Glad I could help, thanks for watching. If you’re looking for additional examples/videos make sure to check out my website adampanagos.org where I have a lot of other videos and resources available that you might find helpful. Thanks, Adam.
Thanks for watching! I've only taught EE382 one time at UAH (about a year ago). I teach EE383 regularly in the summer though. Maybe I'll see you in EE383. Thanks for the nice comment!
Yes, that's correct. At time t = 1.8 we have a slope of +1. At time t = 2 we need to have a slope of -1. So, we need CHANGE in slope of -2. Thus, the use of -2r(t-2). Hope that helps. Adam
Might be a stupid question, but what is the unit of the signal x(t)? I am having trouble understanding the unit of a digital signal and other types of signals. I do know that sometimes the square of x(t) is power
Not a stupid question at all. For the most part, in classes/examples like this the signals are unitless (i.e. they don't have any units). In some more practical application you might be working with specific signals that have units, but for the most part you can work this type of example using a unitless signal and still convey/learn the relevant information. Hope that helps, Adam
Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam
I use an iPad app called Doceri (www.doceri.com) for most of my videos. This app lets you record all your handwriting ahead of time and use "breakpoints" to pause as needed. Once all the writing is down you can "play" the handwriting back while recording audio over it. I find this works much better than trying to write and talk at the same time. I'd definitely recommend checking out the app, I've found it very useful. Hope that helps!
Great question. The unit step function u(t) as defined here is equal to 1 at time t = 0. So, when the unit step function "turns on" it is equal to one when it's argument is zero. So, consider what happens at time t=5 for example. The term we're adding in there is -2u(t-5). This means, when t = 5, we have -2u(0) = -2, so the OVERALL signal is equal to zero at time t =5. Similarly, x(0) = 1 and x(4) = 2. As you noted though, the way I drew the original sketch was a little ambiguous. It might have been better if I'd use open circles and closed circles in the original sketch to indicate which value x(t) was taking on. Nice work! Adam
For most of my videos, I use Doceri (an app for my iPad, apps.apple.com/us/app/doceri-remote/id412443803?ls=1) that lets me pre-record all the writing. Then, I can "play back" the writing while I record the audio. It’s a very useful tool for me. Hope that helps, Adam
Hi! great video! May I ask, why is the checking for x(5) = 0 is correct? when x(5) magnitude is 2 according to the figure? I think in -2u(0) it should be equal to (-2*0), please kindly check if this is correct! Thank you!
The funtion in the plot is equal to zero at t = 5. It's equal to 2 when t is slightly less than 5 (i.e. t = 5-epsilon). I maybe should have put closed circles (or open circles) at the jumps to indicate the value. The unit step is always on a time zero so u(0) = 1, thus -2u(0) = -2*1 = -2. Hope that helps.
Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam
The ramp function is defined in every linear systems textbook. r(t) = t*u(t) where u(t) is the unit step function. The unit ramp function r(t) is equal to t for all positive t.
We're evaluating x(t) at t = 2. That means we need to replace all t's with the value 2. So, u(t) becomes u(2). The unit step u(t) is defined as one any time the argument of u(t) is greater than or equal to zero, and zero otherwise. So, u(2) = 1 since the argument of u(t) (in this case 2) is positive. Hope that helps, Adam
Yes, that's right. To be perfectly precise, u(t) = 1 when t>=0 and u(t) = 0 when t < 0. Note my inequalities include t=0 when u(t) = 1. But in general, I think you have the idea.
That's exactly what happens by adding in the u(t-4). This unit step is located at t = 4 and causes the signal to "jump up" a value of 1 at time t = 4. I hope that helps. Make sure to check out my website adampanagos.org for lots of additional content that you might find helpful. Thanks.
It is. So, when it turns "on" at time t=3, it cancels out the negative slope before it. From time t = 2 to t=3, the signal is going down at a slope of -1. Once we add in the signal r(t-3), it adds a slope of 1 at that point, so we end up with a constant value starting at t = 3. -1 + 1 = 0. So, no slope, just holding the value. Hope that helps, Adam
You can think of it as a vector quantity. The slope is suddenly increased by 2 units (from 1 to -1) at t = 2. (In other words, at t = 2, the signal rotates clockwise by pi over 2). The slope only dictates the direction. The total vertical drop of the unit ramp function from t = 2 to t = 3 is only 1 which is obviously not long enough to cross the horizontal axis.
