Menoufia University Faculty of Electronic Engineering Industrial Electronics and Control Engineering program Digital Control Systems– ACE 314 Lecture 5 Modeling and Stability of Digital Control Systems Course Lecturer Fall 2024 Sessions policy تذكر دائما 2 Assessment Schedule 3 CHAPTER 3 Modeling of Digital Control Systems 4 𝑹 𝒔 Obtain the closed loop pulse transfer function of the system shown in the following figure, where the plant has a transfer function as 𝟏 𝑮𝒑 𝒔 = 𝒔+𝟏 𝒁𝑶𝑯 𝑮𝒑 𝒔 𝒀 𝒔 - s olution 𝒀 𝒔 = 𝑮𝒑 𝒔 𝑮𝒉 𝒔 𝑬∗ (𝒔) = 𝑮𝒑 𝑮𝒉 𝒔 𝑬∗ (𝒔) 𝑬 𝐬 =𝑹 𝒔 −𝒀 𝒔 𝑬(𝒔) + Taking the starred transform 1 𝑬∗ (𝒔) = 𝑹∗ (𝒔) − 𝒀∗ (𝒔) Substituting by 𝑬∗ 𝒔 into equation 1 𝒀 𝒔 = 𝑮𝒑 𝑮𝒉 𝒔 (𝑹∗ (𝒔) − 𝒀∗ (𝒔)) = 𝑮𝒑 𝑮𝒉 𝒔 𝑹∗ 𝒔 − 𝑮𝒑 𝑮𝒉 𝒔 𝒀∗ (𝒔) 5 𝒀∗ (𝒔) = 𝑮𝒑 𝑮𝒉 ∗ 𝒔 𝑹∗ 𝒔 − 𝑮𝒑 𝑮𝒉 ∗ 𝒔 𝒀∗ (𝒔) 𝒀∗ (𝒔)(𝟏 + 𝑮𝒑 𝑮𝒉 ∗ 𝒔 ) = 𝑮𝒑 𝑮𝒉 ∗ 𝒔 𝑹∗ 𝒔 𝒀∗ 𝐬 = 𝑮𝒑 𝑮𝒉 ∗ 𝒔 𝟏+𝑮𝒑 𝑮𝒉 ∗ 𝒔 𝑹∗ 𝒔 𝑮𝒑 𝑮𝒉 𝒛 𝒀 𝐳 = 𝑹 𝒛 𝟏 + 𝑮𝒑 𝑮𝒉 𝒛 𝟏 − 𝒆−𝑻𝒔 𝑮𝒉 𝒔 = 𝑠 𝟏 𝑮𝒑 𝒔 = 𝒔+𝟏 𝑮𝒑 𝑮𝒉 𝒔 Pulse Transfer Function 𝟏 𝟏−𝒆−𝑻𝒔 = × 𝒔+𝟏 𝑠 𝑮𝒑 𝑮𝒉 𝒔 𝟏−𝒆−𝑻𝒔 = 𝒔(𝒔+𝟏) 6 𝑮𝒑 𝑮𝒉 𝒛 𝑮𝒑 𝑮𝒉 𝒔 𝟏−𝒆−𝑻𝒔 = 𝒔(𝒔+𝟏) 𝒛 = 𝒆𝑻𝒔 𝟏−𝒛−𝟏 [𝒔(𝒔+𝟏)] 𝑮𝒑 𝑮𝒉 𝒛 = −𝟏 = (𝟏 − 𝒛 𝟏 [𝒔(𝒔+𝟏)] ) 𝟏 [𝒔(𝒔+𝟏)] 𝒛−𝟏 𝒛 = 𝒛−𝟏 = 𝒆−𝑻𝒔 𝟏 To find Z transform of 1 A B = + 𝑠 𝑠+1 𝑠 𝑠+1 1 1 1 = − 𝑠 𝑠+1 𝑠 𝑠+1 [𝒔(𝒔+𝟏)] Using partial fraction A=1 B=-1 𝒛 𝒛 − 𝒛 − 𝟏 𝒛 − 𝒆−𝑻 𝒛−𝟏 𝒛 𝒛 𝑮𝒑 𝑮𝒉 𝒛 = [ − ] −𝑻 𝒛 𝒛−𝟏 𝒛−𝒆 7 𝒛−𝟏 𝑮𝒑 𝑮𝒉 𝒛 = 𝒛 𝒛 𝒛−𝒆−𝑻 −𝒛(𝒛−𝟏) [ (𝒛−𝟏)(𝒛−𝒆 ) ] −𝑻 𝒛−𝟏 𝑮𝒑 𝑮𝒉 𝒛 = 𝒛 𝑮𝒑 𝑮𝒉 𝒛 = 𝒛 𝟏−𝒆−𝑻 [(𝒛−𝟏)(𝒛−𝒆 )] −𝑻 𝟏−𝒆−𝑻 (𝒛−𝒆−𝑻 ) 𝑮𝒑 𝑮𝒉 𝒛 𝒀 𝐳 = 𝑹 𝒛 𝟏 + 𝑮𝒑 𝑮𝒉 𝒛 𝟏−𝒆−𝑻 𝒛+𝟏−𝟐𝒆−𝑻 𝟏 + 𝑮𝒑 𝑮𝒉 𝒛 =1+ = −𝑻 (𝒛−𝒆 ) (𝒛−𝒆−𝑻 ) 𝒀 𝐳 𝟏 − 𝒆−𝑻 = 𝑹 𝒛 𝒛 + 𝟏 − 𝟐𝒆−𝑻 8 For the system shown in the previous example, if the input signal 𝒓(𝒕) is a unit step input and sampling period 𝑇 =1 Sec. determine: 𝑹 𝒔 a) The output response at the sampling instants. b) The final value of 𝑦(𝑘𝑇) s 𝑬(𝒔) + 𝒁𝑶𝑯 𝒀 𝒔 𝑮𝒑 𝒔 - olution 𝑹 𝒛 = 𝒓(𝒕) is a unit step 𝒛 𝒛−𝟏 The sampling period 𝑻 =1 Sec. Hence, 𝒀 𝐳 𝟏 − 𝒆−𝑻 𝟏 − 𝒆−𝟏 𝟎. 