# Problem 12.1: The set point of the control system shown in the figure below is given a step change of 0.

1unit in set point. a) Develop an overall transfer function
C (s) R( s)

b) Determine the order of the overall system. If it is second order system, determine the values of τ and ζ and state whether the system is underdamped, critically damped or overdamped. c) Calculate the ultimate value of the response C (t ) d) Calculate the offset e) Calculate the maximum value of C (t ) and the time at which it occurs f) Find out the period of oscillation Draw sketch of C(t) as a function of time.

Solution:
5 C (s) ( s + 1)(2 s + 1) = 5 R ( s ) 1 + 1.6 ⋅ ( s + 1)(2 s + 1) 1.6 ⋅

8 8 2 C (s) 8 ( s + 1)(2s + 1) s + 3s + 1 = = = 2 2 2 8 R(s) 1 + 2 s + 3s + 1 + 8 2 s + 3s + 9 ( s + 1)(2s + 1) 2 s 2 + 3s + 1
C (s) 8 = R ( s ) 2 s 2 + 3s + 9

[solution to the part (a)]

(1)

For second order system, we know
Y ( s) K = 2 2 X ( s ) τ s + 2ζτs + 1

(2)

Eq. (1) can also be written as

1

8 = 0. 1. it may be shown that Eq. [solution to part (b)] 0. it may shown that s Giving a step function of R ( s ) = C ( s) = 0. the system is underdamped system.state value − ultimate value or C (∞ ) = Offset = ( Final steady .1 in Eq. Now using final value theorem.8  0.C ( s) 8 8/9 8/9 8/9 = 2 = 2 = = 2 3 9 2 1 R( s ) 2 s + 3s + 9 2 s + 3s + 9 s2 + s + s2 + s + 1 9 9 9 9 3 9 (3) Comparing Eq. and C (t ) →C (∞) .state value) − ultimate value 2 . lim [ f (t )] = [ sf ( s )] t →∞ s →0 lim sC ( s ) = 0. 3.0889 [solution to part (c)]  2 = s → 0  2 s + 3s + 9  9 Offset = Final steady . and we may write K = 8 9 2 9 τ= 2 3 τ2 = ⇒ [solution to part (b)] (4) 2ζτ = 1 3 1 1 3 1 ⋅ = = 0.0 .1 8 ⋅ 2 s 2s + 3s + 9 For the ultimate value.8 2 s + 3s + 9 2 lim  0 . 1 is a transfer function of a second order system [solution to part (b)].state value − initial steady .354 3 2 2 2 2 ⇒ ζ= ⋅ [solution to part (b)] (5) As ζ <1. t → ∞ . 1 and Eq.

Offset = 0.1 − 0.0111 [solution to part (d)] −ζ ⋅t    1−ζ 2  Y (t ) = K ⋅ A ⋅ 1 − e τ  cos   τ      1−ζ 2 ζ t+ sin    τ 1−ζ 2    t    Time to first peak: t p = Overshoot: Decay OS = π ⋅τ 1 −ζ 2      C − 2 ⋅π ⋅ζ 2 ratio: DR = A = (OS ) = exp  1 −ζ 2   −π ⋅ ζ A = exp  1 −ζ 2 B  Time period: T = 2 ⋅ π ⋅τ     1 −ζ 2 3 .0889 = 0.

