Impedance Analysis

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20.309: Biological Instrumentation and Measurement

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Overview of impedances

When a circuit contains capacitors and/or inductors, the behavior of the circuit can change with time. Furthermore, often while using these components, we are interested in time-varying input signals, such as a sinusoidal voltage input. Since the current-voltage relationships for capacitors and inductors involve derivatives or integrals, analyzing such circuits often involve differential equations. To simplify our calculations, we can perform the steady-state analysis of circuits with time-varying signals using the frequency domain instead of the time domain. For example, a sine wave constantly varies over time, but only contains one constant frequency. Here we will discuss how to approach circuits using this mind set.

For analyzing the frequency response of circuits with various components, it is convenient to think of a generalized form of resistance, or opposition to current flow, known as impedance ($ Z $). Clearly, we know that a resistor's resistance to current flow is known by its resistance. Although capacitors and inductors don't technically have a "resistance" value, they do resist current flow in their own ways and we can call those their respective impedances. Thus, Ohm's law applies not only to resistances, but to impedances as well:

$ V = I Z $

where $ V $ is the voltage drop across the components of interest, $ I $ is the current, and $ Z $ is the impedance of those components.


The table below indicates the impedance values for different circuit elements. Here, $ j $ is the imaginary number and $ \omega $ is frequency in units of radians. Recall that $ \omega = 2 \pi f $, where $ f $ is frequency in units of Hz.

ImpedanceChart.png

Derivation of impedance

Below is a short demonstration of how we can derive the impedances for the circuit elements listed above. Let's use the capacitor as an example. Since we are interested in transient (time varying) signals, we will let voltage $ V $ be a time varying signal $ Ae^{j \omega t}. $ (Click here for a refresher on complex numbers.)

Since we know that $ I = C {dV\over dt} $, we can take the derivative of V and plug it in:
$ I = C A j \omega e^{j \omega t} $
With some reorganizing, we can see that $ I = j \omega C V $
To get it into the form $ V = I Z $,
$ Z = {1\over j \omega C} $

This impedance value follows what we already know about capacitors--that they act as an open circuit with a constant DC voltage ($ Z \rarr \infin $ when $ \omega = 0 $, also $ I=0 $ because $ {dV\over dt}=0 $). At very high frequencies, as $ \omega \rarr \infin $, the capacitor acts as a short circuit, like a wire without resistance ($ Z \rarr 0 $).

The same exercise can be done for the inductor. It will be left as an exercise for the reader.

Impedances in series and in parallell

When combining the effect of multiple circuit components in series or parallel, their impedances can be treated as resistances. Thus, for elements in series, their impedances add. Meanwhile, for elements in parallel, the inverses of their impedances add. This is summarized in the following table.

AddImpedanceTable.png

Example circuit problem

Find $ {V_{out} \over V_{in}} $ for the following RC circuit.

ExampleRCCircuit.png


We can convert the circuit elements to look at their impedances, as in the diagram on the right, where the resistor has impedance $ Z_1 = R $ and the capacitor has impedance $ Z_2 = {1 \over j \omega C} $

CircuitwithImpedances.png

We can see that this circuit is a voltage divider, where

$ V_{out} = V_{in} {Z_2 \over Z_1 + Z_2} $
Substituting back the impedance values,
$ V_{out} = V_{in} {{1 \over j \omega C} \over R + {1 \over j \omega C}} $
$ {V_{out} \over V_{in}} = {1 \over 1 + j \omega RC} $

We call this type of filter a low-pass filter, because at low frequencies ($ \omega $ small), the output voltage is very close to the input voltage. In other words, at low frequencies, the circuit passes the input voltage to the output. However, at high frequencies ($ \omega $ large), the output voltage is very small compared to the input voltage. Thus, at high frequencies, the circuit does not pass the input voltage to the output; rather, the circuit filters out or blocks the input signal from the output. We will define how high is a high frequency and how low is a low frequency in a following section.


(If you've forgotten about voltage dividers, please visit this page and/or this video.)

Bode plots

To characterize the frequency response of a circuit, it is often helpful to find the transfer function of a circuit, as we did in the example above. A transfer function, usually denoted as $ H(\omega) $ or $ H(f) $, is defined to be the ratio of the output voltage to the input voltage as a function of frequency, i.e. $ H(f) = {V_{out} \over V_{in}} $. (Remember that $ \omega = 2 \pi f $.) Also note that only two things can happen to a sine wave passing through a linear, time-invariant system (like an RLC circuit): it's magnitude can be changed; and the signal can be delayed. The delay and percentage change in the magnitude are a function of frequency. Thus, the transfer function is a complex-valued function of frequency that specifies the magnitude and phase shift of a particular system for all frequencies. The change in amplitude is often called the gain, and the delay is usually thought of in terms of a phase shift of the sine wave.