𝟔𝟑𝟐 = = = −𝑻 −𝟏 𝑹 𝒛 𝒛 + 𝟏 − 𝟐𝒆 𝒛 + 𝟏 − 𝟐𝒆 𝒛 + 𝟎. 𝟐𝟔𝟒 9 a) The output response at the sampling instants. 𝒀 𝐳 = Using long division 𝟎. 𝟔𝟑𝟐 𝒛 = × 𝒛 + 𝟎. 𝟐𝟔𝟒 𝒛 − 𝟏 𝟎. 𝟔𝟑𝟐𝒛 = 𝟐 𝒛 − 𝟎. 𝟕𝟑𝟔𝒛 − 𝟎. 𝟐𝟔𝟒 𝟎.𝟔𝟑𝟐 𝑹 𝒛 𝒛+𝟎.𝟐𝟔𝟒 𝟎. 𝟔𝟑𝟐𝒛−𝟏 +𝟎. 𝟒𝟔𝟓𝒛−𝟐 + 𝟎. 𝟓𝟎𝟗𝒛−𝟑 + 𝟎. 𝟒𝟗𝟖𝒛−𝟒 + …… 𝒛𝟐 − 𝟎. 𝟕𝟑𝟔𝒛 − 𝟎. 𝟐𝟔𝟒 𝟎. 𝟔𝟑𝟐𝒛 −𝟎. 𝟔𝟑𝟐𝒛 − + 𝟎. 𝟒𝟔𝟓 + − 𝟎. 𝟏𝟔𝟕𝒛−𝟏 𝟎. 𝟒𝟔𝟓 + 𝟎. 𝟏𝟔𝟕𝒛−𝟏 − 𝟎. 𝟒𝟔𝟓 + + 𝟎. 𝟏𝟐𝟑𝒛−𝟐 − 𝟎. 𝟑𝟒𝟐𝒛−𝟏 − 𝟎. 𝟓𝟎𝟗𝒛−𝟏 + 𝟎. 𝟏𝟐𝟑𝒛−𝟐 − + + 𝟎. 𝟓𝟎𝟗𝒛−𝟏 − 𝟎. 𝟑𝟕𝟓𝒛−𝟐 − 𝟎. 𝟏𝟑𝟒𝒛−𝟑 𝟎. 𝟒𝟗𝟖𝒛−𝟐 + 𝟎. 𝟏𝟑𝟒𝒛−𝟑 …………….. 𝒀 𝐳 = 𝟎. 𝟔𝟑𝟐𝒛−𝟏 +𝟎. 𝟒𝟔𝟓𝒛−𝟐 + 𝟎. 𝟓𝟎𝟗𝒛−𝟑 + 𝟎. 𝟒𝟗𝟖𝒛−𝟒 + …… 10 𝒀 𝐳 = 𝟎. 𝟔𝟑𝟐𝒛−𝟏 +𝟎. 𝟒𝟔𝟓𝒛−𝟐 + 𝟎. 𝟓𝟎𝟗𝒛−𝟑 + 𝟎. 𝟒𝟗𝟖𝒛−𝟒 + …… 𝑦(0) 𝑦(1) 𝑦(2) 𝑦(3) 𝑦(4) … 0 𝟎. 𝟔𝟑𝟐 𝟎. 𝟒𝟔𝟓 𝟎. 𝟓𝟎𝟗 𝟎. 𝟒𝟗𝟖 … 11 b) Final value of 𝑦(𝑘𝑇) 𝒀(∞) = lim (𝟏 − 𝒛−𝟏 )𝒀 𝐳 𝒁→𝟏 𝒛−𝟏 𝟎. 𝟔𝟑𝟐𝒛 × 𝟐 𝒁→𝟏 𝒛 𝒛 − 𝟎. 𝟕𝟑𝟔𝒛 − 𝟎. 𝟐𝟔𝟒 = lim 𝒛−𝟏 𝟎. 𝟔𝟑𝟐𝒛 = lim × 𝒁→𝟏 𝒛 (𝒛 − 𝟏)(𝒛 + 𝟎. 𝟐𝟔𝟒) 𝟎. 𝟔𝟑𝟐 𝒁→𝟏 (𝒛 + 𝟎. 𝟐𝟔𝟒) = lim 𝟎. 𝟔𝟑𝟐 = (𝟏 + 𝟎. 𝟐𝟔𝟒) = 𝟎. 𝟓 12 Closed loop digital control system 𝑹 𝒔 𝑬(𝒔) + 𝑫∗ (𝒔) 𝑫𝑨𝑪 Microprocessor ZOH 𝑨𝑫𝑪 - Sampler 𝑮𝒑 𝒔 𝒀 𝒔 𝒀∗ (𝒔) DAC, analog system, ADC , controller , sensor MCU combination PLC transfer function 𝑯(𝒔) Sensor/ transducer 𝑮 𝒔 = 𝑮𝒑 𝒔 𝑮𝒉 𝒔 𝑮 𝒔 = 𝑮𝒑 𝑮𝒉 𝒔 𝒀 𝒔 = 𝑮 𝒔 𝑫∗ (𝒔)𝑬∗ (𝒔) 1 𝑬 𝐬 = 𝑹 𝒔 − 𝑯 𝒔 𝒀 𝐬 = 𝑹 𝒔 − 𝑯 𝒔 𝑮 𝒔 𝑫∗ 𝒔 𝑬∗ 𝒔 = 𝑹 𝒔 − 𝑮𝑯 𝒔 𝑫∗ 𝒔 𝑬∗ 𝒔 𝑬∗ (𝒔) = 𝑹∗ (𝒔) − 𝑮𝑯∗ (𝒔) 𝑫∗ 𝒔 𝑬∗ 𝒔 13 𝑬∗ (𝒔) = 𝑹∗ (𝒔) − 𝑮𝑯∗ (𝒔) 𝑫∗ 𝒔 𝑬∗ 𝒔 ∗ 𝒔 𝑹 𝑬∗ 𝐬 = 𝟏 + 𝑮𝑯∗ (𝒔)𝑫∗ 𝒔 From equation 1, ∗ 𝒔 𝑫∗ 𝒔 𝑹∗ 𝒔 𝑮 𝒀∗ 𝒔 = 𝑮∗ 𝒔 𝑫∗ 𝒔 𝑬∗ 𝒔 = 𝟏+𝑮𝑯∗ (𝒔)𝑫∗ 𝒔 𝒀∗ 𝒔 𝑮∗ 𝒔 𝑫∗ 𝒔 = ∗ 𝑹 𝒔 𝟏 + 𝑮𝑯∗ (𝒔)𝑫∗ 𝒔 𝒀 𝒛 𝑮 𝒛 𝑫(𝒛) = 𝑹 𝒛 𝟏 + 𝑮𝑯 𝒛 𝑫(𝒛) 14 Example Find 𝑮𝒁𝑨𝑺 (𝒛) for the series R-L circuit with the inductor voltage as output. s 𝑮𝒁𝑨𝑺 (𝒛) 𝑮𝒑 𝒔 olution means the z-transfer function for ZOH, analog system and sampler 𝑉𝑖 = 𝑖𝑅 + 𝑉𝑜 Take Laplace transform for both sides. 