Skip to main content

Constellation Diagram of ASK in Detail (with MATLAB + Simulator)

A binary bit '1' is assigned a power level of Eb\sqrt{E_b} (or energy EbE_b), while a binary bit '0' is assigned zero power (or no energy).
 

Simulator for Binary ASK Constellation Diagram

Noisy Modulated Signal (ASK)

Original Modulated Signal (ASK)


Energy per bit (Eb) (Tb = bit duration):

We know that all periodic signals are power signals. Now we’ll find the energy of ASK for the transmission of binary ‘1’.

Eb = ∫0Tb(Ac.cos(2П.fc.t))2 dt
= ∫0Tb(Ac)2.cos2(2П.fc.t) dt
Using the identity cos2x = (1 + cos(2x))/2:
= ∫0Tb((Ac)2/2)(1 + cos(4П.fc.t)) dt
= ((Ac)2/2) ∫0Tb(1) dt + ((Ac)2/2) ∫0Tbcos(4П.fc.t) dt
= ((Ac)2/2) * Tb + 0 (The integral of cos(4П.fc.t) over a full period is zero, assuming Tb is an integer multiple of 1/(2fc))
Eb = (Ac2/2).Tb (where Tb is the bit duration)

** where Ac is the amplitude of the carrier signal and fc is the carrier frequency in Hz.

To save transmitter energy, Eb should be small.

** for transmission of binary ‘0’
Eb = ∫0Tb(S2(t))2dt = 0

** Constellation Diagram
First, we define the orthonormal basis function for this system:
φ1(t) = √(2/Tb) cos(2Пfct) for 0 ≤ t ≤ Tb.
The energy of this basis function is 1.

Now, we can represent our signaling waveforms using this basis function:
For binary '1': S1(t) = Ac cos(2Пfct) = [Ac * √(Tb/2)] * φ1(t)
The coordinate for S1(t) in the constellation diagram is g11 = Ac * √(Tb/2).
The energy of S1(t) is Eb = g112 = (Ac2 * Tb)/2.
Therefore, g11 = √(Eb).

For binary '0': S2(t) = 0. The coordinate for S2(t) is g21 = 0.

So, in the constellation diagram:
1 => point at √(Eb) along the φ1 axis
0 => point at 0 (the origin)


High-order Amplitude Shift Keying (ASK) refers to using a large number of amplitude levels to represent digital data. For instance, in binary ASK (BASK), there are two amplitude levels, usually represented as 0 and 1. High-order ASK can have more than two amplitude levels, such as 4, 8, 16, 64, etc.
 

MATLAB Code For Constellation Diagram of ASK  

 
 

Output 

 
 
 

 

Effect of Noise on Constellation Diagram of ASK

At SNR = 5 dB
 
 
 At SNR = 10 dB

 
 
At SNR = 15 dB

 
 
At SNR = 30 dB


 

Read more about 


 


 
 
 



Contact Us

Name

Email *

Message *

Popular Posts

Online Simulator for ASK, FSK, and PSK Signal Generation

Interactive Digital Signal Processing (DSP) Tutorial and Simulator for ASK, FSK, and BPSK modulation techniques. Try our new Digital Signal Processing Simulator!   •   Interactive ASK, FSK, and BPSK tools updated for 2025. Start Now Digital Modulation Visualizer: ASK, FSK, & BPSK Simulator Learn and visualize binary modulation techniques (ASK, FSK, BPSK) in real-time with adjustable carrier and sampling parameters. Perfect for DSP students and engineers. 📡 ASK Simulator 📶 FSK Simulator 🎚️ BPSK Simulator 📚 More Topics ASK Modulator FSK Modulator BPSK Modulator Demodulation More Topics 1. ASK (Ampli...

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 2025 - Question Paper Download PDF June 2025 - Sol...

