online learning Pearon BTC Higher National in lectrical and lectronic ngineering (QCF) Unit 5: lectrical & lectronic Principle Unit Workbook 4 in a erie of 4 for thi unit Learning Outcome: RLC Tranient Page 1 of 20
Content INTRODUCTION... 3 GUIDANC... 3 4.1 Laplace Tranform... 4 4.1.1 Definition of the Laplace Tranform of a Function... 4 4.1.2 Uing a Table of Laplace Tranform... 5 4.2 Tranient Analyi... 6 4.2.1 xpreion for Component and Circuit Impedance in the -plane... 6 4.2.2 Laplace Solution of Firt Order Sytem... 7 4.2.3 Second Order Sytem... 12 4.3 Circuit Repone... 15 4.3.1 Theory... 15 4.3.2 Overdamped Repone following a Step Input... 17 4.3.3 Underdamped Repone following a Step Input... 18 4.3.4 Critically-damped Repone following a Step Input... 19 4.3.5 Zero-damped Repone following a Step Input... 20 Page 2 of 20
INTRODUCTION Thi Workbook guide you through the learning outcome related to: Laplace tranform: definition of the Laplace tranform of a function; ue of a table of Laplace tranform Tranient analyi: expreion for component and circuit impedance in the -plane; firt order ytem mut be olved by Laplace (i.e. RL and RC network); econd order ytem could be olved by Laplace or computer-baed package Circuit repone: over, under, zero and critically damped repone following a tep input; zero initial condition being aumed GUIDANC Thi document i prepared to break the unit material down into bite ize chunk. You will ee the learning outcome above treated in their own ection. Therein you will encounter the following tructure; Purpoe Theory xample Quetion Challenge Video xplain why you need to tudy the current ection of material. Quite often learner are put off by material which doe not initially eem to be relevant to a topic or profeion. Once you undertand the importance of new learning or theory you will embrace the concept more readily. Convey new material to you in a traightforward fahion. To upport the treatment in thi ection you are trongly advied to follow the given hyperlink, which may be ueful document or application on the web. The example/worked example are preented in a knowledge-building order. Make ure you follow them all through. If you are feeling confident then you might like to treat an example a a quetion, in which cae cover it up and have a go yourelf. Many of the example given reemble aignment quetion which will come your way, o follow them through diligently. Quetion hould not be avoided if you are determined to learn. Pleae do take the time to tackle each of the given quetion, in the order in which they are preented. The order i important, a further knowledge and confidence i built upon previou knowledge and confidence. A an Online Learner it i important that the anwer to quetion are immediately available to you. You will find the anwer, upide down, below each et of quetion. Contact your Unit Tutor if you need help. You can really cement your new knowledge by undertaking the challenge. A challenge could be to download oftware and perform an exercie. An alternative challenge might involve a practical activity or other form of reearch. Video on the web can be very ueful upplement to your ditance learning effort. Wherever an online video() will help you then it will be hyperlinked at the appropriate point. Page 3 of 20
4.1 Laplace Tranform 4.1.1 Definition of the Laplace Tranform of a Function If the independent variable of a function i time (t) then the function may be written a f(t). The Laplace Tranform of f(t) i given by... L{f(t)} = e t f(t) dt Here, i a complex number in the frequency domain. We can think of a merely a parameter to manipulate. We may repreent it a jω, and thi equivalence will be ueful to u later on. 0 When we analye circuit containing combination of reitor, capacitor and/or inductor it i quite uual to come up with firt and econd-order differential equation. To eae thi burden of analyi we ue Laplace Tranform, which