anonymous
  • anonymous
Unit 2, Exponential Response, Sinusoidal Input PDF, page 1 at the bottom: I am confused by how they got from 3+16i = sqrt(265)e^i@ (@ = phi) to the solution 9/sqrt(265)cos(2t-@). I feel fairly comfortable with the complex numbers, Eulers Formula etc, and I also understand the Exponential Response Formula. It's just that step that I am confused about. I used complex conjugate to multiply out 9/(3+16i) then put into exponential form (9/sqrt(265))(cos@ - isin@), which makes the solution the real part of ((9/sqrt(265))(cos@ - isin@)(cos2t + isin2t). I just dont understand how that ends up being (
MIT 18.03SC Differential Equations
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I dont understand how that ends up being 9/sqrt(265) * cos(2t - @). and yes I undertand cos(t) + sin(t) = cost(t - @). I am taking ODE at UVA Wise, but my professor isn't teaching complex replacement. Any help would be seriously appreciated. I really want to learn this method.
wrwrwr
  • wrwrwr
The complex solution is \(z_p = \frac{9 e^{2it}}{3 + 16i}\), and what we want is its real part in the so-called amplitude-phase form. -- Start by transforming the denominator to the polar form: \(3 + 16i = \sqrt{3^2 + 16^2} e^{i \, atan2(16, 3)} = \sqrt{265} e^{i \phi}\), where \(\phi = tan^{-1}(16/3)\); -- (Don't yet use Euler's definition); -- Your solution is now: \(z_p = \frac{9}{\sqrt{265}} \frac{e^{2it}}{e^{i \phi}}\); -- Now use a simple exponents law (\(x^n / x^m = x^{n-m}\)) to get \(z_p = \frac{9}{\sqrt{265}} e^{i (2t - \phi)}\); -- Use Euler's formula: \(z_p = \frac{9}{\sqrt{265}} (cos(2t - \phi) + i \, sin(2t - \phi))\); -- And drop the imaginary part: \(x_p = Re(z_p) = \frac{9}{\sqrt{265}} cos(2t - \phi)\). You could also expand the numerator using Euler's, expand the fraction by denominator's conjugate, multiply brackets (leaving the conjugate in Cartesian form and using \(i^2 = -1\)), get the real part and use the trigonometric formula (\(a \, cos(\theta) + b \, sin(\theta) = A \, cos(\theta - \phi)\) with \(a + ib = A \, e^{i \phi}\)).

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