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Stable and Unstable Systems 

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Комментарии : 166   
@lovelykumari6202
@lovelykumari6202 4 года назад
No need of ofline teachers... Your video is enough... Thanks on behalf of all electronics student..
@ranjankumar2359
@ranjankumar2359 3 года назад
But offline rhta to or effective hota
@Abhi_sr99
@Abhi_sr99 3 года назад
@@ranjankumar2359 but
@mosesadventurediaries
@mosesadventurediaries 2 года назад
Comment endorsed 👍🏻
@itsashishkohli
@itsashishkohli Год назад
Offline *
@tejasbr6420
@tejasbr6420 Год назад
👍
@buqetmuolli2330
@buqetmuolli2330 3 года назад
The system in the homework problem is stable. If we take u(t) as the input and take into consideration that sin(t) is a bounded signal (it takes values from -1 to +1), then after multiplying these signals we get an output that is 0 from -infinity to 0 and from 0 to infinity looks like a normal sine function. So, the output of the system is bounded if the input is bounded.
@kautukraj
@kautukraj 3 года назад
Crisp and clear explanation!
@vladsamonin9794
@vladsamonin9794 6 лет назад
It's stable because amplitude of sin(t) is from -1 to 1 and you can do check for bounded input x=2 for example => sin(2)*x(2) is finite.
@astaragmohapatra9
@astaragmohapatra9 6 лет назад
If we take x(t)=r(t) then?
@ravindrathalari8332
@ravindrathalari8332 6 лет назад
We should take only bounded input..
@youtubeonly7271
@youtubeonly7271 5 лет назад
@@astaragmohapatra9 r(t) is an unbounded input as it goes from 0 to infinity
@souravganguly5680
@souravganguly5680 5 лет назад
@@astaragmohapatra9 r(t) is not bounded input signal
@onnumillasummathan2611
@onnumillasummathan2611 2 года назад
@@astaragmohapatra9 give x(t) = u(t) and it becomes u(t)*sin(t) and further simplify it becomes sin(t)✨
@rishitsharma9777
@rishitsharma9777 2 месяца назад
Love the teaching style and how concise and yet precise everything is. I was pretty confused with this when my college professors taught this. But now its crystal clear and that too within minutes. Thanks a lot for these wonderful videos.
@muhammadwaqas209
@muhammadwaqas209 5 лет назад
stable system because the range of sin(t) is from -1 to +1
@ehsanabbasi9405
@ehsanabbasi9405 6 лет назад
that is a Stable system because Sin(t) is an Stable System :-)
@simonbenjamin7917
@simonbenjamin7917 8 месяцев назад
Your lectures are brief and on point. They are helping me alot with my studies. #Zambia
@dyuthig9518
@dyuthig9518 6 лет назад
Your lectures are just the best......Neso academy 😁
@kavintong151
@kavintong151 3 года назад
This is how teaching should be to the point precise at the same time easy to understand
@02_shraddhakotrange87
@02_shraddhakotrange87 3 года назад
I just like the way you write and the different color's you used.teaching...no words
@joshs.7612
@joshs.7612 6 лет назад
You can't prove a system stable by example. If you do that then you have to show that every bounded input results in a bounded output. You can only show that a system is not stable by example .
@baliramchaurasiya3261
@baliramchaurasiya3261 6 лет назад
you are right
@sapankushwaha4069
@sapankushwaha4069 6 лет назад
I am not able to understand 2 nd line of your comment please explore it
@Muxik4k
@Muxik4k 5 лет назад
you should upload a video with your proof Josh..
@Monugupta-qd9ud
@Monugupta-qd9ud 5 лет назад
Bounded i/p Lo ,ydi bounded o/p aya to system stable hoga ...
@kera2403
@kera2403 4 года назад
Make sense
@nickthewinner2194
@nickthewinner2194 3 года назад
its good to have straight A's!
@smritismriti6012
@smritismriti6012 Месяц назад
Thank u so much sir. Couldn't express my gratitude for this❤❤❤
@akshaysingh1914
@akshaysingh1914 5 лет назад
if we take x(t) = sint (test signal) then it becomes y(t)=sin^2 t hence lies -1
@bhaveshacharya1422
@bhaveshacharya1422 5 лет назад
TRUE, BUT ONE CORRECTION: 0
@viswanathboya6847
@viswanathboya6847 5 лет назад
@@bhaveshacharya1422 u r correct. Sin^2 gives positive value only
@Mansoor.BSCEF22
@Mansoor.BSCEF22 7 месяцев назад
stable system because the value the sint is bounded (-1 to 1) Thanks for your great explanation
@lovelykumari6202
@lovelykumari6202 4 года назад
Laconic and lucid explanation
@joeyzhou8273
@joeyzhou8273 6 лет назад
It is stable since sin(t) has value between -1 and 1, therefore by feeding any value to x(t), the output will always be finite.
