# Applying a gain of dB to a signal (function) help?

• asdf12312
In summary, the conversation discusses determining the gain factor in a problem by considering the effect of multiplying a signal by a constant factor on its energy and then using the logarithmic formula to solve for the gain factor. A discrepancy is noted between a 4 dB and 5 dB gain, but it is later confirmed to be a typo and the correct gain is 5 dB. The correct method for finding the constant factor to multiply the signal by is determined to be taking the square root of the ratio of the desired energy to the original energy.
asdf12312

## Homework Equations

The ratio of the parameters $\frac{E_1}{E_2}$ to find the gain (in dB), 10$\times$log10$\frac{E_1}{E_2}$.

"To determine the gain factor in part 8, first think about what happens to the energy when you multiply the signal by a constant factor. Then use the logarithmic formula above to solve for the gain factor."

## The Attempt at a Solution

Since energy is square of integral of a f(x), I am thinking if f(x) is multiplied by constant factor then energy will be square of that factor greater. I already have E1 of the problem, which is the energy of f(x) above = 2.99. I am confused I think because problem says 5dB then 4dB. anyway..if I assume gain of energy is equal to 4dB i have:

4dB = 10$\times$log10$\frac{E_1}{E_2}$
0.4 = log10$\frac{E_1}{E_2}$
$\frac{E_1}{E_2}$ = 2.512 → E2 = $\frac{2.99}{2.512}$ = 1.19

but why is the energy less than the original though? am i doing this right so far??

asdf12312 said:

## Homework Equations

The ratio of the parameters $\frac{E_1}{E_2}$ to find the gain (in dB), 10$\times$log10$\frac{E_1}{E_2}$.

"To determine the gain factor in part 8, first think about what happens to the energy when you multiply the signal by a constant factor. Then use the logarithmic formula above to solve for the gain factor."

## The Attempt at a Solution

Since energy is square of integral of a f(x), I am thinking if f(x) is multiplied by constant factor then energy will be square of that factor greater. I already have E1 of the problem, which is the energy of f(x) above = 2.99. I am confused I think because problem says 5dB then 4dB. anyway..if I assume gain of energy is equal to 4dB i have:

4dB = 10$\times$log10$\frac{E_1}{E_2}$
0.4 = log10$\frac{E_1}{E_2}$
$\frac{E_1}{E_2}$ = 2.512 → E2 = $\frac{2.99}{2.512}$ = 1.19

but why is the energy less than the original though? am i doing this right so far??

Why do you say it is less? The ratio of 2.512 is correct.

I got confused cause the new energy is less then the one I calculated (2.99). So a gain of 4dB in the energy results in E2=1.19. How would I find the constant factor that I need to multiply the signal by? I am guessing it is something like A*x(t) and I have to find A that results in this new energy value.

asdf12312 said:
I got confused cause the new energy is less then the one I calculated (2.99). So a gain of 4dB in the energy results in E2=1.19. How would I find the constant factor that I need to multiply the signal by? I am guessing it is something like A*x(t) and I have to find A that results in this new energy value.
I don't understand the 4dB vs. 5 dB discrepancy, either. Must be a typo.

4dB = 10$\times$log10$\frac{E_1}{E_2}$
You are confusing ##E_1## and ##E_2##. Once you sort that out you'll be right.

OK that's what I thought. So they would be reversed, right? In that case I get E2=7.51. So then A2=2.74. The ratio $\frac{A_2}{A_1}$ = 1.585. So this would be the constant factor I multiply x(t) by. Is this right?

That looks right. Just √(7.51/2.99)

"The text for part 8 refers to both a 5 dB and a 4 dB gain. This is a typo. The gain to be applied is 5 dB. This has been corrected in the PDF.."

Our instructor just send out that message lol. so I guess it was a typo..

asdf12312 said:
"The text for part 8 refers to both a 5 dB and a 4 dB gain. This is a typo. The gain to be applied is 5 dB. This has been corrected in the PDF.."

Our instructor just send out that message lol. so I guess it was a typo..
That gives you the opportunity to redo the problem, this time using 5dB.

## 1. What does it mean to apply a gain of dB to a signal?

Applying a gain of dB to a signal means increasing or amplifying the strength or amplitude of the signal by a certain amount measured in decibels (dB). This can be done using electronic amplifiers or signal processing techniques.

## 2. Why would someone want to apply a gain of dB to a signal?

There are several reasons why someone may want to apply a gain of dB to a signal. One common reason is to increase the volume of an audio signal, making it louder and easier to hear. Another reason is to improve the signal-to-noise ratio, making the signal clearer and easier to detect.

## 3. How is the gain of dB calculated?

The gain of dB is calculated using the following formula: Gain (dB) = 20 log (Vout / Vin) where Vout is the output voltage and Vin is the input voltage. This formula is based on the logarithmic nature of decibels, where a gain of 6 dB represents a doubling of the signal's power.

## 4. Can a gain of dB be applied to any type of signal?

Yes, a gain of dB can be applied to any type of signal, including audio, video, and data signals. However, the specific method of applying the gain may vary depending on the type of signal and the equipment being used.

## 5. Are there any potential drawbacks to applying a gain of dB to a signal?

Yes, there are potential drawbacks to applying a gain of dB to a signal. One common issue is the introduction of distortion or clipping if the gain is set too high. Additionally, increasing the gain can also amplify any existing noise in the signal, reducing its overall quality. It's important to carefully consider the desired outcome and adjust the gain accordingly to avoid these potential drawbacks.

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