Page 1 Digital Control Module 4 Lecture 2 Module 4: Time Response of discrete time systems Lecture Note 2 1 Prototype second order system The study of a second order system is important because many higher order system can be approximated by a second order model if the higher order poles are located so that their contri- butions to transient response are negligible. A standard second order continuous time system is shown in Figure 1. We can write + _ R(s) C(s) E(s) ? 2 n s(s+2??n) Figure 1: Block Diagram of a second order continuous time system G(s) = ? 2 n s(s+2?? n ) Closed loop: G c (s) = ? 2 n s 2 +2?? n s+? 2 n where, ? = damping ratio ? n = natural undamped frequency Roots: -?? n ±j? n p 1-? 2 1.1 Comparison between continuous time and discrete time systems The simpli?ed block diagram of a space vehicle control system is shown in Figure 2. The objective is to control the attitude in one dimension, say in pitch. For simplicity vehicle body is considered as a rigid body. Position c(t) and velocity v(t) are fedback. The open loop transfer function can be calculated I. Kar 1 Page 2 Digital Control Module 4 Lecture 2 Module 4: Time Response of discrete time systems Lecture Note 2 1 Prototype second order system The study of a second order system is important because many higher order system can be approximated by a second order model if the higher order poles are located so that their contri- butions to transient response are negligible. A standard second order continuous time system is shown in Figure 1. We can write + _ R(s) C(s) E(s) ? 2 n s(s+2??n) Figure 1: Block Diagram of a second order continuous time system G(s) = ? 2 n s(s+2?? n ) Closed loop: G c (s) = ? 2 n s 2 +2?? n s+? 2 n where, ? = damping ratio ? n = natural undamped frequency Roots: -?? n ±j? n p 1-? 2 1.1 Comparison between continuous time and discrete time systems The simpli?ed block diagram of a space vehicle control system is shown in Figure 2. The objective is to control the attitude in one dimension, say in pitch. For simplicity vehicle body is considered as a rigid body. Position c(t) and velocity v(t) are fedback. The open loop transfer function can be calculated I. Kar 1 Digital Control Module 4 Lecture 2 + - + _ C(s) 1 Jvs K R Kp E(s) R(s) K 1 s Figure 2: Space vehicle attitude control as G(s) = C(s) E(s) = KK P × 1 K R +J v s × 1 s = KK P s(J v s+K R ) Closed loop transfer function is G c s = G(s) 1+G(s) = KK p J v s 2 +K R s+KK P K P = Position Sensor gain = 1.65×10 6 K R = Rate sensor gain = 3.71×10 5 K = Ampli?er gain which is a variable J v = Moment of inertia = 41822 With the above parameters, G(s) = 39.45K s(s+8.87) C(s) R(s) = 39.45K s 2 +8.87s+39.45K Characteristics equation ? s 2 +8.87s+39.45K = 0 ? n = v 39.45K rad/sec, ? = 8.87 2? n Sincethesystemisof2 nd order,thecontinuoustimesystemwillalwaysbestableifK P , K R , K, J v are all positive. Now, consider that the continuous data system is subject to sampled data control as shown in Figure 3. I. Kar 2 Page 3 Digital Control Module 4 Lecture 2 Module 4: Time Response of discrete time systems Lecture Note 2 1 Prototype second order system The study of a second order system is important because many higher order system can be approximated by a second order model if the higher order poles are located so that their contri- butions to transient response are negligible. A standard second order continuous time system is shown in Figure 1. We can write + _ R(s) C(s) E(s) ? 2 n s(s+2??n) Figure 1: Block Diagram of a second order continuous time system G(s) = ? 2 n s(s+2?? n ) Closed loop: G c (s) = ? 2 n s 2 +2?? n s+? 2 n where, ? = damping ratio ? n = natural undamped frequency Roots: -?? n ±j? n p 1-? 