Page 1 6.4 Small signal analysis For small signal analysis of a multimachine power system, we need to linearise the di?erential equa tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system having ‘m’ generators. Each generator is represented by its classical model. Further, without any loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system. Now, for linearising the di?erential equations, let us recall the swing equations of each generator here for ready reference. dd i dt =? i ? s for i= 1, 2,??m (6.64) 2H i ? s d? i dt =P mi P ei for i= 1, 2,??m (6.65) Linearising equation (6.64) for thei th generator, one can get, d?d i dt = ?? i (6.66) Now, let us de?ne, ?d = [?d 1 , ?d 2 , ?? ?d m ] T (6.67) ?? = [?? 1 , ?? 2 , ?? ?? m ] T (6.68) In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values of rotor angle and machine speed respectively. Please note that the size of each of these two vectors is (m× 1). Now, collecting equation (6.66) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, d dt ?d =F 1 ?? (6.69) In equation (6.69),F 1 is a (m×m) identity matrix. Now, linearising equation (6.65) for thei th generator, one can get, 2H i ? s d?? i dt =?P ei (6.70) For performing linearisation ofP ei , its expression its required. This expression can be derived as follows. From equation (6.25), one can write, ¯ I i = E i ?d i V i ?? i jx / di = E i x / di e j(d i p~2)  V i x / di e j(? i p~2) 278 Page 2 6.4 Small signal analysis For small signal analysis of a multimachine power system, we need to linearise the di?erential equa tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system having ‘m’ generators. Each generator is represented by its classical model. Further, without any loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system. Now, for linearising the di?erential equations, let us recall the swing equations of each generator here for ready reference. dd i dt =? i ? s for i= 1, 2,??m (6.64) 2H i ? s d? i dt =P mi P ei for i= 1, 2,??m (6.65) Linearising equation (6.64) for thei th generator, one can get, d?d i dt = ?? i (6.66) Now, let us de?ne, ?d = [?d 1 , ?d 2 , ?? ?d m ] T (6.67) ?? = [?? 1 , ?? 2 , ?? ?? m ] T (6.68) In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values of rotor angle and machine speed respectively. Please note that the size of each of these two vectors is (m× 1). Now, collecting equation (6.66) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, d dt ?d =F 1 ?? (6.69) In equation (6.69),F 1 is a (m×m) identity matrix. Now, linearising equation (6.65) for thei th generator, one can get, 2H i ? s d?? i dt =?P ei (6.70) For performing linearisation ofP ei , its expression its required. This expression can be derived as follows. From equation (6.25), one can write, ¯ I i = E i ?d i V i ?? i jx / di = E i x / di e j(d i p~2)  V i x / di e j(? i p~2) 278 Or, ¯ E i ¯ I * i =E i e jd i ¯ I * i = E i x / di e jp~2  E i V i x / di e j(p~2+d i ? i ) Therefore, P ei = Re? ¯ E I ¯ I * i ?