Skip to main content

MATLAB Code Constellation Diagrams of M-ary PSK (e.g, 4, 8, 16, 32, 64, 128)



What is the difference between Bit and Symbol in the perspective of transmission?

Symbols use bandwidth more efficiently than bits. For example, in the case of QPSK, one symbol or signal waveform is represented by 2 bits. Hence symbol rate is one-half of the bit rate. As a result, it occupies half bandwidth compared to the BPSK waveform.

We know the primary purpose of modulation [↗] is to multiplex data. Here multiplexing is done so that there is less interference between parallel data streams. Suppose there is a communication channel; we can transmit a single data stream simultaneously. But if we send a symbol instead of a bit, we can send more than 1 bit at a time. In ASK modulation, we assign two amplitude levels to a signal where a higher level is represented by binary '1' and another level as '0'. For BFSK, we apply phase shift in signal (for example, 0 phase shift for consecutive binary '0' bits and 180 phase shift for a binary bit '1'. ASK, FSK, and PSK [↗] - are primary modulation techniques. With the help of those modulation techniques, we derive many other digital modulations capable of carrying more bits thru a channel as a symbol at a time. For example, in QPSK (Quadrature Phase Shift Keying), we can transmit a symbol two bits at a time thru a channel. A total of 4 symbols use 2 bits per symbol and a phase difference of 90 degrees between them. An example of QPSK is shown below. Here you see that the data rate of the channel is getting double when we transmit 2 bits at a time.


1. What is a constellation diagram


A constellation diagram represents a signal modulated by a digital signal, such as quadrature amplitude modulation (QAM) or quadrature phase shift keying (QPSK). [Read More]


QPSK


Assume we need to modulate four signals or symbols with phase differences of π/2 so that the signals can be orthogonal, which will minimize their mutual interference. Then we can modulate those signals in the following way:

s(t)=Acos(2πfct) for 00

= A cos (2πfct + 90) for 01

= A cos (2πfct + 180) for 10

= A cos (2πfct + 270) for 11

Here, the first signal is modulated with a carrier signal. The next signal is modulated with π/2 shifted same carrier signal, the third signal with additional π/2 shifted to the same carrier signal, and so on. The modulated first signal is represented by the symbol '00', the second modulated signal by the symbol '01', and so forth.





In the above figure, we've shown a constellation diagram of 4 QPSK modulations.


Also, read about the Constellation Diagrams of ASK, FSK, and PSK, Constellation Diagrams of M-ary QAM


2. What is the significance of M-ary PSK?


In Mary PSK, given data bits are modulated with any of the M numbers of phase-shifted carrier signals. Let's send M number of data bits modulated with M number of phase-shifted carriers. Theoretically, there will be no interference (theoretically) between them, and we will achieve 8 times the previous data rate (without modulation).

The RF carrier's phase (or frequency) varies instead of only varying the RF signal's phase, frequency, or amplitude. Mary modulation algorithms transfer baseband data into four or more alternative RF carrier signals since the envelope and phase provide two degrees of freedom. We are talking about four carrier signals because here, 2 or more bits form a symbol, and from 2 bits, we can represent 2^(2) or 4 different signals. M-ary modulation is the name given to such modulation schemes. Two or more bits are joined together to create symbols in the M-ary modulation scheme, and one of the available signals S1(t), S2(t),..., Sm(t) is sent during each symbol period Ts. M = 2^n, where n is an integer that defines the number of bits/symbols, the total number of possible signals.

The modulation is called M-ary ASK, M-ary PSK, or M-ary FSK, depending on whether the amplitude, phase, or frequency is altered. M-ary modulation techniques are appealing for application in bandlimited channels because they improve bandwidth efficiency while sacrificing power efficiency. For example, an 8-PSK system utilizes the channel log8 (base 2) = 3 times more efficiently than a 2-PSK (also known as BPSK) system, as the bandwidth of a physical channel is always limited. M-ary signaling, on the other hand, has lower error performance due to the reduced distances between signals in the constellation diagram. The following sections go through a few of the most common M-ary signaling methods.