convert circuit analyi into a purely algebraic problem. Laplace Tranform may be deduced by performing the integration, a above, but thi oon become a very tediou exercie. Virtually all of the Laplace Tranform we meet in engineering have already been evaluated (i.e. the integration are done for u). We imply need to ue a table containing thee evaluated tranform. It i informative at thi point to try a couple of thee integration for a unit tep ource (cloing a witch to connect to a DC upply) and a unit ramp ource (linearly increaing to 1 V DC ource). Thee exemplify the reaon why we wih to avoid the additional mathematic. Find the Laplace Tranform of f(t) = 1 (unit tep function) L{1} = e t (1) dt = e t dt = [ e t ] 0 0 0 = 1 (e e 0 ) = 0 1 = 1 = 1 That function wa jut the number 1, o any number can be evaluated in the ame way i.e. L{8} = 8. Find the Laplace Tranform of f(t) = t (unit ramp function) L{t} = e t (t) dt 0 Integration by part i required here, a it i for virtually all function for which we need to find the Laplace Tranform. Let do that then... Page 4 of 20
L{t} = e t (t) dt 0 = [ e = [ te t e t 0 dt ] = [ te t 0 e t 2 ] 0 e 2 ] [0e 0 e 0 2 ] = [0 0] [0 1 2] = 1 2 We have managed to find the right anwer to thee mall problem. You could perform all the neceary integration for many other function, but why would you need to do that when they have already been done for you? The reult are found in a table of Laplace Tranform, a previouly mentioned, which we now turn to. 4.1.2 Uing a Table of Laplace Tranform Here i a hort table of Laplace Tranform which will be very ueful when analying circuit. Have a look through it (two of them you already know). We hall then dicu how it i ued. Function of time, f(t) Laplace Tranform of f(t), L{f(t)} 1 δ (unit impule) 1 2 1 (unit tep function) 3 t (unit ramp) 4 e at (exponential growth) 5 e at (exponential decay) 6 in (ωt) 7 co (ωt) 8 e at in (ωt) (decaying ine wave) 9 e at co (ωt) (decaying coine wave) 10 d f(t) dt (firt differential) 1 1 2 1 a 1 + a ω 2 + ω 2 2 + ω 2 ω ( + a) 2 + ω 2 + a ( + a) 2 + ω 2 F() f(0) d 2 f(t) 11 (econd differential) d f(0) dt 2 2 F() f(0) dt 1 12 f(t) dt (integral) F() + 1 f(0) Page 5 of 20
We have already mentioned the fact that Laplace Tranform allow u to convert differential equation into eay algebraic problem that their beauty. Here i a procedure for analying a circuit and uing the table to find olution to circuit parameter. 1) Repreent the circuit in term of an expreion involving voltage, current and component value 2) Tranform each term in the expreion eparately, uing the table 3) Simplify the expreion a bet a you can. If there are any initial value then put them in. nter the component value given in the circuit. nure that the implification repreent expreion which the table can handle. 4) Having found an expreion which look like one of the term on the right-hand-ide of the table, look to the left to find the invere Laplace Tranform (i.e. the circuit olution). Thi whole topic and proce never look too undertandable the firt time you meet it. Soon enough you hall begin to ee the light by way of example circuit. 4.2 Tranient Analyi 4.2.1 xpreion for Component and Circuit Impedance in the -plane It wa mentioned earlier that we can repreent a jω. When we analye the reactance of an inductor we have the uual expreion... +jx L = +j2πfl = +jωl Notice the jω in that expreion. We are aying that jω can be repreented by, o let do that... +jωl L When we analye the reactance of a capacitor we have the uual expreion... jx C = j 1 2πfC = j 1 ωc = 1 jωc Notice the jω in that expreion alo. We are aying that jω can be repreented by, o let do that... jx C 1 C We know that inductive and capacitive reactance are dependent upon the frequency ued. Their value will vary with frequency. Pure reitor are different, their value do not vary with frequency, o we jut think of reitor a contant in our expreion. If we had a reitor and inductor in erie then we can evaluate the overall impedance of the combination in the -domain a follow... Page 6 of 20