= E i V i x / di sin(d i ? i ) (6.71) Linearisation of equation (6.71) yields, ?P ei =k 1i ?d i +k 2i ?? i +k 3i ?V i (6.72) Where, k 1i = E i V i x / di cos(d i ? i ); k 2i = E i V i x / di cos(d i ? i ); k 3i = E i x / di sin(d i ? i ); (6.73) In equation (6.73), the constantsk 1i ,k 2i andk 3i are evaluated using the values ofE i ,V i ,d i and ? i at the current operating point fori= 1, 2,??m. Now, again let us de?ne, ?? g = [?? 1 , ?? 2 , ?? ?? m ] T (6.74) ?V g = [?V 1 , ?V 2 , ?? ?V m ] T (6.75) ?P e = [?P e1 , ?P e2 , ?? ?P em ] T (6.76) In equations (6.74) and (6.75), the vectors ?? g and ?V g denote the vectors of perturbed values of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P e denotes the perturbed values of the generator real powers. Please note that the size of each of these three vectors is also (m× 1). Now, collecting equation (6.72) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, ?P e =K 1 ?d+K 2 ?? g +K 3 ?V g (6.77) In equation (6.77), the size of each of the matricesK 1 ,K 2 andK 3 is (m×m). Moreover, all these three matrices are diagonal matrices and are given by; K 1 = diag(k 11 , k 12 , ??k 1m ) K 2 = diag(k 21 , k 22 , ??k 2m ) K 3 = diag(k 31 , k 32 , ??k 3m ) ? ? ? ? ? ? ? ? ? ? ? (6.78) 279 Page 3 6.4 Small signal analysis For small signal analysis of a multimachine power system, we need to linearise the di?erential equa tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system having ‘m’ generators. Each generator is represented by its classical model. Further, without any loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system. Now, for linearising the di?erential equations, let us recall the swing equations of each generator here for ready reference. dd i dt =? i ? s for i= 1, 2,??m (6.64) 2H i ? s d? i dt =P mi P ei for i= 1, 2,??m (6.65) Linearising equation (6.64) for thei th generator, one can get, d?d i dt = ?? i (6.66) Now, let us de?ne, ?d = [?d 1 , ?d 2 , ?? ?d m ] T (6.67) ?? = [?? 1 , ?? 2 , ?? ?? m ] T (6.68) In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values of rotor angle and machine speed respectively. Please note that the size of each of these two vectors is (m× 1). Now, collecting equation (6.66) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, d dt ?d =F 1 ?? (6.69) In equation (6.69),F 1 is a (m×m) identity matrix. Now, linearising equation (6.65) for thei th generator, one can get, 2H i ? s d?? i dt =?P ei (6.70) For performing linearisation ofP ei , its expression its required. This expression can be derived as follows. From equation (6.25), one can write, ¯ I i = E i ?d i V i ?? i jx / di = E i x / di e j(d i p~2)  V i x / di e j(? i p~2) 278 Or, ¯ E i ¯ I * i =E i e jd i ¯ I * i = E i x / di e jp~2  E i V i x / di e j(p~2+d i ? i ) Therefore, P ei = Re? ¯ E I ¯ I * i ?