@shiveshsatyakam7468
@shiveshsatyakam7468 3 года назад
What if x(t)=ramp(t)
@kundurupavan1429
@kundurupavan1429 3 года назад
@@shiveshsatyakam7468 ramp(t) is unbounded input We cannot use unbounded input
@burgerking220
@burgerking220 5 лет назад
Wow! Thank you for the explanation! Amazing.
@arpitmishra4319
@arpitmishra4319 5 лет назад
please make playlist on electrical machines also
@saniyabanshkar5463
@saniyabanshkar5463 2 года назад
You got the best voice.
@krismemolah4193
@krismemolah4193 8 месяцев назад
Let the input be a dc of a value 4. After passing it through the system, we obtain the output to be 4sin(t) which is bounded from amplitude -4 to 4 in the interval -infinity to positive infinity and is therefore BIBO stable.
@mariama8157
@mariama8157 6 лет назад
Great explanation by great teacher thanks a lot
@hectorgatica1026
@hectorgatica1026 4 года назад
yeah great video Bro!!!!!!!!!
@afaqahmad4518
@afaqahmad4518 Год назад
it is bounded , cuz sin is bounded b/w -1 and 1 , and any applied bounded I/P x(t) will just get multiplied to this limit , thus giving us finite output
@barsilgen120
@barsilgen120 Год назад
Thank you
@ritvikshekhar4900
@ritvikshekhar4900 Месяц назад
Stable
@rorytobin1492
@rorytobin1492 8 месяцев назад
You are a GOAT thank you soooo much !!!!
@arpitmishra4319
@arpitmishra4319 5 лет назад
keep doing great work
@dayalsati9874
@dayalsati9874 7 лет назад
Sir you are one of the best teacher of the word....can we see your your face in the lectures....at least for few minutes
@vrushabh1816
@vrushabh1816 6 лет назад
Kirk Hammer 2010 so curde of u. and ill thinking
@Pratip.
@Pratip. 6 лет назад
@@KirkHammer-fj2of 🤣🤣
@rajankarn90
@rajankarn90 7 лет назад
a kind request. please make a video on digital counters states counting problem.
@ohmakademi
@ohmakademi 6 лет назад
*thank you very much*
@rajankarn90
@rajankarn90 7 лет назад
which one of the following are the STABLE discrete time system. 1. y(n)=x(4n) 2. y(n)=x(-n) 3. y(n)=ax(n)+8 4.y(n)=cosx(n) Please ans. and upload a snapshot of the solution.
@rajeshwararaju9865
@rajeshwararaju9865 7 лет назад
all the four are stable systems... let x(n) = u(n) # we can also chose any one of the bounded i/p i.e., finite amplitude signals 1. Y(n)= u(4n) #just time compression amplitude will not effected so it is stable 2. Y(n)= u(-n) # agian it is time reversal so amplitude is not going to effect so it is stable 3. Y(n)= au(n)+8 # here i've chosen a as finite number ... let a=2; Y(n)= 8 for t=0 so that the signal Y(n) amplitude is finite amplitde over the entire range of time axis..so it is also stable 4. as we know that whatever the angle may be the cos fucntion value is always lies in between (-1,1) so it gives finite magnitude that means it is also stable. :-)
@akshaysingh1914
@akshaysingh1914 5 лет назад
@@@rajeshwararaju9865 very fine , so always go with some test signal is a better choice : very thanks.
@razzazz
@razzazz 5 лет назад
thanks sir, you help me a lot
@ahmedzanoon6330
@ahmedzanoon6330 Год назад
thank u sir
@bipulkumar6637
@bipulkumar6637 3 года назад
I study in one of the most premier engineering institution of India and still my prof uses the SS of neso academy
@shribas087
@shribas087 8 месяцев назад
HW was very easy, where if x(t) will considered as u(t) and unit step function is bounded similarly sint will also bounded as its range varies from -1 to 1 so the multiplication of two bounded system should be bounded one.
@AliHamza-oy3li
@AliHamza-oy3li 5 лет назад
The signal is stable because when we take x(t) as impulse function than the answer will be 0 so it means 0 is the amplitude from this it is clear that it is a stable signal
@asalafanclub
@asalafanclub 5 лет назад
To put it in a more clear way: The system is stable if it maps a any bounded function to a bounded function. Otherwise, It is called unstable.
@rashmikapur2801
@rashmikapur2801 7 лет назад
thanks sir please upload more videos on sns syllabus
@-A-Ruchitha
@-A-Ruchitha 3 года назад
Are these lectures useful for EEE?
@nesoacademy
@nesoacademy 3 года назад
Yes they are.
@samarththakor1210
@samarththakor1210 Год назад
The system is Stable. As Sint can be given as bounded input .
