Logarithms problem

  • Thread starter Rectifier
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Rectifier
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Hey!
I have a question regarding a statement in my physics book. I dont see how
[tex] |H(f)|_{dB} = - 10log (1+( \frac{f}{f_B})^2) [/tex]

approaches this equation below for big values on f.

[tex] |H(f)|_{dB} = - 20log ( \frac{f}{f_B}) [/tex]

Could you please help me out?

Thanks in advance.

EDIT: I am sorry if this is posted to a wrong sub-forum.
 
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Answers and Replies

  • #2
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When f is large, you can ignore 1 and so there only remains [itex] -10 \log{(\frac{f}{f_B})^2} [/itex]. Now you can use the property [itex] \log a^n=n\log a [/itex] to get what you want.
 
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  • #3
Rectifier
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When f is large, you can ignore 1 and so there only remains [itex] -10 \log{(\frac{f}{f_B})^2} [/itex]. Now you can use the property [itex] \log a^n=n\log a [/itex] to get what you want.
Oh gosh. Thank you for your help :)
 

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