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Logarithmic Ratios

  • Thread starter agata78
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The current input of a system is 250mA. If the current ratio of the system is 15dB, determine:
(i) the current output and
(ii) the current ratio expressed in nepers

I have the answer to the questions which are (i) 1.406 A (ii) 1.7272 Np

I am just struggling with the complexity of the logarithmic ratio equations. Can you please show me how to go about showing the workings out?
 
139
0
1. The problem statement, all variables and given/known data

The current input of a system is 250mA. If the current ratio of the system is 15dB, determine:
(i) the current output and
(ii) the current ratio expressed in nepers

I have the answer to the questions which are (i) 1.406 A (ii) 1.727 Np

I am just struggling with the complexity of the logarithmic ratio equations. Can you please show me how to go about showing the workings out?

2. Relevant equations

(ii) Current Ratio in Nepers = In (I2 / I1)

I2 = Current Output
I1 = Current Input

3. The attempt at a solution
 
Last edited:

UltrafastPED

Science Advisor
Gold Member
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15 dB amplification means

15 = 10 x log10(I2/I1)


Correction:

15 = 20 x log10(I2/I1)
 
Last edited:

berkeman

Mentor
54,772
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15 dB amplification means

15 = 10 x log10(I2/I1)
No, when talking about current or voltage, it is db = 20log(ratio).

If you are taking the ratio of powers, then it is dB = 10log(ratio).
 

berkeman

Mentor
54,772
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I've never used the units of Nepers, but wikipedia explains the forumulas to use with them:

http://en.wikipedia.org/wiki/Nepers

:smile:
 
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A text book example I have is:

Current Output = 50 mA
Current Input = 187.2 mA

Current Ratio in dB = 20log(I2/I1)
= 20log(50/187.2)
= -114dB
**Which is all correct for dB, however the formula for Nepers is still unclear. If I use the same equation for example for my initial problem I am not getting the magic anwser for question (ii) 1.727 Np.

Does anyone know the equation for current ratio expressed in Nepers for the following problem:

Current Output = 250 mA
Current Input = 1.406 A
 

collinsmark

Homework Helper
Gold Member
2,723
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If I use the same equation for example for my initial problem I am not getting the magic anwser for question (ii) 1.727 Np.
You should be getting the 1.727 Np. Show your calculation and maybe we can figure out what's going wrong.

Does anyone know the equation for current ratio expressed in Nepers for the following problem:

Current Output = 250 mA
Current Input = 1.406 A
I think you have those swapped (according to the original post). I think you mean

Current Input = 250 mA
Current Output = 1.406 A

---------------------
In the mean time, allow me to make a few comments on decibels.
  • Decibels are commonly used by many people in many fields.
  • Decibels are defined in terms of power ratios -- not current ratios, not voltage ratios, but power ratios. And there is an assumption that whenever you use decibels, the ratio involved is a ratio of powers.
  • With decibels, logarithms are base 10.
  • The definition for value in units of decibels (I'll just call it [itex] L_{\mathrm{dB}} [/itex]) is [itex] L_{\mathrm{dB}} = 10 \ \log_{10} \left( \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \right). [/itex] (although technically, [itex] P_{\mathrm{out}} [/itex] and [itex] P_{\mathrm{in}} [/itex] don't have to be output and input powers, they can be any powers. The important point is that they be powers. Defining it this way, as input and output, is how one would calculate the gain of a circuit.)

So where does changing the "10" to "20" come from when working with currents and voltages? It starts with an assumption: The input resistance of the circuit is the same as the output resistance of the circuit. This is a big assumption. But it is also a common one. Normally you want the input resistance of one stage in a circuit to equal the output resistance of the previous stage for maximum power transfer.* So usually, the input and output resistances of all stages are the same, typically something like 50 Ohms (some systems use 75 Ohms like cable TV).

*(Technically you want the impedance of one to the complex conjugate of the impedance of the other, but I'm getting more complicated than I need to right now.)