It was confusing but then I realize that -2r(t-2) is -r(t-2) - r(t-2). The graph of r(t-1) will cancel out the graph of -r(t-2) at t>=2, leaving v(t) = 2. Then, we add -r(t-2) again to make downward slope of -1 at t=2. You can read the comment of Mr. Panagos above to understand next step or watch example #2 first.
@@SonPham-jy2qithis is what I’ve been searching for. It’s not because of the position compared to the point where it’s last traveled. But the position on the Cartesian plane.
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (540+ videos) you might find helpful. Thanks, Adam
hey i am just wondering if you could answer my question regarding a signal that looks like this: →2 /| / | →1 | | _____| | ← 2→-3- get the energy of this signal in one period. the pulse width of the combined triangle over a square wave is 3 (amp of square is 1 while amp of triangle is also 1). this is a periodic wave. note : what i have drawn is one period of the signal already. hope you could answer me cause i am really confused.
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
I'm not sure how you came up with that answer but it's definitely not correct. For instance, if we compute your answer at time t = 6 we get: x(6) = u(6) +r(6-1) - 2r(6-2) + u(6-3) + 2u(6-4) - 2u(6-5) = u(6) +r(5) - 2r(4) + u(3) + 2u(2) - 2u(1) = 1 + 5 - 2*4 + 1 + 2*1 - 2*1 = 1 + 5 - 8 + 1 + 2 - 2 = -1 which obviously isn't correct since from the plot of x(t) we know that x(6) should equal zero. The correct answer for the signal is worked out in the video.
No, the video is correct. During the time interval 2 < t < 3 we are decreasing with a slope of 1. Since we want to be a constant value during the time interval 3 < t < 4, we need to increase our slope by 1 at time t = 3. Thus, we add in r(t-3). Hope that helps, Adam
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
Thank you so much sir,......at first I didn't get how to draw by watching this video........but I understand after seeing your reply for the comments....... thanks for responding others comments
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (450+ videos) you might find helpful. Thanks, Adam
No, r(t) is the unit ramp function. r(t-3) is the unit ramp function that turns on at time t = 3. r(t) has a slope of 1, so does r(t-3) (when on). When r(t-3) turns on at time t = 3, it cancels out previous slope of -1 to yield a slope of -1+1 = 0. The function r(t) is never a unit step function.
During the time interval t = 2 to t = 3 the signal x(t) is decreasing with a slope of -1 (i.e. it is going down 1 units for every one unit it moves over). At time t = 3, x(t) stops decreasing and holds the constant value of 1 for the time interval t = 3 to t = 4. A constant value has a slope of zero. Since we were previously decreasing with a slope of -1, and we need to stop decreasing, we need to increase the slope by +1 at time t = 3. -1+1 = 0, resulting in a constant value being held. This is why we add in the signal r(t-3). This unit ramp signal has a slope of +1 and turns "on" at time t = 3. Hope that helps, Adam
The video is correct, it should be +r(t-3). From time t = 2 to t = 3 we are going down with a slope of -1. At time t = 3 we "level off" and hold a constant value until t = 4. So, at time t = 3 we need to go from a slope of -1 to a slope of 0. Adding in a unit ramp signal with a slope of 1 does exactly that. Hope that helps. If you found the video useful make sure to check out my website (adampanagos.org) where I have a ton of other resources available and it's easier to watch my videos in a more organized fashion. Thanks, Adam.
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
I'm not sure that I understand your comment. We're working with continuous-time signals here so the unit step function would be u(t) which is equal to 1 for all t greater than or equal to zero, and equal to zero otherwise.
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
You cleared a huge doubt out of my mind just before the exam. Extremely Helpful
Glad I could help, thanks for watching. If you’re looking for additional examples/videos make sure to check out my website adampanagos.org where I have a lot of other videos and resources available that you might find helpful. Thanks, Adam.
Thank you so much for posting this. You made it incredibly simple.
Glad to help, thanks!
Current EE382 student at UAH and this is pure gold. Thanks.
Thanks for watching! I've only taught EE382 one time at UAH (about a year ago). I teach EE383 regularly in the summer though. Maybe I'll see you in EE383. Thanks for the nice comment!
at t=3, we can write u(t-3),
plz help me out to write it in terms of a ramp function.
why are we doing -2r(t-2) when its slope is -1
akshat bhatt my question as well. Adam can you explain?/
akshat bhatt it should be -r (t-2) i think. how to calculator slope????
