Digital Control Module 4 Lecture Module 4: ime Repone of dicrete time ytem Lecture Note ime Repone of dicrete time ytem Abolute tability i a baic requirement of all control ytem. Apart from that, good relative tability and teady tate accuracy are alo required in any control ytem, whether continuou time or dicrete time. ranient repone correpond to the ytem cloed loop pole and teady tate repone correpond to the excitation pole or pole of the input function.. ranient repone pecification In many practical control ytem, the deired performance characteritic are pecified in term of time domain quantitie. Unit tep input i mot commonly ued in analyi of a ytem ince it i eay to generate and repreent a ufficiently dratic change thu providing ueful information on both tranient and teady tate repone. he tranient repone of a ytem depend on the initial condition. It i a common practice to conider the ytem initially at ret. Conider the digital control ytem hown in Figure. r(t) e(t) Digital + Hold Plant Controller c(k) Figure : Block Diagram of a cloed loop digital ytem Similar to the continuou time cae, tranient repone of a digital control ytem can alo be characterized by the following.. Rie time (t r ): ime required for the unit tep repone to rie from 0% to 00% of it final value in cae of underdamped ytem or 0% to 90% of it final value in cae of overdamped ytem. 2. Delay time (t d ): ime required for the the unit tep repone to reach 50% of it final value. I. Kar
Digital Control Module 4 Lecture 3. Peak time (t p ): ime at which maximum peak occur. 4. Peak overhoot (M p ): he difference between the maximum peak and the teady tate value of the unit tep repone. 5. Settling time (t ): ime required for the unit tep repone to reach and tay within 2% or 5% of it teady tate value. However ince the output repone i dicrete the calculated performance meaure may be lightly different from the actual value. Figure 2 illutrate thi. he output ha a maximum value c max wherea the maximum value of the dicrete output i c max which i alway le than or equal to c max. If the ampling period i mall enough compared to the ocillation of the repone then thi difference will be mall otherwie c max may be completely erroneou. c max c max.0 Figure 2: Unit tep repone of a dicrete time ytem t.2 Steady tate error he teady tate performance of a table control ytem i meaured by the teady error due to tep, ramp or parabolic input depending on the ytem type. Conider the dicrete time ytem a hown in Figure 3. r(t) e(t) e (t) + _ e G p() H() Figure 3: Block Diagram 2 From Figure 2, we can write E() = R() H()C() I. Kar 2
Digital Control Module 4 Lecture We will conider the teady tate error at the ampling intant. From final value theorem lim e(k) = lim ( z )E(z) ] k z ] G(z) = ( z Gp () )Z ] GH(z) = ( z Gp ()H() )Z C(z) G(z) = R(z) +GH(z) Again, E(z) = R(z) GH(z)E(z) or, E(z) = +GH(z) R(z) ] e = lim ( z ) z +GH(z) R(z) he teady tate error of a ytem with feedback thu depend on the input ignal R(z) and the loop tranfer function GH(z)..2. ype-0 ytem and poition error contant Sytem having a finite nonzero teady tate error with a zero order polynomial input (tep input) are called ype-0 ytem. he poition error contant for a ytem i defined for a tep input. r(t) = u (t) unit tep input R(z) = z e = lim z +GH(z) = +K p where K p = lim z GH(z) i known a the poition error contant..2.2 ype- ytem and velocity error contant Sytem having a finite nonzero teady tate error with a firt order polynomial input (ramp input) are called ype- ytem. he velocity error contant for a ytem i defined for a ramp input. r(t) = u r (t) unit ramp z R(z) = = Z (z ) 2 ( Z ) 2 e = lim = z (z )GH(z) K v where K v = lim (z )GH(z)] i known a the velocity error contant. z I. Kar 3
Digital Control Module 4 Lecture.2.3 ype-2 ytem and acceleration error contant Sytem having a finite nonzero teady tate error with a econd order polynomial input (parabolic input) are called ype-2 ytem. he acceleration error contant for a ytem i defined for a parabolic input. R(z) = 2 z(z +) = 2 (+z )z 2(z ) 3 2( z ) 3 e = 2 2 lim (z +) = z (z ) 2 +GH(z)] (z ) lim 2 z GH(z) = K a 2 (z ) 2 where K a = lim GH(z) i known a the acceleration error contant. z 2 able how the teady tate error for different type of ytem for different input. able : Steady tate error Sytem Step input Ramp input Parabolic input ype-0 +K p ype- 0 K v ype-2 0 0 K a Example : Calculate the teady tate error for unit tep, unit ramp and unit parabolic input for the ytem hown in Figure 4. R() E() E () C() + ZOH 000 + 0 500 Figure 4: Block Diagram for Example Solution: he open loop tranfer function i: G() = C() E () = G ho()g p () = e 000/0 ( + 500/0) I. Kar 4
Digital Control Module 4 Lecture aking Z-tranform G(z) = 2( z ) Z 0 ] 2 500 + 0 500( + 5000) ] = 2( z z ) (z ) 0z 2 500(z ) + 0z 500(z e 50 ) = ] (500 0+0e 50 )z (500 +0)e 50 +0 250 (z )(z e 50 ) Steady tate error for tep input = +K p where K p = lim z G(z) =. e tep = 0. Steady tate error for ramp input = K v where K v = lim z (z )G(z)] = 2. e ramp = 0.5. Steady tate error for parabolic input = K a where K a = 2 lim z (z ) 2 G(z)] = 0. e para =. I. Kar 5