Question

• 2.1. Obtain process response curve in time domain for a step input of a magnitude of 0.5. (6)
• 2.2. Plot the response curves in a single figure. (use EXCEL, submit the EXCEL file too) (20)
• 2.3. Explain how the values of the transfer function zeros affect the shape of the response curves? (4)

A process has a transfer function with the following equation ?(?)=?(???+1)(?1?+1)(?2?+1) with =2,⁡ ⁡?1=10,?2=2. If ?? has the following case values

Case i: ??=20

Case ii: ??=4

Case iii: ??=1

Case iv:??=−2

For each case (i-iv)

[30

Answer 1

Plotting the response in Excel

DEclaring the function in Ms-Excel

- O X Microsoft Visual Basic for Applications - New Microsoft Excel Worksheet.xlsx Eile Edit View Insert Format Debug Run Tools Add-Ins Window - B Da K Project - VBAProject X Help Ln 8. Col 18 o x New Microsoft Excel Worksheet.xlsx - Module1 (Code) (General) 6-8 VBAProject (New Microsi 6- Microsoft Excel Objects 1 Sheet1 (Sheet1) ThisWorkbook 66 Modules Module 1 Public Function funl(t As Double, taoA As Double) funl = 1 - ((Exp(-t / taoА))/ ((1 - (10 / taoA)) * (1 - (2 / taoА)))) + ((1.25 * Exp(-0.1 * t)) / (1 - (0.1 * End Function Public Function Yt As Double, taoA As Double) A = Exp(-t / taoA) / ((1 - (10 / taoA)) * (1 - (2 / taoА))) B = (1.25 * Exp(-0.1 * t)) / (1 - (0.1 * taoA)) C = (0.25 * Exp(-0.5 * t)) / (1 - (0.5 * taoA)) Y = 1 - A + B + a End Function Properties - Module1 Module1 Module Alphabetic Categorized (Name) Module 1 A » POENG 17:38 25-03-20202

Sign in REVIEW 2 – FILET HOME INSERT PAGE LAYOUT FORMULAS DATA % Cut Calibri -11 YA À == LE Copy Paste B IU - A E Format Painter EE Clipboard Font VIEW COMSOL 5.3 & Wrap Text General Merge & Center, , % Σ AutoSum ΑΣ X 209 Conditional Format as Cell Formatting Table Styles Styles Insert Delete Format - Cells Fill Clear Sort & Find & Filter, Select Alignment Number Editing R18 A D F G H I K L M O P Q R S T B taoA E Case 2 N Case 4 Case 3 20 taoA taoA taoA 0 C Y(t,taoA) -2.5 20 -2.26174 20 -2.04438 20 -1.84491 20 -1.66106 20 -1.49111 20 -1.33366 20 -1.18754 20 -1.05177 20 -0.92547 20 -0.80788 20 -0.69831 20 -0.59614 20 -0.50081 20 -0.41179 20 -0.32863 20 -0.25089 20 -0.17817 20 -0.11012 20 -0.04639 20 0.01332 Y(t,taoA) 4 4.166667 4 3.771846 4 3.422427 4 3.117411 4 2.853172 4 2.625091 4 2.428418 4 2.258702 4 2.11197 4 1.984775 4 1.874178 4 1.777697 4 1.693251 4 1.619098 4 1.553779 4 1.496073 4 1.444955 4 1.399559 4 1.359154 4 1.323118 Y(t) 1 2.777778 1 2.519108 1 2.306029 1 2.134947 1 1.996633 1 1.882698 1 1.786857 1 1.704699 1 1.633189 1 1.570221 1 1.514308 1 1.464362 1 1.419564 1 1.379268 1 1.342952 1 1.31018 1 1.28058 1 1.253829 1 1.229643 1 1.207772 Y(t) -22.083333 -2 1.880962 -2 1.672306 -2 1.426103 -2 1.099412 -2 0.626856 -2 -0.09589 -2 -1.23857 -2 -3.07951 -2 -6.07653 -2 -10.9837 -2 -19.0437 -2 -32.305 -2 -54.1444 -2 -90.1291 -2 -149.438 -2 -247.203 -2 -408.374 -2 -674.085 -2 -1112.15 Sheet1 READY M U - + 100% 17:38 25-03-20202 13 A lla 1) - ENG

The plot is given as

Response curve - Y(t) Y(t) vs t case 3 Y(t)v/st case 1 Y(t) vs t case 2 Y(t) vs t case 4 - 10

There is no Zeroes here but with the introduction of zeroes

1) Adding left hand plane(-ve) zero would increase stability and increase overshoot of the response

2)Adding right hand plane(+ve) zero to the transfer function would slow down the response

Case 4 here is unstable