I just want to thank neso academy for helping engineers to this extent....your videos have proved beneficial not only for students but also for many teachers as well. I therefore want to send my warm regards for reaching and helping engineers in building their base
A lot of love and respect to you and the whole team of neso Academy for the great job you are doing guys keep supporting engineers like this this will give our country well qualified engineers
Wow I must say am so grateful for your lessons in this unit. Am taking signals and systems this semester and have found your content very much understandable much thanks. If I score an A I owe it to you Neso Academy and the entire team
You are a life saver! I can't appreciate your efforts in words. Thank you sir. My ans. For the H.W problem I am getting 2 ramp( t + 2) - 4 ramp( t) - 4 ramp ( t - 2)
@@PoluBalaji If you see the diagram clearly, at t=3 the diagram is not going down to the 4th quadrant but -4r(t-2) is representing the diagram towards down(4th quad). So, we have to change the slope to 0 at t=3 by adding +4r(t-3). Actually '+4' is the slope value. If you don't add this 4r(t-3) term at t=3 the diagram would be different.
already the sign is taken negative as the signal is undergoing downward shift . if you want to follow the "final-initial" then do not take the negative sign beforehand. so either way you will get the negative sign. so do whatever you are comfortable at. good luck.
I want to watch all vidoe from this accademy can any one tell me how can I get it .....this vidoe are superb i can't express in words...dear neso can u help me i want to study every subject full course from your channel.....i belong to ECE branch and preparing for gate pls help me ... eagerly waiting for your reply..
Thank you sir for this clear and nice explanation :), I need to plot the following function: Hw(tau)=H[Cw(t-tau)-r],could you please help me to plot it?Thanks again :)
In my opinion... 2r(t+2) - 4r(t) - 4r(t-2) + 4r(t-3).. there will be 2r(t+2) in place of 4r(t+2). reason............... for the calculation of slope, we use>>>> (4-0)/(0-{-2}) = 4/2 = 2
the correct answer is x(t) = 2r(t+2) - 2r(t) - 4r(t-3) you can plot it on Matlab and check it is right. here's the code: t0=-2:0.01:0; x0=2*(t0+2); t1=0:0.01:2; x1=4*ones(size(t1)); t2=2:0.01:3; x2=-4*(t2-3); t=[t0 t1 t2]; X=[x0 x1 x2]; plot(t,X,'linewidth',2) xlabel('time') ylabel('amplitude') grid on
@@yurifan2537 Without that term the signal continues in the -ve direction after crossing 0 at t=3, which isn't the case in the ques, so to terminate the signal we add the 4r(t-3) part.
* -4r(t-2) because at pont 2 singal is downword turn and down turn consider to be - ve and slope is 4 = 4-0/(3-2) , and the pont of turning is 2 so we take r(t-2) and not r(t-3)
I just want to thank neso academy for helping engineers to this extent....your videos have proved beneficial not only for students but also for many teachers as well. I therefore want to send my warm regards for reaching and helping engineers in building their base
A lot of love and respect to you and the whole team of neso Academy for the great job you are doing guys keep supporting engineers like this this will give our country well qualified engineers
You are saving a career, thanks so much for these wonderful classes
Wow I must say am so grateful for your lessons in this unit. Am taking signals and systems this semester and have found your content very much understandable much thanks. If I score an A I owe it to you Neso Academy and the entire team
thank you so much neso academy...... lots of love.... I ll recommend this channel to all of my frnds definitely...
The answer to the homework problem is:
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
NESO ACADEMY IS GREAT💖
@sarthak gupta at t = 3 the slope tends to upward when compared with 4r(t-2), so should we add the 4r(t-3) signal?
@meharj bro, last term its -4r(t-3)
I liked your video so much because these are really helpful for me to pass my semester examination😊😊
You are a life saver! I can't appreciate your efforts in words. Thank you sir.
