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## Homework Statement

## Homework Equations

The ratio of the parameters [itex]\frac{E_1}{E_2}[/itex] to find the gain (in dB), 10[itex]\times[/itex]log

_{10}[itex]\frac{E_1}{E_2}[/itex].

"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 E

_{1}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[itex]\times[/itex]log

_{10}[itex]\frac{E_1}{E_2}[/itex]

0.4 = log

_{10}[itex]\frac{E_1}{E_2}[/itex]

[itex]\frac{E_1}{E_2}[/itex] = 2.512 → E

_{2}= [itex]\frac{2.99}{2.512}[/itex] = 1.19

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