(-2) is the change in slope. See it is something like this-
(final slope)-(initial slope)
= (-1)-(1)
Yes, that's correct. At time t = 1.8 we have a slope of +1. At time t = 2 we need to have a slope of -1. So, we need CHANGE in slope of -2. Thus, the use of -2r(t-2).
Hope that helps.
Adam
He said why in the video.
i dont get the part 'r(t-3)'....i think it should be 'u(t-3)'...please explain...
same here, did you get any answer to it?
During the time interval 2
Adam Panagos thanks for the quick response, I get it now. Thanks for helping Adam.
@@AdamPanagos thank you for the explanation
Might be a stupid question, but what is the unit of the signal x(t)? I am having trouble understanding the unit of a digital signal and other types of signals. I do know that sometimes the square of x(t) is power
Not a stupid question at all. For the most part, in classes/examples like this the signals are unitless (i.e. they don't have any units). In some more practical application you might be working with specific signals that have units, but for the most part you can work this type of example using a unitless signal and still convey/learn the relevant information. Hope that helps,
Adam
Thank you for all your videos!
You're welcome, thanks for watching.
Great video!
Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam
What is the equation for the output signal, x(t), when the input is a unit ramp?
Thank you so much, Sir.❤❤❤
Clear and easy explanation.
The notes is excellent. What software did you use ?
I use an iPad app called Doceri (www.doceri.com) for most of my videos. This app lets you record all your handwriting ahead of time and use "breakpoints" to pause as needed. Once all the writing is down you can "play" the handwriting back while recording audio over it. I find this works much better than trying to write and talk at the same time. I'd definitely recommend checking out the app, I've found it very useful. Hope that helps!
How to construct this signal sir x(t)=tu(t)-2(t-2)u(t-2)+2(t-4)u(t-4) i have confusion in it.
At t (0) x has values 0 and 1. And at t(5) x values are 2 and 0. So which of the twe values are to be considered - the 0 or 1,2 ?
Great question. The unit step function u(t) as defined here is equal to 1 at time t = 0. So, when the unit step function "turns on" it is equal to one when it's argument is zero.
So, consider what happens at time t=5 for example. The term we're adding in there is -2u(t-5). This means, when t = 5, we have -2u(0) = -2, so the OVERALL signal is equal to zero at time t =5.
Similarly, x(0) = 1 and x(4) = 2. As you noted though, the way I drew the original sketch was a little ambiguous. It might have been better if I'd use open circles and closed circles in the original sketch to indicate which value x(t) was taking on.
Nice work!
Adam
May i know what software you are using here, thank you
For most of my videos, I use Doceri (an app for my iPad, apps.apple.com/us/app/doceri-remote/id412443803?ls=1) that lets me pre-record all the writing. Then, I can "play back" the writing while I record the audio. It’s a very useful tool for me. Hope that helps,
Adam
Hi! great video! May I ask, why is the checking for x(5) = 0 is correct? when x(5) magnitude is 2 according to the figure?
I think in -2u(0) it should be equal to (-2*0), please kindly check if this is correct! Thank you!
The funtion in the plot is equal to zero at t = 5. It's equal to 2 when t is slightly less than 5 (i.e. t = 5-epsilon). I maybe should have put closed circles (or open circles) at the jumps to indicate the value. The unit step is always on a time zero so u(0) = 1, thus -2u(0) = -2*1 = -2. Hope that helps.
Thanks Dude this video will last for genertions
I LOVE YOU MAN
Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam
This is my first time seeing the ramp function. How is it defined? Where is a video where you introduce it?
The ramp function is defined in every linear systems textbook. r(t) = t*u(t) where u(t) is the unit step function. The unit ramp function r(t) is equal to t for all positive t.
Good Explaination
Informative.. thanks :).. one doubt, in your check, at t=2, then how U(2) = 1, when there is no unit step function @ t=2. Please explain..
We're evaluating x(t) at t = 2. That means we need to replace all t's with the value 2. So, u(t) becomes u(2). The unit step u(t) is defined as one any time the argument of u(t) is greater than or equal to zero, and zero otherwise. So, u(2) = 1 since the argument of u(t) (in this case 2) is positive. Hope that helps,
Adam
OK. So it means that in u(argument) whenever argument > 0 we take u(t) = 1 and if argument
Yes, that's right. To be perfectly precise, u(t) = 1 when t>=0 and u(t) = 0 when t < 0. Note my inequalities include t=0 when u(t) = 1. But in general, I think you have the idea.
