ELEC 242 Lab

Background

Frequency Response

In Lab 5 we measured the step response of some first order systems. Although it only describes the response to piecewise constant signals, the step response in fact completely characterizes the behavior of a linear system.

The step response is a time domain characterization. It describes what the output looks like as a function of time. We can also characterize a linear system by its frequency response, which describes its behavior in the frequency domain.

The frequency response is a special case of the transfer function where is replaced by $j\omega$ (where $\omega = 2\pi f$ ): $\displaystyle H(j\omega)=\frac{V_{out}(j\omega)}{V_{in}(j\omega)}$ . Note that since $V_{in}$ and $V_{out}$ are complex, $H(j\omega)$ is complex

We usually express $H(j\omega)$ in polar form, in terms of its magnitude ( $\displaystyle \vert H(j\omega)\vert=\frac{\vert V_{out}(j\omega)\vert}{\vert V_{in}(j\omega)\vert}$ ). and phase ( $\displaystyle \angle H(j\omega)=\angle V_{out}(j\omega) - \angle V_{in}(j\omega)$ ).

For our generic first order system:

$\includegraphics[scale=0.650000]{ckt7.2.ps}$

we have

$\displaystyle V_{out}=\frac{1/sC}{R+1/sC} V_{in}=\frac{1}{1+RCs} V_{in}$ or $\displaystyle H(s)=\frac{V_{out}}{V_{in}}=\frac{1}{1+RCs}=\frac{1}{1+\tau s}$ For a sinusoidal input voltage, $v_{in}=V_i \cos(\omega t)=Re\{V_i e^{j\omega t}\}$ so we can find the output voltage, $v_{out}=V_o \cos(\omega t)$ by setting $s=j\omega$ in the transfer function: $\displaystyle H(j\omega)=\frac{1}{1+j\omega \tau}=\frac{1}{1+j\frac{\omega}{\omega_0}}$ where $\omega_0 = \frac{1}{\tau}$ is called the cutoff frequency of the network. For the magnitude of the frequency response, we have $\displaystyle \vert H(j\omega)\vert=\frac{1}{\vert 1+j\frac{\omega}{\omega_0}\vert}=\frac{1}{\sqrt{1+\frac{\omega^2}{\omega_0^2}}}$ which looks like this: $\includegraphics[scale=0.350000]{fr1.ps}$ Note that for $\omega = \omega_0$ , $\vert H(j\omega)\vert=\frac{1}{\sqrt{2}}=0.707$ .