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I believe there are both non avertible; many to one maping. for the first function you plug -1 and +1 you will get the same answerr which is one. for the second question, when you take the derivative of any constant it could be always zero.
HW PROBLEMS: 1. Non-Invertible 2. DEPENDS on wether dx(t)/dt is STRICTLY MONOTONIC or not EXPLANATION OF 2: dx(t)/dt is the slope of the function x(t). If this slope is STRICTLY MONOTONIC (i.e. if it is constantly increasing in value OR if it is constantly decreasing in value), then it follows 1 to 1 mapping ("1-1"). And thus it is INVERTIBLE. Otherwise, it is Non-Invertible. * How to find if dx(t)/dt is STRICTLY MONOTONIC: We have to check wether d2x(t)/dt2 maintains the SAME sign throughout its range (i.e. if constantly > 0 OR if constantly < 0). If it does, then dx(t)/dt is strictly monotonic, and thus "1-1", and thus INVERTIBLE.
Sir Thank you so much. I couldn't believe that I'm getting a world class tutoring for free. It would be nice if you do a series on microcontrollers microprocessors and embedded systems. Since you have given us a solid background in digital electronics
Sir,can we make a generalization here......that the systems which produce the output as the scaled version of input is always invertible...If not give me an example..@nesoacademy
Bro! To solve such problems you should assume the system to be invertible and then try to find out two inputs for which the output is same if there exist two such inputs then the system is non-invertible. You can apply the same approach to solve system stability problems
Hi, Could you please explain if the below system is invertible or not . And if it is invertible , what is the inverse system in this case ? y[n] = x[n]x[n-1]
take x1[n]=0 and x2[n]=delta[n] , for both the input signals the output signal is y[n]= 0 hence two distinct input signals lead to same output signal implies that the signal is non invertible
Both r non invertible, bcoz many to one mapping. 1. Take x(t) = u(t) and u(-t) : output is u(t) for both. 2. Take x(t) = u(t) and -u(-t) : output is impulse(t) for both.
both HW problems are non-invertible . in 1st problem if we take t=2 and t=-2 , then input will be x(2) and x(-2) and output for both will be y(t)=x (t^2) i.e x(4), so it is many to one mapping . in 2nd question if we take x(t) = u(t) and x(t)=u(-t) then output for both will be zero , so many to one mapping
You told three properties to check if the system is dynamic or not? Can we say that those properties hold true for the invertible system too? Like if time scaling or time shifting is happening then the system is invertible?
you need to first of all understand that a system maps a signal to another signal i.e the domain is a set of signals and the codomain itself is a set of signals, so you need to put different values of input signal to check whether you are getting same output signal , if yes then the system is non invertible if not then you need to check the invertibility.
1. Non invertible, since for negative and positive of some time 't' we get same value of y 2. Invertible, since for different values of x we get different values of y
@@DeepshikhaKumariBEE yes...if the system is non invertibe..this gives two similar output for two different input...and y1 and y2 are straighe lines..and its passes throught origin so m will be equal
1.invertible 2.non invertible For the first one take u(t) & -u(t) as input will get one to one mapping For the second case take input 1 & 2 will get zero because Differentiation of any contast is zero so will get many to one mapping hence its non-invertible system
@@makindeolalekanmonsuru5349 yeah we can use it.....But make sure that if u have another possibility of inputs that proves a particular maping was wrong.
for the first question if we give dirac (t) and -dirac(t) one will become dirac t square and other will be - dirac t square . so if t goes to 2 for both wouldn't to result be 0 because dirac (2)= 0 ? and dirac(4) is also 0 other than t =0 dirac would be 0 right or am i confused?
How is the last example a proof of invertibility? How can you show only two examples of different inputs giving different outputs? Of course it is invertible, but isn't that not sufficient proof?
I think it is non invertible as in case if we assume x(t) to be constant,then in that case time shifting does not alter the value of x(t).thus y(t) remain constant for different values of x(t)
@@ujjawalvats5169 no bro it will be always a constant value that i accept but for every different x(t) u will get different y(t) so one-one mapping and hence it will be invertible. Just put u(t) first then u(-t) u will get different constant values
prithvi krishna..... i think it is invertible because when we put u(t) and -u(t) it will give the result u(t^2) and -u(t^2) respectively it is just like a time scaling as previous ex. In this presentation
let input be x(t)+c where c is a constant and x(t) can be any input, d/dt (input) = x'(t); you have your system performing many to one operation. Derivative operator is non invertible