when the u(t-1), its indicates that t = 1, while when u(t+1), t = -1, assume that t + 1 = 0, use t as the left side and move 1 into the right side. simple algebra.
Thanks for the comment Pratheesh. Yes, I did check again, but there is nothing wrong. Fourth signal is u(t-3), its an unit step signal shifted to the right by 3.
THANKS A LOOOOT you explained it perfectly and easily, i was suffering to understand it
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Love the way you teach sir, nice content in short video☺️
when the u(t-1), its indicates that t = 1, while when u(t+1), t = -1, assume that t + 1 = 0, use t as the left side and move 1 into the right side. simple algebra.
for 3rd signal and 4th signal how you got -1
How do you get amplitude 1+ time signal 3=0 how do you write 4th signal as zero
Hey appreciate your effort can you do with discrete signal as well
Thank you Safwan. Yes those videos are also in the pipeline.
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thx so much for this lecture
excillent for this information and more introduction i wish for you
Awesome bro
In 3rd signal it is about amplitude -2 but u made it -1 how come sir
Thank you
4th signal how's you get 0 amplitude please reply
New amplitude at a time=previous amplitude on graph at that time plus amplitude at which switching starts at that time for the new graph.
Thank you very much
Thank you!
Perfect
I think the x(t) graph is wrong🙄🙄🙄u took amplitude 1 and -1 instead of 2 and -2
Thankyou sir
Is U(-t)= -U(t). Plz reply
Nope.! In U(-t) time is getting reflected and in -U(t) amplitude is becoming negative.
Both are different.
Hope this helps.
Ok. It means in U(-t) line with amplitude 1 is on the left side of x axis . In -U(t)below y axis.
Exactly.! You got it correct.
3rd signal amplitude was -2 you sketched -1
F(t) = t^2[ u (t-1) - u (t-2)]
❤
Your addition for this signal is not clear..🤧
Fourth signal is wrong...please do check that...
Thanks for the comment Pratheesh. Yes, I did check again, but there is nothing wrong. Fourth signal is u(t-3), its an unit step signal shifted to the right by 3.