= E i V i x / di sin(d i ? i ) (6.71) Linearisation of equation (6.71) yields, ?P ei =k 1i ?d i +k 2i ?? i +k 3i ?V i (6.72) Where, k 1i = E i V i x / di cos(d i ? i ); k 2i = E i V i x / di cos(d i ? i ); k 3i = E i x / di sin(d i ? i ); (6.73) In equation (6.73), the constantsk 1i ,k 2i andk 3i are evaluated using the values ofE i ,V i ,d i and ? i at the current operating point fori= 1, 2,??m. Now, again let us de?ne, ?? g = [?? 1 , ?? 2 , ?? ?? m ] T (6.74) ?V g = [?V 1 , ?V 2 , ?? ?V m ] T (6.75) ?P e = [?P e1 , ?P e2 , ?? ?P em ] T (6.76) In equations (6.74) and (6.75), the vectors ?? g and ?V g denote the vectors of perturbed values of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P e denotes the perturbed values of the generator real powers. Please note that the size of each of these three vectors is also (m× 1). Now, collecting equation (6.72) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, ?P e =K 1 ?d+K 2 ?? g +K 3 ?V g (6.77) In equation (6.77), the size of each of the matricesK 1 ,K 2 andK 3 is (m×m). Moreover, all these three matrices are diagonal matrices and are given by; K 1 = diag(k 11 , k 12 , ??k 1m ) K 2 = diag(k 21 , k 22 , ??k 2m ) K 3 = diag(k 31 , k 32 , ??k 3m ) ? ? ? ? ? ? ? ? ? ? ? (6.78) 279 6.4.1 Linearisation of network equations (at the generator buses) To illustrate the procedure for linearisation of the network equations, let us ?rst consider the ?rst equation of the equation set (6.23) below. ? ¯ Y 11 + ¯ y 1 ? ¯ V 1 + ¯ Y 12 ¯ V 2 +?+ ¯ Y 1m ¯ V m +?+Y 1n ¯ V n = ¯ y 1 ¯ E 1 (6.79) Now, let, ¯ Y ij =Y ij exp(ja ij ); ¯ V i =V i exp(j? i ); for i, j = 1, 2,??n; (6.80) and ¯ y k =y k exp(jß k ); ¯ E k =E k exp(jd k ); for k = 1, 2,??m; (6.81) Utilising equations (6.80) and (6.81) in equation (6.79), we get, Y 11 V 1 exp(j(a 11 +? 1 ))+y 1 V 1 exp(j(ß 1 +? 1 ))+Y 12 V 2 exp(j(a 12 +? 2 ))+?+ Y 1m V m exp(j(a 1m +? m ))+?+Y 1n V n exp(j(a 1n +? n ))=y 1 E 1 exp(j(d 1 +ß 1 )) (6.82) Taking the real part of both the sides of equation (6.82), we get, Y 11 V 1 cos(a 11 +? 1 )+y 1 V 1 cos(ß 1 +? 1 )+Y 12 V 2 cos(a 12 +? 2 )+?+ Y 1m V m cos(a 1m +? m )+?+Y 1n V n cos(a 1n +? n )=y 1 E 1 cos(d 1 +ß 1 ) (6.83) Linearising equation (6.83), we get, Y 11 cos(a 11 +? 1 )?V 1 Y 11 V 1 sin(a 11 +? 1 )?? 1 +y 1 cos(ß 1 +? 1 )?V 1  y 1 V 1 sin(ß 1 +? 1 )?? 1 +Y 12 cos(a 12 +? 2 )?V 2 Y 12 V 2 sin(a 12 +? 2 )?? 2 +?+Y 1m cos(a 1m +? m )?V m Y 1m V m sin(a 1m +? m )?? m +?+ Y 1n cos(a 1n +? n )?V n Y 1n V n sin(a 1n +? n )?? n =y 1 E 1 sin(d 1 +ß 1 )?d 1 (6.84) Equation (6.84) can be rewritten as, a 11 ?V 1 +a 12 ?V 2 +?