Z() = R + L If we had a reitor and capacitor in erie then we can evaluate the overall impedance of the combination in the -domain a follow... Z() = R + 1 C If we had a reitor and inductor in parallel then we can evaluate the overall impedance of the combination in the -domain a follow... Z() = R L R + L = LR R + L If we had a reitor and capacitor in parallel then we can evaluate the overall impedance of the combination in the -domain a follow... Z() = R 1 C R R + 1 = CR + 1 C If we had a reitor, inductor and capacitor all in erie then we can evaluate the overall impedance of the combination in the -domain a follow... Z() = R + L + 1 C If we had a reitor, inductor and capacitor all in parallel then we can evaluate the overall impedance of the combination in the -domain a follow... 1 Z() = 1 R + 1 L + 1 = 1 1 R + 1 L + C C After finding the common denominator, manipulating and flipping the expreion we get... Z() = LR 2 RLC + L + R It doen t matter how complicated the circuit i. There may be erie element and parallel element. Jut analye them a you ee them, then write down the expreion in the -domain. 4.2.2 Laplace Solution of Firt Order Sytem We finally come to the real nitty-gritty of all thi Laplace theory applying it to real circuit to ee if it can horten our olution by not having to olve differential equation. Let look at a couple of worked example. Page 7 of 20
Worked xample 1 Conidering the circuit below, ue Laplace Tranform to find an expreion for the current flowing after the witch i cloed. Aume i = 0 when t = 0. We tart our olution by uing Kirchhoff Voltage Law (KVL) around the circuit, once the witch i cloed... = V R + V L We then form a firt order differential equation involving the current in the circuit. The current i a function of time o we hall denote it by i(t)... d i(t) = Ri(t) + L dt Our next tep i to take Laplace Tranform (LT ) for the term in the expreion we currently have. Our table of Laplace Tranform now become ueful. i a unit tep function, ince the battery voltage i introduced in a udden tep by the cloing of the witch. We can then write... Some point to note here: = Ri() + L[F() f(0)] When taking the LT of i(t) we imply write i() F() i our function for current, o we imply replace it with i() f(0) i the current when time i zero. Thi wa tated a zero in the quetion, o we omit it. = Ri() + L i() i() = = i()[r + L] (R + L) We are trying to get thi lat expreion into a format whereby we can read off the anwer in the table from right to left. It doen t look like anything in the table at the moment. What we mut do i to take partial fraction (remember thoe from your Analytical Method unit)... Page 8 of 20
Multiplying both ide by (R + L) give... (R + L) A + B R + L = A(R + L) + B We now need to find value for A and B. The implet tarting point i to let be zero... Let = 0 = AR A = R Now that we have found A we mut think of an eay way to find B. If we make the bracket equal to zero would iolate B for u. In order to make the bracket equal to zero we mut make equal to R L. Let = R L = B ( R L ) B = L R We may now put thee expreion for A and B into our main developed expreion... (R + L) A + i() = B R + L = R R L + R R + L L + R R + L = i() We now try to reolve thi expreion into a format which i table-friendly... i() = R (1 i() = R (1 L L + R ) L L ( L ) L ) + (R L ) i() = R (1 1 + ( R L ) ) There are two term in the big bracket here. Both of them look like entrie in our table of LT. The firt term i from row 2 of the table and the econd term i from row 5 of the table. Notice that R/L i taken to Page 9 of 20
