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A sectional method of analysis of feedback systems containing deadtime
Chemical Engineering Science, 1968, Vol. 23, pp. 1504-1506.
Pergamon Press.
Printed in Great Britain.
A sectional method of analysis of feedback systems containing deadtime 16 April 1968)
(Received
A NUMBER of writers have considered the difficulties involved in feedback control of systems which contain deadtime (transportation lag); and the difficulties involved in analysing deadtime feedback systems, in general, because of the infinite number of roots of the characteristic equation[l]. The response of such a system is easily determined using an analogue computer with a deadtime simulator. However, it does not appear to have been noted that there is also an easy analytical method, which can also contribute to an intuitive understanding of the effect of the deadtime.
sum at each section, increases from section to section. However, quite often it is only the early part of the response which is of interest. If the later part of the response is required, then this is often obtained most easily from the characteristic equation, since by this time it is only the response term corresponding to the smallest root which is significant. It is easily seen that a similar technique applies for the case of positive feedback. In this case, all the terms in the series expansion in Eq. (2) have positive sign. If there is no deadtime in the loop, then the advantage of this series expansion is lost, since all the terms in Eq. (2) then start at f = 0. 1 Consider a negative feedback system where
Example
I
I
Fig. 1. Regulator containing deadtime. Consider a regulation system, as shown in Fig. 1, with deadtime in the feedback loop. (The method applies equally well to servo systems.) The transfer function is y G, --=___ X l+Gle-r8
(1)
where T is the deadtime in the loop, and G1 contains no deadtime. This can be expanded as: $ = G2( 1+ G1e-r8)-’ = Gz- GzG,eeT*+ G,G,2e-eTa- GzG,3e-3T*+. . .
(2)
This corresponds to the proportional control of a system described by a first order plus deadtime transfer function, a transfer function which is commonly used to approximate plant dynamics. Let
C&=1.
l+Ts’
and this makes the system the same as one which has been analysed by Harriott [ 11. The step response up to t = 3T is shown in Fig. 2 together with the individual terms in the sectional expansion. The calculation of this took about 15 min, compared to “an hour or two” using the finite difference method employed by Han-ion. Also shown in Fig. 2 is the response term corresponding to the smallest root of the characteristic equation. The calculation of this term is rather laborious in this case[2], since it involves finding a complex root. The expression obtained for the response is:
This means that, for a given input signal, the response consists of: The response of GZ Minus the response of G2Gl, starting at r = T, Plus the response of G,G,‘, starting at t = 2T,
~~=0.454+0.432e+“~‘~~‘~sin
(1_86(r/T)-1.14)+.
..
(3)
It is seen from Fig. 2 that all response terms, other than the two shown in this expression, have decayed after about two deadtimes.
and so on. Thus, the response divides up naturally into time sections of length T; and, at least to obtain the response over the first few sections, it is only necessary to plot a few ordinary response curves, which can be combined graphically. The existence of an infinite number of roots of the characteristic equation of the system is of no concern. As the number of time sections included in the calculation increases, the labour involved increases. This is because the order of the transfer function, whose response is added to the
Example 2 Now consider the positive feedback case with
G = 0.5 e-Ts 1 1 + Ts and
1504
Gz=&
, I
/1’
I’
1’
I
/’
.’
t/r
1
z
Fig. 2. Step response; Example 1. complete response terms in the Sectional Series dominant term from the characteristic steady-state response.
This gives a transfer function which is the same as considered by Sinclair[3] for a stirred vessel, and of the type considered by Haddad, et al. [4] for multistage systems with recycle. In this case, the step response is as shown in Fig. 3 (which is plotted in the same way as Fig. 2). It is seen that the terms in the sectional expansion, starting at t = T and later, start with zero slope, since they are of order greater than unity. Hence, there is no discontinuity in the slope of the complete response curve at the end of each section; which is contrary to the conclusions of Haddad, et al. In this case, the calculation of the dominant response term
/ ,*’ J
equation
is much easier, since it corresponds to a real root of the characteristic equation. The response is obtained as: v = , - 0.942 e-O.BlS(NT) + ... (4) and the response rapidly converges onto these first two terms (so that, on a plot of log (I -y) vs. r, it approaches a straight line). School of Chemical Engineering University of New South Wales Australia
1505
D. C. DIXON
.. .... ------
Fig. 3. Step Response; Example 2. complete response terms in the Sectional Series dominant term from the characteristic equation steady-state response
REFERENCES HARRIOTT P., Process Conrrol, pp. 82-85, 187-191. McGraw-Hill 1964. COHEN G. H. and COON G. A., Trans. Am Sot. Me&. Engrs 1953 75 827. SINCLAIR C. G.,A.I.Ch.E.J/7 709 1961. HADDAD A., WOLF D. and RESNICK W., Can. J. Chem. Engng 1964 42 2 16.