My ans. For the H.W problem I am getting
2 ramp( t + 2) - 4 ramp( t) - 4 ramp ( t - 2)
Bro starting lo 2 mundu minus vastadhi ga (4/-2=-2)
best video on waveform synthesis,thanks
2r(t+2)-2r(t)+4r(t-2)💯
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
i guess
@@abdelkadermoussidene9139 This is the correct answer 👍
@@magjoy_6 at t=3 in 4r(t-3), I guess we hve to take amplitude as 4/3 right (slope)..????but y you hve taken 4..??
@@abdelkadermoussidene9139 can u explain y u hve taken 4 in 4r(t-3).???
@@PoluBalaji If you see the diagram clearly, at t=3 the diagram is not going down to the 4th quadrant but -4r(t-2) is representing the diagram towards down(4th quad). So, we have to change the slope to 0 at t=3 by adding +4r(t-3). Actually '+4' is the slope value.
If you don't add this 4r(t-3) term at t=3 the diagram would be different.
4:11 is it correct to say : +ve sign when turning anticlockwise, -ve sign when turning clockwise for all 360 degrees?
the change in slope would be zero so it wouldn't matter.
Thank You, Sir, That was great ! But I think ..... i ) 0 + 2r(t) - 2r(t-1) + 2u(t-1) ,
I finally got the correct and easiest method only from u.Tnq sir.
Thank you so much sir. Hope one day I can be as good as you :)
X(t) = 2r(t+2) - 2r(t) -4/3 r(t-2) - 4/3 r(t-3)
Thank you so much for this wonderful explanation 🙏
H.W. answer: 2r(t+2) - 2r(t) - 4r(t-2) + 4r(t-3)
thank you so much, this was very eloquently explained :)
HomeWork Answer is 2r(t+2)-2r(t-0)-4r(t-2)
5:40 horizontal lines don't have slope why do use 2??
thank you so much for this video. i have been waiting for this...
sir please make a video of laplace transform of waveform
while going from 5 to -2 U just substract 5-(-2) means initial -final but in other cases U just substract final - initial
already the sign is taken negative as the signal is undergoing downward shift . if you want to follow the "final-initial" then do not take the negative sign beforehand.
so either way you will get the negative sign.
so do whatever you are comfortable at.
good luck.
You saved me , thnx a lot NESOoo
I want to watch all vidoe from this accademy can any one tell me how can I get it .....this vidoe are superb i can't express in words...dear neso can u help me i want to study every subject full course from your channel.....i belong to ECE branch and preparing for gate pls help me ... eagerly waiting for your reply..
This channel is great
as usual ,life saveur
2r(t-2)-2r(t)-4r(t-2)-4r(t-3)
2r(t+2) - 2r(t)-4r(t-2)
@@sagarshrestha5453 this look incomplete there's no term with r(t-3) already
Thank you for this video
Please which textbook are you using for signal... I really love your videos
God bless you sir
its very helpful, thx very match
Thank you sir for this clear and nice explanation :), I need to plot the following function: Hw(tau)=H[Cw(t-tau)-r],could you please help me to plot it?Thanks again :)
0+2r(t+2)-2r(t-0)-4/3r(t-2)+4/3r(t-3)
no need of zero at start. and after the 2 term all are having wrong amplitudes
2r(t+2)-r(t)-4r(t-2)
+4r(t-3)
thanks
👍👍👍
Love from J.Yan ELEC 221 section :D
2r(t+2) -2r(t) -r(t-2) +r(t-3)
THANKS
In the 1st example also the signal is changing from 0 to 2,then why it is not discountinuity
I need Fourier series representation of continuous periodic signals
thanks sir
Grade saver! God bless y'all ✌
2r(t+2)-2r(t)-4\3(r-3)
Wrong
thank you!
Homework problem answer:
2r(t+2)-2r(t)-4/3r(t-2)
Magnitude of 3rd term is -4 not -4/3 and also there will be one more term at t=3 bcs slope is changing there also
Beautiful How I wish I can give you a big hug right now.
Thanks
10:45 why (5-(-2))?
New level - old level is (-2-5)
At 10.13 . How do u simplify the last two terms as 2t u(t-1)
pls reply as soon as possible..
Since it's downward, so -ve sign is used, then original level - new level, 5-(-2) =7
2r(t+2)-2r(t-0)-4/3(t-2)+4/3(t-3)
right or wrong ???????
