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Forums
Engineering
Electrical Engineering
Bode Plot Method For a Transfer Function
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[QUOTE="NascentOxygen, post: 5485206, member: 336970"] Yes, they factorized the denominator from (jω + 10) to 10(jω/10 + 1) but accidently lost that first 10. The reason for standardising on the form (jω/N + 1) is so that for ω<<N this term contributes a value of approx unity to the T.F. so can be "ignored" until ω→N. I don't like their use of 10‧log |(jω)[sup]2[/sup] + 1| I think the j should not be included like that; we don't want to [I]square [/I]the j. The full expression for the amplitude of each term does correctly involve 20‧log but because each frequency-dependent expression is of the form, e.g., 20‧log √((1/10)[SUP]2[/SUP] + 1[SUP]2[/SUP]) the rules of mathematics[SUP]※[/SUP] allow them to combine the 20 and the √ and write it as 10 [U]without[/U] the √. [SUP]※[/SUP] you should satisfy yourself of the validity of this. A little confusing! [emoji84] [/QUOTE]
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Bode Plot Method For a Transfer Function
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