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

Constellation Diagram of FSK in Detail

📘 Overview 🧮 Simulator for constellation diagram of FSK 🧮 Theory 🧮 MATLAB Code 📚 Further Reading 📚 BER vs SNR from Constellation   Binary bits '0' and '1' can be mapped to 'j' and '1' to '1', respectively, for Baseband Binary Frequency Shift Keying (BFSK) . Signals are in phase here. These bits can be mapped into baseband representation for a number of uses, including power spectral density (PSD) calculations. For passband BFSK transmission, we can modulate signal 'j' with a lower carrier frequency and signal '1' with a higher carrier frequency while transmitting over a wireless channel. Let's assume we are transmitting carrier signal fc1 for the transmission of binary bit '1' and carrier signal fc2 for the transmission of binary bit '0'. Simulator for 2-FSK Constellation Diagram Simulator for 2-FSK Constellation Diagram ...

UGC NET Electronic Science Previous Year Question Papers with Solutions

Home / Engineering & Other Exams / UGC NET 2026 PYQ ⬇️ Download Papers and Solutions 📋 Exam Pattern 💡 Preparation Tips ❓ FAQs 📊 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 - Solved Paper + Explanation ...

FM Bandwidth and FM Band Explained

FM radio uses the frequency band from 88 MHz to 108 MHz , which is a 20 MHz-wide spectrum . This is the range of carrier frequencies available to stations. 108 MHz − 88 MHz = 20 MHz However, a single FM station occupies only about 200 kHz . This is the bandwidth of the modulated FM signal. 1. Why One FM Station Needs ~200 kHz FM uses frequency modulation . The bandwidth depends on how far the carrier swings. Carson's Rule gives the approximate FM bandwidth: B = 2 ( Δf + f m ) ...

BER vs SNR for M-ary QAM, M-ary PSK, QPSK, BPSK, ...(MATLAB Code + Simulator)

Bit Error Rate (BER) & SNR Guide Analyze communication system performance with our interactive simulators and MATLAB tools. 📘 Theory 🧮 Simulators 💻 MATLAB Code 📚 Resources BER Definition SNR Formula BER Calculator MATLAB Comparison 📂 Explore M-ary QAM, PSK, and QPSK Topics ▼ 🧮 Constellation Simulator: M-ary QAM 🧮 Constellation Simulator: M-ary PSK 🧮 BER calculation for ASK, FSK, and PSK 🧮 Approaches to BER vs SNR What is Bit Error Rate (BER)? The BER indicates how many corrupted bits are received compared to the total number of bits sent. It is the primary figure of merit f...

Coherence Bandwidth and Coherence Time (with MATLAB + Simulator)

🧮 Coherence Bandwidth 🧮 Coherence Time 🧮 MATLAB Code s 📚 Further Reading For Doppler Delay or Multi-path Delay Coherence time T coh ∝ 1 / v max (For slow fading, coherence time T coh is greater than the signaling interval.) Coherence bandwidth W coh ∝ 1 / τ max (For frequency-flat fading, coherence bandwidth W coh is greater than the signaling bandwidth.) Where: T coh = coherence time W coh = coherence bandwidth v max = maximum Doppler frequency (or maximum Doppler shift) τ max = maximum excess delay (maximum time delay spread) Notes: The notation v max −1 and τ max −1 indicate inverse proportionality. Doppler spread refers to the range of frequency shifts caused by relative motion, determining T coh . Delay spread (or multipath delay spread) determines W coh . Frequency-flat fading occurs when W coh is greater than the signaling bandwidth. Coherence Bandwidth Coherence bandwidth is...

Intel 8086 Transistor Count: Architecture, Specifications, and Comparison with Other Microprocessors

Intel 8086 Transistor Count: Architecture, Specifications, and Comparison with Other Microprocessors Intel 8086 Transistor Count: Complete Guide with Architecture and Processor Comparison The Intel 8086 microprocessor is one of the most important processors in computer history. Released in 1978 , it introduced the x86 architecture that still influences modern CPUs. One of the most frequently asked questions in computer architecture and microprocessor courses is: How many transistors are present in the Intel 8086? The commonly accepted answer is approximately 29,000 transistors . However, reverse-engineering studies have shown that the actual number of physical transistors is closer to 19,618 , while Intel's published figure includes programmable transistor locations used in ROM and PLA structures. Intel 8086 Transistor Count Metric Value Published transistor count ~29,000 Physical transistor count ~19,618 Release year 1978 Word ...

Online Simulator for ASK, FSK, and PSK

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 More Topics 1. ASK (Amplitude Shift Keying) Simulat...