𝑉𝑖 (𝑠) = 𝑅𝑖(𝑠) + 𝑉𝑜 (𝑠) Note: 𝒅𝒊 𝑉𝐿 = 𝑉𝑜 = 𝑳 𝒅𝒕 1 𝑖 = න 𝑉𝑜 𝐿 1 𝑖(𝑠) = 𝑉𝑜 (𝑠) 𝐿𝑠 15 𝑅 𝑅 𝑅 + 𝐿𝑠 𝑉𝑖 𝑠 = 𝑉𝑜 𝑠 + 𝑉𝑜 𝑠 = + 1 𝑉𝑜 𝑠 = 𝑉𝑜 𝑠 𝐿𝑠 𝐿𝑠 𝐿𝑠 𝑉𝑜 𝑠 𝐿𝑠 (𝐿/𝑅)𝑠 𝜏𝑠 = = = 𝑉𝑖 𝑠 𝑅 + 𝐿𝑠 1 + (𝐿/𝑅)𝑠 1 + 𝜏𝑠 where 𝜏 = (𝐿/𝑅) 𝜏𝑠 ∴ 𝑮𝒑 𝒔 = 1 + 𝜏𝑠 𝑮𝒑 𝒔 [ 𝒔 ] = (𝟏 − 𝒛 ) 𝑮𝒁𝑨𝑺 (𝒛) = (𝟏 − 𝒛−𝟏 ) = (𝟏 − 𝒛−𝟏 ) = −𝟏 𝜏𝑠 [𝒔(1+𝜏𝑠)] 𝜏 [𝜏(𝑠+1Τ )] 𝜏 1 𝑠 + 1ൗ𝜏 -1 𝑡 𝑒 − ൗ𝜏 𝒛 𝒛−𝟏 × −𝑻ൗ𝝉 𝒛 − 𝒆 𝒛 𝑮𝒁𝑨𝑺 (𝒛) = 𝒛−𝟏 𝑻ൗ − 𝒛−𝒆 𝝉 16 Modeling of a physical system ⎯ A simple mechanical system غري مطالب باستنتاج Model 17 18 Recall that the mechanical system in the cruise-control problem can be represented by : 𝑴𝒗ሶ 𝒕 + 𝒃𝒗 𝒕 = 𝒖 𝒕 Find 𝐺𝑍𝐴𝑆 (𝑧) for cruise control system for the vehicle, where 𝒖 is the input force, 𝒗 is the velocity of the car, and 𝑏 is the viscous friction coefficient. s Take Laplace transform for both sides of olution 𝑴𝒗ሶ 𝒕 + 𝒃𝒗 𝒕 = 𝒖 𝒕 𝑀𝑠𝑣 (𝑠) + 𝑏𝑣(𝑠) = 𝑢(𝑠) 𝑣 𝑠 1 (1/𝑀) 𝐾/𝜏 𝑮𝒑 𝒔 = = = = 𝑢 𝑠 𝑀𝑠 + 𝑏 𝑠 + (𝑏/𝑀) 𝑠 + 1/𝜏 𝑮𝒁𝑨𝑺 (𝒛) = (𝟏 − 𝒛−𝟏 ) 𝑮𝒑 𝒔 where 𝐾 = 1/𝑏 and 𝜏 = 𝑀/𝑏 [ 𝒔 ] = (𝟏 − 𝒛 ) −𝟏 𝐾/𝜏 𝒔(𝑠 + 1/𝜏) 19 Using partial fraction 𝐾/𝜏 𝑲 = 𝝉 𝒔(𝑠+1/𝜏) 𝝉 𝝉 − 𝒔 𝒔+𝟏/𝝉 𝟏 𝟏 =𝑲 − 𝒔 𝒔 + 𝟏/𝝉 𝟏 [𝑲 𝟏𝒔 − 𝒔+𝟏/𝝉 ] 𝑮𝒁𝑨𝑺 (𝒛) = (𝟏 − 𝒛−𝟏 ) 𝒛−𝟏 𝒛 𝑮𝒁𝑨𝑺 (𝒛) = 𝑲 𝒛 𝒛−𝟏 [ 𝑮𝒁𝑨𝑺 (𝒛) = 𝑲 [𝟏 − 𝑮𝒁𝑨𝑺 (𝒛) = 𝑲 [ − 𝒛 𝑻ൗ − 𝒛−𝒆 𝝉 𝒛−𝟏 𝑻 𝒛−𝒆− ൗ𝝉 ] ] 𝑻ൗ − 𝟏−𝒆 𝝉 ] 𝑻 𝒛 − 𝒆− ൗ𝝉 20 Example Find 𝑮𝒁𝑨𝑺 (𝒛) for the vehicle position control system, where u is the input force, y is the position of the car, and b is the viscous friction coefficient. Where, the relation between the input and speed is described by the equation 𝑀𝑦(𝑡) ሷ + 𝑏𝑦(𝑡) ሶ = 𝑢(𝑡) 𝒚 𝒔 𝑮𝒑 𝒔 = 𝒖 𝒔 s olution 21 22 Steady-state error and error constants TIME RESPONSE ➢ In continuous system C(t) Response Unit step input Maximum Overshoot Transient 1.02 Cf 0.98 Cf 0.9 Cf 0.63 Cf 0.5 Cf Steady state error t→ ∞ Time Constant tr, ts, tp, td, overshoot, time constant ,𝜼 Steady state ess Delay time td 0.1 Cf Rising time ts Peak time tp