be the quantity a mentioned in the table. We may now take invere Laplace Tranform (uually written a L 1 {F()})... i(t) = R (1 e Rt L) We have managed to find an LT olution for our firt circuit. We were given component value in the quetion o let put them into our new expreion... i(t) = 10 1 (1 e (1)t 0.001) i(t) = 10(1 e 1000t ) [A] So much for an RL circuit then. Let now have a look at an RC circuit. Worked xample 2 Conidering the circuit below, ue Laplace Tranform to find an expreion for the current flowing after the witch i cloed. Aume i = 0 when t = 0. We tart our olution by uing Kirchhoff Voltage Law (KVL) around the circuit, once the witch i cloed... = V R + V C We then form a firt order differential equation involving the current in the circuit... = Ri(t) + 1 i(t) dt C Our next tep i to take Laplace Tranform (LT ) for the term in the expreion we currently have... = Ri() + 1 C (1 F() + 1 f(0)) The initial condition given are: i = 0 when t = 0, o we ignore f(0). A before, we take F() to be i()... Page 10 of 20
i() = R + 1 C = Ri() + i() C i() = = i() (R + 1 C ) = C RC + 1 = (R + 1 C ) C RC ( RC RC ) + ( 1 RC ) = i() = R ( 1 + ( 1 ) RC ) R + ( 1 RC ) We now need to take the invere LT for the expreion in the large bracket. Notice that 1/RC i the ame quantity meant by a in our table. Our anwer for thi circuit i therefore, again, taken from row 5 of the table... i(t) = R e ( 1 RC )t We were given component value in the quetion o let put them into our expreion for current... i(t) = 10 1 1 10 6 e ( 1 10 6 1 10 6)t = 10 5 e t [A] Page 11 of 20
4.2.3 Second Order Sytem Firt order ytem are thoe where the highet differential i of order 1 for example, they feature di/dt a the highet differential. You aw an example of thi in the RL circuit analyed previouly. Second order ytem feature a differential of order 2 i.e. d 2 i dt 2. Thee ytem feature an inductor and a capacitor. Let look at a worked example involving a erie RLC combination. Worked xample 3 Conidering the circuit below, ue Laplace Tranform to find an expreion for the current flowing after the witch i cloed. Aume i = 0 when t = 0. After the witch i cloed we can apply Kirchhoff Voltage Law... Take Laplace Tranform... 1 1 = v R + v L + v C = R1 i() + L1 i() + i() C At thi point it i helpful to put in the component value (implifie the algebra)... Now divide top and bottom by 0.1 20 = 10 i() + 0.1 i() + i() 20 10 5 = i() (10 + 0.1 + 1 10 5 ) 20 20 i() = (10 + 0.1 + 1 = 10 5 ) 0.1 2 + 10 + 10 5 200 i() = 2 + 100 + 10 6 We now need to get the expreion into a form uitable for table ue. We wih to expre it in the following form... Page 12 of 20
ω ( + a) 2 + ω 2... which i from row 8 of the table. The way to do thi i to look at that 100 in the denominator, divide it by 2, which give u 50, then quare that 50 to give 2,500 which we then add and ubtract to the denominator. Thi proce give u... i() = 200 2 + 100 + 10 6 = 200 2 + 100 + 2500 2500 + 10 6 Adding and ubtracting 2500 doe not change it, of coure. The value in doing thi, however, i that we may factorie the firt three term in the denominator, a follow... 2 + 100 + 2500 ( + 50) 2...which look a bit more like the expreion in row 8 of the table. The lat two term evaluate to... 2500 + 1,000,000 = 997,500 We need an ω 2 to repreent 997,500, o the value of ω mut be the quare root of 997,500 i.e. We can now bring thee development together... ω 2 = 997,500 ω = 997,500 998.75 i() = 200 2 + 100 + 10 6 = 200 ( + 50) 2 + (998.75) 2 Now the denominator look very much like what we need. We till have a bit of work to do on the numerator though. We d like ω to appear there. The value of ω i 998.75 but the numerator mut evaluate to 200. We work out the numerator a follow... We may now write... 998.75 ome number = 200 ome number = 200 998.75 0.2 i() = 0.2 998.75 ( + 50) 2 + (998.75) 2 Which can be written a... 998.75 i() = 0.2 ( ( + 50) 2 + (998.75) 2) The expreion in the large bracket i now ready for an Invere Laplace Tranform (row 8 of the table remember)... i(t) = 0. 2e 50t in(998. 75 t) [A] Page 13 of 20
