first of all thx for your awesome videos, they are very easy to understand without being boring. in the very end you put the 5 in front of the integral, but you don't use it in your final answer. As far as i know, your final answer should be something like 5u(t-2)
it feels so amazing to understand it finally..!! thank you so much... i was very confused with all the symbols before watching this video. now it is crystal clear..!!!!
Thanks a lot Iman, your videos have helped me a lot in understanding basic signal processing. I watched your videos a year ago to help me get through my intro signals and systems course, and I am now re-watching to refresh before a real time dsp lab course this fall.
Thanks a bunch for your kind comment, Cooper! It is such a great pleasure to hear my videos have been helpful for you. Please recommend them to your colleagues! Cheers!
thanks 4 this clear easy to grasp lecture ,my question is at 11:00 u didnt multiply by 5, another question if u have half circle instead of "(t-2)^2" at 3;20 ,whats the equation 4 half a circle
Hi Evan, thanks for the feedback. Yes, you are right, I forgot to multiply it by 5. The equation for circle is x^2 + y^2 = r^2 where r is the constant radius. Cheers!
Dirac delta function is equal to infinity and not to 1. Because it's integral is equal to 1 and it's width is approaching zero (a width of a single point) it's hight need to be infinity. Thank you very much for the videos !
Hey Alon, thanks for your comment. Technically impulse function is a rectangular function with the width of Delta and the amplitude of 1/Delta when Delta goes to zero. Thus the area under the curve is one. This definition is technically true but hard to digest. It is common to model this with the unit impulse function with the amplitude of 1 at the origin. Hope this clear things up.
I didn't got one thing, it's OK that integral us a summation but in the above cases the integrand should be summation of d(t) multiplied by dt. I got that d(t) has value 1 at origin but what about dt???
iman hi sir, I am talking about all the examples which involve integration of some function f with d(t), moreover in our college we have been taught that d(t) function is 1/E at t=0 where E is a very infinitesimally small no.
Thank you for the awesome explanation. I have a doubt, at 8:10 in the video. we have in 2nd example cos(2t)delta(t-1) with integral. delta function has t but cos function has 2t. is the 't' in x(t) and delta(t) independent?
iman thank you! I have almost watched most of your videos and it was really helpful. Currently I'm trying to work on non linear signal processing. Could I email you my query? Please let me know your email. Thanks in advance..
Unique Ramya thanks for watching my videos. Please email your questions to imanmoaz@uvic.ca I will do my best to answer them as soon as possible. Best!
Hey. Sorry for the confusion! The value of delta function is actually 1/delta for a width of delta when delta approaches zero. This is a rectangle with the area equal to one. Hope this makes sense.
The impulse is 1 and not value of the function. ..meaning the area of this function is one....this point cannot be over emphasized and is source of confusion in beginning for most.
can I ask a question iman? in the video at 3:20 , when the equation in the below adding together. the value in the origin is 2 , not as shown in the triangular which u plot in the begin?
Hey, sorry for my late reply! I have been super busy with life, my friend. I just wonder how you ended up with two. Assuming unit step is 1 at the origin, here is what we get: (t+1) x (u(t+1) - u(t)) + (-t+1) x (u(t) - u(t-1)) ---> (replace t by zero) 1 x (u(1) - u(0)) + 1 x (u(0) - u(-1)) = 1 x (1-1) + 1 x (1-0) = 1 Hope this helps.
Hey, thank you for the question. Note that "y = constant" is a line parallel to the x-axis. No matter what x is, the y value is always constant. Hope this makes sense.
Hi Iman. Thanks for this (again very good explained) video. I still have a basic question on the equivalence and sifting properties of the delta functions. The "mechanics" of the examples are clear, but where I am not 100% clear is what are these properties of the delta function good for. Referring to your examples at ca 11:00 . What are you proving by multiplying sin(t) with d(t-pi/6) and resulting in 1/2d(t-pi/6)? Thanks again for the time doing these videos!
Hi again Jonathan. Thank you for your question. The sifting and equivalence properties are super useful in signal processing. That's why I decided to dedicate one lecture to this topic. If you keep watching my tutorials, you will see the applications of these properties in convolution, Fourier series/transform and sampling. For instance, multiplying sin(t) with d(t-pi/6) means sampling the sinusoidal function at pi/6. Please let me know if you still have more questions. Cheers!
