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.
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.
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 .
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.
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
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.
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. :-)
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.
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
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?
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
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
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.
