Response to Complex Exponential

MIT RES.18-009 Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler, Fall 2015
View the complete course: ocw.mit.edu/RES-18-009F15
Instructor: Gilbert Strang
For linear equations, the solution for a cosine input is the real part of the solution for a complex exponential input. That complex solution has magnitude G (the gain).
License: Creative Commons BY-NC-SA
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[@SedoKai]

The value of these video MIT OCW series is beyond comprehension. You're given the opportunity to develop marketable skill sets based on formal education for nominally no cost, where other people are paying 6 or high 5 figures for it. And since it comes from MIT, you're essentially assured of it's quality. I have to offer my sincere thanks and appreciation to the professors who volunteered their time, and to MIT for supporting them and allowing them to do it.

[@questionablemathematician3902]

Agreed

[@alimehrabifard1830]

Perfect

[@CatsBirds2010]

Thanks......

[@rylanschaeffer3248]

What happened to the initial condition y(0) and the null solution i.e. y(0)e^{at}? This seems to contradict his textbook's Equation 19.

[@ashutoshacharya8]

Gil only solves for particular solution in his video

[@wimthiels638]

what about the initial conditions ? is the implicit assumption made that they are transitory because a< 0, so we can leave them out of the final solution ?

[@rakhimovv]

Null solution still exists and it is the same for different q(t). The professor is just considering the particular solution for particular q(t).

[@armpit1648]

What's the reason he kept alpha instead of converting to atan2(w/-a). And MIT could do something to improve the complex response of the audio tracks of their videos. Bet they can afford that.

[@elyepes19]

Simply for convenience of brevity and don't have the final solution looking too cumbersome

[@sainandan65]

Why there is no constant in the assumed solution of form? like yc = Ye^iwt + C

[@physicalgraffitti]

aint it that the constant is already there?

[@Chrisgintz88]

Is r supposed to equal sqrt(a^2 - w^2)? I am assuming (iw)^2= i^2*w^2 =-w^2

[@chnegsi]

no, it's the Modulus Amplitude Of A Complex Number

[@suharsh96]

no, its just the length of the hypotenuse that is calculated by the basic Pythagorean theorem.

[@weinihao3632]

@@ak47tetris84 In that case the complex value 1 + i would have a radius r = sqrt(1 - 1) = 0 which is obviously wrong. The correct calculation is r = sqrt ((a)^2 + (w)^2) as shown in the video

[@Dark-tk9xu]

Yes, it ought to be sqrt(a2-w2).

[@sossupummi]

no, increasing omega in that case would decrease the length r of a vector in polar coordinates, but as you can see from the picture there, increasing omega increases the length

[@T4l0nITA]

so the imaginary part would be i*sin(ωt-alpha) ?

[@ing.erickosorio2887]

yes, it comes from Euler great formula, e^(ix)=cosx+isinx , just replace your x with the angle you have.

[@lazywarrior]

Please reorder the videos in the playlist. The current order is not correct.

[@mitocw]

The playlist matches the course on MIT OpenCourseWare. Maybe you are looking at a different playlist? The official playlist for this series is kzbin.info/aero/PLUl4u3cNGP63oTpyxCMLKt_JmB0WtSZfG.

[@fredpelogia]

@@mitocw I believe that this complex exponential video is supposed to be after the Oscillating input video and before the "Solution to any input" video.

[@hathuytu]

Do you believe there exist anti-imaginary number?

[@user-ox4tp4qs5w]

I am the 555th who click the like button. I am so inspired by these responses and thank you very professor.

[@XiaosChannel]

ahh the sound quality...

[@biolinux2307]

Like 👍😀

[@robertw2930]

way overmodulated