Settling time ts Transient response Steady state response 23 ➢ Steady state — Continuous systems 𝑹 𝒔 𝑬(𝒔) + - 𝑪(𝒔) 𝑼(𝒔) 𝒀 𝒔 𝑮(𝒔) 𝒆𝒔𝒔 = 𝒆(∞) = 𝐥𝐢𝐦 𝒔𝑬(𝒔) 𝒔→𝟎 ➢ Steady state — Discrete systems 𝒆𝒔𝒔 = 𝒆(∞) = 𝐥𝐢𝐦(𝟏 − 𝒛−𝟏 )𝑬(𝒛) 𝒛→𝟏 𝑹 𝒛 + - 𝑬(𝒛) 𝑪 𝒛 𝑼(𝒛) 𝒀 𝒛 𝑮𝒁𝑨𝑺 (𝒛) 𝐸 𝑧 = 𝑅 𝑧 − 𝑌(𝑧) 𝐸 𝑧 = 𝑅 𝑧 − 𝐸 𝑧 𝐶(𝑧)𝐺𝑍𝐴𝑆 (𝑧) 𝐸 𝑧 (1 + 𝐶(𝑧)𝐺𝑍𝐴𝑆 (𝑧)) = 𝑅 𝑧 𝑅 𝑧 𝑅 𝑧 𝐸 𝑧 = = 1 + 𝐶(𝑧)𝐺𝑍𝐴𝑆 (𝑧) 1 + 𝐿(𝑧) Where 𝑳 𝒛 = 𝑪(𝒛)𝑮𝒁𝑨𝑺 (𝒛) denotes the loop gain of the system 24 Case 1: 𝒛 If 𝒓(𝒌) is a unit step, 𝑅 𝑧 = 𝒛−𝟏 , by the final value theorem of z-Transform 𝒛−𝟏 𝑅 𝑧 𝒛−𝟏 𝒛 1 = 𝐥𝐢𝐦 × × 𝒛−𝟏 1+𝐿(𝑧) 𝒛→𝟏 𝒛 1+𝐿(𝑧) 𝒛→𝟏 𝒛 𝒆𝒔𝒔 = 𝒆(∞) = 𝐥𝐢𝐦 𝟏 − 𝒛−𝟏 𝑬 𝒛 = 𝐥𝐢𝐦 𝒛→𝟏 1 𝒆 ∞ = 𝐥𝐢𝐦 𝒛→𝟏 1 + 𝐿(𝑧) 𝒆 ∞ 1 1 = = 1+𝐿(1) 1+𝐾𝑝 Type of system is determined by Where 𝐾𝑝 is the position error constant. u𝐧𝐭𝐢𝐥 𝒆 ∞ = 𝟎 𝑲𝒑 =∞ 𝑳 𝒛 = 𝑪(𝒛)𝑮𝒁𝑨𝑺 (𝒛) 𝟏/(𝒛– 𝟏)𝒏 has at least a factor of 𝟏/(𝒛– 𝟏) (TYPE I SYSTEM) in order to make 𝒆(∞) zero. If𝑲𝒑 is a finite scalar, the open-loop system 𝑳 𝒛 = 𝑪(𝒛)𝑮𝒁𝑨𝑺 (𝒛) is said to be a TYPE 0 system. 25 Case 2 If 𝑟(𝑘) is a unit ramp, 𝑅 𝑧 = 𝑻𝒛 , by the final value theorem 𝒛−𝟏 𝟐 𝒛−𝟏 𝑅 𝑧 𝒛−𝟏 𝑻𝒛 1 = 𝐥𝐢𝐦 × × 𝒛−𝟏 𝟐 1+𝐿(𝑧) 𝒛→𝟏 𝒛 1+𝐿(𝑧) 𝒛→𝟏 𝒛 𝒆𝒔𝒔 = 𝒆(∞) = 𝐥𝐢𝐦 𝟏 − 𝒛−𝟏 𝑬 𝒛 = 𝐥𝐢𝐦 𝒛→𝟏 𝒆 ∞ 𝑇 1 = 𝐥𝐢𝐦 = 𝒛→𝟏 (𝒛−𝟏)(1+𝐿 𝑧 ) 𝐾𝑉 1 𝐾𝑉 = 𝐥𝐢𝐦 (𝒛 − 𝟏)(1 + 𝐿 𝑧) 𝒛→𝟏 𝑇 Where, 𝑲𝑽 is the velocity error constant. We need also 𝑲𝑽 = ∞, until 𝒆 ∞ =0 𝑳 𝒛 = 𝑪(𝒛)𝑮𝒁𝑨𝑺 (𝒛) has at least a factor of 𝟏/(𝒛– 𝟏)𝟐 (TYPE II SYSTEM) in order to make 𝒆(∞) zero. 26 Example Find the steady-state position error for the digital position control system with unity feedback and with the transfer functions 1. For a sampled unit step input 2. For a sampled ramp input s olution The loop gain of the system is given by 𝐾𝐾𝑐 (𝑧 + 𝑎) 𝑳 𝒛 = 𝑪 𝒛 𝑮𝒁𝑨𝑺 𝒛 = (𝑧 − 1)(𝑧 − 𝑐) TYPE I SYSTEM 1. For a sampled unit step input 1 1 1 𝒆 ∞ = 𝐥𝐢𝐦 = = 1+𝐿(𝑧) 1+𝐿(1) 1+𝐾𝑝 𝒛→𝟏 𝐾𝐾𝑐 (1+𝑎) 𝐾𝑝 = =∞ (1−1)(1−𝑐) Hence, 𝒆 ∞ =0 27 2. For a sampled unit ramp input 𝒆 ∞ 𝑇 = 𝐥𝐢𝐦 = 𝐾1𝑉 𝒛→𝟏 (𝒛−𝟏)(1+𝐿 𝑧 ) 1 1 𝐾𝐾𝑐 (𝑧+𝑎) 𝐾𝑉 = 𝐥𝐢𝐦 (𝒛 − 𝟏)(1 + 𝐿 𝑧) = 𝐥𝐢𝐦 (𝒛 − 𝟏)(1+ ) 𝑇 𝑇 (𝑧−1)(𝑧−𝑐) 𝒛→𝟏 𝒛→𝟏 1 𝒛→𝟏 𝑇 = 𝐥𝐢𝐦 ((𝒛 − 𝟏) + (𝒛 − 𝟏) 𝐾𝐾𝑐 (𝑧+𝑎) ) (𝑧−1)(𝑧−𝑐) 𝐾𝐾𝑐 (1+𝑎) = + (1−𝑐) ) 1 𝐾𝐾𝑐 (1+𝑎) = ( ) 𝑇 (1−𝑐) 1 ((𝟏 − 𝟏) 𝑇 𝑇 (1−𝑐) 𝒆 ∞ = ( ) 𝐾𝐾𝑐 (1+𝑎) Note: • The steady-state error is reduced by increasing the controller gain. • It is also affected by the choice of controller pole and zero. 28 CHAPTER 4 Stability of Digital Control Systems - Mapping between s-plane and z-plane - Study the stability of digital control systems using jury method 29 Mapping between s-plane and z-plane ➢ Poles and Zeros of a Transfer Function in Continuous Control Systems s-plane 𝒔 = 𝝈 + 𝒋𝝎 𝒋𝝎 • The transfer function is a rational function 𝝈 𝒃𝒎 𝒔𝒎 + 𝒃𝒎−𝟏 𝒔𝒎−𝟏 + ⋯ + 𝒃𝟏 𝒔 + 𝒃𝟎 𝑮 𝒔 = 𝒂𝒏 𝒔𝒏 + 𝒂𝒏−𝟏 𝒔𝒏−𝟏 + ⋯ + 𝒂𝟏 𝒔 + 𝒂𝟎 • It is often convenient to factorize 𝐹 𝑠 s-plane 𝑵 𝒔 𝒔 − 𝒛𝟏 𝒔 − 𝒛𝟐 … 𝒔 − 𝒛𝒎−𝟏 𝒔 − 𝒛𝒎 𝑮 𝒔 = =𝑲 𝑫 𝒔 𝒔 − 𝒑𝟏 𝒔 − 𝒑𝟐 … 𝒔 − 𝒑𝒏−𝟏 𝒔 − 𝒑𝒏 30 𝑵 𝒔 𝑮 𝒔 = 𝑫 𝒔 ✓ Poles are the roots of denominator polynomials 𝑫 𝒔 which makes 𝑫 𝒔 = 𝟎 → lim 𝐺 𝑠 = ∞ 𝑠→𝑝𝑖 ✓ Zeros are the roots of numerator polynomials 𝑵 𝒔 which makes 𝑵 𝒔 = 𝟎 → lim 𝐺 𝑠 = 0 𝑠→𝑧𝑖 The ch’s equation is 𝐷 𝑠 = 𝑠 − 𝑝1 𝑠 − 𝑝2 … 𝑠 − 𝑝𝑛−1 𝑠 − 𝑝𝑛 = 0 𝒋𝝎 𝒔𝟐 + 𝟑𝑠 + 𝟐 = 0 𝒔+𝟐 𝒔+𝟏 =𝟎 𝝈 𝒔𝟏 = −𝟐 𝒔𝟐 = −𝟏 31 Poles & Zeros real → 𝑠 = ±𝜎 complex → 𝑠 = ±𝜎 ± 𝑗𝜔 𝒋𝝎 𝝈 Positive integers 1, 2, 3, 4, 5,… Zero Negative integers -1, -2, -3,- 4, -5,… Fractional 1/2, 5/3, …. Complex Real + imaginary 32 Natural response of LS 𝒚𝒏 𝒕 = σ𝒏𝒊=𝟏 𝑪𝒊 𝒆𝒑𝒊𝒕 𝑒 𝜎𝑇 𝒋𝝎 𝑒 − 𝜎𝑇 𝝈 𝑒 −𝜎𝑇±𝑗𝜔 𝑒 𝜎𝑇±𝑗𝜔 33 • A real pole 𝑝𝑖 = −𝜎 (in LHS) defines an 𝒋𝝎 exponentially decaying 𝐶𝑒 −𝜎𝑡 o poles near the origin → decay slowly. o far poles → decay rapidly • A real pole 𝑝𝑖 = 𝜎 (in RHS) defines an 𝝈 exponentially