8-PSK 

 

16-PSK

 

 
 

MATLAB Code for M-ary PSK (e.g, 4, 8, 16, 32, 64, 128)

%The code is developed by SalimWireless.com
% M-ary PSK Modulation and Demodulation
clc;
clear;
close all;

% Parameters
M = 32;  % Order of PSK (M-PSK)
N = 1000;  % Number of symbols
SNR = 10;  % Signal-to-Noise Ratio in dB

% Generate random data symbols
dataSymbols = randi([0 M-1], N, 1);

% Modulate using M-PSK
txSignal = pskmod(dataSymbols, M);

% Add AWGN noise
rxSignal = awgn(txSignal, SNR, 'measured');

% Demodulate
demodulatedSymbols = pskdemod(rxSignal, M);

% Calculate symbol error rate
symbolErrors = sum(dataSymbols ~= demodulatedSymbols);
SER = symbolErrors / N;

% Display results
disp(['Symbol Error Rate (SER): ', num2str(SER)]);

% Plot constellation diagrams
figure;
subplot(2, 1, 1);
plot(real(txSignal), imag(txSignal), 'o');
grid on;
title('Transmitted Signal Constellation');
xlabel('In-Phase');
ylabel('Quadrature');

subplot(2, 1, 2);
plot(real(rxSignal), imag(rxSignal), 'o');
grid on;
title('Received Signal Constellation');
xlabel('In-Phase');
ylabel('Quadrature');

Output






Copy the MATLAB Code above from here



3. What can we conclude from the above M-ary PSK


Both QPSK and QAM are used to send signals in the form of symbols and to increase the bit rate. If you send a symbol instead of a single bit at a time, then multiple prior data rates will be achieved. Those mary modulation techniques are used to multiplex data.

If you are using simple ASK, FSK, or 2-PSK, and if the data rate is N

Then, the following modulation techniques increase data rates further.

4-PSK, 4-QAM ==>2N

Because here 2 bits are sent as a symbol once

8-PSK, 8-QAM ==>3N

Because here 3 bits are sent as a symbol once

Read More about OFDM, QAM, QPSK, BPSK, FSK, etc.


# constellation diagram of qpsk  # qpsk constellation diagram  # Constellation diagram of ask psk fsk


Further Reading

📊 MATLAB & ONLINE SIMULATIONS (Web-Based)
🎓 GATE & UGC - NET SOLUTIONS
📡 WIRELESS & SIGNAL PROCESSING


Contact Us

Name

Email *

Message *

Popular Posts

UGC NET Electronic Science Previous Year Question Papers with Solutions

Download Papers and Solutions Exam Pattern Preparation Tips FAQs More Home / Engineering & Other Exams / UGC NET 2026 PYQ 📊 Exam Highlights: Electronic Science (88) Feature Details Junior Research Fellowship (JRF) ₹37,000 + HRA per month Eligibility M.Sc/M.Tech in Electronics (55%) Validity of Certificate JRF (3 Years) | Lectureship (Lifetime) 📥 Download UGC NET Electronics PDFs Complete collection of previous year question papers, answer keys and explanations for Subject Code 88. Start Downloading 📂 View All Question Papers June 2026 - Question Paper Download PDF June 202...
Read more

GMSK BER vs SNR (with Simulation)

  GMSK Performance: Why the BER vs. SNR Curve Matters If you’ve ever wondered why GMSK (Gaussian Minimum Shift Keying) became the backbone of the global GSM revolution, the answer lies in its unique balance of spectral efficiency and power performance. But there is no free lunch in engineering. To get that sleek, narrow spectrum, you have to pay a price in Bit Error Rate (BER) . In this guide, we break down the theoretical BER vs. SNR performance of GMSK, the "magic" of the BT product, and how to optimize your link budget for real-world conditions. What is GMSK and Why Should You Care? GMSK is a derivative of MSK (Minimum Shift Keying) that uses a Gaussian pre-modulation filter. This filter "smooths" the phase transitions, significantly reducing sideband power. The result? A constant envelope signal that allows you to use high-efficiency Class-C amplifiers without distorting the signal. This is the secr...
Read more