Thee calculation are fairly complex, o it i alway valuable to ue the MicroCap imulator (perform a Tranient Analyi and make ure you turn OFF the Operating Point checkbox). The imulator produce the following waveform for thi circuit... Notice that waveform ha a decaying exponential envelope. The yellow marker i there o that a check can be made on our calculation. The imulator i aying that after 20 m the circuit current ha a value of around 66 ma. Let ue our derived expreion to ee if thi i o... i(t) = 0.2e 50t in(998.75 t) [A] Let t = 0.02 i(0.02) = 0.2e 50 0.02 in(998.75 0.02) = 66.4 ma Look like we re ok! If you zoom right in on the imulator reult at 20 m the anwer will be exactly right (it jut a matter of mathematical reolution). Page 14 of 20
4.3 Circuit Repone Theory 4.3.1 Theory For a erie RLC circuit there are number of way that the circuit repone may behave. We explore thee behaviour here with the initial aid of ome reviion from the Further Analytical Method unit Ordinary Differential quation (worth a quick reviion of Workbook 4 for that unit). Given a erie RLC circuit connected to ome upply voltage V we may employ Kirchhoff Voltage Law and write... We know that... V = v R + v L + v C [1] We alo know that... If we ubtitute [3] into [2] we get... v L = L di dt i = C dv C dt [2] [3] Furthermore, we know... v L = L d dt {C dv C dt } = LC d2 v C dt 2 [4] v R = ir = (C dv C dt ) R v R = RC dv C dt Subtituting equation [5] and [4] into equation [1] give... [5] Rearranging... V = RC dv C dt + LC d2 v C dt 2 + v C LC d2 v C dt 2 + RC dv C dt + v C = V [6] Page 15 of 20
quation [6] i a linear 2 nd order differential equation. To olve it we turn it into a homogeneou type by forcing V to zero... The tandard olution here i... LC d2 v C dt 2 + RC dv C dt + v C = 0 [7] Let v C = Ae mt [8] dv C dt = Amemt [9] Subtituting [8], [9] and [10] into [7] give... Taking out the common factor... We can then ay that Ae mt i a olution only if... d 2 v C dt 2 = Am2 e mt [10] LC(Am 2 e mt ) + RC(Ame mt ) + Ae mt = 0 Ae mt (m 2 LC + mrc + 1) = 0 [11] m 2 LC + mrc + 1 = 0 [12] quation [12] i known a the Auxiliary quation (remember?) Our auxiliary equation i in quadratic form, which we mut olve uing the quadratic formula... m = RC ± [(RC)2 4(LC)(1)] 2LC = RC ± R2 C 2 4LC 2LC m = RC 2LC ± C 2 4LC R2 (2LC) 2 = R 2L ± R2 C 2 4L 2 C 2 4LC 4L 2 C 2 m = R 2L ± R2 4L 2 1 LC m = R 2L ± ( R 2L ) 2 1 LC [13] We hall next invetigate the poible propertie of equation [13] to determine the characteritic of an RLC erie circuit. Page 16 of 20
4.3.2 Overdamped Repone following a Step Input m = R 2L ± ( R 2L ) 2 1 LC If m ha two different real root then the erie RLC circuit i aid to be overdamped. For thi to be the cae then... ( R 2L ) 2 > 1 LC An overdamped erie RLC circuit i hown below, along with it tranient repone... The tranient current die away very lowly with time. Page 17 of 20
4.3.3 Underdamped Repone following a Step Input m = R 2L ± ( R 2L ) 2 1 LC If m ha two complex root then the erie RLC circuit i aid to be underdamped. For thi to be the cae then... ( R 2L ) 2 < 1 LC An underdamped erie RLC circuit i hown below, along with it tranient repone... The current ocillate around a teady value and die away lowly. Page 18 of 20
4.3.4 Critically-damped Repone following a Step Input m = R 2L ± ( R 2L ) 2 1 LC If m ha two real equal root then the erie RLC circuit i aid to be critically-underdamped. For thi to be the cae then... ( R 2L ) 2 = 1 LC A critically-damped erie RLC circuit i hown below, along with it tranient repone... Notice that thi tranient repone look imilar to the overdamped cae. The difference i that the repone die away much more quickly in the critically-damped circuit. Page 19 of 20
4.3.5 Zero-damped Repone following a Step Input In thi cae the reitance i zero. The circuit i alo known a being undamped. A zero-damped erie RLC circuit i hown below, along with it tranient repone... Auming an ideal ource (no internal reitance) then the ocillation continue indefinitely. Page 20 of 20