An impulse vector, also known as Kang vector, is a mathematical tool used to graphically design and analyze input shapers that can suppress residual vibration. The impulse vector can be applied to both undamped and underdamped systems, as well as to both positive and negative impulses in a unified manner. The impulse vector makes it easy to obtain impulse time and magnitude of the input shaper graphically.[1] A vector concept for an input shaper was first introduced by W. Singhose[2] for undamped systems with positive impulses. Building on this idea, C.-G. Kang introduced the impulse vector (or Kang vector) to generalize Singhose's idea to undamped and underdamped systems with positive and negative impulses.
For a vibratory second-order system
2 | |
\omega | |
n |
/(s2+2\zeta\omegan+
2 | |
\omega | |
n |
)
\omegan
\zeta
Ii
\thetai
Ii
Ai\delta(t-ti)
i=1,2,...,n
Ii=Ai
\zeta\omeganti | |
e |
\thetai=\omegadti
where
Ai
ti
\omegad
\omegan\sqrt{1-\zeta2}
Ai>0
Ai<0
In this definition, the magnitude
Ii
Ai
ti
Ai
\thetai
\delta(t-ti)
t=ti
\omegan
\zeta
Consider two impulse vectors
I1
I2
I1
I1(>0)
\theta1
A1>0
I2
I2=-I1
\theta2=\pi+\theta1
A2<0
I1
I2
t2
I1
I2
The magnitude of the impulse vector determines the magnitude of the impulse, and the angle of the impulse vector determines the time location of the impulse. One rotation,
2\pi
If it is an undamped system (
\zeta=0
Ii=Ai
\thetai=\omeganti
The impulse response of a second-order system corresponding to the resultant of two impulse vectors is same as the time response of the system with a two-impulse input corresponding to two impulse vectors after the final impulse time regardless of whether the system is undamped or underdamped. □
If the resultant of impulse vectors is zero, the time response of a second-order system for the input of the impulse sequence corresponding to the impulse vectors becomes zero also after the final impulse time regardless of whether the system is undamped or underdamped. □
4\pi2/(s2+0.4\pis+4\pi2)
\omegan=2\pi
\zeta=0.1
I1
I2
IR1
IR2
IR1
AR1=IR1/
\zeta\omegantR1 | |
e |
tR1=\thetaR1/\omegad
IR2
AR2=IR2/
\zeta\omegantR2 | |
e |
tR2=\thetaR2/\omegad
The resultants
IR1
IR2
Rx=I1+I2\cos\theta2, Ry=I2\sin\theta2
IR1=-
2 | |
\sqrt{R | |
x |
+
2}, | |
R | |
y |
\thetaR1=\pi+\tan-1(Ry/Rx)
IR2=
2 | |
\sqrt{R | |
x |
+
2}, | |
R | |
y |
\thetaR2=\tan-1(Ry/Rx)
Note that
-\pi/2<\tan-1(a)<\pi/2
yR1
yR2
IR1
IR2
y1+y2
Now, place an impulse vector
I3
I1+I2
I3
I3=
2 | |
\sqrt{R | |
x |
+
2}, | |
R | |
y |
\theta3=\pi+\tan-1(Ry/Rx)
When the impulse sequence corresponding to three impulse vectors
I1,I2
I3
t3
' | |
I | |
3 |
I3
I3
Using impulse vectors, we can redesign known input shapers [3] such as zero vibration (ZV), zero vibration and derivative (ZVD), and ZVDn shapers. The ZV shaper is composed of two impulse vectors, in which the first impulse vector is located at 0°, and the second impulse vector with the same magnitude is located at 180° for
I1+I2=0
\theta1=0, \theta2=\pi
I1=I2=I
Therefore,
t1=0, t2=\pi/\omegad
Since
A1+A2=1
A1=I1, A2=I2/
\zeta\omegant2 | |
e |
I1+
I2 | ||||
|
=I+
I | |
K |
=1,
\zeta\pi/\sqrt{1-\zeta2 | |
K=e |
Therefoere,
I=K/(K+1)
Thus, the ZV shaper
A1\delta(t)+A2\delta(t-t2)
\begin{bmatrix} ti\\ Ai\end{bmatrix} =\begin{bmatrix} 0,&\pi/\omegad\\ K/(K+1),&1/(K+1) \end{bmatrix}
The ZVD shaper is composed of three impulse vectors, in which the first impulse vector is located at 0 rad, the second vector at
\pi
2\pi
I1:I2:I3=1:2:1
I1+I2+I3=0
\theta1=0, \theta2=\pi, \theta3=2\pi
Therefore,
t1=0, t2=\pi/\omegad, t3=2\pi/\omegad
Also from the impulse vector diagram,