(τ m s + 1) = K c ⋅ (1 + τ D s) ⋅ (τ m s + 1) τ 1τ m s + τ 1s + τ m s + 1 + K c + K c ⋅ τ D s 2 = K c ⋅ (1 + τ D s) ⋅ (τ m s + 1) τ 1τ m s + (τ 1 + τ m + K c ⋅ τ D ) s + ( K c + 1) 2 1 ( K c + 1) C ( s) = τ 1τ m 2 (τ 1 + τ m + K c ⋅ τ D ) R( s) s + s +1 ( K c + 1) ( K c + 1) K c ⋅ (1 + τ D s) ⋅ (τ m s + 1) ⋅ This is resembling to a second order system. therefore 4 .Solution: a)  1  K c ⋅ (1 + τ D s) ⋅  τ s + 1  C ( s)  1  = R( s )  1   1  1 + K c ⋅ (1 + τ D s) ⋅  τ s +1 ⋅ τ s + 1   1   m   1  K c ⋅ (1 + τ D s ) ⋅  τ s + 1  K c ⋅ (1 + τ D s ) ⋅ (τ m s + 1)  1  = = (τ 1s + 1) ⋅ (τ m s + 1) + K c ⋅ (1 + τ D s ) (τ 1s + 1) ⋅ (τ m s + 1) + K c ⋅ (1 + τ D s ) (τ 1 s + 1).

7 = 1 1  70 + 3K c  ⋅ ⋅  2 1  60  ⋅ ( K c + 1) 6 5 .7 = 1 1 3   10 ⋅ ⋅ 1 + + Kc ⋅  2 1 60   60 ⋅ ( K c + 1) 6 0.7 = 1 ⋅ 2 1 1⋅ 10 ⋅ ( K c + 1) 60 ⋅ (1 + 10 ) 60 0.τ2 = τ 1τ m τ 1τ m ⇒τ = Kc + 1 Kc + 1 2ζτ = (τ 1 + τ m + K c ⋅ τ D ) ( K c + 1) 1 1 (τ + τ + K c ⋅ τ D ) ζ = ⋅ ⋅ 1 m 2 τ ( K c + 1) 1 ζ = ⋅ 2 (τ + τ + K c ⋅ τ D ) 1 ⋅ 1 m ( K c + 1) τ 1τ m Kc + 1 ζ = 1 1 ⋅ ⋅ (τ 1 + τ m + K c ⋅ τ D ) 2 τ 1τ m ⋅ ( K c + 1) b) When τ D = 0 s 0 .167 b) When τ D = 3 s 0.7 = 1 1  60 + 10  ⋅ ⋅ = 2 1  60  ⋅ ( K c + 1) 6 35 / 60 1 ⋅ ( K c + 1) 6 K c = 3.

6 . we can use Solver® (in MicroSoft Excel). K c = 5.5*(1/SQRT((1/6)*(D5+1)))*((70+3*D5)/60))-0. Click Solver.255 Write the following formula in D7 =(0.7 = 0  60  Here.1 1 ⋅ 2 1 ⋅ ( K c + 1) 6  70 + 3K c  ⋅  − 0.7 Go to Tools then Solver.

7 . The solution is there. By Changing Cells: D5 Click Solve button. Select Equal To: value of: 0.In the Solver window below. do the following: Set Target Cell: D7.

167 = 0. Now using final value theorem.24 1 + 3.760 and Offset = 1 − 0. t → ∞ .840 and Offset = 1 − 0. lim [ f (t )] = [ sf ( s )] t →∞ s →0 lim C (∞) = Kc 1 + Kc When K c = 3.167 When K c = 5.255 Please compare the above two offsets and also find periods and compare.255 = 0. Period is greater when τ D = 0 .c)  1  K c ⋅ (1 + τ D s ) ⋅  τ s + 1  C ( s)  1  = R( s )  1   1   1 + K c ⋅ (1 + τ D s ) ⋅  τ s +1 ⋅    1  τ ms + 1 For a unit step change: R(s) = 1 s  1  K c ⋅ (1 + τ D s ) ⋅  τ s +1  1  1  C (s) = ⋅ s  1   1   1 + K c ⋅ (1 + τ D s ) ⋅  τ s +1 ⋅    1  τ ms + 1 For the ultimate value. Stability of a Control Loop 8 .255 C (∞) = 5.160 1 + 5.760 = 0. and C (t ) →C (∞) .840 = 0.167 C (∞) = 3.