Dirac delta function is NOT equal to 1 at t=0: it equals to infinity. But an integral of Dirac delta function is 1. It is very important to understand, but it is shown wrong in this video and some next too.
Hey, thanks for your comment. Technically impulse function is a rectangular function with the width of Delta and the amplitude of 1/Delta when Delta goes to zero. Thus the area under the curve is one. This definition is technically true but hard to digest. It is common to model this with the unit impulse function with the amplitude of 1 at the origin. Hope this clear things up.
Hi Iman, appreciate the videos, they are of great help! I'm a bit confuse at 8:00, example 2, as you went a bit too fast, this is my understanding of what happen, and I think it's missing, hope you could provide some clarification: cos(2t)∆(t-1) = cos(2 * t)∆(t-1) = cos(2 * 1) = cos(2)
Hey! Thanks for your comment and sorry for the late reply. Life has been hectic recently! You got it all right. I just replaced t by 1. Hope it makes sense.
I guess you did wrong at 4:27 , impulse function is never 1, it is infinity. The area is 1. unit impulse function and unit impulse sequence ( DT ) are not same. There is an abstract "impulse to sequence" conversion. It will be helpful if you make a video about CT to DT conversion, about what exactly the process is. Also, your solutions could be slower. Thanks
Hi iman, I like your video very much and I hope you can continue to make it. Besides, at 6:54 in this video, it seems that you forgot the dt in the Integral formula.
Your vids are amazing I LOVE IT but can u also please teach about Energy and power signal please and also LCDE and State variable. Thx for your great vids 10000 times better at explaining compare to my professor at university and I'm also sharing this great content to my friends Pls keep up the good work XD and btw I'm wondering why won't u teach discrete time since they are similar but important too
Hi, the way you defined the impulse function, it's integral is always zero, because you chose a finite value for t=0, and therefore that finite number over an infinitesimal width has zero area. You can multiply it by any finite function and it will remain zero. For a proper definition refer to wiki under the definition section: en.wikipedia.org/wiki/Dirac_delta_function#Definitions
I think it would be clearer it would be wrong to leave out the delta if you wrote: x(t)=sin(t)*∆(t-t0)=>x(t)=sin(t0)*∆(t-t0) Because then withouy delta you get: x(t)=sin(t0) So the signal x(t) is equal to a constant.
first of all thx for your awesome videos, they are very easy to understand without being boring.
in the very end you put the 5 in front of the integral, but you don't use it in your final answer.
As far as i know, your final answer should be something like 5u(t-2)
Hey Andi, thank you very much for your feedback. You are absolutely right, the final answer is 5u(t-2). I forgot to write it down. cheers!
I think he made a similar mistake in this example here: 7:50
it feels so amazing to understand it finally..!! thank you so much... i was very confused with all the symbols before watching this video.
now it is crystal clear..!!!!
Glad it helped!
How did you arrive at the line equations for the signals in 1:43 and 3:32?
Thanks a lot Iman, your videos have helped me a lot in understanding basic signal processing. I watched your videos a year ago to help me get through my intro signals and systems course, and I am now re-watching to refresh before a real time dsp lab course this fall.
Thanks a bunch for your kind comment, Cooper! It is such a great pleasure to hear my videos have been helpful for you. Please recommend them to your colleagues! Cheers!
Joy to watch....Feeling good after understanding. Thank you
Joy to see you watching all my videos ;) Thank you!
you are saving me right now! thank you so much for this greatly made tutorial man. Keep up the good work! (thumbs up)
Thank you very much for the feedback ;)
AMAZING MAN
7 years have passed but still very current
Thanks for your kind words and pleasure to help!
Dear brother, thanks very much, learn more from you, happy every day!
Thanks a lot ;)
Merci Iman jan , you are the best!
Thanks a bunch!
thanks 4 this clear easy to grasp lecture ,my question is at 11:00 u didnt multiply by 5,
another question if u have half circle instead of "(t-2)^2" at 3;20 ,whats the equation 4 half a circle
Hi Evan, thanks for the feedback. Yes, you are right, I forgot to multiply it by 5. The equation for circle is x^2 + y^2 = r^2 where r is the constant radius. Cheers!