@Msdhonisir07
@Msdhonisir07 Год назад
The ans of the last qn is stable system Is it correct or not
@piaknow3881
@piaknow3881 2 года назад
Pls do more videos on subjects in ece
@bimalkumar4137
@bimalkumar4137 7 лет назад
most easy way ever i read
@kasunjayasekara2737
@kasunjayasekara2737 2 года назад
system is stable system
@Msdhonisir07
@Msdhonisir07 Год назад
One of the best video to learn about this topic Thank you sir 💖
@Mustafajwd
@Mustafajwd 3 года назад
Hi .. if we have this equation Y(t)=SUM(x(t)gama(t-Ts) how can we find the linearty thank you
@sandipansarkar3717
@sandipansarkar3717 2 года назад
H.W :- Stable System
@anjanikumari-vj1cc
@anjanikumari-vj1cc 2 года назад
Thnx alots
@anjanikumari-vj1cc
@anjanikumari-vj1cc 2 года назад
Nice
@srajansoni27
@srajansoni27 6 лет назад
Sir it is stable system because amplitude of sine wave is unity positive
@belwizdadimed3967
@belwizdadimed3967 5 лет назад
having finite value a each instant t doesn't mean bounded! should be there exit a constant C independent of t such as abs(x(t)) < C for any t.
@bingisainath5731
@bingisainath5731 4 года назад
Answer is Stable sir
@ranjanagupta9487
@ranjanagupta9487 2 года назад
the system is stable.
@rahulgarg5144
@rahulgarg5144 6 лет назад
Sir , shouldn’t we check all inputs ? Like if we do example 1 by taking sin(t) .. it will come as stable ..
@nischalrawat1468
@nischalrawat1468 9 месяцев назад
What about x(t)=π/2 then it's stable or not..?
@sushmanaik2228
@sushmanaik2228 3 года назад
Sir,What happens when a system becomes unstable? unstable system has infinite magnitude is it possible to transmit a signal with infinite magnitude??Do we use in realtime applications?
@arunpanse9853
@arunpanse9853 3 года назад
The answer of home work is stable
@siddharthmishra8012
@siddharthmishra8012 3 года назад
When it was mentioned that input should be bounded then why you are taking input till infinity in example 1
@damnfarooq
@damnfarooq 4 года назад
Nice explanation but i think you can't prove a system to be stable by just considering a single test bounded signal.... Forexample y(t) = x(t)/(1-x(t)) For all DC values the system is stable, except of x(t) = 1
@dynamicguy4969
@dynamicguy4969 4 года назад
i dont listen to our dept sir 's i understand your videos only
@jahansaid6382
@jahansaid6382 5 лет назад
Stable system, because the amplitude of sin function is between 1 and -1.
@sounavapal762
@sounavapal762 6 лет назад
Is there digital Fourier transform videos on neso academy?? I badly need those
@hichambenbrika4211
@hichambenbrika4211 Год назад
in your first lecture you said that DC is not signal ! but now you said it is bounded signal how ?
@mbhagyarajareddy4639
@mbhagyarajareddy4639 6 лет назад
It's stable
@naginaprasad8279
@naginaprasad8279 6 лет назад
stable system
@dynamicguy4969
@dynamicguy4969 4 года назад
sir can u explain feedback and non feedback systems
@shubhamgoyal1988
@shubhamgoyal1988 5 лет назад
Sir please tell me one question What is the range of 'a' for which h(t)=(exp(-a+3))*u(-t) Is BIBO static LTI system
@agnivachatterjee220
@agnivachatterjee220 5 лет назад
a
@iamhere2125
@iamhere2125 7 лет назад
nyce sir...
@vandanamehra5524
@vandanamehra5524 4 года назад
In ex y(t)= t x(t) The system is stable or unstable If we take unit step signal we get ramp signal but what about the delta signal.
@avantuksingh
@avantuksingh 4 года назад
Delta signal is not bounded, its amplitude is infinite at time equals zero.
@aishwaryabuyya3793
@aishwaryabuyya3793 3 года назад
stable
@adrenochromeaddict4232
@adrenochromeaddict4232 7 месяцев назад
a bounded system is bounded??!!?! damn what a phenomenal definition!!
@ertugrul98835
@ertugrul98835 5 лет назад
Stable system
@amar_1234_paul
@amar_1234_paul Год назад
In the given homework problem, according to me, it should be an unstable system coz x(t) may have any value from -infinity and +infinity .. although sin() will have values from -1 to +1 but x(t) may have any value
@shailendhera6512
@shailendhera6512 Год назад
No the input x(t) should be a bounded signal so the answer would be bounded system
@rockyyadav5112
@rockyyadav5112 5 лет назад
The given network is stable
@AjayPatel-ii8ud
@AjayPatel-ii8ud 7 лет назад
In stable system u use one word DC value what is it... fullform..
