Frequency-Domain Response of First-Order Systems

For first-order systems, we typically have:

$$m\dot{v}(t)+cv(t)=f(t)$$

where $f(t)=A\sin \omega t$. This is $f(t)$ because we are interested in the system frequency response, which deals with $s=j\omega$ along the imaginary axis of the complex plane; cosine term deals with real part so it is excluded.

Then we have:

$$H(s)=\frac{V(s)}{F(s)}=\frac{1}{ms+c}$$

Then, since $s=j\omega$, we have:

$$H(j\omega )=\frac{1}{mj\omega +c}$$

and

$$\begin{array}{rl}|H(j\omega )|& =\frac{|1|}{|mj\omega +c|}=\frac{1}{\sqrt{m^2{\omega }^2+c^2}}=\frac{1}{c\sqrt{{\tau }^2{\omega }^2+1}}\\ \angle H(j\omega )& =\angle 1-\angle (mj\omega +s)=0-\arctan \left(\frac{m\omega }{c}\right)=-\arctan (\omega \tau )\end{array}$$

where $\tau =\frac{m}{c}$ is a time constant.

Then, the steady-state response, $V(s)=H(s)F(s)$ in the Laplace domain, can be found in the time domain by multiplying the input by the magnitude of the frequency response and adding the phase shift, such that:

$$\begin{array}{rl}v(t{)}_{\text{steady-state}}& =A|H(j\omega )|\sin (\omega t+\angle H(j\omega ))\\ & =A\frac{1}{c\sqrt{{\tau }^2{\omega }^2+1}}\sin [\omega t+(-\arctan (\omega \tau ))]\end{array}$$