Glad it was helpful and good luck on your exam! I continue to add videos to my channel as time permits....hoping to get some more completed on the upcoming holiday break. Best, Adam.
Thank you for the kind words, I appreciate you watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
3:55, How the blue and red give -4 and how the green and purple give -3? Should not the Blue and Red give -2 like below: 3u(3.5) - 5u(3.5-2) = 3u(3.5) - 5u(1.5) = 3 x 1 - 5 x 1 = -2 and should not the Green and Purple give 1 instead: r(3.5) - r(3.5-1) = r(3.5) - r(2.5) = 3.5 - 2.5 = 1 we sum them up and get -2 + 1 = -1 as a final value of x(3.5).
Yes, what you've written is true for just the "red" and "blue" but x(t) is the sum of ALL parts at each point in time. At time t = 3.5 we have 3u(3.5) + r(3.5) -r(3.5-1) -5u(3.5-2) = 3u(3.5) + r(3.5) -r(2.5) -5u(1.5) = 3*1+3.5-2.5-5 =3+3.5-2.5-5 = -1. Recall, r(t) = t for all t>0.
Adam Panagos Thank you, that makes a perfect sense, however, on the actual video I reckon you have mistakenly said that "the blue curve and the red curve give -4 and green and the purple curve give -3". So I was confused but not anymore. Thanks once again, I am learning a lot from your videos :)
Ah yes. If I said that I definitely misspoke. Too many different colors to keep track of in this video. =) Sorry for the confusion. Glad you're getting something out of the videos!
Sir I think we can plot the signal without drawing individual signals.........by just doing opposite of what we did in the "signal construction" and for a check we can write the signal expression for the signal made by us and check if it is matching with the given signal expression.Hope I'm right.
You can try and do it that way, but I think it will be much more difficult. What if you had a signal x(t) that was a summation of 30 different terms? Sure, you could draw your best guess at the signal expression and check different points to see if it matches, but the odds of just being able to write down an expression that is correct for all time seem pretty slim! I think the method presented in the video is the best systematic procedure to use. Thanks!
We can arrange all the terms in ascending values of "t" viz (t+15),(t+14),(t+13)...(t-15) and do the procedure taught by you and (I think) we will get the signal pretty easily .Also as the number of signal components increases it will be slightly cubersome to determine signals at larger values of "t" because we have to look for a lot of signals for that time but if one uses the method ,which I'm claiming ,we will get the signal straightway.Hope I made my point. Thanks a lot for the reply.
Thanks a lot. Could you tell how r(2t-1) looks like in the plot? i struggle a lot when the t is not alone within the ramp function (within in the argument). Many thanks
r(t) = t*u(t) is just the ramp function. If you have r(2t-1), then r(2t-1) = (2t-1)u(2t-1). This function is zero anytime that 2t-1 < 0 --> t < 1/2. For values of t greater than this, it is a line with slope 2, since 2t-1 has a slope of 2. So, starting at t = 1/2, just draw a line with slope 2. Hope that helps, Adam
@@aashalash40 Do you have a specific time you can tell me where this is said? I'm guessing I maybe misspoke but I'm not sure exactly what you're asking about.
Ok, I found it. Yes, I definitely just misspoke. The total of all signals at that point in time is -1, but when I mentioned the subset of individual curves adding up I said a value that was a little off. Thanks for catching that.
No, you are incorrect. No mistake in the video. If not for the last term in the expression for x(t), the signal would hold a constant value of 4 for all time. At t =2, a unit step with magnitude with -5 turns on, which changes the value to 4 - 5 = -1. This value of -1 (the gray line) then continues for all time.
@@kunalsharma6496 Yes, I did plot in Matlab and it's exactly what I have in the video. If you could provide some explanation for what you think is wrong I might be able to help you figure out where you've made your mistake. But just saying, "this is wrong" doesn't help me figure out where you've made an error.......
@@AdamPanagos I am really really very sorry, I did a mistake while plotting the graph in MATLAB. That mistake was in Expression of X(t) in which I wrote 5*r(t-2) instead of 5*u(t-2). That's why I was confused and I also bother my course teacher in it. I think you are a *great learner*. I want to have your Email, if you can share.(kunalsharma2482000@gmail.com this is mine).
@@kunalsharma6496 No worries at all. I've posted hundreds of videos and have definitely made mistakes that have needed correction from time-to-time. I was pretty certain things here were correct though. Glad we got everything figured out. Thanks much. Adam
Amazing work! So easy to understand when you explain it! Bravo!
please do more videos on this plotting stuff, I have my exams few days after so it will be very helpful
Glad it was helpful and good luck on your exam! I continue to add videos to my channel as time permits....hoping to get some more completed on the upcoming holiday break. Best, Adam.
subscribe for you bro!!! glad to see your videos, quite easy to understand.
This is the difference between an engineer and a math teacher .. pls make more videos on this
Thank you for the kind words, I appreciate you watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
ideas and exampleas are so clear to understand.ı want appreciate for this video collage.I will keep going for other video series
3:55, How the blue and red give -4 and how the green and purple give -3?
