Question
Download Solution PDFMatch List 1 and List 2
List 1 Location of roots of characteristic equation |
List 2 System characteristics |
||
A |
(-1 + j), (-1 - j) |
D |
मामूली तौर पर स्थिर |
B |
(-2 + j), (-2 - j), (2), (-2j) |
E |
अस्थिर |
C |
(-j), (j), (1), (-1) |
F |
स्थिर |
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Characteristics of Roots:
- Stable System: All roots lie in the left half of the s-plane (negative real parts). The system will return to equilibrium after a disturbance.
- Marginally Stable System: Roots lie on the imaginary axis, but no root has a positive real part. The system neither grows unbounded nor settles completely but oscillates perpetually.
- Unstable System: At least one root lies in the right half of the s-plane (positive real part) or there are repeated roots on the imaginary axis. The system grows unbounded in response to disturbances.
Analysis of Given Lists:
The characteristic equations and their respective roots provide insight into the system's behavior:
- List 1: Represents the location of roots of the characteristic equation.
- List 2: Represents the corresponding system characteristics.
Matching List 1 and List 2:
Option (A): (-1 + j), (-1 - j): The roots lie in the left half of the s-plane (negative real part), indicating a marginally stable system. Hence, A matches with D.
Option (B): (-2 + j), (-2 - j), (2), (-2j): Here, one root lies in the right half of the s-plane (positive real part, i.e., 2). This makes the system unstable. Hence, B matches with E.
Option (C): (-j), (j), (1), (-1): The root (1) lies in the right half of the s-plane (positive real part), making the system unstable. Hence, C matches with E. However, if (-j) and (j) were the only roots on the imaginary axis, the system would be marginally stable. But the presence of (1) shifts the system to instability.
The correct match from the options provided is: Option 3:
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