@@Kuchdelan thanks man, love u so much already
you are very kind, my friend ;)
Dirac delta function is equal to infinity and not to 1. Because it's integral is equal to 1 and it's width is approaching zero (a width of a single point) it's hight need to be infinity.
Thank you very much for the videos !
Hey Alon, thanks for your comment. Technically impulse function is a rectangular function with the width of Delta and the amplitude of 1/Delta when Delta goes to zero. Thus the area under the curve is one. This definition is technically true but hard to digest. It is common to model this with the unit impulse function with the amplitude of 1 at the origin. Hope this clear things up.
At 4:45, how is the function x(t) defined?
Hi Libby, x(t) is a general function. So, it can be anything such as cosine, exponential, linear, etc. Hope this helps!
@@Kuchdelan Got it. Thanks!
Doroood iman joon aalie so clear❤❤❤
Mamnon az lotfet and pleasure to help!
I didn't got one thing, it's OK that integral us a summation but in the above cases the integrand should be summation of d(t) multiplied by dt. I got that d(t) has value 1 at origin but what about dt???
Hi Yogesh. Sorry for confusion. Please let me know which example you are talking about.
iman hi sir, I am talking about all the examples which involve integration of some function f with d(t), moreover in our college we have been taught that d(t) function is 1/E at t=0 where E is a very infinitesimally small no.
please read page 60 & 61 here:
www.ece.uvic.ca/~frodo/sigsysbook/downloads/lecture_slides_for_signals_and_systems-2016-01-25.pdf
Hope this helps!
Thank you for the awesome explanation. I have a doubt, at 8:10 in the video. we have in 2nd example cos(2t)delta(t-1) with integral.
delta function has t but cos function has 2t. is the 't' in x(t) and delta(t) independent?
Thank you very much for the feedback. Regarding your question, no both functions (x and cos) depend on the same variable (t). Hope this helps.
iman thank you! I have almost watched most of your videos and it was really helpful. Currently I'm trying to work on non linear signal processing. Could I email you my query? Please let me know your email. Thanks in advance..
Unique Ramya thanks for watching my videos. Please email your questions to imanmoaz@uvic.ca I will do my best to answer them as soon as possible. Best!
iman sure. thank you for the kind response:) God bless!
Hey Iman thanks for this course, it is really very helpful. Btw is there any chance of getting your notes on blackboard slides in any format.
I will release them sometimes in the future. Thanks for your feedback!
this is a brilliant video! double thumb up!
You are awesome ;)
You are a life saver!!!!!!!!!!!!!!!!
Glad I could help!
Thank you for the video but seems weird... how can integral of a point of value 1 (no width) be equal to 1?
Hey. Sorry for the confusion! The value of delta function is actually 1/delta for a width of delta when delta approaches zero. This is a rectangle with the area equal to one. Hope this makes sense.
The impulse is 1 and not value of the function. ..meaning the area of this function is one....this point cannot be over emphasized and is source of confusion in beginning for most.
can I ask a question iman? in the video at 3:20 , when the equation in the below adding together. the value in the origin is 2 , not as shown in the triangular which u plot in the begin?
Hey, sorry for my late reply! I have been super busy with life, my friend.
I just wonder how you ended up with two. Assuming unit step is 1 at the origin, here is what we get:
(t+1) x (u(t+1) - u(t)) + (-t+1) x (u(t) - u(t-1)) ---> (replace t by zero)
1 x (u(1) - u(0)) + 1 x (u(0) - u(-1)) = 1 x (1-1) + 1 x (1-0) = 1
Hope this helps.
ohhhhhhh~! I see! . I misunderstand the calculate process, but now I get it. tks for ur kind reply iman~
Not a problem at all ;)
5:29 x(t0) is a value of the function x(t) at (t0) which I think it is just a point. why is it a line?
Hey, thank you for the question. Note that "y = constant" is a line parallel to the x-axis. No matter what x is, the y value is always constant. Hope this makes sense.