+a 1m ?V m +?+a 1n ?V n + b 11 ?? 1 +b 12 ?? 2 +?+b 1m ?? m +?+b 1n ?? n =g 1 ?d 1 (6.85) In equation (6.85), a 11 =Y 11 cos(a 11 +? 1 )+y 1 cos(ß 1 +? 1 ); a 12 =Y 12 cos(a 12 +? 2 ); a 1m =Y 1m cos(a 1m +? m ); a 1n =Y 1n cos(a 1n +? n ); b 11 =Y 11 V 1 sin(a 11 +? 1 )y 1 V 1 sin(ß 1 +? 1 ); b 12 =Y 12 V 2 sin(a 12 +? 2 ); b 1m =Y 1m V m sin(a 1m +? m ); b 1n =Y 1n V n sin(a 1n +? n ); g 1 =y 1 E 1 sin(d 1 +ß 1 ); (6.86) Again, taking the real part of both sides of the second equation of the equation set (6.23) (after 280 Page 4 6.4 Small signal analysis For small signal analysis of a multimachine power system, we need to linearise the di?erential equa tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system having ‘m’ generators. Each generator is represented by its classical model. Further, without any loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system. Now, for linearising the di?erential equations, let us recall the swing equations of each generator here for ready reference. dd i dt =? i ? s for i= 1, 2,??m (6.64) 2H i ? s d? i dt =P mi P ei for i= 1, 2,??m (6.65) Linearising equation (6.64) for thei th generator, one can get, d?d i dt = ?? i (6.66) Now, let us de?ne, ?d = [?d 1 , ?d 2 , ?? ?d m ] T (6.67) ?? = [?? 1 , ?? 2 , ?? ?? m ] T (6.68) In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values of rotor angle and machine speed respectively. Please note that the size of each of these two vectors is (m× 1). Now, collecting equation (6.66) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, d dt ?d =F 1 ?? (6.69) In equation (6.69),F 1 is a (m×m) identity matrix. Now, linearising equation (6.65) for thei th generator, one can get, 2H i ? s d?? i dt =?P ei (6.70) For performing linearisation ofP ei , its expression its required. This expression can be derived as follows. From equation (6.25), one can write, ¯ I i = E i ?d i V i ?? i jx / di = E i x / di e j(d i p~2)  V i x / di e j(? i p~2) 278 Or, ¯ E i ¯ I * i =E i e jd i ¯ I * i = E i x / di e jp~2  E i V i x / di e j(p~2+d i ? i ) Therefore, P ei = Re? ¯ E I ¯ I * i ?= E i V i x / di sin(d i ? i ) (6.71) Linearisation of equation (6.71) yields, ?P ei =k 1i ?d i +k 2i ?? i +k 3i ?V i (6.72) Where, k 1i = E i V i x / di cos(d i ? i ); k 2i = E i V i x / di cos(d i ? i ); k 3i = E i x / di sin(d i ? i ); (6.73) In equation (6.73), the constantsk 1i ,k 2i andk 3i are evaluated using the values ofE i ,V i ,d i and ? i at the current operating point fori= 1, 2,??m. Now, again let us de?ne, ?? g = [?? 1 , ?? 