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Error constants and K range for K/[s(s+1)(s²+2s+2)]

Short answer

Type 1 system: Kp=∞, Kv=K/2, Ka=0. For r(t)=10+50t, ess=100/K. ess<0.5 demands K>200, but Routh stability limits K to 0<K<2.22, so no K satisfies the error condition with a stable system.

The open loop transfer function of a unity feedback system is given by G(s) = K/[s(s+1)(s²+2s+2)]. Calculate the static error constants for this system and find the range of K if static error is less than 0.5 for input r(t) = 10 + 50t.

G(s)G(s) has one pole at the origin, so the system is Type 1: the step error is zero and the ramp error is 1/Kv1/K_v.

Static error constants

Kp=lim⁡s→0G(s)=∞Kv=lim⁡s→0sG(s)=K(1)(2)=K2Ka=lim⁡s→0s2G(s)=0\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{K}{(1)(2)} = \frac{K}{2} \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}

Steady state error for r(t)=10+50tr(t) = 10 + 50t

ess=101+Kp+50Kv=0+50K/2=100Ke_{ss} = \frac{10}{1 + K_p} + \frac{50}{K_v} = 0 + \frac{50}{K/2} = \frac{100}{K}

Condition ess<0.5e_{ss} < 0.5

100K<0.5  ⇒  K>200\frac{100}{K} < 0.5 \;\Rightarrow\; K > 200

Stability check (needed for the result to be valid)

Characteristic equation: s4+3s3+4s2+2s+K=0s^4 + 3s^3 + 4s^2 + 2s + K = 0.

Row
s4s^414KK
s3s^332
s2s^210/310/3KK
s1s^12−0.9K2 - 0.9K
s0s^0KK

For stability: 2−0.9K>02 - 0.9K > 0 and K>0K > 0, so 0<K<2.220 < K < 2.22.

Answer: the error condition needs K>200K > 200, but the system is stable only for 0<K<2.220 < K < 2.22. The two ranges do not overlap, so no value of K meets ess<0.5e_{ss} < 0.5 with a stable system. A compensator (for example lag compensation) is needed to raise KvK_v without causing instability.

✓ Answer checked by Claude Opus 5.5 (high effort)

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