Should not the Blue and Red give -2 like below:
3u(3.5) - 5u(3.5-2) = 3u(3.5) - 5u(1.5) = 3 x 1 - 5 x 1 = -2
and should not the Green and Purple give 1 instead:
r(3.5) - r(3.5-1) = r(3.5) - r(2.5) = 3.5 - 2.5 = 1
we sum them up and get -2 + 1 = -1 as a final value of x(3.5).
Yes, what you've written is true for just the "red" and "blue" but x(t) is the sum of ALL parts at each point in time.
At time t = 3.5 we have
3u(3.5) + r(3.5) -r(3.5-1) -5u(3.5-2) = 3u(3.5) + r(3.5) -r(2.5) -5u(1.5) = 3*1+3.5-2.5-5 =3+3.5-2.5-5 = -1.
Recall, r(t) = t for all t>0.
Adam Panagos Thank you, that makes a perfect sense, however, on the actual video I reckon you have mistakenly said that "the blue curve and the red curve give -4 and green and the purple curve give -3". So I was confused but not anymore. Thanks once again, I am learning a lot from your videos :)
Ah yes. If I said that I definitely misspoke. Too many different colors to keep track of in this video. =) Sorry for the confusion. Glad you're getting something out of the videos!
thank you very much....this video helps me a lot
Sir I think we can plot the signal without drawing individual signals.........by just doing opposite of what we did in the "signal construction" and for a check we can write the signal expression for the signal made by us and check if it is matching with the given signal expression.Hope I'm right.
You can try and do it that way, but I think it will be much more difficult. What if you had a signal x(t) that was a summation of 30 different terms? Sure, you could draw your best guess at the signal expression and check different points to see if it matches, but the odds of just being able to write down an expression that is correct for all time seem pretty slim! I think the method presented in the video is the best systematic procedure to use. Thanks!
We can arrange all the terms in ascending values of "t" viz (t+15),(t+14),(t+13)...(t-15) and do the procedure taught by you and (I think) we will get the signal pretty easily .Also as the number of signal components increases it will be slightly cubersome to determine signals at larger values of "t" because we have to look for a lot of signals for that time but if one uses the method ,which I'm claiming ,we will get the signal straightway.Hope I made my point.
Thanks a lot for the reply.
Thanks a lot. Could you tell how r(2t-1) looks like in the plot? i struggle a lot when the t is not alone within the ramp function (within in the argument). Many thanks
r(t) = t*u(t) is just the ramp function. If you have r(2t-1), then r(2t-1) = (2t-1)u(2t-1). This function is zero anytime that 2t-1 < 0 --> t < 1/2. For values of t greater than this, it is a line with slope 2, since 2t-1 has a slope of 2. So, starting at t = 1/2, just draw a line with slope 2. Hope that helps,
Adam
ευχαριστω μαγκα,ακριβως αυτο που εψαχνα
At the end of the video you say the blue curve and red curve add up to -4...how do you get -4 from that plot???
sorry, -3
@@aashalash40 Do you have a specific time you can tell me where this is said? I'm guessing I maybe misspoke but I'm not sure exactly what you're asking about.
Ok, I found it. Yes, I definitely just misspoke. The total of all signals at that point in time is -1, but when I mentioned the subset of individual curves adding up I said a value that was a little off. Thanks for catching that.
one part of equation has 3u(t)u(t-4), how to plot this?
u(t) is zero for all t =4 it's equal to 3. Hope that helps.
Adam
@@AdamPanagos oh great! got it
thank u so much sir
You're welcome, thanks for watching.
add more examples on signal sketching
Thanks
You're welcome, thanks for watching!
xe(t)= 0.5[x(t)+x(-t)]
xo(t)= 0.5[x(t)+x(-t)] help
Sir Koncham Telugu looo Chepandi plz
You did mistake at t=2
Plot this function in MATLAB you will find your mistake.
Indeed you are doing a great job.🙏🙏🙏🙏
Namaskaram
No, you are incorrect. No mistake in the video. If not for the last term in the expression for x(t), the signal would hold a constant value of 4 for all time. At t =2, a unit step with magnitude with -5 turns on, which changes the value to 4 - 5 = -1. This value of -1 (the gray line) then continues for all time.
@@AdamPanagos
Plot this X(t) in MATLAB and see please....
@@kunalsharma6496 Yes, I did plot in Matlab and it's exactly what I have in the video. If you could provide some explanation for what you think is wrong I might be able to help you figure out where you've made your mistake. But just saying, "this is wrong" doesn't help me figure out where you've made an error.......
@@AdamPanagos
I am really really very sorry,
I did a mistake while plotting the graph in MATLAB. That mistake was in Expression of X(t) in which I wrote 5*r(t-2) instead of 5*u(t-2). That's why I was confused and I also bother my course teacher in it.
I think you are a *great learner*.
I want to have your Email, if you can share.(kunalsharma2482000@gmail.com this is mine).
@@kunalsharma6496 No worries at all. I've posted hundreds of videos and have definitely made mistakes that have needed correction from time-to-time. I was pretty certain things here were correct though. Glad we got everything figured out. Thanks much.
Adam