2 , ?? ?? m ] T (6.74) ?V g = [?V 1 , ?V 2 , ?? ?V m ] T (6.75) ?P e = [?P e1 , ?P e2 , ?? ?P em ] T (6.76) In equations (6.74) and (6.75), the vectors ?? g and ?V g denote the vectors of perturbed values of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P e denotes the perturbed values of the generator real powers. Please note that the size of each of these three vectors is also (m× 1). Now, collecting equation (6.72) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, ?P e =K 1 ?d+K 2 ?? g +K 3 ?V g (6.77) In equation (6.77), the size of each of the matricesK 1 ,K 2 andK 3 is (m×m). Moreover, all these three matrices are diagonal matrices and are given by; K 1 = diag(k 11 , k 12 , ??k 1m ) K 2 = diag(k 21 , k 22 , ??k 2m ) K 3 = diag(k 31 , k 32 , ??k 3m ) ? ? ? ? ? ? ? ? ? ? ? (6.78) 279 6.4.1 Linearisation of network equations (at the generator buses) To illustrate the procedure for linearisation of the network equations, let us ?rst consider the ?rst equation of the equation set (6.23) below. ? ¯ Y 11 + ¯ y 1 ? ¯ V 1 + ¯ Y 12 ¯ V 2 +?+ ¯ Y 1m ¯ V m +?+Y 1n ¯ V n = ¯ y 1 ¯ E 1 (6.79) Now, let, ¯ Y ij =Y ij exp(ja ij ); ¯ V i =V i exp(j? i ); for i, j = 1, 2,??n; (6.80) and ¯ y k =y k exp(jß k ); ¯ E k =E k exp(jd k ); for k = 1, 2,??m; (6.81) Utilising equations (6.80) and (6.81) in equation (6.79), we get, Y 11 V 1 exp(j(a 11 +? 1 ))+y 1 V 1 exp(j(ß 1 +? 1 ))+Y 12 V 2 exp(j(a 12 +? 2 ))+?+ Y 1m V m exp(j(a 1m +? m ))+?+Y 1n V n exp(j(a 1n +? n ))=y 1 E 1 exp(j(d 1 +ß 1 )) (6.82) Taking the real part of both the sides of equation (6.82), we get, Y 11 V 1 cos(a 11 +? 1 )+y 1 V 1 cos(ß 1 +? 1 )+Y 12 V 2 cos(a 12 +? 2 )+?+ Y 1m V m cos(a 1m +? m )+?+Y 1n V n cos(a 1n +? n )=y 1 E 1 cos(d 1 +ß 1 ) (6.83) Linearising equation (6.83), we get, Y 11 cos(a 11 +? 1 )?V 1 Y 11 V 1 sin(a 11 +? 1 )?? 1 +y 1 cos(ß 1 +? 1 )?V 1  y 1 V 1 sin(ß 1 +? 1 )?? 1 +Y 12 cos(a 12 +? 2 )?V 2 Y 12 V 2 sin(a 12 +? 2 )?? 2 +?+Y 1m cos(a 1m +? m )?V m Y 1m V m sin(a 1m +? m )?? m +?+ Y 1n cos(a 1n +? n )?V n Y 1n V n sin(a 1n +? n )?? n =y 1 E 1 sin(d 1 +ß 1 )?d 1 (6.84) Equation (6.84) can be rewritten as, a 11 ?V 1 +a 12 ?V 2 +?+a 1m ?V m +?+a 1n ?V n + b 11 ?? 1 +b 12 ?? 2 +?+b 1m ?? m +?+b 1n ?? n =g 1 ?d 1 (6.85) In equation (6.85), a 11 =Y 11 cos(a 11 +? 1 )+y 1 cos(ß 1 +? 1 ); a 12 =Y 12 cos(a 12 +? 2 ); a 1m =Y 1m cos(a 1m +? m ); a 1n =Y 1n cos(a 1n +? n ); b 11 =Y 11 V 1 sin(a 11 +? 1 )y 1 V 1 sin(ß 1 +? 1 ); b 12 =Y 12 V 2 sin(a 12 +? 2 ); b 1m =Y 1m V m sin(a 1m +? m ); b 1n =Y 1n V n sin(a 1n +? n ); g 1 =y 1 E 1 sin(d 1 +ß 1 ); (6.86) Again, taking the real part of both sides of the second equation of the equation set (6.23) (after 280 substituting equations (6.80) and (6.81) into it), we get, Y 21 V 1 cos(a 21 +? 1 )+y 2 V 2 cos(ß 2 +? 2 )+Y 22 V 2 cos(a 22 +? 2 )+?+ Y 2m V m cos(a 2m +? m )+?+Y 2n V n cos(a 2n +? n )=y 2 E 2 cos(d 2 +ß 2 ) (6.87) Linearising equation (6.87), one can get, a 21 ?V 1 +a 22 ?V 2 +?+a 2m ?V m +?