# 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 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.

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.

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.

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.

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.

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.

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

Hence unit circle lie between 0.4 and 2.

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