Insertion Loss Calculator for Signal Power

Insertion Loss Calculator:

Enter the values of Input Voltage, V_1(V) and output Voltage, V_2(V) to determine the value of Insertion Loss, IL_(dB).

Enter Input Voltage:		V
Enter Output Voltage:		V
Result -Insertion Loss:		dB

Insertion Loss:

Insertion Loss is a measure of the reduction in signal power resulting from the insertion of a device or component (like a cable, connector, filter, or other passive network element) into a transmission line or optical path. It quantifies how much signal is lost when the device is placed in the signal path.

Insertion loss is typically expressed in decibels (dB), which is a logarithmic unit used to describe ratios of power or amplitude. A lower insertion loss indicates a more efficient system with minimal signal degradation, which is crucial in high-performance communication systems, audio systems, and radio frequency (RF) applications.

Insertion loss may result from resistance, impedance mismatch, dielectric loss, or radiation. It’s especially important in RF, microwave, and optical systems where even small losses can affect system performance.

Manufacturers often specify insertion loss ratings for components, and ensuring minimal IL is key to maintaining signal strength and clarity. Testing is done using signal generators and measuring instruments to determine the input and output voltage (or power) before and after insertion.

Insertion Loss, IL_(dB) in decibels equals 10 multiplied by the base-10 logarithm of Input Voltage, V_1(V) in volts, divided by Output Voltage, V_2(V) in volts.

Insertion Loss, IL_(dB)= 20 * log₁₀(V_1(V)/ V_2(V))

IL_(dB)= insertion loss in decibels, dB.

V_1(V)= input voltage in volts, V.

V_2(V) = output voltage in volts, V.

Insertion Loss Calculation:

1. An input signal has a voltage of 5V, and the output voltage after passing through a filter is 4V. Find the insertion loss.

Given:V_1(V)= 5V, V_2(V) = 4V.

Insertion Loss, IL_(dB)= 20 * log₁₀(V_1(V)/ V_2(V))

IL_(dB)= 20 * log₁₀(5 / 4)

IL_(dB)= 20 * log₁₀(1.25)

IL_(dB)= 20 * 0.0969

IL_(dB)= 1.94dB.

2. An input voltage of 10V results in an insertion loss of 6 dB. What is the output voltage V2?

Given:V_1(V)= 10V, IL_(dB)= 6dB.

Insertion Loss, IL_(dB)= 20 * log₁₀(V_1(V)/ V_2(V))

V_2(V) = V_1(V)/ 10^{(IL(dB) / 20)}

V_2(V) = 10 / 10^{(6 / 20)}

V_2(V) = 10 / 1.995

V_2(V) = 5.01V.