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P1- Complex Algebra. 4. Solving Differential Equations with Complex Numbers (I). Summary of previous lecture. The derivative of a complex function with respect to a real variable follows the same rules of differentiation as for real functions. We simply treat i as a constant.
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P1- Complex Algebra 4. Solving Differential Equations with Complex Numbers (I)
Summary of previous lecture • The derivative of a complex function with respect to a real variable follows the same rules of differentiation as for real functions. We simply treat i as a constant. • Complex functions are only differentiable wrt a complex variable zif they are analytic. All the known rules of differentiation then apply, and again we treat i as a constant.
Summary cont. • In integrating complex functions along the real line, it is useful to bear in mind that
Summary cont. • Substitution of trial solutions of the form can be used to solve homogenouslinear ordinary differential equations, i.e. of the form • This produces an auxiliary polynomial with n complex solutions.
Summary cont. • Each of those solutions sk, substituted into produces one valid solution to the differential equation. • The value of the constants Akis determined by the initial conditions. • The general solution to the original differential equation is given by
Contents 4.1Inhomogenous ordinary differential equations 4.2 Introduction to phasors 4.3 Solution of 1st order differential equations using phasors 4.4 Solution of 2nd order differential equations using phasors
4.1 Inhomogeneous ODEs • Many differential equations describing physical systems are inhomogeneous, i.e. they contain terms that are not functions of the dependent variable, e.g. • The inhomogeneous term (the non-zero RHS) is known as the forcing function.
Inhomogeneous ODEs cont. • The solution to an inhomogenous linear ODE will consist of 2 parts: • The complementary function or transient solution • The particular integral or steady state solution
Inhomogeneous ODEs cont. • The complementary function is found by ignoring the forcing function and solving the corresponding homogenous equation (as in section 4.2). • The particular integral requires the use of a different trial solution, whose form will be determined by the forcing function.
Inhomogeneous ODEs cont. • Complex numbers can be used to simplify finding the particular integral. • First the forcing function must be rewritten in complex form. • We use the fact that (cosθ + isinθ) = eiθ and so cosθ = Re(eiθ).
Inhomogeneous ODEs cont. • Thus the forcing function in this case can be rewritten as Note that C = ceiφis complex and denotes both the magnitude (c) and phase(φ) of the forcing function with respect to the reference function eiωt.
4.2 Phasors • For this reason, C = ceiφis called a phasor. • Our trial solution for the steady state response of the system x will be of the same form where X is again complex and is known as the response phasor
Phasorscont. • If we differentiate , we obtain the phasor component of which is • Note that differentiating xmultiplies the magnitude of the phasor by ωand rotated the phasor by π/2. • Differentiating again repeats the effect giving the phasor
Phasorscont. • are complex numbers and so can be plotted on an Argand diagram, which is then called a phasor diagram.
Phasors cont. • Notation convention • Responses that have phasorsclockwise of another phasor are said to lag it. • Responses that have phasorsanticlockwise of another phasor are said to lead it.
Phasors cont. • So if there are 2 functions • x2 is said to lag x1 by ϕ or alternatively ϕis the phase lag between x1 and x2
4.3 Solution of 1st order differential equations using phasors • We can use this type of diagram to solve differential equations • Consider again • Converting to phasor form
Solution of differential equations using phasors cont. • Substituting gives • We can drop the Re() notation for convenience as this is implicit in equating real numbers.
Solution of differential equations using phasors cont. • We can obtain the solution either algebraically or graphically using a phasor diagram. • Algebraically:
Solution of differential equations using phasors cont. • Thus • And hence • With phase difference
Solution of differential equations using phasors cont. • Graphically: • To plot the phasor diagram we choose one function to be our reference function. • Let us choose xin this case and Ψ as the phase lead of the forcing function phasorC.
Solution of differential equations using phasors cont. • We can now use Pythagoras’ theorem to find |X| • and trigonometry gives
Solution of differential equations using phasors cont. • We therefore have the steady state solution
4.4 Solution of 2nd order differential equations using phasors • e.g. Consider a mass mounted on a spring and a damper. We assume that the spring and the damper are linear, the spring providing a force proportional to its extension, and the damper a force proportional to the rate of extension. There is an additional sinusoidally varying force f acting on the mass. Describe the steady state motion of the mass.
Solution of 2nd order differential equations using phasors cont. • The equation of motion of the mass is: • As before we make the substitutions and giving
Solution of 2nd order differential equations using phasors cont. • f varies sinusoidally so we can represent it using a phasorF • Similarly the steady state response will be of the form • And thus and
Solution of 2nd order differential equations using phasors cont. • Substituting the above into the equation of motion gives: • Cancelling the exponentials and rearranging the equation gives
Solution of 2nd order differential equations using phasors cont. • Which can be written in polar form as
Solution of 2nd order differential equations using phasors cont. • The steady state solution is then
Solution of 2nd order differential equations using phasors cont. • We can plot the magnitude and phase of x as a function of ωand c (i.e. the driving frequency and damping in the system)
Solution of 2nd order differential equations using phasors cont. • For low levels of damping we can see a peak at around the frequency of free vibrations (resonance). • The lower the damping, the higher the peak and the closer the resonance frequency to the natural frequency. • As damping increases, the peak disappears.
Solution of 2nd order differential equations using phasors cont.
Solution of 2nd order differential equations using phasors cont. • At low frequencies, the displacement is in phase with the force: this is the response of a spring, and the magnitude is then given by Hooke’s law. • At resonance, the response is out of phase with the force. • At high frequencies, the response is in anti-phase
Mass phasor Damper phasor Spring phasor Solution of 2nd order differential equations using phasors cont. • We can also predict these trends from the phasor diagram:
Solution of 2nd order differential equations using phasors cont. • At low frequencies, the phase lag is small and the force phasor is almost equal to the spring force phasor. • Near resonance, the spring force and the mass force phasors cancel each other out and the response is dominated by damping (the damping phasor). • At high frequencies, the mass phasor grows to be much larger than all others.
