Learn how to apply the graphical "flip and slide" interpretation of the convolution integral to convolve an input signal with a system's impulse response.
Memo Pony And it could be explained even shorter - At least the fact HOW convolution works - However, WHY convolution works and why it is so powerful, is quite hard to understand in my opinion - At least, I am not able to find any good information on that. So either it's too hard to understand, so people are just fine with understanding HOW it works - or I am simply too dumb to understand why convolution works and is so useful.
Surely!!! Even in many first world countries they will charge you thousands of dollars to teach you, when knowledge must be free and free! Luckily that is one of the good things about my country, and that in some time I will receive my degree as an Electronic Engineer without paying anything but my taxes. Greetings from Argentina.
In despite the clear explanation, which is great by the way, the video totally succeed due the use of a non ordinary example! Most of the books explain continuos convolution using step and exponential functions. This is the first time that I can see a convolution example using a ramp function as h(t). This is really a good job and really helped me a lot! Thank you very much!
Out of several videos I have watched since the morning (6:00AM) until now (7:00PM) hands down to this one. It explained the concept crystal clear. Well done, sir!
The "flip" is part of the mathematical definition of convolution, and generally makes a difference in the result. An exception is when the "flipped" waveform already has even symmetry, for example a triangle ramp that goes up and comes back down at the same rate (even symmetry means you can flip the waveform about its center value and it looks the same).
At 8:13... we need the equation of the downward-sloping line between t = 1 and 2 seconds. Projecting this line back at t = zero means that the line hits the vertical axis at 4 (this is the y-intercept). Even though the line does not actually extend up to this value (because x(t) is constant in the 0 to 1 range), you need to write the function that describes x(t) during the 1 to 2 second range. Hope that helps!
Hi man, I just wanted to know, of what all to know before learning convolution pls let me know and your. Videos are really helpful for my college thank you so much ans God bless
can u please explain why did u not change the eqn while ur h(-taw) is positive slope.. for me while h(-taw), eqn of line is = (1/3)taw+1 and after shifting it while h(t-taw)= (1/3)(t-taw)+1. please correct me if i am wrong...
Hi, thank you for the explanation. It was would have been better if you described how to mathematically obtain the interval, rather than just showing it graphically.
Thank you very much for this video, great! I am wondering: Can you tell me please which tool you are using to show the graphical convolution? Did you program it in Mathematica or an other program? And if Mathematica: Is there already a tool to show this type of analysis?
Tell me please, what program do you use to get the convolution when you move the inverted triangle along the axis? Is there a program in free access? Thank you in advance for the answer.
Really appreciate your work! It has helped immensely! I have a question, if you don't mind... At 5:23, you needed to find an equation for h(t-tau). You developed this equation from the original, unmodified version of h(t), that is a ramp with negative slope. Why did you use the original h(t) instead of the modified version, h(t-tau)?
+Thomas Turkington He just came to a point, where he needed to insert the function of h. Using h(t) or the "modified version" h(t-tau) is just playing around with the argument of the function. So it's straight forward to just look at the shape of h(t) and think about the function definition, which will just give you a simple linear function as you are used to work with. Without any other constants or variables to take care of. Once you have the function of h(t), you get the function h(t-tau) by inserting "t-tau" for every occurence of "t" in h(t). That's it :)
Imagine that section from t=1 to 2 of x(t) alone, and then picture it extending in all the domain of t. This new line function would be x1(t)=-2t+4 when extended, but if you took only a piece of it, from 1 to 2, it'd still take the same equation to describe it. Hope it helps!
hi ! please tell me why we must turn around h(-t)? for what this is ? please explane..... what happened if dont this is .... ( iim sorry if i did same mistakes in my message ... just im not English speaking person)
i just had this one in my test exactly same , and i did it wrong ,despite of having watched it before because i only watched and didnt repeat it on paper.
So you must first slide, and only after you ahve slite it by t you will reverse it, right? If you do this, a very big negative t will bring your signal to the far left(first you bring it to far right, then flip it and it will go to far left) insted of brnging it to far right , is it correct?
in the 0,1 interval we get Parabolic function 7:15. but we can see on the simulator that it is straight line and i allso draw the function and im getting parabolic line
Hi Jan Be -- The red curve straight line at 3:44 shows only the *product* of the two functions, not the convolution result. The parabolic function at 7:15 is the *area* under the red curve which is the convolution result in this interval.
