Problem 51 - Algebra and Trigonometry with Analytic Geometry

Question

When the volume control on a stereo system is increased, the voltage across a loudspeaker changes from \(V_{1}\) to \(V_{2}\), and the decibel increase in gain is given by $$ \mathrm{db}=20 \log \frac{V_{2}}{V_{1}} . $$ Find the decibel increase if the voltage changes from 2 volts to \(4.5\) volts.

Step-by-Step Solution

Answer

The decibel increase is approximately 7.044 dB.

Step 1: Identify Known Values

First, identify the given variables in the problem. Here, we have:\( V_1 = 2 \) volts and \( V_2 = 4.5 \) volts. We need to find the decibel increase in gain.

Step 2: Substitute Into the Formula

Insert the identified values into the formula for decibel gain. The formula given is:\[ \text{db} = 20 \log \left( \frac{V_2}{V_1} \right) \]Substitute \( V_1 = 2 \) and \( V_2 = 4.5 \) into this formula.

Step 3: Calculate the Voltage Ratio

Calculate the ratio \( \frac{V_2}{V_1} \):\[ \frac{4.5}{2} = 2.25 \]

Step 4: Apply the Logarithm

Apply the logarithm to the voltage ratio:\[ \log(2.25) \approx 0.3522 \]

Step 5: Compute the Decibel Increase

Substitute the logarithm value back into the formula to find the decibel increase:\[ \text{db} = 20 \times 0.3522 \approx 7.044 \]

Key Concepts

Voltage Ratio

In the context of electronics and sound systems, understanding the voltage ratio can be very enlightening. The voltage ratio is simply the comparison of two voltage levels. When you adjust the volume on a stereo, you are actually altering the voltage applied across the loudspeaker. This change in voltage from an initial value \( V_1 \) to a new value \( V_2 \) is expressed as a ratio \( \frac{V_2}{V_1} \).
This concept is crucial because the human perception of sound is more responsive to proportional changes rather than absolute ones. Thus, knowing how much the voltage has increased or decreased helps in assessing how the perceived volume will change.

• For example, if the voltage changes from 2 volts to 4.5 volts, the voltage ratio will be \( \frac{4.5}{2} = 2.25 \).

Understanding the voltage ratio forms the foundational step in calculating the corresponding change in sound level, usually measured in decibels.

Logarithmic Function

Logarithms simplify multiplicative relationships into additive ones, making them very handy in various scientific applications. When dealing with audio signals, a logarithmic scale is often used because it reflects the way our ears perceive changes in sound intensity.
The logarithmic function lets us calculate how significant a change in voltage (like from 2 volts to 4.5 volts) is, in terms of sound level perceptions. The formula we're dealing with uses base 10, written as \( \log \), which converts the voltage ratio into a number more understandable in the context of sound volume.

• In this specific example, when you calculate \( \log(2.25) \), you get approximately 0.3522.

This value then gets further processed in the decibel formula, but the concept of logarithms is important because it ties the physical change in electricity to the more abstract auditory change.

Decibel Formula

The decibel formula is a standard method to quantify sound level changes, specifically changes in electrical signals like voltage across a loudspeaker. This formula reflects the logarithmic nature of sound perception and builds on the voltage ratio.
The equation is \( \text{db} = 20 \log \left( \frac{V_2}{V_1} \right) \), where \( V_2/V_1 \) is the voltage ratio. The factor of 20 scales the logarithmic value to make it compatible with human hearing sensitivity.

• After computing the logarithm for 2.25, which is 0.3522, you then multiply this by 20, giving you approximately 7.044 decibels.

This result tells us that the sound gained about 7 decibels of loudness when increasing the voltage from 2 volts to 4.5 volts, highlighting how the formula connects changes in physical voltage to changes in perceived sound.