+a 2n ?V n + b 21 ?? 1 +b 22 ?? 2 +?+b 2m ?? m +?+b 2n ?? n =g 2 ?d 2 (6.88) In equation (6.88), a 21 =Y 21 cos(a 21 +? 1 ); a 22 =Y 22 cos(a 22 +? 2 )+y 2 cos(ß 2 +? 2 ); a 2m =Y 2m cos(a 2m +? m ); a 2n =Y 2n cos(a 2n +? n ); b 21 =Y 21 V 1 sin(a 21 +? 1 ); b 22 =Y 22 V 2 sin(a 22 +? 2 )y 2 V 2 sin(ß 2 +? 2 ); b 2m =Y 2m V m sin(a 2m +? m ); b 2n =Y 2n V n sin(a 2n +? n ); g 2 =y 2 E 2 sin(d 2 +ß 2 ); (6.89) Similarly, continuing with linearisation of the real parts of ?rst ‘m’ equations (corresponding to the generator buses) of the equation set (6.23), we get, A 1 B 1 ? ? ? ? ? ? ? ? ? ? ? ?V g ?V L ?? g ?? L ? ? ? ? ? ? ? ? ? ? ? = G?d (6.90) In equation (6.90), A 1 = ? ? ? ? ? ? ? ? ? ? ? ? ? a 11 a 12 ? a 1m a 1,(m+1) ?? a 1n a 21 a 22 ? a 2m a 2,(m+1) ?? a 2n ? ? ? ? ? ? ? ? ? ? ? ? ? ? a m1 a m2 ? a mm a m,(m+1) ?? a mn ? ? ? ? ? ? ? ? ? ? ? ? ? ? is a (m×n) matrix (6.91) B 1 = ? ? ? ? ? ? ? ? ? ? ? ? ? b 11 b 12 ? b 1m b 1,(m+1) ?? b 1n b 21 b 22 ? b 2m b 2,(m+1) ?? b 2n ? ? ? ? ? ? ? ? ? ? ? ? ? ? b m1 b m2 ? b mm b m,(m+1) ?? b mn ? ? ? ? ? ? ? ? ? ? ? ? ? ? is a (m×n) matrix (6.92) 281 Page 5 6.4 Small signal analysis For small signal analysis of a multimachine power system, we need to linearise the di?erential equa tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system having ‘m’ generators. Each generator is represented by its classical model. Further, without any loss of generality, it is assumed that the generators are connected at the ?rst ‘m’ buses of the system. Now, for linearising the di?erential equations, let us recall the swing equations of each generator here for ready reference. dd i dt =? i ? s for i= 1, 2,??m (6.64) 2H i ? s d? i dt =P mi P ei for i= 1, 2,??m (6.65) Linearising equation (6.64) for thei th generator, one can get, d?d i dt = ?? i (6.66) Now, let us de?ne, ?d = [?d 1 , ?d 2 , ?? ?d m ] T (6.67) ?? = [?? 1 , ?? 2 , ?? ?? m ] T (6.68) In equations (6.67) and (6.68), the vectors ?d and ?? denote the vectors of perturbed values of rotor angle and machine speed respectively. Please note that the size of each of these two vectors is (m× 1). Now, collecting equation (6.66) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, d dt ?d =F 1 ?? (6.69) In equation (6.69),F 1 is a (m×m) identity matrix. Now, linearising equation (6.65) for thei th generator, one can get, 2H i ? s d?? i dt =?P ei (6.70) For performing linearisation ofP ei , its expression its required. This expression can be derived as follows. From equation (6.25), one can write, ¯ I i = E i ?d i V i ?? i jx / di = E i x / di e j(d i p~2)  V i x / di e j(? i p~2) 278 Or, ¯ E i ¯ I * i =E i e jd i ¯ I * i = E i x / di e jp~2  E i V i x / di e j(p~2+d i ? i ) Therefore, P ei = Re? ¯ E I ¯ I * i ?