Consider the following block diagram Taking τ1 =1 . 9 .e. R ( s ) = . and τ3 = . find the response C(t) for a step change in the set 1 s 1 2 1 3 point i. τ 2 = .

Or A process is said to be unstable if its output becomes larger and larger (either positively or negatively) as time increases. Also a saturated or near saturated value of a variable may affect the quality of a product. working at saturation is of course not a satisfactory control. Step function is a bounded input while the function f (t ) = t is unbounded. Therefore physical systems saturate and do not go beyond bounds. Saturation If we keep on opening the valve. 10 .Definition of stability A control system is said to be stable when the output response is bounded for all bounded inputs. A bounded input is a function of time that always falls within certain bounds during the course of time. However. With some exothermic reactions. a point will reach where further pressure to the diaphragm will not change the flowrate through the valve. the saturation point may be highly unsatisfactory and may damage the equipment and in some cases the human life. such type of limitation is called “saturation”.

G1 ( s ) = K c G2 ( s ) = 1 (τ1s +1)(τ 2 s +1) H (s) = 1 τ 3 s +1 Overall transfer function for set point change C ( s) G1 ( s ) ⋅ G2 ( s ) = R ( s ) 1 + G1 ( s ) ⋅ G2 ( s ) ⋅ H ( s ) Overall transfer function for load point change 11 .Figure 1 Stable and unstable responses Characteristic equation Consider the following block diagram Where.

R ( s ) = . 12 . when equal to zero is called the characteristic equation. or to the right of the imaginary axis. A linear control system is unstable if any roots of its characteristic equation are On.C (s) G2 ( s) = U ( s ) 1 + G1 ( s ) ⋅ G2 ( s ) ⋅ H ( s ) The denominator is same in both cases. Roots of the characteristic equation For unit step change in set point. and r3 are the roots of the following characteristic equation. it may shown that 1 s C (s) = K c ⋅ (τ 3 s + 1) s τ 1 ⋅ τ 2 ⋅ τ 3 ⋅ s + (τ 1 ⋅ τ 2 + τ 1 ⋅ τ 3 + τ 2 ⋅ τ 3 ) s 2 + (τ 1 + τ 2 + τ 3 ) s + (1 + K c ) ( 3 ) C ( s) = K c ⋅ (τ 3 s + 1) /(τ1 ⋅τ 2 ⋅τ 3 ) s( s − r1 )( s − r2 )( s − r3 ) Where r1. The term G1 ( s ) ⋅ G2 ( s ) ⋅ H ( s ) in the denominator is called the open loop transfer function. τ1 ⋅τ 2 ⋅τ 3 ⋅ s 3 + (τ1 ⋅τ 2 + τ1 ⋅τ 3 + τ 2 ⋅τ 3 ) s 2 + (τ 1 + τ 2 + τ 3 ) s + (1 + K c ) = 0 Criteria of stability and roots of a characteristic equation The feedback control system is stable if and only if all roots of the characteristic equation are negative or have negative real parts. r2. Otherwise system is stable.

Figure: Stability criteria and roots of the characteristic equation 13 .

Example 13.5s + 1) =0 s (2 s + 1) s 2 + 3s + 5 = 0 One can solve the quadratic equation by hand as well. However.1 For the figure below. solving by Matlab Since real parts of s1 and s2 are negative. the system is stable. 14 . a control system has the transfer functions Characteristic equation: 1+ 10 ⋅ (0.