2 1.1 Comparison between continuous time and discrete time systems The simpli?ed block diagram of a space vehicle control system is shown in Figure 2. The objective is to control the attitude in one dimension, say in pitch. For simplicity vehicle body is considered as a rigid body. Position c(t) and velocity v(t) are fedback. The open loop transfer function can be calculated I. Kar 1 Digital Control Module 4 Lecture 2 + - + _ C(s) 1 Jvs K R Kp E(s) R(s) K 1 s Figure 2: Space vehicle attitude control as G(s) = C(s) E(s) = KK P × 1 K R +J v s × 1 s = KK P s(J v s+K R ) Closed loop transfer function is G c s = G(s) 1+G(s) = KK p J v s 2 +K R s+KK P K P = Position Sensor gain = 1.65×10 6 K R = Rate sensor gain = 3.71×10 5 K = Ampli?er gain which is a variable J v = Moment of inertia = 41822 With the above parameters, G(s) = 39.45K s(s+8.87) C(s) R(s) = 39.45K s 2 +8.87s+39.45K Characteristics equation ? s 2 +8.87s+39.45K = 0 ? n = v 39.45K rad/sec, ? = 8.87 2? n Sincethesystemisof2 nd order,thecontinuoustimesystemwillalwaysbestableifK P , K R , K, J v are all positive. Now, consider that the continuous data system is subject to sampled data control as shown in Figure 3. I. Kar 2 Digital Control Module 4 Lecture 2 + - C(s) + - ZOH K R E(s) R(s) KK P E * (s) 1 Jvs 1 s Figure 3: Discrete representation of space vehicle attitude control For comparison purpose, we assume that the system parameters are same as that of the con- tinuous data system. G(s) = C(s) E * (s) = G ho (s)G p (s) = 1-e -Ts s . KKp/J v s(s+K R /J v ) G(z) = (1-z -1 ) KKp K R Z 1 s 2 - J v K R s + J v K R (s+K R /J v ) = (1-z -1 ) KKp K R Tz (z-1) 2 - J v z K R (z-1) + J v z K R (z-e -K R T/Jv ) = KKp K 2 R (TK R -J v +J v e -K R T/Jv )z-(TK R +J v )e -K R T/Jv +J v (z-1)(z-e -K R T/Jv ) Characteristic equation of the closed loop system: z 2 +a 1 z+a 0 = 0, where a 1 = f 1 (K,Kp,K R ,Jv) a 0 = f 0 (K,Kp,K R ,Jv) Substituting the known parameters: a 1 = 0.000012K(3.71×10 5 T -41822+41822e -8.87T )-1-e -8.87T a 0 = e -8.87T +0.000012K 41822-(3.71×10 5 T +41822)e -8.87T For stability (1) |a 0 | < 1 (2) P(1) = 1+a 1 +a 0 > 0 = 1-e -8.87T > 0 always satis?ed since T is positive (3) P(-1) = 1-a 1 +a 0 > 0 Choice of K and T: If we plot K versus T then according to conditions (1) and (3) the stable region is shown in Figure 4. Pink region represents the situation when condition (1) is satis?ed but the (3) is not. Red region depicts the situation when condition (1) is satis- ?ed, not the (3). Yellow is the stable region where both the conditions are satis?ed. If we I. Kar 3 Page 4 Digital Control Module 4 Lecture 2 Module 4: Time Response of discrete time systems Lecture Note 2 1 Prototype second order system The study of a second order system is important because many higher order system can be approximated by a second order model if the higher order poles are located so that their contri- butions to transient response are negligible. A standard second order continuous time system is shown in Figure 1. We can write + _ R(s) C(s) E(s) ? 2 n s(s+2??n) Figure 1: Block Diagram of a second order continuous time system G(s) = ? 2 n s(s+2?? n ) Closed loop: G c (s) = ? 2 n s 2 +2?? n s+? 2 n where, ? = damping ratio ? n = natural undamped frequency Roots: -?? n ±j? n p 1-? 2 1.1 Comparison between continuous time and discrete time systems The simpli?ed block diagram of a space vehicle control system is shown in Figure 2. The objective is to control the attitude in one dimension, say in pitch. For simplicity vehicle body is considered as a rigid body. Position c(t) and velocity v(t) are fedback. The open loop transfer function can be calculated I. Kar 1 Digital Control Module 4 Lecture 2 + - + _ C(s) 1 Jvs K R Kp E(s) R(s) K 1 s Figure 2: Space vehicle attitude control as G(s) = C(s) E(s) = KK P × 1 K R +J v s × 1 s = KK P s(J v s+K R ) Closed loop transfer function is G c s = G(s) 1+G(s) = KK p J v s 2 +K R s+KK P K P = Position Sensor gain = 1.65×10 6 K R = Rate sensor gain = 3.71×10 5 K = Ampli?er gain which is a variable J v = Moment of inertia = 41822 With the above parameters, G(s) = 39.45K s(s+8.87) C(s) R(s) = 39.45K s 2 +8.87s+39.45K Characteristics equation ? s 2 +8.87s+39.45K = 0 ? n = v 39.45K rad/sec, ? = 8.87 2? n Sincethesystemisof2 nd order,thecontinuoustimesystemwillalwaysbestableifK P , K R , K, J v are all positive. Now, consider that the continuous data system is subject to sampled data control as shown in Figure 3. I. Kar 2 Digital Control Module 4 Lecture 2 + - C(s) + - ZOH K R E(s) R(s) KK P E * (s) 1 Jvs 1 s Figure 