One closed loop, five transfer functions. Each one answers a different question: where does the signal go in, and where do you put the meter? Example plant: the DC motor from the first practical exam, .
4.1
Absolute vs relative stability estabilidad absoluta · relativa
Absolute
Yes or no: does the response settle? Yes when every closed-loop pole has Re < 0.
Relative
How much: how far is the loop from losing stability? A glass in the middle of a table and one 1 cm from the edge both “stand”, only one is safe.
Symptoms
Small relative stability shows as big overshoot, long ringing, poles high up on the root locus (small ζ). Margins and the peak Ms put one number on it (Step 05).
4.2
Input and meter funciones de sensibilidad
Takeaway
Same loop, different question. Pick where the signal enters (green dot) and where you measure (orange square). Click each function.
One rule
Numerator = blocks on the straight path from input to meter. Denominator = always 1 + L, because the signal always circles the same loop. So all of them share the same poles.
Names
Plain “sensibilidad” almost always means S. Si = sensibilidad de entrada (FR-05 p. 5). The subscript says where: i = plant input, u = control signal u.
Quick check
You want to know how much the motor voltage u jumps when you step the reference r. Which function?
4.3
Where 1 + L comes from derivación de S
Recipe
One equation per block, substitute until only the wanted signal and the input remain, the wanted signal appears on both sides (it came back round the loop), pull it out: that bracket is 1 + L.
4.4
Steady-state error and system type error en régimen permanente · tipo
When
Unity feedback, step on the reference, stable loop, error after a long time. “Long time” = put s = 0.
Rule
— if L has a pure 1/s, L(0) = ∞ and ep = 0.
Type ≠ order
Type (tipo) = number of pure integrators 1/s in L. Order (orden) = number of poles.
L
Order
Type
Step error ep
Ramp error ev
oven 6K/((s+1)(s+2)(s+3))
3
0
1/(1+K)
∞
motor + P: 500K/(s(s+500))
2
1
0
1/K
motor + PI: 500K(s+100)/(s²(s+500))
3
2
0
0
S = 1/(1+L);
isstable(S) % 1 = stable. Check this first!
ep = dcgain(S) % error after a step r = 1
ev = dcgain(S/s) % error after a ramp r = t
Part
What it does
Steady-state error
P
pushes in proportion to the error
can leave an error (it stops pushing when e = 0)
I
adds the error up over time, keeps pushing while e ≠ 0
removes it
D
reacts to how fast the error changes
none (derivative of a constant = 0); D adds damping, not accuracy
4.5
A constant load: P vs PI rechazo de perturbaciones
Function
Input d, meter y →
Result
d = 5 on the motor. P leaves an offset 5/K. PI brings y back to 0 for every K.
Why
What counts is where the integrator sits. The motor's own integrator is after d: it integrates the load, it doesn't help. The PI integrator is before d: it keeps raising the voltage until u = −5 and e = 0.
P control: offset stays
PI control: back to zero
K = 50K = 125K = 1000
Quick checks
L = 500K(s+100)/(s²(s+500)). What is its type?
P control only, d = 5. Smallest K for an offset below 0.01?
FR-03 p. 5–11 · FR-05 p. 5, p. 11 (tipo ante perturbaciones) · examen práctico 1, apartado 2.2