Skip to main content

Pulse Amplitude Modulation and Demodulation



 Pulse Amplitude Modulation (PAM)

Sampling allow us to represent real world continuous signal, such as audio or video, in a format suitable for digital processing and storage. This sampled discrete-time signal is inherently digital. A digital signal is a discrete-time signal that is further quantized in amplitude.

Pulse Amplitude modulation (PAM) is the modulation technique in which amplitude of carrier pulses is made to vary in accordance with the input message signal, similar to Amplitude Modulation (AM). But here we use pulse generator as carrier signal. So, Pulse–amplitude modulation (PAM) is a form of signal modulation where the message information is encoded in the amplitude of a series of signal pulses.


 Fig 1: Pulse Amplitude Modulation




The basic idea in PAM for communication over a Continuous Time (CT) channel is to transmit a sequence of Continuous Time pulses of some per-specified pulse shape, with the sequence of pulse amplitudes carrying the information. 


 Pulse Amplitude Demodulation

To demodulate a PAM signal, pass it through a reconstruction filter. As here, the amplitude of the pulse carrier is varied according to the amplitude of the message signal, we only need to pass this received pulse signal through a low-pass filter with a cut-off frequency the same as the message signal or slightly higher.


If you perform quantization at the transmitter side and assign some levels (amplitude levels), then demodulation is performed by detecting the amplitude level of the carrier at every single period. The number of possible pulse amplitudes in analog PAM is theoretically infinite. Digital PAM reduces the number of pulse amplitudes to some power of two. For example, in 4-level PAM there are (2^2 = 4) possible discrete pulse amplitudes; in 8-level PAM there are (2^3 = 8) possible discrete pulse amplitudes; and in 16-level PAM there are (2^4 = 16) possible discrete pulse amplitudes.
 

Instructions for Pulse Amplitude Modulation (PAM)

  • Note: Use the input fields to enter the message frequency and the square pulse carrier frequency.
  • Step 1: Click the "Generate Message" button to generate the input message signal.
  • Step 2: Click the "Generate Carrier" button to generate the carrier signal. Carrier must be > Message.
  • Step 3: Click the "Generate PAM Signal" button to generate the Modulated signal.

PAM Modulation Control Center

Perform Pulse Amplitude Modulation by interacting with the signal generators below.

Ready for Signal Recovery?

After generating your PAM signal, proceed to the Demodulation section to recover the original message using a reconstruction filter.

Go to Demodulation

Instructions for Pulse Amplitude Demodulation

  • The reconstruction filter recovers the original message from the sampled PAM signal.
  • A low-pass filter (LPF) is used with a cutoff frequency at or above the message frequency.
  • Click 'Demodulate' to view the recovered baseband signal.

Technical Definition: PAM Signal

MATLAB Mathematical Representation Copy Snippet
% Ideal Sampling with Pulse Train t = 0:0.001:1; fm = 2; fc = 10; m_t = cos(2*pi*fm*t); c_t = square(2*pi*fc*t); s_pam = m_t .* (c_t > 0); % Low Pass Reconstruction [b,a] = butter(4, fm/(fs/2)); demod = filter(b,a, s_pam);

Further Reading 



Contact Us

Name

Email *

Message *

Popular Posts

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...

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...

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...

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...

Flat vs Frequency Selective Online Simulator

Flat vs Frequency Selective Online Simulator Channel Type Without Fading Flat Fading Multipaths Nakagami m SNR(dB) Run Simulation Input Signal Signal After Fading Constellation Diagram BER vs SNR Explore Advanced Flat vs Frequency-Selective Fading Simulator Want to see these equations in action? Visualize it. Launch Simulator Tool Flat Fading and Frequency-Selective Fading Effects of Multipath Components and the Nakagami-m Parameter 1. Flat Fading Flat fading occurs when the signal bandwidth is smaller than the channel's coherence bandwidth, meaning that all frequency components of the transmitted signal experience approximately the same fading. It is typically associated with narrowband signals, where the multipath delay spread is small relative to the symbol duration. ...

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 ...