3: Discrete representation of space vehicle attitude control For comparison purpose, we assume that the system parameters are same as that of the con- tinuous data system. G(s) = C(s) E * (s) = G ho (s)G p (s) = 1-e -Ts s . KKp/J v s(s+K R /J v ) G(z) = (1-z -1 ) KKp K R Z 1 s 2 - J v K R s + J v K R (s+K R /J v ) = (1-z -1 ) KKp K R Tz (z-1) 2 - J v z K R (z-1) + J v z K R (z-e -K R T/Jv ) = KKp K 2 R (TK R -J v +J v e -K R T/Jv )z-(TK R +J v )e -K R T/Jv +J v (z-1)(z-e -K R T/Jv ) Characteristic equation of the closed loop system: z 2 +a 1 z+a 0 = 0, where a 1 = f 1 (K,Kp,K R ,Jv) a 0 = f 0 (K,Kp,K R ,Jv) Substituting the known parameters: a 1 = 0.000012K(3.71×10 5 T -41822+41822e -8.87T )-1-e -8.87T a 0 = e -8.87T +0.000012K 41822-(3.71×10 5 T +41822)e -8.87T For stability (1) |a 0 | < 1 (2) P(1) = 1+a 1 +a 0 > 0 = 1-e -8.87T > 0 always satis?ed since T is positive (3) P(-1) = 1-a 1 +a 0 > 0 Choice of K and T: If we plot K versus T then according to conditions (1) and (3) the stable region is shown in Figure 4. Pink region represents the situation when condition (1) is satis?ed but the (3) is not. Red region depicts the situation when condition (1) is satis- ?ed, not the (3). Yellow is the stable region where both the conditions are satis?ed. If we I. Kar 3 Digital Control Module 4 Lecture 2 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 5 10 15 20 25 30 35 40 45 T K T=0.08 Figure 4: K vs. T for space vehicle attitude control system want a comparatively large T, such as 0.2, the gain K is limited by the range K < 5. Similarly ifwewantacomparativelyhighgainsuchas25, wehavetogoforT assmallas0.08orevenless. From studies of continuous time systems it is well known that increasing the value of K gen- erally reduces the damping ratio, increases peak overshoot, bandwidth and reduces the steady state error if it is ?nite and nonzero. 2 Correlation between time response and root locations in s-plane and z-pane The mapping between s-plane and z-plane was discussed earlier. For continuous time systems, thecorrelationbetweenrootlocationins-planeandtimeresponseiswellestablishedandknown. • A root in negative real axis of s-plane produces an output exponentially decaying with time. • Complex conjugate pole pairs in negative s-plane produce damped oscillations. • Imaginary axis conjugate poles produce undamped oscillations. • Complex conjugate pole pairs in positive s-plane produce growing oscillations. Digital control systems should be given special attention due to sampling operation. For ex- ample, if the sampling theorem is not satis?ed, the folding e?ect can entirely change the true response of the system. I. Kar 4 Page 5 Digital Control Module 4 Lecture 2 Module 4: Time Response of discrete time systems Lecture Note 2 1 Prototype second order system The study of a second order system is important because many higher order system can be approximated by a second order model if the higher order poles are located so that their contri- butions to transient response are negligible. A standard second order continuous time system is shown in Figure 1. We can write + _ R(s) C(s) E(s) ? 2 n s(s+2??n) Figure 1: Block Diagram of a second order continuous time system G(s) = ? 2 n s(s+2?? n ) Closed loop: G c (s) = ? 2 n s 2 +2?? n s+? 2 n where, ? = damping ratio ? n = natural undamped frequency Roots: -?? n ±j? n p 1-? 2 1.1 Comparison between continuous time and discrete time systems The simpli?ed block diagram of a space vehicle control system is shown in Figure 2. The objective is to control the attitude in one dimension, say in pitch. For simplicity vehicle body is considered as a rigid body. Position c(t) and velocity v(t) are fedback. The open loop transfer function can be calculated I. Kar 1 Digital Control Module 4 Lecture 2 + - + _ C(s) 1 Jvs K R Kp E(s) R(s) K 1 s Figure 2: Space vehicle attitude control as G(s) = C(s) E(s) = KK P × 1 K R +J v s × 1 s = KK P s(J v s+K R ) Closed loop transfer function is G c s = G(s) 1+G(s) = KK p J v s 2 +K R s+KK P K P = Position Sensor gain = 1.65×10 6 K R = Rate sensor gain = 3.71×10 5 K = Ampli?er gain which is a variable