increasing 𝐶𝑒 𝜎𝑡 • A complex pole at −𝜎 ± 𝑗𝜔 → a decaying sinusoid. • A complex pole at 𝜎 ± 𝑗𝜔 → an increasing sinusoid • A pole 𝑝𝑖 = 0 → constant in amplitude. • Poles on ±𝒋𝝎 → constant amplitude sinusoid 𝐶 sin 𝜔𝑡 + 𝜑 34 Summary An 𝑛𝑡ℎ order linear system is asymptotically stable only if all of the components in the natural response from a finite set of initial conditions decay to zero as time increases, or 𝒍𝒊𝒎 σ𝒏𝒊=𝟏 𝑪𝒊 𝒆𝒑𝒊 𝒕 = 𝟎 𝒕→∞ • If all poles have −𝑣𝑒 real parts → stable system. • If any pole has a +𝑣𝑒 real part → unstable system. • If one or more poles lying on the imaginary axis → marginally or critical stable system. 35 Mapping • The stability of digital systems is related to the stability in 𝒔-plane as follows: s-plane Generally, 𝒔 = 𝝈 ± 𝒋𝝎. z-plane 𝒛 = 𝒆𝒔𝑻 • We map the imaginary axis 𝒔 = ±𝒋𝝎 (marginally SS) to the unit circle. 𝒛 = 𝒆𝒔𝑻 = 𝒆𝒋𝝎𝑻 = 𝐜𝐨𝐬 𝝎𝑻 +𝒋 𝒔𝒊𝒏 𝝎𝑻 = 𝟏∠𝝎𝑻 𝒛 = 𝒆𝒔𝑻 = 𝒆−𝒋𝝎𝑻 = 𝐜𝐨𝐬 𝝎𝑻 −𝒋 𝒔𝒊𝒏 𝝎𝑻 = 𝟏∠ − 𝝎𝑻 𝐼𝑚 𝒋𝝎 𝟏∠𝟑𝝅/𝟒 𝝈 s-plane 𝑧 =1 𝟏∠𝝅 𝟏∠𝝅/𝟐 𝟏∠𝝅/𝟒 𝟏∠𝟎 𝑹𝒆 z-plane 36 • LHS of s-plane, we map the vertical line 𝒔 = −𝝈 ± 𝒋𝝎 (Stable system) 𝑠 = −𝜎 ± 𝑗𝜔 → 𝑧 = 𝑒 −𝜎±𝑗𝜔 𝑇 → 𝒆−𝝈𝑻 𝒆𝒋𝝎𝑻 = 𝒆−𝝈𝑻 ∠𝝎𝑻 ( 𝒛 = 𝒆−𝝈𝑻 < 𝟏). → 𝒆−𝝈𝑻 𝒆−𝒋𝝎𝑻 = 𝒆−𝝈𝑻 ∠ − 𝝎𝑻 ( 𝒛 = 𝒆−𝝈𝑻 < 𝟏). 𝒆−𝟏 =0.368 ∠𝝎𝑻 𝐼𝑚 𝒆−𝟐 =0.135 ∠𝝎𝑻 𝒋𝝎 𝒆−𝟑 =0.0498 ∠𝝎𝑻 𝑹𝒆 -3 -2 𝝈 -1 z-plane s-plane 37 • RHS of s-plane, we map the vertical line 𝒔 = 𝝈 ± 𝒋𝝎 (Unstable system) 𝑠 = 𝜎 ± 𝑗𝜔 → 𝑧 = 𝑒 (𝜎±𝑗𝜔)𝑇 → 𝒆𝝈𝑻 𝒆𝒋𝝎𝑻 = 𝒆𝝈𝑻 ∠𝝎𝑻 ( 𝒛 = 𝒆𝝈𝑻 > 𝟏). → 𝒆𝝈𝑻 𝒆−𝒋𝝎𝑻 = 𝒆𝝈𝑻 ∠ − 𝝎𝑻 ( 𝒛 = 𝒆𝝈𝑻 > 𝟏). 𝒆𝟏 =2.718 ∠𝝎𝑻 𝒆𝟐 =7.389 ∠𝝎𝑻 𝒋𝝎 𝒆𝟑 =20.086 ∠𝝎𝑻 1 s-plane 𝐼𝑚 2 3 𝑹𝒆 𝝈 z-plane 38 𝐼𝑚 Critical Stable 𝑹𝒆 𝟏 z-plane 39 Stability analysis of closed loop systems in the z-plane • The stability of LTI SISO discrete control systems 𝑹 𝒛 𝒀 𝒛 𝑮 𝒛 = 𝑹 𝒛 𝟏+𝑮 𝒛 𝑯 𝒛 + - 𝑬(𝒛) 𝑮(𝒛) 𝒀 𝒛 𝑯 𝒛 • It may be determined by the locations of closed loop poles