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mginaction
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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 (
 2 years ago
 2 years ago
mginaction Group Title
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 (
 2 years ago
 2 years ago

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mginaction Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

wrwrwr Group TitleBest ResponseYou've already chosen the best response.0
The complex solution is \(z_p = \frac{9 e^{2it}}{3 + 16i}\), and what we want is its real part in the socalled amplitudephase 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 (\(x^n / x^m = x^{nm}\)) 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 \(a + ib = A \, e^{i \phi}\)).
 2 years ago
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