System Behavior and stability
The behavior and stability of a system is obtained from analyzing the system characteristic equation roots:
r2 + (B/M)r + (K/M) = 0
The roots to the 2nd order equation are
r1 = (-B/M +
sqrt((B/M)2-4K/M))/2
r2
= (-B/M - sqrt((B/M)2-4K/M))/2
| System Stability | |
| Stable | All roots are on the left-hand side of the complex plane |
| Unstable | Real roots or roots in the right half-plane or double-root at origin of complex plane |
| Marginally Stable |
One or more distinct roots on imaginary axis and all other roots in left half-plane |
Case 1
The graph below describes the system behavior when the applied force
is the unit step function. (car displacement on a bumpy road). Comparing
the graphs we can see that a change of value on the system friction element
B causes the system to take longer to reach its equilibrium value
System elements values:
M = 2560 lbs
K = 2000
B = 720
r1=-0.14+j0.87
r2=-0.14-j0.87
Graph 1
System elements values:
M = 2560 lbs
K = 2000
B = 80
Graph 2
Case
2
The graph below describes the system behavior in time
when the applied force is a periodic function. (very bumpy road)
System elements values:
M = 2560 lbs
K = 2000
B = 720
Graph 3
Previous topics:
- Case study Analysis
- Definition of system equations
- Definition of system state variables, output and input-output equations
- System solution
- Laplace Transforms
All graphs are included in the above pages, to see graphs ONLY (no solutions) go here.