in z-plane or the roots of the characteristic equation 𝟏 + 𝑮 𝒛 𝑯 𝒛 =0 closed loop poles Unit circle inside Stable outside Unstable Closed loop zeros do not affect the absolute stability. If a pole on circle Critical stable 41 For K=1 determine the closed loop stability of the system shown below. Example 1 𝑹 𝒔 s 𝑬(𝒔) + 𝑲 - 𝒁𝑶𝑯 𝑻=𝟏 𝟏 𝒔 𝒔+𝟏 𝒀 𝒔 olution 𝑮 𝒔 𝒀 𝒛 𝑹 𝒛 𝑮 𝒛 =ℤ 𝑲 𝟏−𝒆−𝑻𝒔 𝒔 𝑮 𝒛 = 𝟏+𝑮 𝒛 𝟏 𝒔 𝒔+𝟏 = ℤ 𝟏 − 𝒆−𝑻𝒔 𝑮 𝒛 = 𝟏 − 𝒛−𝟏 ℤ 𝟏 𝒔𝟐 𝒔+𝟏 𝑮 𝒛 = 𝟏 − 𝒛−𝟏 𝑻𝒛 𝒛 𝒛 − + 𝒛−𝟏 𝟐 𝒛−𝟏 𝒛−𝒆−𝑻 = 𝟏 − 𝒛−𝟏 ℤ 𝑲 𝒔𝟐 𝒔+𝟏 𝟏 𝟏 𝟏 − + 𝒔𝟐 𝒔 𝒔+𝟏 , 𝑲 = 𝟏 we obtain Using partial fraction 42 𝒛−𝟏 𝑮 𝒛 = 𝒛 𝑮 𝒛 ȁ𝑻=𝟏 𝒔𝒆𝒄. = 𝑻𝒛 𝒛 𝒛 − + 𝟐 𝒛−𝟏 𝒛 − 𝟏 𝒛 − 𝒆−𝑻 𝒛 𝟏−𝟏+𝒆−𝟏 −𝒆−𝟏 −𝒆−𝟏 +𝟏 𝒛−𝟏 𝒛−𝒆−𝟏 𝟎.𝟑𝟔𝟖𝒛+𝟎.𝟐𝟔𝟒 = 𝟐 𝒛 −𝟏.𝟑𝟔𝟖𝒛+𝟎.𝟑𝟔𝟖 Thus, the transfer function of the closed loop system is 𝑮 𝒛 𝟏+𝑮 𝒛 𝟎.𝟑𝟔𝟖𝒛+𝟎.𝟐𝟔𝟒 = 𝟐 𝒛 −𝒛+𝟎.𝟔𝟑𝟐 The characteristic equation is 𝑭 𝒛 = 𝟏 + 𝑮 𝒛 = 𝒛𝟐 − 𝒛 + 𝟎. 𝟔𝟑𝟐 = 𝟎 → 𝒛𝟏,𝟐 = 𝟎. 𝟓 ± 𝒋𝟎. 𝟔𝟐 𝒛𝟏,𝟐 = 𝟎. 𝟓𝟐 + 𝟎. 𝟔𝟐𝟐 = 𝟎. 𝟕𝟗 < 𝟏 i.e. the system is stable. 43 Example 2 An isothermal chemical reactor where the product concentration is controlled by manipulating the feed flow rate is modeled by the following transfer function Determine 𝑮𝒁𝑨𝑺 (𝒛) with a sampling rate T= 0.1 s, and then study the absolute stability of the closed-loop system with the feedback controller s olution 44 The discretized process transfer function is The transfer function from the reference input to the system output is given by The system is stable where all its poles inside the unit circle. NOTE: the stability of the closed-loop transfer function is not always sufficient for proper system operation because some of the internal variables may be unbounded. 