## 3 Numerical Analysis
### Harmonic Force with a Zero Mean Value  
The first example in this study uses a harmonic excitation with a zero mean on a single-degree-of-freedom (SDOF) system. The harmonic excitation with a zero mean can be expressed as a sine force, \(f(t)=\sin\theta t\), with \(\theta/\omega=1\), \(\eta=0.1\), \(m=1\) kg, \(k=4\pi^{2}/1.28^{2}\) N/m. The time history and Fourier spectrum of sine function of the applied excitation are shown in Figure 2. The calculation duration used in this example is 20.48 s and the time interval \(\Delta t\) is 0.01 s.  
Figure 1: Flowchart of the developed fast frequency domain method.  
Figure 2: The time history and Fourier spectrum of zero mean harmonic force.  
The exact steady solution of such a dynamic problem can be expressed as:  
\[u=\frac{-i}{-\theta^{2}m+(1+i\eta)k}e^{i\theta t}, \tag{29}\]  
However, if this problem is solved without dual forces, the governing equation of motion can be expressed as:  
\[\begin{array}{c}\dot{m}\dot{u}_{R}+k(u_{R}-\eta u_{I})=\sin\theta t\\ \dot{m}\dot{u}_{I}+k(\eta u_{R}+u_{I})=0\end{array}\Bigg{\}}, \tag{30}\]  
To express conveniently, Equation (30) is called the direct time domain method (Direct TDM). The dynamic response calculated by the Direct TDM can be expressed as:  
\[\begin{array}{c}u_{R}=\frac{\left(k-m\theta^{2}\right)^{2}+\left(k\right)^{ 2}}{\left(k-m\theta^{2}\right)^{2}+\left(\eta k\right)^{2}}\sin\theta t\\ u_{I}=\frac{-\kappa\eta}{\left(k-m\theta^{2}\right)^{2}+\left(\eta k\right)^ {2}}\sin\theta t\end{array}\Bigg{\}}, \tag{31}\]  
When comparing Equation (31) with Equation (29), it can be seen that they are not equal to each other. Thus, to compare the difference between different methods, this given example was solved by using five different methods, including the Direct TDM, the CFDM, the DTDM, the new accurate FFDM and the new approximate FFDM. Figure 3 compares the dynamic displacement responses from the exact solution as in Equation (29), with the solutions from the developed new FFDM and the other three existing methods.  
The relative error \(e\) of the amplitude of displacements can be expressed as  
\[e=\frac{|u-u^{*}|}{u*}\times 100\%, \tag{32}\]  
in which \(u\) and \(u^{*}\) are the approximate and exact solutions, respectively. The relative errors of the amplitude of displacements are also shown in Table 1.  
Figure 3: Displacement response time history under zero mean sine force (\(\theta/\omega=1\)).  
It can be seen from Figure 3 and Table 1 that the real part and imaginary part of displacement obtained by the DTDM, the accurate FFDM and the approximate FFDM are identical with the exact solution. This is the reason that the DTDM and the accurate FFDM are identical and equal to the discrete values of exact solution. As for the approximate FFDM, the \(D_{N/2}\) is equal to zero for sine function, therefore, ignoring of the last term of Equation (23) would not cause errors.  
As for the CFDM, the real part obtained by CFDM is identical with the exact solution. The Fourier series of \(U_{j}\) is conjugated symmetric on the Nyquist frequency in Equation (8), therefore the imaginary part of displacement is approximately equal to zero after summation which is different with the exact solution. Fortunately, the real part of solution is the real response of system, and the imaginary part is a dual term for the real part and not used to decide the status of real system. Therefore, the CFDM can be used to solve the response of system with complex damping. For the direct time domain solution, the real part of the solution is equal to zero, which is obviously incorrect because the dual relationship of the load is ignored. It means that the direct time domain solution is not analyzed in the subsequent examples.