@udayasrimaddi2907
@udayasrimaddi2907 5 лет назад
It is ntg but supply voltage that is constant
@youtubeonly7271
@youtubeonly7271 5 лет назад
direct current, meaning supply voltage is constant as a certain voltage
@venkateshpolisetty5624
@venkateshpolisetty5624 6 лет назад
y(t)=sinx(t) is a stable system as i considered x(t)=impulse function
@Adhi0209
@Adhi0209 3 года назад
Is impulse function bounded input?
@Rohitkumar-IIMRaipur
@Rohitkumar-IIMRaipur 6 лет назад
What about discrete signals
@raviraj582
@raviraj582 7 лет назад
stable system...
@salemellid96
@salemellid96 6 лет назад
What about this system y(t)=x(t)/(1+x(t-1)) is stable on non-stable
@mathematicsphobia5495
@mathematicsphobia5495 6 лет назад
stable
@kito323
@kito323 6 лет назад
I believe it is unstable system because if x(t) is such bounded system that x(t-1) would equal -1 then we have a devision by 0 and that would reach to infinity or minus infinity and it means that the output is unbounded.
@rajeevsingh2694
@rajeevsingh2694 7 лет назад
Sir Delta(t) is bounded input or not.
@raviraj582
@raviraj582 7 лет назад
rajeev singh &(t) is unbounded.... &(n) is bounded...
@nidhimanhar9878
@nidhimanhar9878 7 лет назад
how delta[n] is unbounded
@raviraj582
@raviraj582 7 лет назад
delta(n) is bounded...
@raviraj582
@raviraj582 7 лет назад
delta(t) is unbounded in amplitude..
@raviraj582
@raviraj582 7 лет назад
NIDHI MANHAR it is bounded...
@prabhakardas4261
@prabhakardas4261 7 лет назад
stable system as because the sint amplitude lies between 1 and -1...
@siddharthsaurabh8157
@siddharthsaurabh8157 6 лет назад
Bounded op so stable system
@uniqueversalunicorn2412
@uniqueversalunicorn2412 6 лет назад
Bounded output for the bounded input
@dynamicguy4969
@dynamicguy4969 4 года назад
and its a stable system
@kishanbaranwal8331
@kishanbaranwal8331 4 года назад
Why ramp signal is unbounded because we put infinity make me angry where is BIBO gone
@Monugupta-qd9ud
@Monugupta-qd9ud 5 лет назад
Bounded i/p Lo,ydi bounded o/p aya to system stable hoga ......
@rajankarn90
@rajankarn90 7 лет назад
stable sys
@shivaniyadav5243
@shivaniyadav5243 5 лет назад
what if x(t) is a ramp?
@abhijeetpathak2032
@abhijeetpathak2032 5 лет назад
To check stability u have to provide bounded i/p. But ramp is not bounded. So u can't put x(t) as ramp.
@shivaniyadav5243
@shivaniyadav5243 5 лет назад
@@abhijeetpathak2032 got it
@aqqq4097
@aqqq4097 Год назад
Isnt u(t) step signal and r(t) ramp?
@ujjawal1366
@ujjawal1366 4 года назад
*stable system *Put X(t)=4 ... 4×(+-1)=+-4... BIBO satisfied
@rajankarn90
@rajankarn90 7 лет назад
u dont reply to us sir. please repond
@holly1190
@holly1190 5 лет назад
Can any one recommend me a book to follow for signal and system
@devanshgarg6298
@devanshgarg6298 5 лет назад
You can follow A. Anand kumar
@Michelle-rb6nl
@Michelle-rb6nl 4 года назад
Simon Haykin and Barry Van Veen
@acfake7858
@acfake7858 5 лет назад
show i cant find the unstable system example , someone help me please ...
@pokpikchan
@pokpikchan 5 лет назад
Tan (x) is an unstable system I guess
@sushmanaik2228
@sushmanaik2228 3 года назад
Ramp signal
@riseabovehate9476
@riseabovehate9476 6 лет назад
Guys need help. Tell me whether log|x(n)| is a stable or unstable system
@refreshment567
@refreshment567 6 лет назад
unstable
@hafizhamza3747
@hafizhamza3747 6 лет назад
its a constant signal so stable
@riseabovehate9476
@riseabovehate9476 6 лет назад
Hafiz Hamza how is logarithmic function a constant function ?
@ravikant1845
@ravikant1845 6 лет назад
A
@ashishtiwary7947
@ashishtiwary7947 Год назад
stable systummm
@arpitmishra4319
@arpitmishra4319 5 лет назад
please reply quickly
@Miss___Imam
@Miss___Imam 2 года назад
I don’t get it
@bayzidhossain7422
@bayzidhossain7422 Год назад
stable system because the value the sint is bounded (-1 to 1) Thanks for your great explanation
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