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
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
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
u(t) is the unit step function. It's "on" for all t >=0 and "off" for all t < 0. When the unit step is "on", it has a value of 1. So, u(5) = 1 since the argument of the unit step function 5 > 0 and thus it's on. Hope that helps, Adam
Say we have a positive ramp function with slope = 1 starting at t=0, then at t=1 the signal falls and commences as a horizontal line from t=1 to t=2. How would you write this when there's two things that happen at t=1?( adding a positive slope to make it horizontal then subtracting a step function??)
Yes, what you said sounds perfectly right. If "two things" happen at the same time, just put two signals at that time. You could put a unit step to shift things, and a ramp to change the slope. Hope that helps, Adam
No, I'm pretty sure the videos are correct. Can you tell me what time in the video you think there's an error? u(t) is the unit step function and is equal to 1 for all t >=0 and zero otherwise So, u(2) = 1. r(t) is the unit ramp function and is equal to r(t) =tu(t). So, r(t) is equal to t for all t>=0 and zero otherwise. So r(2) = 2*u(2) = 2*1 = 2. Did you maybe mix up the r(t) and u(t)?
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
please add some more problems on signal construction to clear the concept....this was awesome...but still needs to watch some more to get into d concept...plz
That's what the final ramp term r(t-2) does. If the last term was not present, we would continue to decrease with a slope of -1 for all time. Adding in r(t-2) adds a slope of +1 at time t = 2 to cancel it out. So, the final slope at t = 2 is -1+1 = 0. Thus, we hold a constant value for the rest of time. This constant value happens to be zero. Hope that helps.
you said that the ramp function is equal to 0 if t is less than or equal to 0! Is that true? I checked it online and my professor told us that the ramp function is equal to 0 only if t is less than 0. if ramp of 0 is t then the x(t) won't be equal to 1 at t-1.
Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (400+ videos) you might find helpful. Thanks much, Adam
During times 1 < t < 2, the signal is decreasing with a slope of -1. Once the ramp function at time t= 2 turns on with slope, the overall signal now has slope 0. Hope that helps, Adam
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (435+ 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’re welcome, thanks for watching. If you found the video useful make sure to check out my website adampanagos.org where I have a variety of other resources available that you might find helpful. Thanks, Adam.
As an EE student, Thank YOU!
yes you are an ass
😂🤣
thank you so much!!! just 1 video and it's worth than my lecture at university. This is what I'm trying to understand for days!!!
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 was struggling with these types of questions, I can't thank you enough!❤
Super useful as i'm taking a signal processing course... THANK YOU ADAM!
I'm watching this for the sake of online class during the pandemic. Thanks!
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
You have made this really simple. Thank you!
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
اتمنى ان تستمر بشرح هاذه الماده ❤️
Thank you! Very concise and informative.
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.
A big thanks for making everything clear!
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
Didn’t understand the checking part. How is u(5) =1?
u(t) is the unit step function. It's "on" for all t >=0 and "off" for all t < 0. When the unit step is "on", it has a value of 1. So, u(5) = 1 since the argument of the unit step function 5 > 0 and thus it's on. Hope that helps,
Adam
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.
Say we have a positive ramp function with slope = 1 starting at t=0, then at t=1 the signal falls and commences as a horizontal line from t=1 to t=2. How would you write this when there's two things that happen at t=1?( adding a positive slope to make it horizontal then subtracting a step function??)
Yes, what you said sounds perfectly right. If "two things" happen at the same time, just put two signals at that time. You could put a unit step to shift things, and a ramp to change the slope. Hope that helps,
Adam
Explained beautifully... helpful indeed...Thanks!
Sir, how you consider 1 at u(2) , because according to your previous video it should be 2 yup.
No, I'm pretty sure the videos are correct. Can you tell me what time in the video you think there's an error?
u(t) is the unit step function and is equal to 1 for all t >=0 and zero otherwise So, u(2) = 1.
r(t) is the unit ramp function and is equal to r(t) =tu(t). So, r(t) is equal to t for all t>=0 and zero otherwise. So r(2) = 2*u(2) = 2*1 = 2.
Did you maybe mix up the r(t) and u(t)?
what is u(-t) if u(t)=1 for t>=0?
It's just the time-reversed version of u(t). The signal u(-t) is equal to 1 for all t
great video thank you
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
please add some more problems on signal construction to clear the concept....this was awesome...but still needs to watch some more to get into d concept...plz
x(t) =(3t+1) (u(t +2)+u(t-2)) can you in this question?
x(t) =(3t+1) (u(t +2)+u(t-2)) how to solve this question? What to do with 3t+1 part?
I'm not sure exactly what you're asking. Are you trying to plot this signal?
@@AdamPanagos yeah.. But i did it myself. Thanks ^^
Don't you have to create a ramp function that cancels the first where the value is held from -1?
That's what the final ramp term r(t-2) does. If the last term was not present, we would continue to decrease with a slope of -1 for all time. Adding in r(t-2) adds a slope of +1 at time t = 2 to cancel it out. So, the final slope at t = 2 is -1+1 = 0. Thus, we hold a constant value for the rest of time. This constant value happens to be zero. Hope that helps.
Thank you it really helped !!💯🙏
Glad I could help, thanks for watching.
Adam
you said that the ramp function is equal to 0 if t is less than or equal to 0! Is that true? I checked it online and my professor told us that the ramp function is equal to 0 only if t is less than 0. if ramp of 0 is t then the x(t) won't be equal to 1 at t-1.
Either definition is the same.
You can define it as:
r(t) = t for t>=0 and 0 for t < 0
Or you can define it as:
r(t) = t for t>0 and 0 for t
Very useful!
Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (400+ videos) you might find helpful. Thanks much, Adam
how can we get -1 slope
confused how the last one is r(t-2) isn't that 0*r(t-2)
During times 1 < t < 2, the signal is decreasing with a slope of -1. Once the ramp function at time t= 2 turns on with slope, the overall signal now has slope 0. Hope that helps,
Adam
thank you !!
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (435+ videos) you might find helpful. Thanks, Adam.
great, THANKS!
Thanks a lot man!!
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.
Oh god, the train. I'm getting fred Harris war flashbacks.
r(0) is equal to 0
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..
I'm sorry, but this video example doesn't have anything to do with Matlab so there's not any Matlab code to go along with it.
@@AdamPanagos my lab examination is knocking at the door.I need matlab code like that problem... Can you make me a solution, where I get this code??
Very helpful , Thanks a lot man
You’re welcome, thanks for watching. If you found the video useful make sure to check out my website adampanagos.org where I have a variety of other resources available that you might find helpful. Thanks, Adam.