I1=I3=I, I2=2I
Since
A1+A2+A3=1
I1+
I2 | ||||
|
+
I3 | ||||
|
=I+
2I | |
K |
+
I | |
K2 |
=1,
\zeta\pi/\sqrt{1-\zeta2 | |
K=e |
Therefore,
I=K2/(K+1)2
Thus, the ZVD shaper
A1\delta(t)+A2\delta(t-t2)+A3\delta(t-t3)
\begin{bmatrix} ti\\ Ai\end{bmatrix} =\begin{bmatrix} 0,&\pi/\omegad,&2\pi/
2/(K+1) | |
\omega | |
d\\ K |
2,&2K/(K+1)2,&1/(K+1)2 \end{bmatrix}
The ZVD2 shaper is composed of four impulse vectors, in which the first impulse vector is located at 0 rad, the second vector at
\pi
2\pi
3\pi
I1:I2:I3:I4=1:3:3:1
I1+I2+I3+I4=0
\theta1=0, \theta2=\pi, \theta3=2\pi, \theta4=3\pi
Therefore,
t1=0, t2=\pi/\omegad, t3=2\pi/\omegad, t4=3\pi/\omegad
Also, from the impulse vector diagram,
I1=I4=I, I2=I3=3I
Since
A1+A2+A3+A4=1
I1+
I2 | ||||
|
+
I3 | ||||
|
+
I4 | ||||
|
=I+
3I | |
K |
+
3I | |
K2 |
+
I | |
K3 |
=1,
\zeta\pi/\sqrt{1-\zeta2 | |
K=e |
Therefore,
I=K3/(K+1)3
Thus, the ZVD2 shaper
A1\delta(t)+A2\delta(t-t2)+A3\delta(t-t3)+A4\delta(t-t4)
\begin{bmatrix} ti\\ Ai\end{bmatrix} =\begin{bmatrix} 0,&\pi/\omegad,&2\pi/\omegad,&3\pi/\omegad\\ K3/(K+1)3,&3K2/(K+1)3,&3K/(K+1)3,&1/(K+1)3 \end{bmatrix}
Similarly, the ZVD3 shaper with five impulse vectors can be obtained, in which the first vector is located at 0 rad, the second vector at
\pi
2\pi
3\pi
4\pi
I1:I2:I3:I4:I5=1:4:6:4:1
(i-1)\pi
I1:I2:I3: … :In+2=\tbinom{n+1}{0}:\tbinom{n+1}{1}:\tbinom{n+1}{2}: … :\tbinom{n+1}{n+1}
\tbinom{m}{k}
Now, consider equal shaping-time and magnitudes (ETM) shapers, with the same magnitude of impulse vectors and with the same angle between impulse vectors. The ETMn shaper satisfies the conditions
\theta1=0, \theta2=
2\pi | |
n-1 |
, … ,\thetan-1=
(n-2)2\pi | |
n-1 |
I2=I3= … =In-1=I1+In, In=mI1 (m>0)
n | |
\sum | |
i=1 |
Ai=1
Thus, the resultant of the impulse vectors of the ETMn shaper becomes always zero for all
n\ge2
\begin{bmatrix} ti\\ Ai\end{bmatrix} =\begin{bmatrix} 0,&(2\pi/3)/\omegad,&(4\pi/3)/\omegad,&2\pi/\omegad\\ I/(1+m),&I/K2/3,&I/K4/3,&mI/[(1+m)K2] \end{bmatrix}
I= | (1+m)K2 |
K2+(1+m)(K4/3+K2/3)+m |
,
\zeta\pi/\sqrt{1-\zeta2 | |
K=e |
The ETM5 shaper with five impulse vectors is obtained similarly as
\begin{bmatrix} ti\\ Ai\end{bmatrix} =\begin{bmatrix} 0,&0.5\pi/\omegad,&\pi/\omegad,&1.5\pi/\omegad,&2\pi/\omegad\\ I/(1+m),&I/K1/2,&I/K,&I/K3/2,&mI/[(1+m)K2] \end{bmatrix}
I= | (1+m)K2 |
K2+(1+m)(K3/2+K+K1/2)+m |
,
\zeta\pi/\sqrt{1-\zeta2 | |
K=e |
In the same way, the ETMn shaper with
n\ge6
m=1
Moreover, impulse vectors can be applied to design input shapers with negative impulses. Consider a negative equal-magnitude (NMe) shaper, in which the magnitudes of three impulse vectors are
I1=I(>0), I2=-I, I3=I
\theta1=0, \theta2=\pi/3, \theta3=2\pi/3
t2,t3
t2=(\pi/3)/\omegad, t3=(2\pi/3)/\omegad
A1=I,A2=
\zeta\omegant2 | |
-I/e |
, A3=
\zeta\omegant3 | |
I/e |
A1+A2+A3=1
The resulting NMe shaper
A1\delta(t)+A2\delta(t-t2)+A3\delta(t-t3)
\begin{bmatrix} ti\\ Ai\end{bmatrix} =\begin{bmatrix} 0,&(\pi/3)/\omegad,&(2\pi/3)/\omegad\\ I,&-I/K1/3,&I/K2/3\end{bmatrix}
I=K/(K-K2/3+K1/3),
\zeta\pi/\sqrt{1-\zeta2 | |
K=e |
The NMe shaper has faster rise time than the ZVD shaper, but it is more sensitive to modeling error than the ZVD shaper. Note that the NMe shaper is the same with the UM shaper[5] if the system is undamped (
\zeta=0
Figure (a) in the right side shows a typical block diagram of an input-shaping control system, and figure (b) shows residual vibration suppressions in unit-step responses by ZV, ZVD, ETM4 and NMe shapers.
Refer to the reference for sensitivity curves of the above input shapers, which represent the robustness to modeling errors in
\omegan
\zeta