of error (Pe) is given as;

\(P_e=Q\sqrt{\frac{E_b}{N_0}}\)

Pe is minimum when energy (Eb) is maximum. 

Now,

Eb ∝ d2 (distance between the symbol)

i.e., Pe is minimum when the distance is maximum. 

Calculation:

For M-ary PSK, Energy is calculated as:

E_s = E_blog₂M

\(d=2\sqrt{{E_s}\frac{sin\pi}{M}}\)

Where E_s gives the distance between adjacent signaling points.

1. For 8-PSK

Es = Eblog28

E_s = 3E_b

\(d=2√{3E_b}\frac{sin\pi}{8}\)

\(d=0.67√{E_b}\)  ---(1)

2.  For QPSK

M = 4

Es = Eblog24

Es = 2Eb

\(d=2√ {2E_b}\frac{sin\pi}{4}\)

d = 2√E_b   ---(2) 

For QAM:

\(d=\sqrt {0.2E_s}\)

Where E_s = 4E_b 

\(d=2\sqrt{0.2E_b}\)  ----(3)

So from all equations, we get that:

d is maximum for QPSK

then Pe is minimum when the distance is maximum. 

Hence Minimum power is required to transmit QPSK

Hence option (1) is the correct answer.

Important Points

1. QPSK is a modulation scheme that allows one symbol to transfer two bits of data. There are four possible two-bit numbers (00, 01, 10, 11).

2. \(Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}\)

QPSK modulation represents symbols by a constellation of four-phase angles of the carrier signal, orthogonal to each other.

There are two bits per symbol. So for example 00 = 0°, 01 = 90° , 10 = 180°m and 11= 270°. 

QPSK transmits twice the data rate in a given bandwidth compared to BPSK - at the same BER. 

The constellation diagram for 4 different symbols is as shown:

Quadrature Amplitude Modulation:

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:

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:

The correct answer is 1) QPSK (Quadrature Phase Shift Keying).

Explanation

• QPSK (Quadrature Phase Shift Keying):
  • QPSK is a multilevel modulation technique that uses four distinct phase shifts to represent four different symbols.
  • These four phases are typically separated by 90 degrees (e.g., 45°, 135°, 225°, and 315°).
  • Each symbol represents two bits of information, allowing for twice the data rate compared to BPSK.
• Other options:
  • BFSK (Binary Frequency Shift Keying): Uses two different frequencies.
  • BPSK (Binary Phase Shift Keying): Uses two phase shifts.
  • 8-PSK: Uses eight phase shifts.

Concept:

Probability of error (Pe) is given as;

\(P_e=Q\sqrt{\frac{E_b}{N_0}}\)

Pe is minimum when energy (Eb) is maximum. 

Now,

Eb ∝ d2 (distance between the symbol)

i.e., Pe is minimum when the distance is maximum. 

Calculation:

For M-ary PSK, Energy is calculated as:

E_s = E_blog₂M

\(d=2\sqrt{{E_s}\frac{sin\pi}{M}}\)

Where E_s gives the distance between adjacent signaling points.

1. For 8-PSK

Es = Eblog28

E_s = 3E_b

\(d=2√{3E_b}\frac{sin\pi}{8}\)

\(d=0.67√{E_b}\)  ---(1)

2.  For QPSK

M = 4

Es = Eblog24

Es = 2Eb

\(d=2√ {2E_b}\frac{sin\pi}{4}\)

d = 2√E_b   ---(2) 

For QAM:

\(d=\sqrt {0.2E_s}\)

Where E_s = 4E_b 

\(d=2\sqrt{0.2E_b}\)  ----(3)

So from all equations, we get that:

d is maximum for QPSK

then Pe is minimum when the distance is maximum. 

Hence Minimum power is required to transmit QPSK

Hence option (1) is the correct answer.

Important Points

1. QPSK is a modulation scheme that allows one symbol to transfer two bits of data. There are four possible two-bit numbers (00, 01, 10, 11).

2. \(Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}\)

QPSK modulation represents symbols by a constellation of four-phase angles of the carrier signal, orthogonal to each other.

There are two bits per symbol. So for example 00 = 0°, 01 = 90° , 10 = 180°m and 11= 270°. 

QPSK transmits twice the data rate in a given bandwidth compared to BPSK - at the same BER. 

The constellation diagram for 4 different symbols is as shown:

Quadrature Amplitude Modulation:

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:

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:

Given 64 kbps PCM coding is used to code each channel, hence bandwidth of QPSK modulation is

BW_QPSK=R_S(1+0.4)=1.4R_s

_{\({R_s} = \frac{{{R_B}}}{{{{\log }_2}M}} = \frac{{{R_b}}}{{{{\log }_2}4}} = \frac{{{R_b}}}{2} = 32kbps\)}

BW _QPSK = 1.4 X 32 = 44.8 kHz

Number of channels supported by 36 MHz bandwidth is \( = \frac{{36 \times {{10}^6}}}{{44.8\times {{10}^3}}}\;\)

= 803 channels

The correct answer is 1) QPSK (Quadrature Phase Shift Keying).

Explanation

• QPSK (Quadrature Phase Shift Keying):
  • QPSK is a multilevel modulation technique that uses four distinct phase shifts to represent four different symbols.
  • These four phases are typically separated by 90 degrees (e.g., 45°, 135°, 225°, and 315°).
  • Each symbol represents two bits of information, allowing for twice the data rate compared to BPSK.
• Other options:
  • BFSK (Binary Frequency Shift Keying): Uses two different frequencies.
  • BPSK (Binary Phase Shift Keying): Uses two phase shifts.
  • 8-PSK: Uses eight phase shifts.