Neso Academy u are the best I've prepared for my interview 2 days back using ur digital electronics tutorials 👏👌 can u start a new series about Microprocessors Microcontrollers and embedded systems
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
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
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
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
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?
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.
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?
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)
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?
@@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
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Neso Academy u are the best
I've prepared for my interview 2 days back using ur digital electronics tutorials 👏👌
can u start a new series about Microprocessors Microcontrollers and embedded systems
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.
tks
Hello. When proving that a system is invertible, when should we use the step function instead of the impulse function?
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
Did you find some good resource of them plz reply
Both are non invertable
Please give me the solution
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
how can we gurantee the invertiblity of system by only considering impulse or unit step function?
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
@@AJitKumar-vg6no yr tu que you can get some years ago I love❤❤
can we use u(-t) as input to hw problem 1 such that its output in u(t^2)
Vaidesh Iyer here the system is replacing t by t^2 so ur going to get the ans.as u(-t^2)brother...
@@kirankumar2432 cant we see it like system is squaring input
When i/p is u(-t) then sys will square I/p and make it to u(t²).?
Q2.Differentiation of any constant is zero....so many to one .. so non-invertible
Q1. When you put u(t) then you will get invertible
Q2 we cant take constant as i/p becoz i/p must be function of t as given in question x(t)
@@gyatika8902 ....
Madam, we can take constant as I/P ,see example 1 of this video....
He has taken 2, -2 as I/p.
Thanks for response.
@@gyatika8902
Mam..
Constant × t^0 = constant
So it's function of t.
@@ujjawal1366
Thanks to correct my concept 😊👍yes u r right we can put constant and answer is noninvertible..
@@gyatika8902
Your welcome.
plz answer for HW problem
U have stopped ece branch lecture videos please continue networks and other subjects also sir???
Can we use 0&π to prove sine terms as non invertible
H. W.
1.non invertible
2.non invertible
1.Invertible 2.Non Invertible
Learn New
1 Invertible, 2 non Invertible
revanth Vatturi... in problem 1 check the value at t=-1, and t=1...you got problem 1 is also non invertible...
mr. unique but we are actually putting values for x(t) not for 't'. right..?
2 pblm ans is invertible
1).Invertible 2). Non-invertible
Glt h
@@dharmendrachoudhary5568how?
1. Put x(t2) =t2 ... (-1)2 =(1)2. Hence many to one
2. Put x(t)= t. Diff we get y(t)=1. Hence many to one
plz elaborate
Right bro both are many to one
sir please give answer of HW problems
Both non invertible 👍
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
first is invertible, since we have to check for x(t^2) and not t itself.
@@atmajyotighosh2049 it is non invertible, put x(t) = u(-t) and u(t) both will give u(t^2)
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]
It is non invertible
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
Hello sir
Will plz explain it with numerical Value than it is better way to understand for us
The third question is many to one mapping I guess.
Because as we put different values of t we get same answer i.e. 1.
Correct me if I am wrong.
we do not put value of t we put value of xt
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.
Sir please share the answers in next video
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
We need to put different values of input not for different values of time..so the system is invertible
2. Non-invertible... x1(t)=mt+c1 and x2=mt+c2.. this implies y1(t)=y2(t)=m
@@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
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?
No
Sir, If in the 1st home work question, we take input as u(t) & u(-t) , will the output be x(t^2) ?, system is non - Invertible?
yes second is invertible
@@ClickBait123 bro from my logic it's non-invertible , because if u put u(t) and u(-t) then answer will be 0 .
@@mayankdewan7793 bro if you diff unit step funcytion u get impulse function
please add convoloution, fourier transform
1.Non invertible
2.Non-Invertible
1 Non invertible & 2.Invertible
2nd ex. is invertible coz for x(t) = r(t) nd u(t) we gt different o/p. so 1-1 mapping . hence I.V.
No
take x(t) = 1, y(t) = d/dt(1) = 0
take x(t) = 2, y(t) = d/dt(2) = 0
Hence 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.
these are two diffrent signals !
you are wrong
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
Why don’t you use a u(t) for the second assignment
@@makindeolalekanmonsuru5349 yeah we can use it.....But make sure that if u have another possibility of inputs that proves a particular maping was wrong.
