next up previous
Next: Second order system Up: Evaluation of Fourier Transform Previous: Evaluation of Fourier Transform

First order system

Consider the RL circuit with output taken from $R$. The transfer function is

\begin{displaymath}H(s)=\frac{1/\tau}{s+1/\tau} \end{displaymath}

and frequency response function is

\begin{displaymath}H(j\omega)=\frac{1/\tau}{j\omega+1/\tau}
=\frac{1/\tau}{j\omega-(-1/\tau)}=\frac{1/\tau}{v} \end{displaymath}

where $v\stackrel{\triangle}{=}j\omega-(-1/\tau)$ is a vector representing the difference between two points $s_p=-1/\tau$ and $s=j\omega$ in the s-plane. Note that $\vert H(j\omega)\vert=(1/\tau)/\vert v\vert$ and $\angle H(j\omega)=-\angle v$. Consider $H(j\omega)$ at three particular frequencies: As $\vert H(j\omega)\vert=(1/\tau)/\vert v\vert \rightarrow 0$ when $\omega \rightarrow \infty$, the system is a low-pass filter.

When the output is taken from $L$, the transfer function is

\begin{displaymath}H(s)=\frac{s}{s+1/\tau} \end{displaymath}

and frequency response function is

\begin{displaymath}H(j\omega)=\frac{j\omega}{j\omega+1/\tau}=\frac{u}{v} \end{displaymath}

where $u$ is a vector from $s=j\omega$ to the origin, and $v$ is the same vector as before. When $\omega=0$, $\vert u\vert=0$ and $\vert H(j\omega)\vert=0$. When $\omega \rightarrow \infty$, both $\vert u\vert \rightarrow \infty$ and $\vert v\vert \rightarrow \infty$, and $\vert H(j\omega)\vert \rightarrow 1$. The system is a high-pass filter.


next up previous
Next: Second order system Up: Evaluation of Fourier Transform Previous: Evaluation of Fourier Transform
Ruye Wang 2012-01-28