Posted by Rajeshwara Raju unstable if u take x(t) as u(t) in this case it is stable only but it has to satisfy for all the bounded inputs even if atleast for one signal it is unbounded means it is unstable system let us take x(t) as cost then it will become cos square of t INTEGRATION { cos square of t }= (1+cos2t)/2 here we get integration of 1/2 which is equal to { 1/2 * t } i.e., ramp with slope 0.5 as we alredy know ramp is not a stable signal it is unbounded so the system is unstable
System is always stable for finite input, But if we take input is cos(t) we get integration 1/2*(t) + something, & 1/2*(t) is a ramp signal , & ramp signal is a unbounded signal, so for a any one bounded input the output is reaches to infinite of amplitude so the system considering as a unstable system.
These examples only show if the system is stable *for a particular input.* But the definition of a BIBO-stable system requires that the system be stable for *any* input.
@SP step signal - it is a constant but the integration of step signal or constant is the variable to which the constant is integrated and which is nothing but the ramp signal right
Thank you so much for the clear explanation!! But still wondering whether there are some general methods instead of trying functions according to the experience and intuition.
For the home work, i used integration by part. at the end i had: Y(t)=sin(minus inifinty)*x(minus infinite) So if we assume that x(t) is a bounded input, the output automatically is gonna be bounded. (Because the sin(minus infinity) becomes an amplitude scaling)
This way you can never know if a system is stable or not since you have to try all possible inputs to make sure that the system is stable, and if you happen to find that one of the inputs make the system unstable then you sa yit is unstable so I think that this is not the most efficient way to do this. Thank you for your efforts.
I have a doubt In complex analysis, we studied that sine function is an unbounded function when we use sin(z). So in the first example, should we consider the function to be stable or unstable ? And if stable how should we know where to consider x(t) as real function and where as complex function.
Hi there, if you would take x(t)= invsin(t),then you are confined to take values of t between -1 and +1,so due to domain restrictions input will always be bounded and so will be output.But i think we should avoid using such functions so that we can't vary our input for a broader range.Hope you will get what i am trying to convey.
Ty Nishant Gupta But invsin(t) isn't confined between -1 and +1. Sint(t) is confined between -1and +1. There are no limitations for invsin. Please check
If we assume u(t) as x(t), everytime we're getting correct answer....So will it be safe to say it an ideal bounded function n can surely assume it in every example.. correct me if I'm wrong Sir ...?
you cannot take so as it is said in the video that we should check for all possible bounded input. for example in the above question consider x(t)=cos(t). therefore integrating cost*cost will be an unstable system
jai prakash narottam das meena multiplication of 2 bounded sgn z stable..I guess d case here z abt d integration of the sng..nd though cos is the part of the integration(it has fun of t) it can't be considered as constant...so I guess again diz system fun must b integrated nd I don't whether it is the crt method..is there any 2 explain diz clearly.???
unstable if u take x(t) as u(t) in this case it is stable only but it has to satisfy for all the bounded inputs even if atleast for one signal it is unbounded means it is unstable system let us take x(t) as cost then it will become cos square of t INTEGRATION { cos square of t }= (1+cos2t)/2 here we get integration of 1/2 which is equal to { 1/2 * t } i.e., ramp with slope 0.5 as we alredy know ramp is not a stable signal it is unbounded so the system is unstable
Posted by Rajeshwara Raju
unstable
if u take x(t) as u(t) in this case it is stable only but it has to satisfy for all the bounded inputs
even if atleast for one signal it is unbounded means it is unstable system
let us take x(t) as cost
then it will become cos square of t
INTEGRATION { cos square of t }= (1+cos2t)/2
here we get integration of 1/2 which is equal to { 1/2 * t } i.e., ramp with slope 0.5
as we alredy know ramp is not a stable signal it is unbounded
so the system is unstable
Limits should be submitted
Good work 😊
Thank you sir for ur ans
🙏🙏🙏
good explanation:)
System is always stable for finite input, But if we take input is cos(t) we get integration 1/2*(t) + something, & 1/2*(t) is a ramp signal , & ramp signal is a unbounded signal, so for a any one bounded input the output is reaches to infinite of amplitude so the system considering as a unstable system.
Sir, you have great capability of teaching. Thank you sir.
please reveal the correct answer for HW problems
These examples only show if the system is stable *for a particular input.* But the definition of a BIBO-stable system requires that the system be stable for *any* input.
as we know that cost is finite(bounded) signal -1
can u pls explain me in detail about the process of proving unstable system
this helped me a lot,thanks sir
@@poulamim06 What exactly you are not getting it...
Can you push it here?
@SP step signal - it is a constant but the integration of step signal or constant is the variable to which the constant is integrated and which is nothing but the ramp signal right
The system is stable
Isnt there a common procedure for this kind of questions?do we need to check for that many number of inputs??
Thank you so much for the clear explanation!! But still wondering whether there are some general methods instead of trying functions according to the experience and intuition.
UNSTABLE
if x(t) = cost
y(t) = [1/2t +sin2t/4] with limits
and after limits y(t) = infinite
Integral of Cos[Tau] does not converge on {-Infinity,t}. Hence unstable
For the home work, i used integration by part. at the end i had:
Y(t)=sin(minus inifinty)*x(minus infinite)
So if we assume that x(t) is a bounded input, the output automatically is gonna be bounded. (Because the sin(minus infinity) becomes an amplitude scaling)
This way you can never know if a system is stable or not since you have to try all possible inputs to make sure that the system is stable, and if you happen to find that one of the inputs make the system unstable then you sa yit is unstable so I think that this is not the most efficient way to do this.