= E i V i x / di sin(d i ? i ) (6.71) Linearisation of equation (6.71) yields, ?P ei =k 1i ?d i +k 2i ?? i +k 3i ?V i (6.72) Where, k 1i = E i V i x / di cos(d i ? i ); k 2i = E i V i x / di cos(d i ? i ); k 3i = E i x / di sin(d i ? i ); (6.73) In equation (6.73), the constantsk 1i ,k 2i andk 3i are evaluated using the values ofE i ,V i ,d i and ? i at the current operating point fori= 1, 2,??m. Now, again let us de?ne, ?? g = [?? 1 , ?? 2 , ?? ?? m ] T (6.74) ?V g = [?V 1 , ?V 2 , ?? ?V m ] T (6.75) ?P e = [?P e1 , ?P e2 , ?? ?P em ] T (6.76) In equations (6.74) and (6.75), the vectors ?? g and ?V g denote the vectors of perturbed values of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ?P e denotes the perturbed values of the generator real powers. Please note that the size of each of these three vectors is also (m× 1). Now, collecting equation (6.72) for all the ‘m’ generators and rewriting them in matrix notation we can obtain, ?P e =K 1 ?d+K 2 ?? g +K 3 ?V g (6.77) In equation (6.77), the size of each of the matricesK 1 ,K 2 andK 3 is (m×m). Moreover, all these three matrices are diagonal matrices and are given by; K 1 = diag(k 11 , k 12 , ??k 1m ) K 2 = diag(k 21 , k 22 , ??k 2m ) K 3 = diag(k 31 , k 32 , ??k 3m ) ? ? ? ? ? ? ? ? ? ? ? (6.78) 279 6.4.1 Linearisation of network equations (at the generator buses) To illustrate the procedure for linearisation of the network equations, let us ?rst consider the ?rst equation of the equation set (6.23) below. ? ¯ Y 11 + ¯ y 1 ? ¯ V 1 + ¯ Y 12 ¯ V 2 +?+ ¯ Y 1m ¯ V m +?+Y 1n ¯ V n = ¯ y 1 ¯ E 1 (6.79) Now, let, ¯ Y ij =Y ij exp(ja ij ); ¯ V i =V i exp(j? i ); for i, j = 1, 2,??n; (6.80) and ¯ y k =y k exp(jß k ); ¯ E k =E k exp(jd k ); for k = 1, 2,??m; (6.81) Utilising equations (6.80) and (6.81) in equation (6.79), we get, Y 11 V 1 exp(j(a 11 +? 1 ))+y 1 V 1 exp(j(ß 1 +? 1 ))+Y 12 V 2 exp(j(a 12 +? 2 ))+?+ Y 1m V m exp(j(a 1m +? m ))+?+Y 1n V n exp(j(a 1n +? n ))=y 1 E 1 exp(j(d 1 +ß 1 )) (6.82) Taking the real part of both the sides of equation (6.82), we get, Y 11 V 1 cos(a 11 +? 1 )+y 1 V 1 cos(ß 1 +? 1 )+Y 12 V 2 cos(a 12 +? 2 )+?+ Y 1m V m cos(a 1m +? m )+?+Y 1n V n cos(a 1n +? n )=y 1 E 1 cos(d 1 +ß 1 ) (6.83) Linearising equation (6.83), we get, Y 11 cos(a 11 +? 1 )?V 1 Y 11 V 1 sin(a 11 +? 1 )?? 1 +y 1 cos(ß 1 +? 1 )?V 1  y 1 V 1 sin(ß 1 +? 1 )?? 1 +Y 12 cos(a 12 +? 2 )?V 2 Y 12 V 2 sin(a 12 +? 2 )?? 2 +?+Y 1m cos(a 1m +? m )?V m Y 1m V m sin(a 1m +? m )?? m +?+ Y 1n cos(a 1n +? n )?V n Y 1n V n sin(a 1n +? n )?? n =y 1 E 1 sin(d 1 +ß 1 )?d 1 (6.84) Equation (6.84) can be rewritten as, a 11 ?V 1 +a 12 ?V 2 +?+a 1m ?V m +?+a 1n ?V n + b 11 ?? 1 +b 12 ?? 2 +?+b 1m ?? m +?+b 1n ?? n =g 1 ?d 1 (6.85) In equation (6.85), a 11 =Y 11 cos(a 11 +? 1 )+y 1 cos(ß 1 +? 1 ); a 12 =Y 12 cos(a 12 +? 