Lecture 11 | The Fourier Transforms and its Applications

Lecture by Professor Brad Osgood for the Electrical Engineering course, The Fourier Transforms and its Applications (EE 261). Professor Osgood lectures on confronting the convergence of intervals.

The Fourier transform is a tool for solving physical problems. In this course the emphasis is on relating the theoretical principles to solving practical engineering and science problems.

Complete Playlist for the Course:
http://www.youtube.com/view_play_list?p=B24BC7956EE040CD

EE 261 at Stanford University:
http://eeclass.stanford.edu/ee261/

Stanford University:
http://www.stanford.edu

Stanford University Channel on YouTube:
http://www.youtube.com/stanford

Category:

Education

Tags:

• Electrical
• engineering
• computers
• math
• physics
• geometry
• algebra
• calculus
• technology
• functions
• linear
• operations
• sin
• cosin
• Fourier
• transformations
• series
• central
• limit
• theorem
• repeated
• convolutions
• random
• variables
• CLT
• convergence

[nkdevde]

I did wonder about the /s².

---

In lecture 8 (or something) you mentioned "How mathematics works."

In the past I indeed had plenty of such lecturers who just defined what seemed a random crazy operation and afterwards gave a list(!) of what seemed to be random properties of that operation.

Another annoying habit: Not mentioning when something important happened, assuming the student will somehow "see it".

This is definitely one of the best lectures I ever heard. It avoids all those pitfalls.

[noobmartin]

Absolutely brilliant.

[grunder20]

Good man, know all about engineering.

[AllOtherNamesTaken2]

38:20 is solved easily with LaHopital's rule

[jombo222]

Dualities in Lecture 7 (40:00). "Good Luck" at 45:38.

[Rildays]

scary guy...

[bitrex]

@cartmansuperstar The duality identities are that F F f = f^-1, and F^-1 f = F f^-1, where f^-1 is f(-x). In words, that's the Fourier transform of the Fourier transform of a function is the inverse of the function, and the Fourier transform of the inverse of a function is the inverse of the Fourier transform of the function. I hope that makes sense!

[cartmansuperstar]

ok, i guess i should be more patient... :-/

[cartmansuperstar]

Part 3:

(...)that he would switch between definitions without explicitly saying this. So a little confused right now.

And by the way: what lecture was the part he spoke about that "duality"? Tried to find it, but i couldn´t and don´t want to watch it all through again. Does anybody know? thanks.

[cartmansuperstar]

Part 2: (...)

In addition at 15:19 he speaks about the values of the integral at certain ranges of s whereas it´s actually supposed to be a function of t (which he also mentions shortly before at 15:06)

Of course in one of the previous lectures he stated, that the definition of the FT varies and i also heard about it earlier that it might be defined without the "minus" in the exponential so that the way he wrote the integral might in fact represent the iFT but i just don´t think ..(end Part 2)

[cartmansuperstar]

a little cofused about what he´s doing beginning at 14:11: first he states, that the problem finding the inverse FT (iFT) of the sinc funtion is equivalent to finding the FT of it.

okay. then he wants to "stick with the problem of Fourier-inversion", so he wants to write down the expression (integral) for the iFT, but actually is writing down the definition of the FT (exp(-(2Pi)ist - with emphasis on the MINUS).

(end Part 1)

[alquiora]

I love his theatrical sense of humor