According to the textbook, the default meaning of “error” for non-unity feedback is the so-called true error E(=R-C). Therefore, if you want to apply Table 8.1 in this case, you will have to convert the non-unity feedback system into its equivalent unity feedback form, then determine the system type and error coefficients based upon G’(s).
But we have learned that a more appropriate definition of error should be Ea(=R-HC), called actuating error in the textbook. To calculate actuating error, there is no need to distinguish between unity feedback and non-unity feedback cases, because system type and error coefficients for Ea are always based upon the open-loop TF, which takes form of G(s) in unity feedback systems, and GH(s) in non-unity feedback systems.
8.3 Solution (For “true error”:)
Method 1: Find the open-loop TF of the equivalent unity-feedback system. G′(s)=
G10(1+s)
=3 (type 0 system) 2
1+G(H−1)s+3s−3s+5
10=25
⇒Kp=limG′(s)=
s→0
11
⇒ess==
1+Kp3
Method 2: Calculate directly using equation 8.59 and FVT (Final Value Theorem). E1+G(H−1)s3+3s2−3s+5==3
s+3s2+7s+15R1+GH
32
s+3s−3s+511
⇒ess=limsE(s)=s⋅3⋅=
s→0s+3s2+7s+15s3(For “actuating error”:) Method 1: The open-loop TF is GH(s)=
10
(type 0 system)
(s2+2s+5)(1+s)
10=25
⇒Kp=limGH(s)=
s→0
11
⇒ess==
1+Kp3
Method 2: Calculate directly using equation 8.59 and FVT (Final Value Theorem). E1s3+3s2+7s+5==3
R1+GHs+3s2+7s+15
32
s+3s+7s+511
⇒ess=limsE(s)=s⋅3⋅=
s→0s+3s2+7s+15s3
8.4 Solution:
If H(s)=1 (unity feedback, for which true error equals actuating error):
Kv=limsGH(s)=lims⋅
s→0
s→0
5
⋅1=5
s(1+s)
44
⇒ess==
Kv5
(For “true error”:) a.
G′(s)=
G5
=2 (type 0 system)
1+G(H−1)s+s+15
⇒ess=∞
5G
(type 0 system)=2
1+G(H−1)s+6s−5
⇒ess=∞
b. G′(s)=
c.
G′(s)=
5(1+s)G
= (type 1 system)2
1+G(H−1)s(s+2s−4)
5
⇒Kv=limsG′(s)=−
s→04416
⇒ess==−=−3.2
5Kv
d. G′(s)=
5(s+2)G
=3 (type 0 system)2
1+G(H−1)s+3s+2s−5
⇒ess=∞
(For “actuating error”:)
a.
GH(s)=
20
(type 1 system)s(1+s)
s→0
⇒Kv=limsGH(s)=20⇒ess=b.
5
(type 0 system)(1+s)
⇒ess=∞GH(s)=c. GH(s)=
41=205
5
(type 1 system)2
s(1+s)
4
⇒Kv=limsGH(s)=5⇒ess=
s→05
d.
5
(type 1 system)
s(s+2)
58
⇒Kv=limsGH(s)=⇒ess=
s→025GH(s)=
8.8 Solution:
Steady-state error due to a unit step change in speed command R (assuming T =0):
K1⋅
CK
=s+1Js=2
K1R1+⋅⋅1Js+Js+Ks+1JsECJs2+Js
⇒=1−=2
RRJs+Js+K
Js2+Js1
⇒ess=limsE(s)=lims⋅2⋅=0
s→0s→0Js+Js+Ks
In fact, this conclusion can be drawn directly since the system is of type 1
(for input R).
Steady-state error due to a unit step change in disturbance torque T (assuming R =0):
1
Cs+1Js==2 T1+1⋅1⋅KJs+Js+K
Jss+1
In this case, the desired output is 0 (which means perfect disturbance rejection).
∴ E=0−C=−C
s+111
⇒ess=−limsC(s)=−lims⋅2⋅=−=−1
s→0s→0Js+Js+KsK
According to the theorem of superposition (叠加原理), we know that the total
steady-sate error of speed is:
11
ess=0−=−=−1
KK
Clearly, to reduce change of speed as required, K should be larger than 10.
【Note: The integrator in feedforward path is located behind the point where the disturbance is added, thus can not eliminate steady-state error due to step-type disturbance.】
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