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

[asdf12312]

Homework Statement

Homework Equations

The ratio of the parameters \frac{E_1}{E_2} to find the gain (in dB), 10\timeslog₁₀\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\timeslog₁₀\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??

 

[berkeman]

Homework Statement

Homework Equations

The ratio of the parameters \frac{E_1}{E_2} to find the gain (in dB), 10\timeslog₁₀\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\timeslog₁₀\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.

 

[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.

 

[NascentOxygen]

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\timeslog₁₀\frac{E_1}{E_2}

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

 

[asdf12312]

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?

 

[NascentOxygen]

That looks right. Just √(7.51/2.99)

 

[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..

 

[NascentOxygen]

"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.