45 Methods for Testing Absolute Stability 1. Jury’s Stability Test ➢ It is possible to investigate the stability of z-domain polynomials directly using the Jury test. ➢ These tests involve determinant evaluations as in the RouthHurwitz test for s-domain polynomials but are more time consuming. ❑ For second order From the ch’s equation Three conditions NO. of conditions n+1 𝑭 𝒛 = 𝒂𝟐 𝒛𝟐 + 𝒂𝟏 𝒛 + 𝒂𝒐 = 𝟎 𝟏 . 𝑭 𝟏 >𝟎 𝟐 . 𝑭 −𝟏 > 𝟎 𝟑 . 𝒂𝒐 < 𝒂𝟐 If the conditions are satisfied the system is stable. IF F(1)=0 OR F(-1)=0 then the system is critical Stable 46 Check the stability of 𝑭(𝒛) = 𝒛𝟐 − 𝒛 + 𝟎. 𝟔𝟑𝟐. Example 3 s 𝒂𝟎 = 𝟎. 𝟔𝟑𝟐 olution 𝒂𝟏 = −𝟏 𝒂𝟐 = 𝟏 𝟏 . 𝑭(𝟏) = 𝟏 − 𝟏 + 𝟎. 𝟔𝟑𝟐 = 𝟎. 𝟔𝟑𝟐 > 𝟎 𝟐 . 𝑭 −𝟏 = 𝟏 + 𝟏 + 𝟎. 𝟔𝟑𝟐= 𝟐. 𝟔𝟑𝟐 > 𝟎 𝟑 . 𝒂𝒐 < 𝒂𝟐 The stability conditions are satisfied. Hence the system is stable. Stability check via the roots of the ch’s equation → 𝑭 𝒛 = 𝒛𝟐 − 𝒛 + 𝟎. 𝟔𝟑𝟐 = 𝟎 𝒛𝟏,𝟐 = 𝟎. 𝟓 ± 𝒋𝟎. 𝟔 → 𝑧1,2 = 0.79 < 1 the system is stable 47 Example 4 𝐹 𝑧 = 𝑧 2 − 0.25 s 𝒂𝟎 = −𝟎. 𝟐𝟓 olution 𝒂𝟏 = 𝟎 𝒂𝟐 = 𝟏 𝐹 −1 = 1 − 0.25 = 0.75 𝐹 1 = 1 − 0.25 = 0.75 𝟏 . 𝐹 1 >0 𝟐 . 𝐹 −1 > 0 𝟑 . 𝑎𝑜 < 𝑎2 Since all the conditions are satisfied, the system is stable. 48 Example 5 Determine the range of gain K for stability using Jury’s test for the system shown below. 𝑹 𝒔 𝑬(𝒔) + 𝑲 - 𝟏 𝒔 𝒔+𝟏 𝒁𝑶𝑯 𝑻=𝟏 𝒀 𝒔 𝑮 𝒔 𝒀 𝒛 𝑹 𝒛 𝑲𝑮 𝒛 = 𝟏+𝑲𝑮 𝒛 s olution 𝑲(𝟎.𝟑𝟔𝟖𝒛+𝟎.𝟐𝟔𝟒) 𝑲𝑮 𝒛 = 𝟐 𝒛 −𝟏.𝟑𝟔𝟖𝒛+𝟎.𝟑𝟔𝟖 𝑲(𝟎. 𝟑𝟔𝟖𝒛 + 𝟎. 𝟐𝟔𝟒) 𝒛𝟐 − 𝟏. 𝟑𝟔𝟖𝒛 + 𝟎. 𝟑𝟔𝟖 + 𝟎. 𝟑𝟔𝟖𝑲𝒛 + 𝟎. 𝟐𝟔𝟒𝑲 𝑭 𝒛 = 𝟏 + 𝑲𝑮 𝒛 = 𝟏 + 𝟐 = =0 𝟐 49 𝒛 − 𝟏. 𝟑𝟔𝟖𝒛 + 𝟎. 𝟑𝟔𝟖 𝒛 − 𝟏. 𝟑𝟔𝟖𝒛 + 𝟎. 𝟑𝟔𝟖 𝑭 𝒛 = 𝒛𝟐 − 𝟏. 𝟑𝟔𝟖𝒛 + 𝟎. 𝟑𝟔𝟖 + 𝟎. 𝟑𝟔𝟖𝑲𝒛 + 𝟎. 𝟐𝟔𝟒𝑲 = 𝒛𝟐 + 𝒛 𝟎. 𝟑𝟔𝟖𝑲 − 𝟏. 𝟑𝟔𝟖 + 𝟎. 𝟑𝟔𝟖 + 𝟎. 𝟐𝟔𝟒𝑲 𝑭 𝟏 = 𝟏 + 𝟎. 𝟑𝟔𝟖𝑲 − 𝟏. 𝟑𝟔𝟖 + 𝟎. 𝟑𝟔𝟖 + 𝟎. 𝟐𝟔𝟒𝑲 = 𝟎. 𝟔𝟑𝟐𝑲 For stable system 𝐹 1 > 0 𝟎. 𝟔𝟑𝟐𝑲 > 0 Hence, 𝑲 > 𝟎 𝐹 −1 = 𝟏 − 𝟎. 𝟑𝟔𝟖𝑲 + 𝟏. 𝟑𝟔𝟖 + 𝟎. 𝟑𝟔𝟖 + 𝟎. 𝟐𝟔𝟒𝑲 = −𝟎. 𝟏𝟎𝟒𝑲 + 𝟐. 𝟕𝟑𝟔 For stable system 𝐹 −1 > 0 −𝟎. 𝟏𝟎𝟒𝑲 + 𝟐. 𝟕𝟑𝟔 > 0 For stable system 𝑎𝑜 < 𝑎2 𝟎. 𝟑𝟔𝟖 + 𝟎. 𝟐𝟔𝟒𝑲 < 1 𝟎. 𝟐𝟔𝟒𝑲 < 1 − 0.368 𝟎 < 𝑲 < 𝟐. 𝟑𝟗 Hence, 𝑲 < 𝟐𝟔. 𝟑𝟏 Hence, 𝑲 < 𝟐. 𝟑𝟗 50 Thank You 51
