Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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 (
 one year ago
 one year 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 (
 one year ago
 one year ago

This Question is Open

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.
 one year 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 \(a + ib = A \, e^{i \phi}\)).
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.