Thank you for your efforts.
If input is u(t) then intergal of u(t) is ramp signal which is unbounded.. Hence it is unstable..
sounds right to me!
All the concepts are very excellent and straight forward.
Very easy to learn here.🎶👌🎇
I have a doubt
In complex analysis, we studied that sine function is an unbounded function when we use sin(z).
So in the first example, should we consider the function to be stable or unstable ? And if stable how should we know where to consider x(t) as real function and where as complex function.
Stable system: when you put u(t) then from -infinity to 0 value will zero and from 0 to t will be some finite value
Bro it is unstable
Unstable ✅
In the integration example why you didn't put x(t) = u(t) and by the way it will show an unbounded output
What if the question is of summation running over some variable from minus infinity to ' t ' instead of integration?
It's for Discrete sgnls...same answr
Sir how to check linearity in split function ??
unstable ..x(t)=u(t)
Homework problem army!
HW ANSWER
STABLE if you do integration by parts.
You should get u(t).sint + cost as a result which is bounded
Put x(t) = Cos t and Solve With Integration Range.
Ans Will be Infinity ♾️. Hence Unstable
UNSTABLE SYSTEM CUZ THE LIMITS ARE RANGING FROM MINUS INFINITY TO T SO THAT LEADS TO UNBOUNDED (INFINITE) OUTPUT
Unstable for unit step thus unstable
👍
in sins case limit is -1 to 1 meanwhile cos has no limit so technically unstable since unbounded right?
thanks
Hello sir y(n)=x(1/2n) is it casula or non casual pls clr that
non causal bcs n
what if the integration is between two finite numbers, lets say t-2 and t?
Sir, In the first problem. If we take x(t) = invsin(t). Then y(t)=t. This is an unstable system. Please check. thanks in advance
Hi there, if you would take x(t)= invsin(t),then you are confined to take values of t between -1 and +1,so due to domain restrictions input will always be bounded and so will be output.But i think we should avoid using such functions so that we can't vary our input for a broader range.Hope you will get what i am trying to convey.
Ty Nishant Gupta
But invsin(t) isn't confined between -1 and +1. Sint(t) is confined between -1and +1. There are no limitations for invsin.
Please check
You should not use inverse sin function, because it is unbounded value function which is against BIBO
hlo sir Kya ye saari vedio gate syllabus ko complete krti Hai.kya sirf ye saari vedio see syllabus complete hoga...plzzz sir rply
better is check gate syllabus
but gate syllabus main to sirf topic likha hota only ...ej topic k ander bhut topic aate
Y(n)=exp[ x(n)], is it stable ?
according to BIBO criterion, yes it is :)
H. W question ---- Unstable
why not just take the limit as t tends to infinity?
Unstable
What's the answar? can anybody give me explanation?
Unstable
we do partial differential than we get biubo , so unstable.
Add links to previous n the next video
Y(n)=x(-n+1) Is stable r unstable
Unstable for cos t
If we assume u(t) as x(t), everytime we're getting correct answer....So will it be safe to say it an ideal bounded function n can surely assume it in every example.. correct me if I'm wrong Sir ...?
you cannot take so as it is said in the video that we should check for all possible bounded input. for example in the above question consider x(t)=cos(t). therefore integrating cost*cost will be an unstable system
stable
can uexplain
Unstable.. Take x(t)=cost, it'll make t+cos(2t) whole by 2. And when you integrate 1 it'll give t which is unbounded. Thus =unstable
It is a stable system
Harshita Mohapatra If you put x(t)=cost and solve it it will produce unbounded output.
how explain plz
If we put x(t) =cost then it will become sin2t/2, so its stable naa
Unstable
@@vinaypant557 x(tau)=cos(tau).....it is not x(t),it is x(tau)...........if we substitute x(tau) =cos(tau).....unstable system
It's unstable
unstable,
take x(t)=u(t),
and integration of u(t) is r(t),which is an unstable signal,
so y(t) is an unstale signal.
unstable.
it is unstable
unstable
Stable
if x(t)=del(t), then also unstable
Hwk unstable system
Stable system
👍
It's a Stable System
Un stable
Worst techniques
your speed is very slow in every video i have to speed up everything try to make it fast
Watch video on 2x speed
@@rajatmajhi9905 😂😂😂LOL
stable
jai prakash narottam das meena how is it stable?
x(t) is bounded input and y(t) is bounded between +x(t) and -x(t) so it is also bounded here value of cos t would be from 0 to 1 hence stable system
jai prakash narottam das meena multiplication of 2 bounded sgn z stable..I guess d case here z abt d integration of the sng..nd though cos is the part of the integration(it has fun of t) it can't be considered as constant...so I guess again diz system fun must b integrated nd I don't whether it is the crt method..is there any 2 explain diz clearly.???
unstable
if u take x(t) as u(t) in this case it is stable only but it has to satisfy for all the bounded inputs
even if atleast for one signal it is unbounded means it is unstable system
let us take x(t) as cost
then it will become cos square of t
INTEGRATION { cos square of t }= (1+cos2t)/2
here we get integration of 1/2 which is equal to { 1/2 * t } i.e., ramp with slope 0.5
as we alredy know ramp is not a stable signal it is unbounded
so the system is unstable
my answer for taking u(t) it is stable
unstable
Stable
Unstable
stable
how?
Unstable
unstable
unstable
Unstable
Unstable
unstable
Unstable