2 ); a 1m =Y 1m cos(a 1m +? m ); a 1n =Y 1n cos(a 1n +? n ); b 11 =Y 11 V 1 sin(a 11 +? 1 )y 1 V 1 sin(ß 1 +? 1 ); b 12 =Y 12 V 2 sin(a 12 +? 2 ); b 1m =Y 1m V m sin(a 1m +? m ); b 1n =Y 1n V n sin(a 1n +? n ); g 1 =y 1 E 1 sin(d 1 +ß 1 ); (6.86) Again, taking the real part of both sides of the second equation of the equation set (6.23) (after 280 substituting equations (6.80) and (6.81) into it), we get, Y 21 V 1 cos(a 21 +? 1 )+y 2 V 2 cos(ß 2 +? 2 )+Y 22 V 2 cos(a 22 +? 2 )+?+ Y 2m V m cos(a 2m +? m )+?+Y 2n V n cos(a 2n +? n )=y 2 E 2 cos(d 2 +ß 2 ) (6.87) Linearising equation (6.87), one can get, a 21 ?V 1 +a 22 ?V 2 +?+a 2m ?V m +?+a 2n ?V n + b 21 ?? 1 +b 22 ?? 2 +?+b 2m ?? m +?+b 2n ?? n =g 2 ?d 2 (6.88) In equation (6.88), a 21 =Y 21 cos(a 21 +? 1 ); a 22 =Y 22 cos(a 22 +? 2 )+y 2 cos(ß 2 +? 2 ); a 2m =Y 2m cos(a 2m +? m ); a 2n =Y 2n cos(a 2n +? n ); b 21 =Y 21 V 1 sin(a 21 +? 1 ); b 22 =Y 22 V 2 sin(a 22 +? 2 )y 2 V 2 sin(ß 2 +? 2 ); b 2m =Y 2m V m sin(a 2m +? m ); b 2n =Y 2n V n sin(a 2n +? n ); g 2 =y 2 E 2 sin(d 2 +ß 2 ); (6.89) Similarly, continuing with linearisation of the real parts of ?rst ‘m’ equations (corresponding to the generator buses) of the equation set (6.23), we get, A 1 B 1 ? ? ? ? ? ? ? ? ? ? ? ?V g ?V L ?? g ?? L ? ? ? ? ? ? ? ? ? ? ? = G?d (6.90) In equation (6.90), A 1 = ? ? ? ? ? ? ? ? ? ? ? ? ? a 11 a 12 ? a 1m a 1,(m+1) ?? a 1n a 21 a 22 ? a 2m a 2,(m+1) ?? a 2n ? ? ? ? ? ? ? ? ? ? ? ? ? ? a m1 a m2 ? a mm a m,(m+1) ?? a mn ? ? ? ? ? ? ? ? ? ? ? ? ? ? is a (m×n) matrix (6.91) B 1 = ? ? ? ? ? ? ? ? ? ? ? ? ? b 11 b 12 ? b 1m b 1,(m+1) ?? b 1n b 21 b 22 ? b 2m b 2,(m+1) ?? b 2n ? ? ? ? ? ? ? ? ? ? ? ? ? ? b m1 b m2 ? b mm b m,(m+1) ?? b mn ? ? ? ? ? ? ? ? ? ? ? ? ? ? is a (m×n) matrix (6.92) 281 G= ? ? ? ? ? ? ? ? ? ? ? ? ? g 1 0 ? 0 ?? 0 0 g 2 ? 0 ?? 0 ? ? ? ? ? ? ? ? ? ? ? ? 0 0 ? 0 ?? g m ? ? ? ? ? ? ? ? ? ? ? ? ? ? is a (m×m) diagonal matrix (6.93) The elements of the matrices A 1 , B 1 and G are given as; a ii =Y ii cos(a ii +? i )+y i cos(ß i +? i ); i= 1, 2,??m; a ij =Y ij cos(a ij +? j ); i= 1, 2,??m; j = 1, 2,??n; i?j; b ii =Y ii V i sin(a ii +? i )y i V i sin(ß i +? i ); i= 1, 2,??m; b ij =Y ij V j sin(a ij +? j ); i= 1, 2,??m; j = 1, 2,??n; i?j; g i =y i E i sin(d i +ß i ); i= 1, 2,??m; (6.94) Further, the vectors ?? L and ?V L are de?ned as; ?? L = [?? m+1 , ?? m+2 , ?? ?? n ] T (6.95) ?V L = [?V m+1 , ?V m+2 , ?? ?V n ] T (6.96) Please note that the size of each of the above two vectors is ((nm)× 1). So far, we have considered only the algebraic equations at the generator buses. However, for completing the small signal model, the algebraic equations at the load buses all need to be linearised. We will discuss this issue in the next lecture. 282Read More
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