Hi Iman. Thanks for this (again very good explained) video. I still have a basic question on the equivalence and sifting properties of the delta functions. The "mechanics" of the examples are clear, but where I am not 100% clear is what are these properties of the delta function good for. Referring to your examples at ca 11:00 . What are you proving by multiplying sin(t) with d(t-pi/6) and resulting in 1/2d(t-pi/6)? Thanks again for the time doing these videos!
Hi again Jonathan. Thank you for your question. The sifting and equivalence properties are super useful in signal processing. That's why I decided to dedicate one lecture to this topic. If you keep watching my tutorials, you will see the applications of these properties in convolution, Fourier series/transform and sampling. For instance, multiplying sin(t) with d(t-pi/6) means sampling the sinusoidal function at pi/6. Please let me know if you still have more questions. Cheers!
Dirac delta function is NOT equal to 1 at t=0: it equals to infinity. But an integral of Dirac delta function is 1. It is very important to understand, but it is shown wrong in this video and some next too.
Hey, thanks for your comment. Technically impulse function is a rectangular function with the width of Delta and the amplitude of 1/Delta when Delta goes to zero. Thus the area under the curve is one. This definition is technically true but hard to digest. It is common to model this with the unit impulse function with the amplitude of 1 at the origin. Hope this clear things up.
Hi Iman, appreciate the videos, they are of great help!
I'm a bit confuse at 8:00, example 2, as you went a bit too fast, this is my understanding of what happen, and I think it's missing, hope you could provide some clarification:
cos(2t)∆(t-1) = cos(2 * t)∆(t-1) = cos(2 * 1) = cos(2)
Hey! Thanks for your comment and sorry for the late reply. Life has been hectic recently! You got it all right. I just replaced t by 1. Hope it makes sense.
@@Kuchdelan shouldn't it be cos2 * U(t-1)
Which program you use for drawing?
I used to use photoshop but now I am using Castofly!
I guess you did wrong at 4:27 , impulse function is never 1, it is infinity. The area is 1.
unit impulse function and unit impulse sequence ( DT ) are not same. There is an abstract "impulse to sequence" conversion. It will be helpful if you make a video about CT to DT conversion, about what exactly the process is.
Also, your solutions could be slower. Thanks
thanx man!! very nice videos!!
Hi iman, I like your video very much and I hope you can continue to make it. Besides, at 6:54 in this video, it seems that you forgot the dt in the Integral formula.
Thank you very much for the feedback, Binshuai. You are very nice. And yes, I forgot dt in the integral. Sorry for the confusion. Best!
sepas Iman!
Mamnonam dooste aziz!
clear af
hahaa, thanks dude ;)
If it doesn't matter what you call the integral variable then why not refer to it as t dt instead of tau dtau?
Because I already used t in the integral which is different from tau
nice video! The example in the end, i am not so clear, how to deal with the 5?
Thank you! You are right, the final answer is 5u(t-2). I forgot to write it down. cheers!
Your vids are amazing I LOVE IT but can u also please teach about Energy and power signal please and also LCDE and State variable. Thx for your great vids 10000 times better at explaining compare to my professor at university and I'm also sharing this great content to my friends Pls keep up the good work XD
and btw I'm wondering why won't u teach discrete time since they are similar but important too
Hey. Thanks a lot for your feedback and kindness. I will cover more videos very soon. Cheers!
Hi, the way you defined the impulse function, it's integral is always zero, because you chose a finite value for t=0, and therefore that finite number over an infinitesimal width has zero area. You can multiply it by any finite function and it will remain zero. For a proper definition refer to wiki under the definition section: en.wikipedia.org/wiki/Dirac_delta_function#Definitions
Why not correct with an annotation?
I think it would be clearer it would be wrong to leave out the delta if you wrote:
x(t)=sin(t)*∆(t-t0)=>x(t)=sin(t0)*∆(t-t0)
Because then withouy delta you get:
x(t)=sin(t0)
So the signal x(t) is equal to a constant.
Yep, that's correct. As explained in the video, the delta function cannot be left out. Cheers!
thank you for the video but I think that the last answer=5 if t>=0
thank you x100000000000000000000000000000000000000000000
You are very welcome Faris ;)
university is really useless....all you need
X(Iman)/teacher.Done
loll! thank you ;)