Skip to main content

How to normalize a highly distorted signal


 

Signal normalization is a common practice in signal processing, especially after a signal has undergone filtering. For example, when using an FIR low-pass filter during the demodulation process of a modulated signal—such as AM or DSB-SC demodulation—you may observe that the first few samples of the demodulated signal exhibit significant transients due to the filtering effect. This occurs because the filter requires approximately N past input samples to produce a steady-state or valid output. To correct for amplitude attenuation, the filtered signal can be normalized to a standard range, such as -1 to 1

 

MATLAB Script

% Parameters for the sine wave
fs = 1000; % Sampling frequency
t = 0:1/fs:1; % Time vector
f = 15; % Frequency of the sine wave

% Example signal: Noisy sine wave
filtered_signal = 0.03*sin(2 * pi * f * t) + 0.05*sin(2 * pi * f * t);
% Step 2: Normalize the filterd signal to the range [-1, +1]
normalized_signal = (filtered_signal - min(filtered_signal)) / (max(filtered_signal) - min(filtered_signal));
normalized_signal = normalized_signal * 2 - 1; % Scale to [-1, +1]


% Original Signal
figure();
plot(t, filtered_signal, 'b', 'LineWidth', 1.5);
title('Filtered Signal');
xlabel('Time (s)');
ylabel('Amplitude');
ylim([-0.1 0.1]);
grid on;

% Normalized Signal
figure();
plot(t, normalized_signal, 'g', 'LineWidth', 1.5);
title('Normalized Signal [-1, +1]');
xlabel('Time (s)');
ylabel('Amplitude');
grid on;

Output








 

 

Copy the MATLAB Code from here 

 


 

Normalize a Highly distorted filtered signal

When normalizing a filtered signal, you may observe that the initial data points are often highly distorted, while the remainder of the signal appears stable. Therefore, it's recommended to discard the first N points (where N is the filter order) before performing normalization.

 

MATLAB script for normalizing a highly distorted filtered signal, where the first few samples are highly transient due to filtering effects

 
 % Parameters for the sine wave
fs = 1000; % Sampling frequency
t = 0:1/fs:1; % Time vector
t1 = N/fs:1/fs:1;
f = 15; % Frequency of the sine wave

% Example signal: Noisy sine wave
filtered_signal = 0.03*sin(2 * pi * f * t) + 0.05*sin(2 * pi * f * t);

N = 10; % N = Filter order
signal1 = filtered_signal(1:N) * 15;
signal2 = filtered_signal(N+1:end);

filtered_signal = [signal1 signal2];


% Normalize the filterd signal to the range [-1, +1] without discarding
% first N-points (N = order of filter)
normalized_signal = (filtered_signal - min(filtered_signal)) / (max(filtered_signal) ...
- min(filtered_signal));
normalized_signal = normalized_signal * 2 - 1; % Scale to [-1, +1]

% Normalize the filterd signal to the range [-1, +1]
normalized_signal1 = (filtered_signal(N+1:end) - min(filtered_signal(N+1:end))) / (max(filtered_signal(N+1:end)) ...
- min(filtered_signal(N+1:end)));
normalized_signal1 = normalized_signal1 * 2 - 1; % Scale to [-1, +1]


% Original Signal
figure();
plot(t, filtered_signal, 'b', 'LineWidth', 1.5);
title('Highly Distorted Filtered Signal');
xlabel('Time (s)');
ylabel('Amplitude');
ylim([-1 1]);
grid on;

% Normalized Signal
figure();
plot(t, normalized_signal, 'g', 'LineWidth', 1.5);
title('Normalized Signal [-1, +1] without discarding first N-points');
xlabel('Time (s)');
ylabel('Amplitude');
grid on;

% Normalized Signal
figure();
plot(t1, normalized_signal1, 'g', 'LineWidth', 1.5);
title('Normalized Signal [-1, +1]');
xlabel('Time (s)');
ylabel('Amplitude');
grid on;

Output 

 
















 Suppose we are using a filter of order N (e.g., 200), which uses N+1 taps (for FIR). The filter requires approximately N past input samples to produce a steady-state or valid output. Therefore, the first N or so samples are based on incomplete data, leading to startup transients. Discarding the first N samples is thus a conservative and practical way to avoid these artifacts.

Further Reading



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...

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 ...

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...

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 ...