6.3.4 Modified Euler s method of integration

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1 6.3.4 Modfed Euler s method of ntegraton Before dscussng the applcaton of Euler s method for solvng the swng equatons, let us frst revew the basc Euler s method of numercal ntegraton. Let the general from of a dfferental equaton s gven by; dx = f(x, y); y(x o) = y o ; (6.28) In equaton (6.28), x and y are ndependent and dependent quanttes respectvely and x o and y o are ntal values of x and y respectvely. For the purpose of numercal ntegraton, the ndependent axs (x-axs) s dvded nto ntervals of length h such that dscrete ponts on the ndependent axs are x o, x o + h, x o + 2h, etc. As ndcated n equaton (6.28), the value of y at x = x o s y o. The task s to calculate the values y 1, y 2, correspondng to the x co-ordnates x o + h, x o + 2h, respectvely. Once these values are obtaned, the smooth curve representng the soluton of the dfferental equaton gven n equaton (6.28) can be plotted. In the modfed Euler s method, the values y 1, y 2, are calculated n two steps: Predctor In ths step, the approxmate value of y 1 (denoted as y (1) 1 ) as; Corrector y (1) 1 = y o + h dx x=x o = y o + hf(x o, y o ) (6.29) a) Wth the calculated value of y (1) 1, calculate the approxmate value of dx at x = x o + h as; dx = f(x o + h, y (1) 1 ) (6.30) x=x o+h b) Wth ths updated value of dx at x = x 1 = x o + h, the fnal value of y 1 s calculated as; y 1 = y o + h 2 [ dx + x=x o dx ] x=x o+h or, y 1 = y o + h 2 [f(x o, y o ) + f(x o + h, y (1) 1 )] (6.31) Wth ths fnal value of y 1 obtaned at x = x o + h = x 1, the above two steps are repeated to calculate y 2 at x = x o + 2h = x 2 and subsequently, ths process s repeated to obtan the complete soluton of the dfferental equaton. Now, for our applcaton, let us note that the ndependent axs (x-axs) denotes tme. Therefore, as alrea dscussed, for solvng the dfferental equatons, the tme axs s dvded nto ntervals of duraton t sec. (.e. h = t). Further, let us also assume that the values of δ and ω ( = 1, 2, m) have alrea been obtaned at t = t o and these values are denoted as δ o and ω o respectvely. Moreover, the ntal values of P e (denoted as P (o) e ) are also assumed to be calculated utlsng the values 261

2 of δ o. Wth thse known values, the values of δ and ω at t = t o + t are calculated as follows. Predctor step In ths step, the approxmate values of δ and ω are calculated as (denoted as δ (1) and ω (1) respectvely); δ (1) = δ o + t dδ dt t=t o = δ o + t(ω o ω s ) (6.32) Corrector step ω (1) = ω o + t dω dt t=t o = ω o + t ω s 2H (P m P (o) e ) (6.33) Wth the new values of δ (1) ( = 1, 2, m) obtaned n the predctor step, the values of P e ( = 1, 2, m) are updated usng equatons (6.23)-(6.25) (after approprately ncorporatng the network condtons n the equaton set (6.23)). Let these updates values of P e be denoted as P (1) e. Thereafter, the dervatves at the end of the present tme step are calculated as follows: dδ dt = ω (1) ω s (6.34) dω dt (P m P (1) e ) (6.35) 2H Wth the above new dervatve values obtaned, the fnal values of δ and ω at t = t o + t (denoted as δ 1 and ω 1 respectvely) are calculated as; δ 1 = δ o + t 2 [ dδ dt + dδ t=t o dt ] (6.36) ω 1 = ω o + t 2 [ dω dt + dω t=t o dt ] (6.37) Proceedng further, for calculatng δ and ω at t = t o +2 t, the quanttes δ o and ω o are replaced by δ 1 and ω 1 respectvely and equatons (6.32)-(6.37) are followed agan Runga Kutta 4 th order method of ntegraton Let us agan consder the same general form of a dfferental equaton as n equaton (6.28): dx = f(x, y); y(x o) = y o ; (6.38) Agan, the meanngs of dfferent notatons used n equaton (6.38) are same as those n equaton 262

