Response to Complex Exponential

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MIT OpenCourseWare

MIT OpenCourseWare

Күн бұрын

Пікірлер: 29
@SedoKai
@SedoKai 7 жыл бұрын
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
@questionablemathematician3902 3 жыл бұрын
Agreed
@rylanschaeffer3248
@rylanschaeffer3248 4 жыл бұрын
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
@ashutoshacharya8 Жыл бұрын
Gil only solves for particular solution in his video
@armpit1648
@armpit1648 4 жыл бұрын
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
@elyepes19 Жыл бұрын
Simply for convenience of brevity and don't have the final solution looking too cumbersome
@CatsBirds2010
@CatsBirds2010 7 жыл бұрын
Thanks......
@alimehrabifard1830
@alimehrabifard1830 8 жыл бұрын
Perfect
@wimthiels638
@wimthiels638 8 жыл бұрын
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
@rakhimovv Жыл бұрын
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).
@sainandan65
@sainandan65 Жыл бұрын
Why there is no constant in the assumed solution of form? like yc = Ye^iwt + C
@physicalgraffitti
@physicalgraffitti 7 ай бұрын
aint it that the constant is already there?
@hathuytu
@hathuytu 3 жыл бұрын
Do you believe there exist anti-imaginary number?
@lazywarrior
@lazywarrior 5 жыл бұрын
Please reorder the videos in the playlist. The current order is not correct.
@mitocw
@mitocw 5 жыл бұрын
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
@fredpelogia 4 жыл бұрын
@@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.
@T4l0nITA
@T4l0nITA 5 жыл бұрын
so the imaginary part would be i*sin(ωt-alpha) ?
@ing.erickosorio2887
@ing.erickosorio2887 5 жыл бұрын
yes, it comes from Euler great formula, e^(ix)=cosx+isinx , just replace your x with the angle you have.
@XiaosChannel
@XiaosChannel 8 жыл бұрын
ahh the sound quality...
@Chrisgintz88
@Chrisgintz88 7 жыл бұрын
Is r supposed to equal sqrt(a^2 - w^2)? I am assuming (iw)^2= i^2*w^2 =-w^2
@chnegsi
@chnegsi 7 жыл бұрын
no, it's the Modulus Amplitude Of A Complex Number
@suharsh96
@suharsh96 7 жыл бұрын
no, its just the length of the hypotenuse that is calculated by the basic Pythagorean theorem.
@weinihao3632
@weinihao3632 4 жыл бұрын
@@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
@Dark-tk9xu Жыл бұрын
Yes, it ought to be sqrt(a2-w2).
@sossupummi
@sossupummi 5 ай бұрын
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
@马晓晓-i2y
@马晓晓-i2y Жыл бұрын
I am the 555th who click the like button. I am so inspired by these responses and thank you very professor.
@biolinux2307
@biolinux2307 2 жыл бұрын
Like 👍😀
@robertw2930
@robertw2930 7 жыл бұрын
way overmodulated
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