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

1. Feb 13, 2014

### asdf12312

1. The problem statement, all variables and given/known data

2. Relevant 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."

3. 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??

2. Feb 13, 2014

### Staff: Mentor

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

3. Feb 13, 2014

### asdf12312

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.

4. Feb 13, 2014

### Staff: Mentor

I don't understand the 4dB vs. 5 dB discrepancy, either. Must be a typo.

You are confusing $E_1$ and $E_2$. Once you sort that out you'll be right.

5. Feb 13, 2014

### asdf12312

OK thats 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?

6. Feb 14, 2014

### Staff: Mentor

That looks right. Just √(7.51/2.99)

7. Feb 14, 2014

### asdf12312

"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..

8. Feb 14, 2014

### Staff: Mentor

That gives you the opportunity to redo the problem, this time using 5dB.