Can you explain how to sketch the derivative of a signal. Just to sketch, not using the equations.
Do you have a video on how to plot this equation in matlab? Thanks in advance.
Nice Explanation, Very helpful.
Don't we need an impulse on 4? Since the value is instantly jumping on 2 amplitude?
That's exactly what happens by adding in the u(t-4). This unit step is located at t = 4 and causes the signal to "jump up" a value of 1 at time t = 4.
I hope that helps. Make sure to check out my website adampanagos.org for lots of additional content that you might find helpful. Thanks.
@@AdamPanagos but the value for u(t) will be 1 .How could that jump to 2. Will that not be 2u(t-4)??
Please do reply me....
Between time 3
@@AdamPanagos wow it's very clear now
Thanks a lot !!! You're the best
You're welcome, thanks for watching!
3:42
I thought *r(t-3)* was diagonal line beginning from the 3 . 🤔
It is. So, when it turns "on" at time t=3, it cancels out the negative slope before it. From time t = 2 to t=3, the signal is going down at a slope of -1. Once we add in the signal r(t-3), it adds a slope of 1 at that point, so we end up with a constant value starting at t = 3. -1 + 1 = 0. So, no slope, just holding the value. Hope that helps,
Adam
I don't understand how adding -2r(t - 2) to u(t)+r(t-1) causes the graph to instantly have an opposite slope and still have a positive of x(t).
You can think of it as a vector quantity. The slope is suddenly increased by 2 units (from 1 to -1) at t = 2. (In other words, at t = 2, the signal rotates clockwise by pi over 2). The slope only dictates the direction. The total vertical drop of the unit ramp function from t = 2 to t = 3 is only 1 which is obviously not long enough to cross the horizontal axis.
It was confusing but then I realize that -2r(t-2) is -r(t-2) - r(t-2).
The graph of r(t-1) will cancel out the graph of -r(t-2) at t>=2, leaving v(t) = 2. Then, we add -r(t-2) again to make downward slope of -1 at t=2.
You can read the comment of Mr. Panagos above to understand next step or watch example #2 first.
@@SonPham-jy2qithis is what I’ve been searching for. It’s not because of the position compared to the point where it’s last traveled. But the position on the Cartesian plane.
Thank you for your quick respponse.
Helped a lot
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
You are a god thanks for helping
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (540+ videos) you might find helpful. Thanks, Adam
hey i am just wondering if you could answer my question regarding a signal that looks like this:
→2 /|
/ |
→1 | |
_____| |
← 2→-3-
get the energy of this signal in one period. the pulse width of the combined triangle over a square wave is 3 (amp of square is 1 while amp of triangle is also 1).
this is a periodic wave.
note : what i have drawn is one period of the signal already. hope you could answer me cause i am really confused.
This was helpful
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
thank u so much 💖
#iraq
You're welcome, thanks for watching!
Duha Almaliki
u r welcome dear
sir why we dont use u(t-3) instead of r(t-3) ?
Because at t = 3, we need the SLOPE to change by 1. u(t-3) just shifts things by 1. r(t-3) starts a ramp with a slope that we need.
best explanation!
thank you so much sir ..but i have badly in need the matlab code of this singal..
please sent the code as soon as possible..
the correct answer is X(t)= u(t) + r(t-1) - 2r(t-2) + u(t-3) + 2u(t-4) - 2u(t-5)
I'm not sure how you came up with that answer but it's definitely not correct. For instance, if we compute your answer at time t = 6 we get:
x(6) = u(6) +r(6-1) - 2r(6-2) + u(6-3) + 2u(6-4) - 2u(6-5)
= u(6) +r(5) - 2r(4) + u(3) + 2u(2) - 2u(1)
= 1 + 5 - 2*4 + 1 + 2*1 - 2*1
= 1 + 5 - 8 + 1 + 2 - 2
= -1
which obviously isn't correct since from the plot of x(t) we know that x(6) should equal zero. The correct answer for the signal is worked out in the video.
Thank you brother ...... I'm Sorry !
No worries! This is some tricky stuff! Thanks for watching.