J v = Moment of inertia = 41822 With the above parameters, G(s) = 39.45K s(s+8.87) C(s) R(s) = 39.45K s 2 +8.87s+39.45K Characteristics equation ? s 2 +8.87s+39.45K = 0 ? n = v 39.45K rad/sec, ? = 8.87 2? n Sincethesystemisof2 nd order,thecontinuoustimesystemwillalwaysbestableifK P , K R , K, J v are all positive. Now, consider that the continuous data system is subject to sampled data control as shown in Figure 3. I. Kar 2 Digital Control Module 4 Lecture 2 + - C(s) + - ZOH K R E(s) R(s) KK P E * (s) 1 Jvs 1 s Figure 3: Discrete representation of space vehicle attitude control For comparison purpose, we assume that the system parameters are same as that of the con- tinuous data system. G(s) = C(s) E * (s) = G ho (s)G p (s) = 1-e -Ts s . KKp/J v s(s+K R /J v ) G(z) = (1-z -1 ) KKp K R Z 1 s 2 - J v K R s + J v K R (s+K R /J v ) = (1-z -1 ) KKp K R Tz (z-1) 2 - J v z K R (z-1) + J v z K R (z-e -K R T/Jv ) = KKp K 2 R (TK R -J v +J v e -K R T/Jv )z-(TK R +J v )e -K R T/Jv +J v (z-1)(z-e -K R T/Jv ) Characteristic equation of the closed loop system: z 2 +a 1 z+a 0 = 0, where a 1 = f 1 (K,Kp,K R ,Jv) a 0 = f 0 (K,Kp,K R ,Jv) Substituting the known parameters: a 1 = 0.000012K(3.71×10 5 T -41822+41822e -8.87T )-1-e -8.87T a 0 = e -8.87T +0.000012K 41822-(3.71×10 5 T +41822)e -8.87T For stability (1) |a 0 | < 1 (2) P(1) = 1+a 1 +a 0 > 0 = 1-e -8.87T > 0 always satis?ed since T is positive (3) P(-1) = 1-a 1 +a 0 > 0 Choice of K and T: If we plot K versus T then according to conditions (1) and (3) the stable region is shown in Figure 4. Pink region represents the situation when condition (1) is satis?ed but the (3) is not. Red region depicts the situation when condition (1) is satis- ?ed, not the (3). Yellow is the stable region where both the conditions are satis?ed. If we I. Kar 3 Digital Control Module 4 Lecture 2 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 5 10 15 20 25 30 35 40 45 T K T=0.08 Figure 4: K vs. T for space vehicle attitude control system want a comparatively large T, such as 0.2, the gain K is limited by the range K < 5. Similarly ifwewantacomparativelyhighgainsuchas25, wehavetogoforT assmallas0.08orevenless. From studies of continuous time systems it is well known that increasing the value of K gen- erally reduces the damping ratio, increases peak overshoot, bandwidth and reduces the steady state error if it is ?nite and nonzero. 2 Correlation between time response and root locations in s-plane and z-pane The mapping between s-plane and z-plane was discussed earlier. For continuous time systems, thecorrelationbetweenrootlocationins-planeandtimeresponseiswellestablishedandknown. • A root in negative real axis of s-plane produces an output exponentially decaying with time. • Complex conjugate pole pairs in negative s-plane produce damped oscillations. • Imaginary axis conjugate poles produce undamped oscillations. • Complex conjugate pole pairs in positive s-plane produce growing oscillations. Digital control systems should be given special attention due to sampling operation. For ex- ample, if the sampling theorem is not satis?ed, the folding e?ect can entirely change the true response of the system. I. Kar 4 Digital Control Module 4 Lecture 2 The pole-zero map and natural response of a continuous time second order system is shown in Figure 5. X X s1 ?1 j? s t (a) (b) Figure 5: Pole zero map and natural response of a second order system If the system is subject to sampling with frequency ? s < 2? 1 , it will generate an in?nite number of poles in the s-plane at s = s 1 ± j? 1 + jn? s for n = ±1,±2,........ The sampling operation will fold the poles back into the primary strip where -? s /2 < ? < ? s /2. The net e?ect is equivalent to having a system with poles at s = s 1 ±j(? s -? 1 ). The corresponding plot is shown in Figure 6. X X X X s1 ?1 j? s t (a) (b) ?s ?s/2 -?s/2 -?s -?1 +?s -?1 ?1-?s Actual With folding Figure 6: Pole zero map and natural response of a second order system Root locations in z-plane and the corresponding time responses are shown in Figure 7. I. Kar 5Read More

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