So what is the power to voltage relationship between input and output voltages and powers? [itex] P_{\mathrm{in}} = \frac{V_{\mathrm{in}}^2}{R_{\mathrm{in}}} [/itex] and [itex] P_{\mathrm{out}} = \frac{V_{\mathrm{out}}^2}{R_{\mathrm{out}}}. [/itex]

But since the input and output resistances are equal, [itex] P_{\mathrm{in}} = \frac{V_{\mathrm{in}}^2}{R} [/itex] and [itex] P_{\mathrm{out}} = \frac{V_{\mathrm{out}}^2}{R}. [/itex]

Plugging that into our decibel equation gives

[tex] L_{\mathrm{dB}} = 10 \ \log_{10} \left( \frac{\frac{V_{\mathrm{out}}^2}{R}}{\frac{V_{\mathrm{in}}^2}{R}} \right) = 10 \ \log_{10} \left( \frac{V_{\mathrm{out}}^2}{V_{\mathrm{in}}^2} \right) = 10 \ \log_{10} \left[ \left( \frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} \right)^2 \right] = 20 \ \log_{10} \left( \frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} \right)[/tex]

You can do the same thing with current ratios, realizing that [itex] P = I^2R [/itex], and following the same idea. And that's where the "20" comes from.

--------------------------------
Now a few comments on Nepers.

  • Nepers are not as commonly used, at least from my experience (I could be wrong, but today is the first I ever remember hearing of them.)
  • Nepers are not defined in terms of powers, so you can simply plug your voltages or currents right in, without having to worry about changing "10" to "20" or anything like that, when working with voltages and currents.
  • With Nepers, logarithms are base e.
  • The definition value in units of Nepers (I'll just call it [itex] L_{\mathrm{Np}} [/itex]) is [itex] L_{\mathrm{Np}} = \ln \left( \frac{X_{\mathrm{out}}}{X_{\mathrm{in}}} \right), [/itex] where [itex] X [/itex] can be a voltage or a current.
 
139
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Thanks for the response. Therefore question (ii) is:

ln([itex]\frac{I2}{I1}[/itex])

ln([itex]\frac{1406}{250}[/itex])

= 1.727 Np

Question (i) is calculating Current Output (UNKNOWN). Even though I know the answer is 1.406 A. My workings out are as follows (Am I along the right lines so far?):

dB Current Ratio = 10log([itex]\frac{I2}{I1}[/itex])

Hence, 15 = 10log([itex]\frac{I2}{I1}[/itex]), where I2 is the Current Output

15 = log([itex]\frac{I2}{250}[/itex])

([itex]\frac{I2}{250}[/itex]) = 10^15
 

collinsmark

Homework Helper
Gold Member
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Thanks for the response. Therefore question (ii) is:

ln([itex]\frac{I2}{I1}[/itex])

ln([itex]\frac{1406}{250}[/itex])

= 1.727 Np
There ya go. :approve:

Question (i) is calculating Current Output (UNKNOWN). Even though I know the answer is 1.406 A. My workings out are as follows (Am I along the right lines so far?):

dB Current Ratio = 10log([itex]\frac{I2}{I1}[/itex])

Hence, 15 = 10log([itex]\frac{I2}{I1}[/itex]), where I2 is the Current Output

15 = log([itex]\frac{I2}{250}[/itex])

([itex]\frac{I2}{250}[/itex]) = 10^15
Try that one again, but this time be sure to use [itex] 20 \ \log_{10} \left( \frac{I_2}{I_1} \right). [/itex]

Remember, when working with currents or voltages in dB land, you need to use the "20" not the "10." Yes, this assumes that the input and output resistances are equal, but that is an assumption we can live with.

The reason for that is because when you communicate with other engineers about the gain of a particular circuit being 15 dB, or some other value in dB, that figure is always in reference to power. Always. No matter what. You might solve the problem working with current, some other engineer might solve the problem working with voltages, and a third engineer might be measuring powers directly. But you should all get the same answer. And that answer is ultimately in terms of power.

In order for that to work out when using currents (and given the assumption about the input and output resistances), you must use [itex] 20 \ \log_{10} \left( \frac{I_2}{I_1} \right) [/itex] when working with currents.
 
