 #1
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 68
Homework Statement:
 Find the maximum depth of a submarine below the surface of the sea which enables it to detect electromagnetic signals launched from the surface? (all the numbers given)
Relevant Equations:

Damped Wave Equation
Definition of skin depth
Hi,
So the main question is: How to deal with power loss in EM waves numerically when we are given power loss in dB's?
The context is that we are dealing with the damped wave equation: [itex] \nabla ^ 2 \vec E = \mu \sigma \frac{\partial \vec E}{\partial t} + \mu \epsilon \frac{\partial ^ 2 \vec E}{\partial t^2} [/itex]. The problem is assuming a good conductor [itex] \left( \sigma >> \omega \epsilon \right) [/itex] and thus we get that the onedimensional spatial solution is:
[tex] \vec E = \vec E_0 \exp \left(  \frac{(1 + j)z}{s} \right) [/tex] where [itex] s = \sqrt \frac{2}{\omega \sigma \mu} [/itex].
The data is given for all these constants, but am not concerned about that aspect of this problem. We are told that the maximum allowable power loss is 60 dB.
Here is my attempt to start the problem, and I was wondering whether I was starting the problem off correctly?
[tex] 60 dB = 10 \log_{10} \left( \frac{P_{sea}}{P_0} \right) [/tex] However, we know that [itex] P_{EM} \propto  \vec E  ^2 [/itex] and thus we can write that [itex] 60 dB = 20 \log_{10} \left(  \frac{E_{sea}}{E_{o}}  \right) [/itex]. Therefore, we have that [itex] 10^{3} =  \exp \left(  \frac{(1 + j)z}{s}  \right) [/itex].
This is where I am slightly confused on how to proceed  I am not fully sure how to evaluate this expression. If I just momentarily ignore the absolute value signs, then I get: [itex] \ln \left( 10^{3} \right)s = 1 + jz [/itex]. I feel as if I have cheated to get to this step. Anyways, that will lead me to the solution: [itex] z = \frac{\ln \left( 10^{3} \right)s}{\sqrt 2} [/itex].
When I substitute the numbers in, then I get the incorrect answer for some reason. I find that I need to have a 2 in the denominator rather than a [itex] \sqrt 2 [/itex]. Is my expression for the power loss in dB correct?
Thanks in advance for the help
So the main question is: How to deal with power loss in EM waves numerically when we are given power loss in dB's?
The context is that we are dealing with the damped wave equation: [itex] \nabla ^ 2 \vec E = \mu \sigma \frac{\partial \vec E}{\partial t} + \mu \epsilon \frac{\partial ^ 2 \vec E}{\partial t^2} [/itex]. The problem is assuming a good conductor [itex] \left( \sigma >> \omega \epsilon \right) [/itex] and thus we get that the onedimensional spatial solution is:
[tex] \vec E = \vec E_0 \exp \left(  \frac{(1 + j)z}{s} \right) [/tex] where [itex] s = \sqrt \frac{2}{\omega \sigma \mu} [/itex].
The data is given for all these constants, but am not concerned about that aspect of this problem. We are told that the maximum allowable power loss is 60 dB.
Here is my attempt to start the problem, and I was wondering whether I was starting the problem off correctly?
[tex] 60 dB = 10 \log_{10} \left( \frac{P_{sea}}{P_0} \right) [/tex] However, we know that [itex] P_{EM} \propto  \vec E  ^2 [/itex] and thus we can write that [itex] 60 dB = 20 \log_{10} \left(  \frac{E_{sea}}{E_{o}}  \right) [/itex]. Therefore, we have that [itex] 10^{3} =  \exp \left(  \frac{(1 + j)z}{s}  \right) [/itex].
This is where I am slightly confused on how to proceed  I am not fully sure how to evaluate this expression. If I just momentarily ignore the absolute value signs, then I get: [itex] \ln \left( 10^{3} \right)s = 1 + jz [/itex]. I feel as if I have cheated to get to this step. Anyways, that will lead me to the solution: [itex] z = \frac{\ln \left( 10^{3} \right)s}{\sqrt 2} [/itex].
When I substitute the numbers in, then I get the incorrect answer for some reason. I find that I need to have a 2 in the denominator rather than a [itex] \sqrt 2 [/itex]. Is my expression for the power loss in dB correct?
Thanks in advance for the help