@@santhoshchandchelli6581 exactly! So for 1st Question : for x(-2) ---> y(-2) = x(4) and x(2)----> y(2) = x(4).Many to one : Non-Invertible.
y(t)=(2+sint)x(t) is invertible or not? plzz ans it
Non invertible..
invertible.... please try for impulse input , negative impulse input and delayed impulse input.
Invertible if we take unit impulse as a input
Invertible
It is non invertible
sir plz explain the homework sums
Both non invertible .. right?
alka kumari wrong
right
@@karindratalukdar9198 why...?
@@pentainamerica First one is invertible
your explanations are the greatest!! I even let the adds run for you! hahah
can we use impulse signal in place of step signal
no never
Sir agar y(t)=x(2t) m x(t)=u(t) aur x(t)=u(2t) lenge to non invert aaga
Ut and U2t are same fns
@@nityanand4581 no it is not ut and u2t are not same time scaling
wt abt the system when y(t)=x(t+1)
It is invertible
Can a system be invertible and non at the same time ?
draw graph and check , both question is non-invertible
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?
y(t) = x(t-5) - x(3-t)
What about this...invertible or not ???
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)
adarsh raj yeah....but given answer is Invertible..
yes it is invertible system
What so if we take x(t) =, u(t) and u(-t), then it will be invertible
Both seems invertible
Both are Invertible
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?
Thanks sir
Delta(t) =1 at t=0
Invertible is first 2nd may or may not be invertible
Both are non invertible
Non -invertible system and invertible system
Both should be non invertible
1) Inv
2) Non Inv
solve plz y(t )= ∫ 𝒙(𝟓𝝉)𝒅𝝉 𝟒
interate from -infinity to 4 is invertable or not??
It is invertible bro
Y(t) will always be constant so it will be many one and non invertible
@@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
@Apoorv gaming for every x(t), y(t) will be independent of x(t). No matter what will be x(t). It is just like y(t) =4
@@ujjawalvats5169 y(t) is equal to integration of x(5t) so how it will be independent of x(t). Bro just clarify this
Both are non invertible
Put x(t)=u(-t) and u(t) in first case
Put x(t)= 2 and 1 in second case
i think you shouldnt substitute for t but substitute for x(t)
@@raghunandan2470 same thing
sir if i take u(t) and u(2t) as two inputs in 3rd problem the system will not be invertible?
Both r invertible ?........reply
Yes..
@@reshmipriya3358 No
take x(t) = 1, y(t) = d/dt(1) = 0
take x(t) = 2, y(t) = d/dt(2) = 0
Hence non invertible
Both the problems having the time invertible property.
Both are invertible
i think 1st one is invertible but 2nd is non invertible
@6:29 smj nii aaya sir kuch bhi
Both are non invertable
1.invertible
2.non invertible
both invertible .
Both noninvertible
Non Invertible both of them.
how?
both are NI easy.
#Neso Academy
Non invertible
Invertiable
1.Non invertible
2.Invertible
1)non-invertible
2)non-invertible
Both are non - invertible
invertible
non invertible
Both invertible...
both are non invertible
if possible give ans to me .
1) NI
2) NI
1.non invertible,2.invertible
How it is
+Ashok Vardineni because differentiation of any constant is zero... implies many to one function
Ashok Vardineni that's why the 2nd one should be non invertible as many to one ...
Eva Humaira : how is the first one non invertible
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
1.Invertible 2.Non Invertible
1.Invertible,2.Non Invertible
Both are non invertible
both are invertible
1.invertible
2.non invertible
1. Non-Invertible
2. Invertible
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
Lets put x(t) =u(t) so d/dt u(t) will give unit impulse function and if x(t) is ramp it will give step function si it is a invertible function
Put DC value and derivative of DC 0 and for also zero value output also be zero so non invertible
1.Non invertible.
2.invertible
1.non invertible
2.invertible
1.Invertible,2.Non Invertible
1. Invertible
2. Non-Invertible
1) for 2 and (-2) both you will get 4 are output... many to one mapping so it is non invertible
both are non invertable