139
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Thanks for the support. But I have spent many hours trying to work out question (i), and I am no further forward. But here is what I have done so far to try and show my working out.

As I reminder here is the original question and the correct answer:

The current input of a system is 250mA. If the current ratio of the system is 15dB, determine:
(i) the current output?

The answer to the questions which is (i) 1.406 A.

I need to show the complete calculations and workings out. I have come to the answer of 1.4067 in my calculations and I know this is incorrect. I am just trying to show that I am racking my brain for the solution and I am struggling now.

ln 250/15 = ln 1.667

= 2.8134

2.8134 / 2

= 1.4067

As you can see, I have come up with the answer instead of using the equation from the previous post. I would like to get to the answer of 1.406A using the equation 20log10(I2/I1).

Can someone please help me?
 

collinsmark

Homework Helper
Gold Member
2,723
1,065
Thanks for the support. But I have spent many hours trying to work out question (i), and I am no further forward. But here is what I have done so far to try and show my working out.

As I reminder here is the original question and the correct answer:

The current input of a system is 250mA. If the current ratio of the system is 15dB, determine:
(i) the current output?

The answer to the questions which is (i) 1.406 A.

I need to show the complete calculations and workings out. I have come to the answer of 1.4067 in my calculations and I know this is incorrect. I am just trying to show that I am racking my brain for the solution and I am struggling now.

ln 250/15 = ln 1.667

= 2.8134

2.8134 / 2

= 1.4067

As you can see, I have come up with the answer instead of using the equation from the previous post. I would like to get to the answer of 1.406A using the equation 20log10(I2/I1).

Can someone please help me?
How about I just set up the equation.

[tex] 15 \ \mathrm{dB} = 20 \log_{10} \left( \frac{I_2}{250 \ \mathrm{mA}} \right). [/tex]
Solve for [itex] I_2 [/itex]. The first step is to divide both sides of the equation by 20. Then pay careful attention that the logarithm involved here is base 10; it is not the natural logarithm (i.e. not base e).
 
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15 =20 log10 ( I2/ 250)
15/20 = log 10 (I2/250)
3/4= log10 (I2)- log10 (250)
0.75= log (I2)- 2.397
log (I2)= 2.397+0.75
log (I2)= 3.147

How to ger rid off log on left side?
 

collinsmark

Homework Helper
Gold Member
2,723
1,065
15 =20 log10 ( I2/ 250)
15/20 = log 10 (I2/250)
3/4= log10 (I2)- log10 (250)
0.75= log (I2)- 2.397
log (I2)= 2.397+0.75
log (I2)= 3.147

How to ger rid off log on left side?
Take each side to the power of 10.

[tex] \log_{10}(x) = y [/tex]
[tex] x= 10^{y} [/tex]
(And be careful, you made a small rounding error in your "2.397" value. You might want to keep more significant figures in your intermediate steps.)
 
Last edited:
139
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COMPLETED

log(I2) = 2.39794 + 0.75

log(I2) = 3.14794

x = 10^3.14794

x = 1405.8532

x = 1.4058

Current Output = 1.406 A

Its easy when you know how to do it! Cheers!!!

Agata
 
32,364
4,147
COMPLETED

log(I2) = 2.39794 + 0.75

log(I2) = 3.14794
In the three lines below, it should be I2, not x. In the next step you're writing each side as the exponent on 10, like so:

10log(I2) = 103.14794

10 raised to the log of something is just that same something, so the left side is now plain old I2.
x = 10^3.14794

x = 1405.8532

x = 1.4058

Current Output = 1.406 A

Its easy when you know how to do it! Cheers!!!

Agata
 
139
0
dB = 20log10 (I2 / I1)

15 = 20log10 (I2 / 250)

15 / 20 = log10 (I2 / 250)

3/4 = log10 (I2) - log10 (250)

0.75 = log(I2) – 2.39794

log(I2) = 2.39794 + 0.75

log(I2) = 3.14794

10log(I2) = 103.14794

I2 = log 3.14794

I2 = 1405.8532

I2 = 1.4058

Current Output = 1.406 A
 

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