3 (6.28). In RK 4 th order method, the value y 1 (correspondng to x = x o + h) s calculated as; y 1 = y o + h 6 (k 1 + 2k 2 + 2k 3 + k 4 ) (6.39) In equaton (6.39), k 1 = hf(x o, y o ) (6.40) k 2 = hf (x o + h 2, y o + k 1 2 ) (6.41) k 3 = hf (x o + h 2, y o + k 2 2 ) (6.42) k 4 = hf (x o + h, y o + k 3 ) (6.43) Now, for solvng the transent stablty problem wth RK 4 th order method, let us agan assume that the value of δ and ω ( = 1, 2, m) are known at t = t o (denoted as δ o and ω o respectvely). Moreover, the values of P e are also assumed to be known (calculated utlsng the values of δ o ). From these ntal values, the procedure of calculaton of δ 1 and ω 1 (values of δ and ω at t = t o + t) s as follows. Calculaton of frst estmate of the dervatves In ths step, the frst estmates of the dervatves for the th calculated as: as: machne ( = 1, 2, m) are dδ = ω o ω s (6.44) dω 2H [P m P (o) e ] (6.45) Wth these frst estmates of the dervatves, the values of δ and ω ( = 1, 2, m) are updated δ (1) = δ o t dδ (6.46) ω (1) = ω o t dω (6.47) Wth these values of δ (1) ( = 1, 2, m), the values of P e ( = 1, 2, m) are updated from equatons (6.23) - (6.25). Let these newly calculated values of P e be denoted as P (1) e ( = 1, 2, m). We now proceed to the next step. Calculaton of second estmate of the dervatves 263

4 The second estmates of the dervatves are calculated as: dδ = ω (1) ω s (6.48) dω 2H [P m P (1) e ] (6.49) Wth these second estmates of the dervatves, the values of δ and ω ( = 1, 2, m) are updated as: δ (2) = δ o t dδ (6.50) ω (2) = ω o t dω (6.51) Wth these values of δ (2), the values of P e are agan updated from equatons (6.23) - (6.25). Let these newly calculated values of P e be denoted as P (2) e ( = 1, 2, m). We now proceed to the next step. Calculaton of thrd estmate of the dervatves The thrd estmates of the dervatves are calculated as; dδ = ω (2) ω s (6.52) as; dω 2H [P m P (2) e ] (6.53) Wth these thrd estmates of the dervatves, the values of δ and ω ( = 1, 2, m) are updated δ (3) = δ o + t dδ (6.54) ω (3) = ω o + t dω (6.55) Wth these values of δ (3), the values of P e are agan updated from equatons (6.23) - (6.25). Let these newly calculated values of P e be denoted as P (3) e ( = 1, 2, m). We now proceed to the next step. Calculaton of fourth estmate of the dervatves 264

5 The fourth estmates of the dervatves are calculated as: dδ = ω (3) ω s (6.56) dω 2H [P m P (3) e ] (6.57) After the fourth estmates are obtaned, we are now n a poston to calculate δ 1 ( = 1, 2, m). and ω 1 Calculaton of fnal values The fnal, updated values are calculated as: δ 1 = δ o + t 6 [ dδ + 2 dδ + 2 dδ + dδ ] (6.58) ω 1 = ω o + t 6 [ dω + 2 dω + 2 dω + dω ] (6.59) Proceedng further, for calculatng δ and ω at t = t o +2 t, the quanttes δ o and ω o are replaced by δ 1 and ω 1 respectvely and equatons (6.44)-(6.59) are followed agan. In the next lecture, we wll llustrate the applcaon of Modfed Euler s method for transent stablty calculaton. 265

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