If any coefficient is negative. the system may or may not be stable. Procedure for examining the roots o Write the characteristic equation in the form a0 s n + a1s n −1 + a2 s n −2 + ⋅ ⋅ ⋅ + an = 0 Where a0 is positive (if a0 is originally negative. o The coefficients of above equation should be positive. o Arrange the coefficients into the first two rows of the Routh array as shown in the next slide. the system is definitely unstable and Routh test is not needed. their proximity to the imaginary axis. Therefore it can not be used to test a control system with a transportation lag. The test is limited to the systems that have polynomial characteristic equations. 15 . the system is stable). o If all the coefficients are positive. It gives no information about the actual location of the roots and. in particular. both sides are multiplied by –1).Routh test for stability It is an algebraic method for determining how many roots of the characteristic equation have positive real pars (No roots with positive real parts.

then all the elements of the nth row will vanish and none of the elements of the preceding row will vanish. equidistant from the origin. o If some of the elements in the first column are negative. o If one pair of roots is on the imaginary axis. Example 13. the number of roots with positive real part (in the right half plane) is equal to the number of sign changes in the first column. respectively.2 Given the characteristic equation s 4 + 3s 3 + 5s 2 + 4s + 2 = 0 . The location of the pair of imaginary roots can be found by solving the equation Cs 2 + D = 0 where the coefficients C and D are the elements of the array in the ( n−1)st row as read from left to right. 16 . determine the stability by the Routh criterion.o For a stable system all elements of the first column of the Routh array should be positive and nonzero. and all other roots are in the left half-plane.

5s + 1)  ⋅ s + 3 = 0          3  1 ⋅ 1 + Kc ⋅   =0 s s +3    ( s +1)( +1)    2      3  2 1 + Kc ⋅   ( s + 1)( s + 2)  ⋅ s + 3 = 0     17 .5s + 1) 2 = = R( s)    3    1    3  1 + Kc ⋅   ( s + 1)(0. Is the system stable for (a) Kc = 9.1 Write the characteristic equation and construct the Routh array for the control system shown in the figure below.5s + 1)  ⋅ s + 3 1  1 + Kc ⋅  ⋅     s s +3   ( s + 1)( + 1)     2  2 2 ⋅ Kc 2 ⋅ Kc ( s + 1)( s + 2) ( s + 1)( s + 2) = = = 6 ⋅ K ( s + 1 )( s + 2 )( s + 3) + 6 ⋅ K c    3  1+ 2 c  1 + Kc ⋅  ⋅    ( s + 1)( s + 2)  s + 3 ( s + 1)( s + 2)( s + 3) ( s + 3)     Kc ⋅ C ( s) 2 ⋅ K c ⋅ ( s + 3) = R ( s ) ( s +1)( s + 2)( s + 3) + 6 ⋅ K c Characteristic equation:    3  1 1 + Kc ⋅   ( s + 1)(0. and (c) Kc = 12? Overall transfer function: 1 Kc ⋅ 1 s Kc ⋅ ( s + 1)( + 1) C ( s) ( s + 1)(0.Problem 13. (b) Kc = 11.5.

5 10 − Kc = 10 − 9.0.5 For K c =11 10 − Kc = 10 − 11 = − 1 For K c =12 10 − Kc = 10 − 12 = − 2 System is stable only for Kc = 2.5 = 0 . 18 .1+ 6 ⋅ Kc =0 ( s + 1)( s + 2)(s + 3) ( s + 1)( s + 2)( s + 3) + 6 ⋅ K c = 0 ( s 2 + 3s + 2)( s + 3) + 6 ⋅ K c = 0 s 3 + 6 s 2 + 11s + 6 + 6 ⋅ K c = 0 s 3 + 6 s 2 + 11s + 6 ⋅ (1 + K c ) = 0 Routh array: Row 1 2 3 1 6 10 − K c 11 6 ⋅ (1 + K c ) Note the calculations of for b1: b1 = 6 × 11 − 1 × 6 ⋅ (1 + K c ) 11 − 1 − K c = = 10 − K c 6 1 For K c = 9.

10  3   1 + 2 ⋅ 1 +  ⋅ 2 ⋅  2 =0  s   2 s + 4s + 10  2 s 3 + 4 s 2 + 50 s + 120 = 0 Row 1 2 3 2 4 − 10 50 120 System is unstable. 19 .0 .For K c = 2.
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