The output voltage V0 (in Volt) for the network given in the Figure is ___

  Find the Output Voltage V o Solve the given circuit using Kirchhoff's Current Law (KCL), Ohm's law, and the supernode method. 1. Given Circuit Parameters Current source 1 mA Resistors 1 kΩ Each resistor Voltage source 6 V Required V o 2. Understand the Circuit The circuit contains a 1 mA current source, four 1 kΩ resistors, and a 6 V source connected between the two midpoint nodes. The output voltage is measured between the top and bottom rails. Reference node Take the bottom rail as ground: 0 V . Let the top rail voltage be V T , and...
Read more

What is a Supernode?

  Circuit Analysis What Is a Supernode? A supernode is formed when an ideal voltage source connects two non-reference nodes. It allows us to apply Kirchhoff's Current Law (KCL) to the combined circuit boundary. 1. Understanding a Supernode In nodal analysis, a voltage source between two unknown-voltage nodes makes the current through that source difficult to express directly using Ohm's law. We combine the two nodes and the voltage source into one extended region called a supernode. Key idea Apply KCL around the outer boundary of the supernode, then use the voltage source to write an additional voltage equation. 2. Supernode Circuit Diagram The voltage source connects nodes V L an...
Read more

A binary ripple counter is designed to count (0)₁₀ to (64)₁₀. Which of the following is/are the number of flip-flops required ...

  Q. A binary ripple counter is designed to count  (0)₁₀ to (64)₁₀ . Which of the following is/are the number of flip-flops required to design the counter?   (A) 6 (B) 7 (C) 4 (D) 5 Solution A binary ripple counter must count from (0)₁₀ to (64)₁₀ , including both endpoints. Step 1: Calculate the total number of states Since both 0 and 64 are included, add 1 to the difference: N = Last value − First value + 1 N = 64 − 0 + 1 = 65 states The counter visits every integer from 0 through 64: 0, 1, 2, 3, ..., 63, 64 There are 65 distinct states in total. Step 2: Calculate the number of flip-flops Each flip-flop stores one bit. With n flip-flops, the counter can represent 2ⁿ distinct states. 2⁶ = 64 states (not enough) 2⁷ = 128 states (eno...
Read more

The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of ...

  Q. The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of inner dimension 0.28 inch×0.14 inch is __. (rounded off to two decimal places). Given Information Waveguide dimensions = 0.28 inch × 0.14 inch Wider dimension (a) = 0.28 inch Smaller dimension (b) = 0.14 inch Medium = Air Dominant mode = TE 10 Speed of light (c) = 3 × 10 8 m/s Step 1: Cutoff Frequency Formula The cutoff frequency of a rectangular waveguide is: f c = (c / 2) √[(m/a) 2 + (n/b) 2 ] For the dominant TE 10 mode: m = 1, n = 0 f c = c / 2a Step 2: Convert Inch into Meter We know that 1 inch = 0.0254 meter. a = 0.28 × 0.0254 a = 0.007112 m ...
Read more

Parametric Down-Conversion of Photons

  Nonlinear Optics One incoming photon is converted into two lower-energy photons, typically in a nonlinear optical medium. Energy is conserved: E in = E 1 + E 2 E in ​ = E 1 ​ + E 2 ​ . Parametric Down-Conversion A nonlinear optical process in which one pump photon is converted into two lower-energy photons, known as the signal and idler photons. 1 Energy Conservation The energy of the pump photon equals the sum of the energies of the signal and idler photons. \[ \boxed{\hbar\omega_p=\hbar\omega_s+\hbar\omega_i} \] \[ \boxed{\omega_p=\omega_s+\omega_i} \] ωₚ Pump angular frequency ...
Read more

SRAM, DRAM, EPROM, MRAM, and DDRAM Explained

Computer Fundamentals Types of Memory and Their Technologies Understand the fundamental relationship between different types of computer memory and the technologies used to implement them. Fundamental Matching Memory type, technology, and key characteristics Memory Type Technology Key Characteristic Cache Memory SRAM Very fast; stores frequently accessed data Main Memory DRAM Holds programs and data currently in use Au...
Read more