Constellation Diagrams of ASK, PSK, and FSK (with MATLAB Code + Simulator)

Constellation Diagrams: ASK, FSK, and PSK Comprehensive guide to signal space representation, including interactive simulators and MATLAB implementations. 📘 Overview 🧮 Simulator ⚖️ Theory 📈 Q-function 📚 Resources BASK Modulation Transmits one of two signals: 0 or $\sqrt{E_b}$, representing binary 0 and 1. Simple but sensitive to noise. BFSK Modulation Transmits one of two signals: $\sqrt{E_b}$ on the Y-axis or $\sqrt{E_b}$ on the X-axis. These are orthogonal signals. BPSK Modulation Transmits $+\sqrt{E_b}$ or $-\sqrt{E_b}$ (antipodal signaling). Most efficient binary scheme. ...

OFDM Symbols and Subcarriers Explained

This article explains how OFDM (Orthogonal Frequency Division Multiplexing) symbols and subcarriers work. It covers modulation, mapping symbols to subcarriers, subcarrier frequency spacing, IFFT synthesis, cyclic prefix, and transmission. Step 1: Modulation First, modulate the input bitstream. For example, with 16-QAM , each group of 4 bits maps to one QAM symbol. Suppose we generate a sequence of QAM symbols: s0, s1, s2, s3, s4, s5, …, s63 Step 2: Mapping Symbols to Subcarriers Assume N sub = 8 subcarriers. Each OFDM symbol in the frequency domain contains 8 QAM symbols (one per subcarrier): Mapping (example) OFDM symbol 1 → s0, s1, s2, s3, s4, s5, s6, s7 OFDM symbol 2 → s8, s9, s10, s11, s12, s13, s14, s15 … OFDM sym...

Direction of Arrival (DoA) Online Simulator (using MUSIC)

Interactive DOA Simulator X-axis XY angle (deg): 45 XZ angle (deg): 30 Noise: 0.05 Y-axis XY angle (deg): 60 YZ angle (deg): 45 Noise: 0.05 Z-axis XZ angle (deg): 60 YZ angle (deg): 30 Noise: 0.05 Estimated DOA (deg): 0 Simulation Workflow and Mathematical Background This simulator demonstrates Direction of Arrival (DOA) estimation using three-axis sensor signals (X, Y, Z), Maximal Ratio Combining (MRC) , and the MUSIC algorithm . It allows interactive control of signal angles and noise for teaching purposes. 1. Signal Generation A pure sinewave signal of frequency f is projected onto three axes using user-defined angles in different planes: X-axis: θ XY , θ XZ Y-axis: θ XY , θ YZ Z-axis: θ XZ , θ YZ Mathematically, for each time sample t : x(t) = s(t) * cos(θ_xy_x) * cos(θ_xz_x) + n_x(t) y(t) = s(t) * sin(θ_xy_y) * cos(θ_yz_y) + n_y(t) z(t) = s(t) * sin(θ_xz_z) * sin(θ_yz_z) + n_z(t) wh...

BER vs. SER: Why Do They Differ?

Below is the derivation for Bit Error Rate (BER) and Symbol Error Rate (SER) at 0 dB \(E_b/N_0\) . 1. The 16-QAM Case (\(E_b/N_0 = 0\) dB) 16-QAM is treated as two independent 4-PAM signals on the Real and Imaginary axes. At 0 dB, the linear ratio \(\gamma_b = 1\). Bit Error Rate (BER) Calculation: \[P_b \approx \frac{3}{4} Q\left( \sqrt{\frac{4}{5} \frac{E_b}{N_0}} \right)\] \[P_b \approx 0.75 \times Q(\sqrt{0.8}) = 0.75 \times Q(0.8944)\] \[P_b \approx 0.75 \times 0.1855 = \mathbf{0.139} \approx \mathbf{0.14}\] Symbol Error Rate (Per Dimension) Calculation: For 4-PAM (one axis of 16-QAM): \[P_{pam} = \frac{3}{2} Q\left( \sqrt{\frac{4}{5} \frac{E_b}{N_0}} \right)\] \[P_{pam} = 1.5 \times Q(0.8944) = 1.5 \times 0.1855 = \mathbf{0.278} \approx \mathbf{0.28}\] ...