1. The problem statement, all variables and given/known data Solvinf the unit time step equation mathematically. 2. Relevant equations x(t) = u(t+1) - 2u(t) + u(t-1) 3. The attempt at a solution How can i solve this signal mathemtically and using plotted signals , i couldn't understand, here is my approach and i'm not sure it is correct, i simply wrote the coordiantes of each signal, u(t+1) = (-1,1) -2ut = (0,-2) u(t-1) = (1,1) and then added them which give answer = '0' is this the way of solving the signals ? please help
Give values to t, 0 1 2 3 and so on. At each value evaluate each term in the equation. If t in u(t) is equal or greater than zero then u(t) equals 1. Remember the definition of a step signal.
i wanna solve it without putting a value for t, what my teacher did is , he simply draw those three signals and cancelled out the opposite signals , the remaining signal is the answer ! what the hell was that ??? is that a way of solving a signal ? he said there is no mathematical approach to solve signals (btw we have just started signals n system and this was my 2nd lecture), i wonder how the hell there is no way to solve the above equation mathematically. there must be a way which my teacher doesn't aware about, can you draw the output waveform of this signal and approach of solving it without assigning a value of t ?
Yes, you can plot each u() component of x(t) individually and then sum them up. Assume the component u(t+1): set t+1=0 and solve for t t=-1 Thus, at t=-1 the signal rises from 0 t 1 (step) Do the same for all u() components.
i already did that as i mentioned in my first post , re-read the first post please. but how to solve further, after this step ?? look at this, u(t+1) = (-1,1) -2ut = (0,-2) u(t-1) = (1,1) you said exactly what i've already posted. can i solve it like u(t+1) - 2u(t) + u(t-1) (-1,1) + (0,-2) + (1,1) = 0 is this correct ? if not then how to solve it ?
The three functions are not continuous over the region from t=-∞ to t=+∞ and cannot be summed as you propose. Your summation only holds true for either t<-1 or t>=1.