Yeah I thought so about the second to the last expression "2u(t-4)" is it correct or wrong?
Dont you think the 4th term should be 2r(t-3) instead of r(t-3)????
No, the video is correct. During the time interval 2 < t < 3 we are decreasing with a slope of 1. Since we want to be a constant value during the time interval 3 < t < 4, we need to increase our slope by 1 at time t = 3. Thus, we add in r(t-3). Hope that helps,
Adam
Ya got it thanks 😄
why is u(5) = 1?
its should be check x(5)=2
u(t) can only be either 1 when t > 0 and 0 otherwise. so u(5), 5 > 0, so it is = 1.
thank you so much sir
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
nice, thank you very much!
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
no entiendo ingles...pero con lo visual pude entender...muchas gracias
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
thank you so much
Thank you so much sir,......at first I didn't get how to draw by watching this video........but I understand after seeing your reply for the comments....... thanks for responding others comments
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (450+ videos) you might find helpful. Thanks, Adam
r(t-3) is unit step function as it goes straight line ri8
No, r(t) is the unit ramp function. r(t-3) is the unit ramp function that turns on at time t = 3. r(t) has a slope of 1, so does r(t-3) (when on). When r(t-3) turns on at time t = 3, it cancels out previous slope of -1 to yield a slope of -1+1 = 0. The function r(t) is never a unit step function.
att time = 3 i didnt get it plss explain again
During the time interval t = 2 to t = 3 the signal x(t) is decreasing with a slope of -1 (i.e. it is going down 1 units for every one unit it moves over). At time t = 3, x(t) stops decreasing and holds the constant value of 1 for the time interval t = 3 to t = 4. A constant value has a slope of zero. Since we were previously decreasing with a slope of -1, and we need to stop decreasing, we need to increase the slope by +1 at time t = 3. -1+1 = 0, resulting in a constant value being held. This is why we add in the signal r(t-3). This unit ramp signal has a slope of +1 and turns "on" at time t = 3.
Hope that helps,
Adam
Adam Panagos at t=4 u have simply written u(t-4).y won't we take the amplitude of 2 into account..??
Why is it -2r(t-2)
At time t = 1.8 we have a slope of +1. At time t = 2 we need to have a slope of -1. So, we need CHANGE in slope of -2. Thus, the use of -2r(t-2).
cool video dude
I second that
I think - r(t-3) not r(t-3)
Can you explain
The video is correct, it should be +r(t-3). From time t = 2 to t = 3 we are going down with a slope of -1. At time t = 3 we "level off" and hold a constant value until t = 4. So, at time t = 3 we need to go from a slope of -1 to a slope of 0. Adding in a unit ramp signal with a slope of 1 does exactly that. Hope that helps.
If you found the video useful make sure to check out my website (adampanagos.org) where I have a ton of other resources available and it's easier to watch my videos in a more organized fashion. Thanks, Adam.
Thank you buddy..😊
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
Thank you
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
god bless you🍁🍁
I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
thank u
Thank you.
I think you got it wrong with the ramp part.
No, I don't think so. What do you think is wrong?
@@AdamPanagos My bad. I didn't notice the ramp next to it. I'm sorry.
@@ronaldrosete4086 No worries! Thanks,
Adam
How is u(5)=1?
u(t) is the unit step function is equal to 1 for all t >=0
I've got tons of questions on this especially with triangles anyway I could get to you
thanks
+Fadhel Alshaikh You're welcome, thanks for watching!
doubts clear about construction of signal..
unit step signal , [x]= n u [n-6]
I'm not sure that I understand your comment. We're working with continuous-time signals here so the unit step function would be u(t) which is equal to 1 for all t greater than or equal to zero, and equal to zero otherwise.
Can you plot this equation
Thank You!
Very difficult and very hard
And one question -3u(t-5) step unit signal
❤️
This seems wrong.
nice
Hindi pls
Honestly, I wish I could. Unfortunately, this is the only language I speak. Definitely jealous of those that speak more than one language. Best,
Adam
u(0)=¿¿
u(t) is the unit step function u(t) = 0 for t= 0. Hope that helps,
Adam
How to be slope -2? I know negative part but how 2
Thank you so much for posting this. You made it incredibly simple.
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
thank you
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
@@AdamPanagos man do you explain Fourier series
I got 0 at FS exam, fuck me