PART III ADVANCED TOPICS IN POWER SYSTEM DYNAMICS
A.3 Linear Ordinary Differential Equations
A.3.6 First-Order Complex Differential Equation
A particular case of a linear first-order differential equation is a homogeneous equation of the form
˙
z−λz=0 whereλis a complex number. The equation can be rewritten as
˙
z=λz. (A.97)
According to the theory developed earlier the solution will be of the form
z(t)=eλtz0, (A.98)
y=Im z
x=Re z
y0 z0
r0
x0 φ0
Figure A.2 Initial condition in the complex plane.
wherez0=z(t0) is an initial condition (a complex number). Assume the following notation:
z(t)=x(t)+jy(t), z0=x0+jy0, λ=α+j. (A.99) The solution will be interpreted in the complex plane of coordinates x=Rez and y=Imz.
Substituting (A.99) into (A.98) gives
x(t)+jy(t)=e(α+j)t(x0+jy0), or
x(t)+jy(t)=eαt(x0+jy0)(cost+j sint).
Multiplying and ordering the terms gives
x(t)=eαt(x0cost−y0sint), (A.100a) y(t)=eαt(y0cost+x0sint). (A.100b) Figure A.2 shows that the initial conditionz0=x0+jy0is a point on the complex plane where
x0=r0cosφ0, y0=r0sinφ0, r0=
x02+y20. (A.101) Substituting (A.101) into (A.100a) gives
x(t)=r0eαt(cosφ0cost−sinφ0sint), (A.102a) y(t)=r0eαt(sinφ0cost+cosφ0sint). (A.102b) Equations (A.102a,b) can be expressed as
x(t)=r0eαtcos(t+φ0), (A.103a) y(t)=r0eαtsin(t+φ0). (A.103b) Obviously the solutionsx(t) andy(t) given by (A.103a, b) are proportional to the sine and cosine and are therefore shifted in time byπ/2. Squaring and adding both sides of (A.103a, b) gives
r(t)=r0eαt where r(t)=
[x(t)]2+[y(t)]2. (A.103) Again creating a complex numberz(t)=x(t)+jy(t) from the solutions (A.103a, b) gives
z(t)=r0eαt[cos(t+φ0)+j sin(t+φ0)]=r0eαtej(t+φ0)=r(t)ãej(t+φ0). (A.104) Figure A.3 shows that function (A.105) describes a logarithmic spiralin the complex plane starting at a point (x0,y0) corresponding to the initial condition. The spiral rotates anticlockwise if =Imλ >0 and clockwise if=Imλ <0. Forα=Reλ <0 the spiral is converging towards the
α < 0 ω > 0
α < 0
ω > 0 α < 0 ω > 0
α < 0 ω > 0
x y
x y
x y
x y
Figure A.3 Logarithmic spirals.
origin of coordinates while forα=Reλ >0 the spiral is diverging. Forα=Reλ=0 the solution z(t) corresponds to a circle in the complex plane.
Obviously for a conjugate valueλ∗=α−j=α+j(−) the imaginary part of λ∗ has the opposite sign toλ. This means that the spiral corresponding toλ∗rotates in the opposite direction to the spiral corresponding toλ. Hence if a function is the sum of solutions for complex conjugate pairsλ,λ∗then the imaginary parts of the solutions will cancel each other out and the only remaining part will be the double real part of the spiral, that is
zi(t)+zj(t)=zi(t)+z∗i(t)=2Rezi(t)=2x(t)=2r0eαtcos(t+φ0). (A.105) This observation is important for the solution of matrix differential equations discussed in Section 12.1.
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