Electronics and Communication Engineering - Signals and Systems - Discussion
Discussion Forum : Signals and Systems - Section 1 (Q.No. 1)
1.
What about the stability of system in 
Answer: Option
Explanation:
ROC include the unit circle hence it is stable.
Discussion:
49 comments Page 5 of 5.
Arun Kumar Gupta said:
1 decade ago
Without ROC one can't say that system is stable or not.
Shamsheer said:
1 decade ago
ROC is |Z|>2 and .4<|Z|<4/3.
Why everyone are concerned with pole Z=2?
Why everyone are concerned with pole Z=2?
Manan said:
1 decade ago
According to me system is unstable.
Shivanand jaiswal said:
1 decade ago
roc of given problem z>2 and z>0.4 so intersection of this will be z>2 that is outside the unit circle tending to infinite this is an casual system but not stable.
Sathiya (ponds) said:
1 decade ago
The pole Z=2 lies outside the unitcircle. The ROC is not mentioned in this question so the system is unstable.
Ajay kumar sharma said:
1 decade ago
Pole doesn't lies b/w the circle so the system is not stable.
Raam said:
1 decade ago
One of the pole z=2 lies outside the unit circle, so the system is unstable.
S.p.Bharath said:
1 decade ago
The pole Z=2 lies outside the unity circle. Then how can we say it as stable ?
Iman said:
1 decade ago
I think the ROC would have to be mentioned in the question to answer the question. Like 0.4 < |z| < 2.
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