1506
Pergamon Press.
Printed in Great Britain.
A sectional method of analysis of feedback systems containing deadtime 16 April 1968)
(Received
A NUMBER of writers have considered the difficulties involved in feedback control of systems which contain deadtime (transportation lag); and the difficulties involved in analysing deadtime feedback systems, in general, because of the infinite number of roots of the characteristic equation[l]. The response of such a system is easily determined using an analogue computer with a deadtime simulator. However, it does not appear to have been noted that there is also an easy analytical method, which can also contribute to an intuitive understanding of the effect of the deadtime.
sum at each section, increases from section to section. However, quite often it is only the early part of the response which is of interest. If the later part of the response is required, then this is often obtained most easily from the characteristic equation, since by this time it is only the response term corresponding to the smallest root which is significant. It is easily seen that a similar technique applies for the case of positive feedback. In this case, all the terms in the series expansion in Eq. (2) have positive sign. If there is no deadtime in the loop, then the advantage of this series expansion is lost, since all the terms in Eq. (2) then start at f = 0. 1 Consider a negative feedback system where
Example
I
I
Fig. 1. Regulator containing deadtime. Consider a regulation system, as shown in Fig. 1, with deadtime in the feedback loop. (The method applies equally well to servo systems.) The transfer function is y G, --=___ X l+Gle-r8
(1)
where T is the deadtime in the loop, and G1 contains no deadtime. This can be expanded as: $ = G2( 1+ G1e-r8)-’ = Gz- GzG,eeT*+ G,G,2e-eTa- GzG,3e-3T*+. . .
(2)
This corresponds to the proportional control of a system described by a first order plus deadtime transfer function, a transfer function which is commonly used to approximate plant dynamics. Let
C&=1.
l+Ts’
and this makes the system the same as one which has been analysed by Harriott [ 11. The step response up to t = 3T is shown in Fig. 2 together with the individual terms in the sectional expansion. The calculation of this took about 15 min, compared to “an hour or two” using the finite difference method employed by Han-ion. Also shown in Fig. 2 is the response term corresponding to the smallest root of the characteristic equation. The calculation of this term is rather laborious in this case[2], since it involves finding a complex root. The expression obtained for the response is:
This means that, for a given input signal, the response consists of: The response of GZ Minus the response of G2Gl, starting at r = T, Plus the response of G,G,‘, starting at t = 2T,
~~=0.454+0.432e+“~‘~~‘~sin
(1_86(r/T)-1.14)+.
..
(3)
It is seen from Fig. 2 that all response terms, other than the two shown in this expression, have decayed after about two deadtimes.
and so on. Thus, the response divides up naturally into time sections of length T; and, at least to obtain the response over the first few sections, it is only necessary to plot a few ordinary response curves, which can be combined graphically. The existence of an infinite number of roots of the characteristic equation of the system is of no concern. As the number of time sections included in the calculation increases, the labour involved increases. This is because the order of the transfer function, whose response is added to the
Example 2 Now consider the positive feedback case with
G = 0.5 e-Ts 1 1 + Ts and
1504
Gz=&
, I
/1’
I’
1’
I
/’
.’
t/r
1
z
Fig. 2. Step response; Example 1. complete response terms in the Sectional Series dominant term from the characteristic steady-state response.
This gives a transfer function which is the same as considered by Sinclair[3] for a stirred vessel, and of the type considered by Haddad, et al. [4] for multistage systems with recycle. In this case, the step response is as shown in Fig. 3 (which is plotted in the same way as Fig. 2). It is seen that the terms in the sectional expansion, starting at t = T and later, start with zero slope, since they are of order greater than unity. Hence, there is no discontinuity in the slope of the complete response curve at the end of each section; which is contrary to the conclusions of Haddad, et al. In this case, the calculation of the dominant response term
/ ,*’ J
equation
is much easier, since it corresponds to a real root of the characteristic equation. The response is obtained as: v = , - 0.942 e-O.BlS(NT) + ... (4) and the response rapidly converges onto these first two terms (so that, on a plot of log (I -y) vs. r, it approaches a straight line). School of Chemical Engineering University of New South Wales Australia
1505
D. C. DIXON
.. .... ------
Fig. 3. Step Response; Example 2. complete response terms in the Sectional Series dominant term from the characteristic equation steady-state response
REFERENCES HARRIOTT P., Process Conrrol, pp. 82-85, 187-191. McGraw-Hill 1964. COHEN G. H. and COON G. A., Trans. Am Sot. Me&. Engrs 1953 75 827. SINCLAIR C. G.,A.I.Ch.E.J/7 709 1961. HADDAD A., WOLF D. and RESNICK W., Can. J. Chem. Engng 1964 42 2 16.
1506