Skip to main content

Simulation of ASK, FSK, and PSK using MATLAB Simulink (with Online Simulator)




Simulation of Amplitude Shift Keying (ASK) using MATLAB Simulink

In Simulink, we pick different components/elements from MATLAB Simulink Library. Then we connect the components and perform a particular operation.

Result

A sine wave source, a pulse generator, a product block, a mux, and a scope are shown in the diagram above. The pulse generator generates the '1' and '0' bit sequences. Sine wave sources produce a specific amplitude and frequency. The scope displays the modulated signal as well as the original bit sequence created by the pulse generator. Mux is a tool for displaying both modulated and unmodulated signals at the same time.

The result section shows that binary '1' is modulated by a certain sine wave amplitude of 1 Volt, and binary '0' is modulated by zero amplitude.

Simulation of Frequency Shift Keying (FSK) using MATLAB Simulink

Result

The diagram above shows two sine wave sources, a pulse generator, a switch, a mux, and a scope. The pulse generator generates the '1' and '0' bit sequences. Two different frequencies are provided by sine wave sources. The result section shows that binary '1' is modulated at a higher frequency than binary '0'.

Simulation of Phase Shift Keying (PSK) using MATLAB Simulink

Result

In the result section, binary '1' is modulated with a 180-degree phase shift (corrected), whereas binary '0' is modulated with the original 0-degree phase.

Simulator for Binary ASK Modulation
Simulator for Binary FSK Modulation
Simulator for Binary PSK Modulation
Mathematical Background

1. Amplitude Shift Keying (ASK)

The signal is represented by changing the amplitude of the carrier wave. For Binary ASK:

$$s(t) = \begin{cases} A \cos(2\pi f_c t) & \text{for bit 1} \\ 0 & \text{for bit 0} \end{cases}$$

2. Frequency Shift Keying (FSK)

The frequency of the carrier is shifted between two discrete values ($f_0$ and $f_1$):

$$s(t) = \begin{cases} \cos(2\pi f_1 t) & \text{for bit 1} \\ \cos(2\pi f_0 t) & \text{for bit 0} \end{cases}$$

3. Phase Shift Keying (PSK)

The phase of the carrier is shifted by 180° ($\pi$ radians) to represent different bits:

$$s(t) = \begin{cases} \cos(2\pi f_c t) & \text{for bit 1 (0 rad)} \\ \cos(2\pi f_c t + \pi) & \text{for bit 0 (\(\pi\) rad)} \end{cases}$$

Read More about

  1. Online Simulator for Amplitude Modulation (AM)
  2. Online Simulator for Frequency Modulation (FM)
  3. Online Simulator for Phase Modulation (PM)
  4. QPSK Modulation and Demodulation using MATLAB Simulink



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...

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...

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...

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...

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...

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 ...

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...

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 ...