8. Solution of Linear Differential Equations Mathematical models of a 1-DOF system become ordinary linear differential equations. In this section, the solution of these differential equations will be given. Let’s consider Example 8.1 for this lesson. Example 8.1 In the differential equation given below, f(t) is the input, u(t) is the response. Let’s find the transfer function at first. f(t) : Input u(t) : Response 8.1 Transfer function (Continued Example 8.1) In order to find the transfer function, if the input is f(t)=est, it is assumed that the response is u(t)=H(s)est. Substituting the formulas (assumpitons) into the differential equations, the following equation is written. s3H(s)est+4s2H(s)est+14sH(s)est+20H(s)est=3est+sest Dividing the equation by est and rearranging the equation (s3+4s2+14s+20)H(s)=3+s Then, the transfer function is found as

When the denominator of the transfer function equals to zero, the eigenvalues are found. The eigenvalues can be calculated with the command “roots” in MATLAB. The eigenvalues for Example 8.1 can be found with these commands. a=[1,4,14,20];roots(a) The eigenvalues are found as -1±3i, -2 8.2. Exponential/Harmonic Input (Continued Example 8.1) If the input has the form of Ae-σtcos(ωt-φ), the input is called as Exponential/Harmonic. Consider Example 8.1. If the input is in the form of We will find the response u(t) as the following. f(t) can be written as Re implies that it is the real part. C and s equal to C=-1.8e-1.6i and s=-0.2+2.7i.

If the input were est, the response would be H(s)est If the input were est, the response would be H(s)est. However, multiplying the input by C, then, the response is multiplied by C. If the input is taken the parenthesis of Re, then the response would be in the parenthesis of Re For the value of s=-0.2+7i, is the amplitude of the transfer function, while is the phase angle of the transfer function. The solution is completed by taking the real parts of the response. For the value of s=-0.2+7i, The amplitude and the phase angle can be found by MATLAB program as follows. s=-0.2+2.7i; hs=(s+3)/(s^3+4*s^2+14*s+20); abs(hs), angle(hs) The amplitude and phase angle are calculated as 0.2437 and -1.2882, respectively. Thus, the response is found as

It is noted that the form of the response is the same as the form of the input. The frequency of the response is same as the excitation frequency. For the response, the amplitude of the response is the product of the amplitudes of the input and the transfer function for the value of s=-0.2+7i. For the response, the phase angle of the response is found by summing the phase angles of the input and the transfer function. In general, It is noted that its amplitude and phase angle of a response will change if the inputs are exponential/harmonic or pure harmonic input. Resonance : If the value of s determined from the input equals to one of the eigenvalues, the denominator of the transfer function equals to zero. Then, the amplitude of the response will equal to infinity. A resonance condition occurs if the inputs are as the following f(t)= ae-tcos(3t-b) or f(t)=ae-2t where a and b are the constants or any number.

Summary of Course 8.a f(t) : Input u(t) : Response Example 8.1 Transfer Function.: If f(t)=est , so u(t)= H(s)est s3H(s)est+4s2H(s)est+14sH(s)est+20H(s)est=3est+sest a=[1,4,14,20];roots(a) (s3+4s2+14s+20)H(s)=3+s Eigenvalues: -1±3i, -2 Exponential / Harmonic Input: s=-0.2+2.7i; hs=(s+3)/(s^3+4*s^2+14*s+20); abs(hs), angle(hs) RESONANCE