Static Error coefficients Kp, Kv, and Ka for an open loop function given below, with unit feedback respectively will be:

26. Static Error coefficients K_p, K_v, and K_a for an open loop function given below, with unit feedback respectively will be:

G(s) = 10 / [s(s+2)(s+5)]

1. K_p = ∞, K_v = 1, K_a = 0
2. K_p = 0, K_v = 1, K_a = ∞
3. K_p = 0, K_v = ∞, K_a = 1
4. K_p = 1, K_v = ∞, K_a = 0

Answer: Option 1

Solution

Step 1: Identify System Type

The Type of a system is determined by the number of poles at the origin (the power of s in the denominator). Since there is one s¹ term outside the brackets, this is a Type 1 System.

Step 2: Calculate Position Error Coefficient (K_p)

The formula for K_p (for step inputs) is:

K_p = lim_{s → 0} G(s) = lim_{s → 0} [ 10 / s(s+2)(s+5) ]

As s approaches 0, the denominator becomes 0. Therefore:
K_p = ∞

Step 3: Calculate Velocity Error Coefficient (K_v)

The formula for K_v (for ramp inputs) is:

K_v = lim_{s → 0} s ⋅ G(s) = lim_{s → 0} [ s ⋅ 10 / s(s+2)(s+5) ]

The s in the numerator and denominator cancel each other out:

K_v = lim_{s → 0} [ 10 / (s+2)(s+5) ] = 10 / (2 × 5) = 10 / 10
K_v = 1

Step 4: Calculate Acceleration Error Coefficient (K_a)

The formula for K_a (for parabolic inputs) is:

K_a = lim_{s → 0} s² ⋅ G(s) = lim_{s → 0} [ s² ⋅ 10 / s(s+2)(s+5) ]

One s cancels, leaving one s in the numerator:

K_a = lim_{s → 0} [ 10s / (s+2)(s+5) ] = 0 / 10
K_a = 0

Final Answer: K_p = ∞, K_v = 1, K_a = 0.
This matches Option 1.

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