Frequency-Domain System Analysis

It can be easier to analyze systems using the frequency domain as a mathematical tool.

Suppose we have a transfer function:

$$H(s)=\frac{X(s)}{F(s)}$$

Then, given an input of $e^{st}$, in the time domain we would have:

$$x(t)=H(s)e^{st}$$

where $s=\sigma +j\omega$ and $x(t)$ is the output in response to the complex exponential $e^{st}$.

If we are on the imaginary axis, such that $s=j\omega$, then $H(j\omega )$ is the system’s frequency response. Then, we have:

$$e^{st}{|}_{s=j\omega }=e^{j\omega t}=\cos \omega t+j\sin \omega t$$

In other words, the frequency response is the [[Time-Domain Response of First-Order Systems#Transient vs Steady-State Response|steady-state response]] to a sinusoid (after [[Time-Domain Response of First-Order Systems#Transient vs Steady-State Response|time response]] disappears).

Thus, we have:

$$\begin{array}{rl}x(t)& =H(j\omega )e^{j\omega t}\\ & =|H(j\omega )||e^{j\omega t}|e^{\angle H(j\omega )}{e}^{j\omega t}\\ & =|H(j\omega t)|e^{[j\omega t+\angle H(j\omega )]}\end{array}$$

What this tell us is that given an input, it is amplified/attenuated by the magnitude of teh frequency response transfer function $H(j\omega t)$ and there is a phase shift of $\angle H(j\omega )$ due to the physics of the system.

If $f(t)=A\sin \omega t$, we would have:

$$x(t)=A|H(j\omega )|\sin (\omega t+\varphi )$$

where $\varphi =\angle H(j\omega )$.