8/9/2019 Complete Solution of Transient Response of Series RLC Circuit for Step Excitation

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Complete solution of transient response of series RLC circuit for step excitation

Partha Sarathi Sengupta To analyze the response of series RLC we must first know two fundamental concepts.

1. General Solution of a 2ndorder O! with constant forcin" function #$% consists ofcomplementary function and particular inte"ral.2. The important physical property concernin" this pro&lem is initial conditions in a

capacitor and inductor' we should know that current in an inductor or (olta"e in acapacitor can)t rise a&ruptly unless there are impulses or their deri(ati(es dri(in" them.

*pplyin" +$L in the a&o(e circuit loop .,e know that the

current throu"h and (olta"e across a capacitor is related &y' . Su&stitutin"

this relation in +$L e-uation a&o(e we "et the differential e-uation/

0or non/homo"eneous 2nd order linear O! with constant coefficients the "eneralsolution is o&tained &y findin" the roots of *uiliary e-uation #m 1' m2% usin" the roots toform the C.0 #complementary function% and addin" the .3 #particular inte"ral% with it.

The roots of *.! are

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ampin" 0actor ; 4atural fre-uency

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/

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The form of C.0 will depend on the nature of the roots of *.! which in turn depends upon the (alue of 5 .

Ran"e of 5 4ature of roots of *.! 0orm of Complementary 0unction

5 61

Real and distinct A1em1x+ A2em2x (m1'm2 roots of *.!%

57 1 Real and e-ual#dou&le root%

#A1 8A2x) emx #mis the dou&le root%

5 91 Comple con:u"ate (p + :q),(p - ep#A1cos qx 8 A2sinqx %

8/9/2019 Complete Solution of Transient Response of Series RLC Circuit for Step Excitation

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:q)

Case 1: 561

The articular inte"ral #.3% of a non/homo"eneous O..! is defined as any function ofindependent (aria&le which satisfies the O..! . 3n case the ecitation function is constant#as in this case% the constant is the .3 . This can &e (erified &y su&stitutin" $;#t%7$ in theO..! and findin" that L.

8/9/2019 Complete Solution of Transient Response of Series RLC Circuit for Step Excitation

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Let)s plot the responses for some (alue of 5 61 to try "au"e the shape and nature of the response/

5 >52/1 #/5 8>5 2/1 % #/5 />52/1 % Response $;#t%

1.? 1.12 /;.@A /2.B2 $ ;#t%7 18 ;.1 e/2.B2D;t / 1.1 e/;.@AD;t i#t%7 /;.E?D;e/2.B2D;t 8 ;.EE? D;e/;.@AD;t

Response $;#t%' i#t% in terms ofxF scale of ais 71;; D;t F $7 1 (olt' D;7 1+hz' C710

0or the a&o(e (alues it can &e seen that transient (olta"e and current ceases in 1.B ms

Case 2: 5 7 1Su&stitutin" root #m1' m2% as #/D;% we can form the C.0 as

(A1+A2t) e-0 t

and G.SV0(t) = V + (A1+A2t) e-0 t///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

Su&stitutin" initial conditions $;#;%7 ; we find that A1= -V

8/9/2019 Complete Solution of Transient Response of Series RLC Circuit for Step Excitation

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sin" and i(0)=0we can deri(eA2= -0V' su&stitutin" (ales of constants

in G.S we find $;#t% and i#t%.

0rom the (alues of this eample it can &e said that transient (olta"e and current ceases at appro 1;ms.

Case 3: 5 9 10or 591 roots are comple con:u"ate' G.S is of the form

Su&stitutin" initial condition of V0(0)=0 we can find A1=-V. Su&stitutin" i(0)=0 we find

let us define =cos-1and sinceA1andA2&oth are

ne"ati(e lies in @rd-uadrant and its (alue is in &etween H and ' the &elow circle will &e an

useful reference actual (alue of calculated usin" the &elow epression' &rackets denote

ma"nitude.

Sol(in" &y su&stitutin" (alue ofA1andA2we "et the (olta"e and current epressions as &elow

V0(t)= V[1- (1 + 0t) e-0 t] and i(t)= CV 02t e-0

8/9/2019 Complete Solution of Transient Response of Series RLC Circuit for Step Excitation

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,we can further deri(e that

4et we plot and i (t)for particular (alues of , ;.2' ;.? and ;.A