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
system is stable
unstable
stable at 0.4
cant say
Answer: Option
Explanation:

ROC include the unit circle hence it is stable.

Discussion:
49 comments Page 1 of 5.

Abdul Wahab said:   2 months ago
The System is unstable,

Since, the pole at z = 0.4 lies inside the unit circle and the other pole at z = 2 lies outside the unit circle.

For a system to be stable, all the poles must be inside the unit circle in the z-plane.

Therefore, the system is unstable.
(4)

Iram said:   6 years ago
H(z) = (-3/4z)/(z-0.4) + (16/5z)/(z-2) this is final transfer function. by sketching its ROC it can be seen that pole at 2 Roc lie outside unity circle while pole at 0.4 Roc includes unity circle by expanding in an outward direction.

So, the system is stable.

Shivanand jaiswal said:   10 years 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.

Moni said:   3 years ago
Find ROC by using partial differentiating it gives two poles both are causal so the greatest pole is roc that means z>2, should lie outside the circle. It is unstable.

Devyanee chandra said:   8 years ago
I think the question is incomplete .

Without ROC, it is difficult to define or at least the signal type i.e. causal/non-causal /anti-causal should have been mentioned.

Hareesh said:   8 years ago
Here, we need to specify the region of convergence (ROC) (ROC:0. 4, <Z<2), then only it's possible to say, it is stable or not.

Other wise its difficult.

Aman said:   9 years ago
For stability we check mod of z which IC greater than 2 and 0.4. Hence we use upper limit that is z=2 which is outside uni circle, so system unstable.

Usama said:   3 years ago
H(z) is actually z transform. Now finding the ROC:
The roots are 0.2, 2.
which makes a circle that encloses the unit circle, so the system is stable.
(2)

Vikas said:   9 years ago
The condition of stable system is that the pole of a transfer function must be lie in the left half of s plane. So the system is unstable.

It's me . . . said:   7 years ago
System is stable.

For the system to be stable, the unit circle should lie between ROC.

Hence unit circle lie between 0.4 and 2.


Post your comments here:

Your comments will be displayed after verification.