Constellation Diagram MCQ Quiz - Objective Question with Answer for Constellation Diagram - Download Free PDF

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° 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:

Since the minimum distance between adjacent point is 2A, so we redraw the constellation diagram shown in figure (I) as:

From, the diagram, we have the distance of four signal points from the origin (amplitude of signals) as 2A.

By using the Pythagorean theorem, we obtain the distance of rest 4-signal point from the origin as:

\(\sqrt {{{\left( {2A} \right)}^2} + {{\left( {2A} \right)}^2}} = 2\sqrt 2 \;A\) 

Thus, the average transmitted power for the constellations is obtained as:

\({P_{av}} = \frac{1}{8}\left[ {4 \times {{\left( {2A} \right)}^2} + 4 \times {{\left( {2\sqrt 2 \;A} \right)}^2}} \right]\) 

P₁ = 6 A²

For constellation shown in figure (II), we have the minimum distance between adjacent points as 2A. So, we redraw the constellations as shown below:

For the above constellation diagram, we have the distance of two signal points from the origin (amplitude of signals) as A.

By using the Pythagorean theorem, we obtain the distance of two signal points at a vertical axis from the origin as:

\(\sqrt {{{\left( {2A} \right)}^2} - {{\left( A \right)}^2}} = \sqrt {4{A^2} - {A^2}} = \sqrt 3 \;A\) 

Now, for determining the distance of rest your points from the origin, consider the triangle shown below:

Again, applying the Pythagorean theorem, we obtain the distance of the four constellation points as:

\(R = \sqrt {{{\left( {\sqrt 3 \;A} \right)}^2} + {{\left( {2A} \right)}^2}} = \sqrt {3{A^2} + 4{A^2}} \) 

\( = \sqrt {7{A^2}} = \sqrt 7 \;A\) 

Thus, the average transmitted power is obtained as:

\({P_{av}} = \frac{1}{8}\left[ {2 \times {{\left( A \right)}^2} + 2 \times {{\left( {\sqrt 3 \;A} \right)}^2} + 4 \times {{\left( {\sqrt 7 \;A} \right)}^2}} \right]\) 

\(= \frac{1}{8}\left[ {2{A^2} + 6{A^2} + 28\;{A^2}} \right]\) 

\({P_2} = \frac{9}{2}{A^2}\) 

So, we have P₂ < P₁

Therefore, the second constellation is more power-efficient.

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° 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:

Since the minimum distance between adjacent point is 2A, so we redraw the constellation diagram shown in figure (I) as:

From, the diagram, we have the distance of four signal points from the origin (amplitude of signals) as 2A.

By using the Pythagorean theorem, we obtain the distance of rest 4-signal point from the origin as:

\(\sqrt {{{\left( {2A} \right)}^2} + {{\left( {2A} \right)}^2}} = 2\sqrt 2 \;A\) 

Thus, the average transmitted power for the constellations is obtained as:

\({P_{av}} = \frac{1}{8}\left[ {4 \times {{\left( {2A} \right)}^2} + 4 \times {{\left( {2\sqrt 2 \;A} \right)}^2}} \right]\) 

P₁ = 6 A²

For constellation shown in figure (II), we have the minimum distance between adjacent points as 2A. So, we redraw the constellations as shown below:

For the above constellation diagram, we have the distance of two signal points from the origin (amplitude of signals) as A.

By using the Pythagorean theorem, we obtain the distance of two signal points at a vertical axis from the origin as:

\(\sqrt {{{\left( {2A} \right)}^2} - {{\left( A \right)}^2}} = \sqrt {4{A^2} - {A^2}} = \sqrt 3 \;A\) 

Now, for determining the distance of rest your points from the origin, consider the triangle shown below:

Again, applying the Pythagorean theorem, we obtain the distance of the four constellation points as:

\(R = \sqrt {{{\left( {\sqrt 3 \;A} \right)}^2} + {{\left( {2A} \right)}^2}} = \sqrt {3{A^2} + 4{A^2}} \) 

\( = \sqrt {7{A^2}} = \sqrt 7 \;A\) 

Thus, the average transmitted power is obtained as:

\({P_{av}} = \frac{1}{8}\left[ {2 \times {{\left( A \right)}^2} + 2 \times {{\left( {\sqrt 3 \;A} \right)}^2} + 4 \times {{\left( {\sqrt 7 \;A} \right)}^2}} \right]\) 

\(= \frac{1}{8}\left[ {2{A^2} + 6{A^2} + 28\;{A^2}} \right]\) 

\({P_2} = \frac{9}{2}{A^2}\) 

So, we have P₂ < P₁

Therefore, the second constellation is more power-efficient.

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° 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: