Question

A certain amplifier is to give a power output of \(1000 \mathrm{W}\) for an input of \(50 \mathrm{W}\) Find the dB gain.The power output of a certain device is half the input. What is the change in decibel level?

Short answer

The dB gain for an input of 50 W and output of 1000 W is approximately 13 dB. When the power output is half the input, there is a change in decibel level of approximately -3 dB.

Step-by-step solution

Understand Decibel Gain

The decibel (dB) gain of an amplifier is given by the formula: \( G_{dB} = 10 \log_{10}\left(\frac{P_{out}}{P_{in}}\right) \), where \(P_{out}\) is the power output and \(P_{in}\) is the power input.

Calculate dB Gain for the First Scenario

Substitute the given power output \(P_{out} = 1000\mathrm{W}\) and power input \(P_{in} = 50\mathrm{W}\) into the formula: \( G_{dB} = 10 \log_{10}\left(\frac{1000 \mathrm{W}}{50 \mathrm{W}}\right) \).

Simplify the Expression for dB Gain

Calculate the ratio and then take the logarithm: \( G_{dB} = 10 \log_{10}(20) \approx 10 \times 1.3010 = 13.01 \) dB. The decibel gain is approximately 13 dB.

Understand Decibel Level Change for Reduced Power Output

When the output is half the input, the ratio of powers \( \frac{P_{out}}{P_{in}} = \frac{1}{2} \). Use this ratio to find the change in the decibel level.

Calculate Decibel Level Change

Use the decibel formula for a power ratio of 1/2: \( \Delta G_{dB} = 10 \log_{10}\left(\frac{1}{2}\right) = 10 \times (-0.3010) = -3.01 \) dB. There is a decrease of approximately 3 dB.

Key concepts

dB gain formula

Understanding the decibel gain formula is essential when dealing with amplifiers or any system that boosts signal strength. It's a logarithmic expression that quantifies the increase in power due to an amplifier or other signal-boosting device. The formula is defined as:
\[ G_{dB} = 10 \log_{10}\left(\frac{P_{out}}{P_{in}}\right) \]
In this expression, \(G_{dB}\) represents the gain in decibels, \(P_{out}\) is the power output, and \(P_{in}\) is the power input. Thanks to the logarithmic nature of the formula, the decibel gain corresponds to a logarithmic scale, which means that each increase by a factor of 10 in the power ratio corresponds to an increase of 10 dB in the gain.

Power Output

The concept of power output is critical in determining an amplifier’s efficiency and capability. It refers to the amount of power, measured in watts (W), that an amplifier can deliver to its load. For instance, in the exercise, the amplifier's power output was indicated to be \(1000\mathrm{W}\). Various factors can affect this output, such as the quality of the amplifier's components, temperature, or even the frequency of the signal. When evaluating the dB gain, the power output directly influences the calculated value, as an increase in output generally leads to a higher dB gain.

Power Input

Power input is the energy in watts (W) provided to the amplifier from an external source, which is then converted into a larger power output. In our given problem, the power input is marked at \(50\mathrm{W}\). This input power acts as a baseline for determining how much an amplifier or similar device amplifies the signal. High-fidelity amplifiers aim to increase the output power without adding too much distortion or noise relative to the input power.

Logarithm

The logarithm is a mathematical operation that is the inverse of exponentiation. It plays a crucial role in dB gain calculation due to its nature to handle multiplicative processes and turn them into additive ones. When calculating dB gain, we use the base-10 logarithm to determine how many times 10 should be multiplied to get the ratio of the output power to the input power. In simpler terms, the logarithm helps us convert the power ratio, which can be a large or small number, into a more manageable scale when expressed as dB gain. This makes comparisons and adjustments in audio and electronic applications much more intuitive. For example, when an output is half the input, using the logarithm simplifies the computation of decibel change significantly, as demonstrated in the solution part of the exercise.