# Frequency Response in AC Circuits

concept

Because the impedance of capacitors and inductors changes based on the frequency of the supply voltage or current AC circuits behave a little differently when stimulated at one frequency over another.
For instance, the normal voltage divider in AC circuits can have one resistor replaced with a capacitor. Now the output voltage will depend not only on the sizes of the two elements, but also the frequency of the input signal.
Using this phenomena we can create filters, these are circuits that boost some frequencies and elminated others, like a mixer for a sound system that allows you to boost some frequencies more than others.
Filters are used to remove unwanted "noise" from sensor inputs in almost all electronic devices.
This topic will cover the basics of frequency response, in a later topic we'll cover actually designing a simple filter to remove certain frequencies.

fact

The voltage divider in AC circuits works just like the voltage divider in DC circuits:
\(V_{out} = V_s\frac{Z_2}{Z_1+Z_2}\)

We'll use some examples to show how frequency affects the output voltage in AC circuits.

example

Find the output voltage for the following circuit with \(V = \cos(2\pi t)\), \(R = 100\Omega\) and \(C = 10\)mF

With \(\omega = 2\pi\) we can find the impedances: \(Z_1 = R = 100\) \(\begin{align} Z_2 = jX_c & = \frac{1]{j2\pi \cdot 10^{-2}}} \\ & = 15.92\angle -90^\circ \\ \end{align}\) \(\begin{align} Z_1+Z_2 & = 100 + \frac{10^2}{j2\pi} \\ & = 100 - j\frac{10^2}{2\pi} \\ & = 101.26\angle -9^\circ \\ \end{align}\) \(\begin{align} V_{out} & = 1\cdot \frac{15.92}{101.26}\angle -81^\circ \\ & = 0.16\angle -81^\circ \\ \end{align}\) So our output voltage is given by: \(V_{out} = 0.16\cos(2\pi t -81^\circ)\) Now what happens if we were to repeat that example with hundreds of different frequencies? We'd get a graph of how much the output is reduced based on the input frequency.

example

Find a graph for the magnitude of the output voltage for the circuit below based on the input frequency

This is going to be much like the last example but we'll be using more algebra since we don't have a numerical value for \(\omega\). \(Z_1 = R\) \(Z_2 = \frac{1}{j\omega C} = \frac{1}{\omega C}\angle -90^\circ\) \(\begin{align} Z_1+Z_2 & = R - j\frac{1}{\omega C} \\ & = \frac{1}{\omega C}\sqrt{1+(\omega RC)^2} \\ \end{align}\) Now for this we only need the magnitude and not the angle so: \(|V_{out}| = \frac{|Z_2|}{|Z_1+Z_2|} = \frac{1}{\sqrt{1+(\omega RC)^2}}\) So we can see that the output voltage changes very strongly depending on the input frequency: When we plot frequency responses we use logarithmic plots for the gain and frequency because we're interested in percentage changes. Increasing the frequency from 90Hz to 95Hz matters less to us than increasing it from 10Hz to 15Hz even though both are an increase of 5Hz.
The gains we plot using the decibel scale.

fact

The gain in decibels is given by:
\(G_{DB} = 20\log_{10}(\frac{|V_o|}{|V_i|})\)
Where \(V_o\) is the output voltage we're measuring and \(V_i\) is the input voltage.
decibels is written "dB" as a unit.

example

A filter provides a gain of 0.707, find the gain in decibels.

We're given that \(\frac{V_o}{V_i} = 0.707\) so using our formula: \(G_{db} = 20\log_{10}(0.707) = -3\)dB So our gain is -3dB. There's nothing really "special" about negative decibels. It just means that our "gain" was less than 1 which will often be the case.

fact

When dealing with powers we use the slightly different formula for decibels:
\(G_{db} = 10\log_{10}(\frac{|P_o|}{|P_i|})\)

That power uses a 10 out front rather than the usual 20 might seem odd at first, just write it down somewhere and you'll get used to it.

So using a logarithmic frequency scale and decibels for the gain we get the following graph (assuming \(RC = 1\))
You can see from the graph above that the magnitude appears to be more or less two straight lines joined at a "knee".
In fact, the graph looks like two line segments so much that we often just pretend that it IS two line segments as follows:

fact

The straight line approximation assumes that for frequencies below the knee point the gain is 1 (0dB), for frequencies greater than the knee point the line drops at a slope of -20dB/decade.
We call this the "corner frequency" (\(\omega_c\))

fact

The corner frequency in the RC circuits we've bee looking at is:
\(\omega_c = \frac{1}{RC}\)

fact

The real gain at the corner frequency is:
\(G_c = \frac{1}{\sqrt{2}} = -3\)dB

We often use the above fact to find the corner frequency. Simply find where the gain is \(\frac{1}{\sqrt{2}}\) and that's our corner frequency.

example

Analyse the following circuit for varying input frequencies.

\(Z_1 = R\) \(Z_2 = j\omega L = \omega L\angle 90^\circ\) \(\begin{align} Z_1 + Z_2 & = \sqrt{R^2 + \omega^2L^2}\angle \tan^{-1}(\frac{\omega L}{R}) \\ & = \omega L\sqrt{\frac{R^2}{\omega^2L^2} + 1}\angle \tan^{-1}(\frac{\omega L}{R}) \\ \end{align}\) \(\begin{align} V_{out} & = V_s\frac{Z_2}{Z_1+Z_2} \\ |V_{out}| & = \frac{1}{\sqrt{(\frac{R}{\omega L})^2+1}} \\ \end{align}\) So a graph of our function for \(R = L = 1\) looks like: We can see that in this case we have a "high pass" filter that removes low frequency components. The corner frequency for this filter is given by: \(\frac{R}{\omega_c L} = 1\) \(\omega_c = \frac{R}{L}\)
practice problems