2.r(t+2)-2.r(t)-4.r(t-2)+4.r(t-3)
2r(t+2)-2r(t-1)-4/3 r(t-2) +4/3 r(t-3)
what if thre are 3 type of signals ? ramp exp and unite step
super
2 (t + 2) u[t + 2] - 2 t u[t] - 4 (t - 2) u[t - 2] + 4 (t - 3) u[t - 3]
4r(t+2) - 4r(t) - 4r(t-2) + 4r(t-3)
Sir please reply is it correct or not??
In my opinion...
2r(t+2) - 4r(t) - 4r(t-2) + 4r(t-3)..
there will be 2r(t+2) in place of 4r(t+2).
reason............... for the calculation of slope, we use>>>> (4-0)/(0-{-2}) = 4/2 = 2
Why does the first signal have no dicontinuity
Why neso academy don't provide the solution of the hw problems???????????????? Website is also not showing the answers, please provide the answers.
2r(t+2)-4r(t)-4r(t-4)+4r(t-3)
4r(t+2) - 4r(t-0) - 4r(t-2) - r(t-3)
whats the answer pleaseee
2r(t+2)-2r(t)-4r(t-2)-2r(t-3)??
the correct answer is x(t) = 2r(t+2) - 2r(t) - 4r(t-3)
you can plot it on Matlab and check it is right.
here's the code:
t0=-2:0.01:0;
x0=2*(t0+2);
t1=0:0.01:2;
x1=4*ones(size(t1));
t2=2:0.01:3;
x2=-4*(t2-3);
t=[t0 t1 t2];
X=[x0 x1 x2];
plot(t,X,'linewidth',2)
xlabel('time')
ylabel('amplitude')
grid on
pls post the answer sjr
2r (t+2)-2r (t)-2r (t-2)-4r (t-3) is is correct or not sir please tell me
if it is 3u(t)u(t-4), how to represent it?
As 3(t-4)
2r(t+2) - 2r(t) - 4/3 r(t-2) *** sir right or wrong
Tu Feku
2r(t+2)-2r(t)-4r(t-2)+4(r-3)
mohammad tawfiq ri8
mohammad tawfiq correct
*4r(t-3)
Wrong
2r+(t+2)-2r(t)-4r(t-3)
its t-2, not t-3
Thanks
2r(t+2)-2r(t)+4r(t-2)
Bhai please bta de wahan 4 kaise aa rha hai mera 4/3 aa rha hai
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
Shi hai tera
Last slope is decreasing so it must be - 4r(t-3) right?
@@keerthanajospin2874 true its mistake.
@@keerthanajospin2874 you are from which country?
@@ujjawal1366 im an indian
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
2r(t+2)-2r(t)-4r(t-2)
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
How is last term 4r(t-3)
@@advikmaniar5783 it changes to zero after t=3
@@sparshkatiyar8909 Hello sir, can you elaborate on that?
Wrogn
@@yurifan2537 Without that term the signal continues in the -ve direction after crossing 0 at t=3, which isn't the case in the ques, so to terminate the signal we add the 4r(t-3) part.
2r(t+2)-2r(t)-4r(t-2)
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
2r(t+2)-2r(t)-2r(t-2)
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
shouldn't it be- x(t)= 2r(t+2)-2r(t)+4r(t-2).
no, because there are 4 transitions.So there must be 4 terms
Santanu Mandal BRO SLOPE MAY COME 4/3
you are right
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
2r(t+2) - 2r(t)-4r(t-2) +4r(t-3)
* -4r(t-2) because at pont 2 singal is downword turn and down turn consider to be - ve and slope is 4 = 4-0/(3-2) , and the pont of turning is 2 so we take r(t-2) and not r(t-3)
2r(t+2)-2r(t)-4r(t-2)+4r(t-3)
Please explain
2r(t+2)-2r(t)-2r(t-2)+2r(t-3)
2r(t+2)-2t-4r(t-2)-4(t-3)
2r(t+2) - 4r(t) - 4r(t-2) + 4r(t-3)