One helpful way to visualize a transfer function is to make two plots, called Bode plots. The first plot shows the magnitude of the transfer function verses frequency on a set of log-log axes. The second plot shows the phase shift versus log frequency. (Traditionally, Bode plots are constructed only using straight-line segments to represent the transfer function by means of straight edges and fine draft pen sets. Since plotting complicated functions is now done by computer, exact plots may be substituted for the traditional methods.)

Plotting the magnitude of the transfer function

Let's first plot the magnitude of the transfer function as a Bode plot. We will take the RC circuit from the example in the previous section to demonstrate how to create the plot. Recall that the transfer function for the low-pass RC circuit above was

$ H(\omega ) = {V_{out} \over V_{in}} = {1 \over 1+ j \omega RC} $

The magnitude of this function is given by (for a refresher on how to find the magnitude of a complex number, click here)

$ |H(\omega )| = {1 \over \sqrt {1 + \omega ^2 R^2 C^2}} $

A magnitude Bode plot typically has the y-axis in units of decibels (dB), which is defined to be 20 times the logarithm of the magnitude. More specifically,

$ |H(\omega )|_{dB} = 20 log |H(\omega )| $

The x-axis is typically in units of radians if you're using $ \omega $ or Hz if you're using $ f $. It is also on a log-scale, so you'll usually see it in increments that increase by ten-fold, known as a decade.

To create the Bode plot, think about the following:

  1. Let's first define the cutoff frequency ($ \omega_c $). This is a convenient frequency to think about, where we will consider $ \omega << \omega_c $ to be a low frequency and $ \omega >> \omega_c $ to be a high frequency. You will see why it is called a cutoff frequency soon.
    Looking at the denominator of your magnitude function, you can see that at small frequencies the 1 dominates, but at large frequencies, the frequency term dominates. We can say that roughly, where the frequency term begins to play a role is when $ 1 = \omega^2 R^2 C^2 $. At this point, we can rearrange the expression to see that $ \omega = {1 \over RC} $.
    This is what we define to be the cutoff frequency:
    $ \omega_c = {1 \over RC}. $
  2. What is the value of $ |H(\omega )| $ at low frequencies ($ \omega << \omega_c $)?
    Plug in $ \omega = 0 $ to your magnitude function and you can see that
    $ |H(\omega )| = 1 $, so $ |H(\omega )|_{dB} = 0 $
    Let's plot what we know so far. With a Bode plot we can approximate most things with straight line segments. We will get into more exact details later.
    BodePlotMag1.png
  3. What is the value of $ |H(\omega )| $ at high frequencies ($ \omega >> \omega_c $)?
    As $ \omega \rarr \infin $, you can see that
    $ |H(\omega )| \rarr 0 $, so $ |H(\omega )|_{dB} \rarr - \infin $ or a very negative number.
    Let's add this information to our plot.
    BodePlotMag2.png

You're Bode plot pretty much looks complete! How did we know the slope of the line at higher frequencies?

Let's look at the equation in a bit more detail.
$ |H(\omega )|_{dB} = 20 log({1 \over \sqrt {1+ \omega^2 R^2 C^2}}) $
$ |H(\omega )|_{dB} = 20 log(1) - 20 log(\sqrt {1+ \omega^2 R^2 C^2}) $
$ |H(\omega )|_{dB} = - 20 log(\sqrt {1+ ({\omega \over \omega_c})^2}) $
$ |H(\omega )|_{dB} = - 10 log( 1+ ({\omega \over \omega_c})^2) $
When we are considering $ \omega >> \omega_c $, the frequency term dominates the 1, such that
$ |H(\omega )|_{dB} \cong - 20 log( {\omega \over \omega_c}) $
Thus, every time the frequency increases ten-fold (a decade), the magnitude decreases -20 dB. We say that the slope of the line at high frequencies is -20 dB per decade.

Guidelines for drawing magnitude Bode plots:

  • for first order systems, with omega in denominator--
  • if the numerator cannot be simplified to

kinda need to know about poles and zeros...or get it into a certain form.

Plotting the phase of the transfer function