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Bode plot
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===Straight-line amplitude plot=== Amplitude decibels is usually done using <math>\text{dB} = 20 \log_{10}(X)</math> to define decibels. Given a transfer function in the form :<math>H(s) = A \prod \frac{(s - x_n)^{a_n}}{(s - y_n)^{b_n}},</math> where <math>x_n</math> and <math>y_n</math> are constants, <math>s = \mathrm{j}\omega</math>, <math>a_n, b_n > 0</math>, and <math>H</math> is the transfer function: * At every value of ''s'' where <math>\omega = x_n</math> (a zero), '''increase''' the slope of the line by <math>20 a_n\ \text{dB}</math> per [[Decade (log scale)|decade]]. * At every value of ''s'' where <math>\omega = y_n</math> (a pole), '''decrease''' the slope of the line by <math>20 b_n\ \text{dB}</math> per decade. * The initial value of the graph depends on the boundaries. The initial point is found by putting the initial angular frequency <math>\omega</math> into the function and finding {{nowrap|<math>|H(\mathrm{j}\omega)|</math>.}} * The initial slope of the function at the initial value depends on the number and order of zeros and poles that are at values below the initial value, and is found using the first two rules. To handle irreducible 2nd-order polynomials, <math>ax^2 + bx + c</math> can, in many cases, be approximated as <math>(\sqrt{a}x + \sqrt{c})^2 </math>. Note that zeros and poles happen when <math>\omega</math> is ''equal to'' a certain <math>x_n</math> or <math>y_n</math>. This is because the function in question is the magnitude of <math>H(\mathrm{j}\omega)</math>, and since it is a complex function, <math>|H(\mathrm{j}\omega)| = \sqrt{H \cdot H^*}</math>. Thus at any place where there is a zero or pole involving the term <math>(s + x_n)</math>, the magnitude of that term is <math>\sqrt{(x_n + \mathrm{j}\omega)(x_n - \mathrm{j}\omega)} = \sqrt{x_n^2 + \omega^2}</math>.
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