# Control Systems/Bode Plots

## Bode Plots

A Bode Plot is a useful tool that shows the gain and phase response of a
given LTI system for different frequencies. Bode Plots are generally
used with the Fourier Transform of a given system.

center\|framed\|An example of a Bode magnitude and phase plot set. The
Magnitude plot is typically on the top, and the Phase plot is typically
on the bottom of the set.

The frequency of the bode plots are plotted against a logarithmic
frequency axis. Every tickmark on the frequency axis represents a power
of 10 times the previous value. For instance, on a standard Bode plot,
the values of the markers go from (0.1, 1, 10, 100, 1000, \...) Because
each tickmark is a power of 10, they are referred to as a **decade**.
Notice that the \"length\" of a decade decreases as you move to the
right on the graph. (note that this description doesn\'t match the chart
above\... there are 10 tickmarks per decade, not one, but since it is a
log chart they are not evenly spaced).

The bode Magnitude plot measures the system Input/Output ratio in
special units called **decibels**. The Bode phase plot measures the
phase shift in degrees (typically, but radians are also used).

### Decibels

A **Decibel** is a ratio between two numbers on a logarithmic scale. To
express a ratio between two numbers (A and B) as a decibel we apply the
following formula for numbers that represent amplitudes (numbers that
represent a power measurement use a factor of 10 rather than 20):

$$dB = 20 \log\left({A \over B}\right)$$

Where dB is the decibel result.

Or, if we just want to take the decibels of a single number C, we could
just as easily write:

$$dB = 20 \log(C)$$

### Frequency Response Notations

If we have a system transfer function T(s), we can separate it into a
numerator polynomial N(s) and a denominator polynomial D(s). We can
write this as follows:

$$T(s) = \frac{N(s)}{D(s)}$$

To get the magnitude gain plot, we must first transit the transfer
function into the frequency response by using the change of variables:

$$s = j\omega$$

From here, we can say that our frequency response is a composite of two
parts, a real part R and an imaginary part X:

$$T(j\omega) = R(\omega) + jX(\omega)$$

We will use these forms below.

### Straight-Line Approximations

The Bode magnitude and phase plots can be quickly and easily
approximated by using a series of straight lines. These approximate
graphs can be generated by following a few short, simple rules (listed
below). Once the straight-line graph is determined, the actual Bode plot
is a smooth curve that follows the straight lines, and travels through
the **breakpoints**.

### Break Points

If the frequency response is in pole-zero form:

$$T(j\omega) = \frac{\prod_n|j\omega + z_n|}{\prod_m|j\omega + p_m|}$$

We say that the values for all z~n~ and p~m~ are called **break points**
of the Bode plot. These are the values where the Bode plots experience
the largest change in direction.

Break points are sometimes also called \"break frequencies\", \"cutoff
points\", or \"corner points\".

## Bode Gain Plots

**Bode Gain Plots**, or **Bode Magnitude Plots** display the ratio of
the system gain at each input frequency.

### Bode Gain Calculations

The magnitude of the transfer function T is defined as:

$$|T(j\omega)| = \sqrt{R^2 + X^2}$$

However, it is frequently difficult to transition a function that is in
\"numerator/denominator\" form to \"real+imaginary\" form. Luckily, our
decibel calculation comes in handy. Let\'s say we have a frequency
response defined as a fraction with numerator and denominator
polynomials defined as:

$$T(j\omega) = \frac{\prod_n|j\omega + z_n|}{\prod_m|j\omega + p_m|}$$

If we convert both sides to decibels, the logarithms from the decibel
calculations convert multiplication of the arguments into additions, and
the divisions into subtractions:

$$Gain = \sum_n20\log(|j\omega + z_n|) - \sum_m20\log(|j\omega + p_m|)$$

And calculating out the gain of each term and adding them together will
give the gain of the system at that frequency.

### Bode Gain Approximations

The slope of a straight line on a Bode magnitude plot is measured in
units of **dB/Decade**, because the units on the vertical axis are dB,
and the units on the horizontal axis are decades.

The value ω = 0 is infinitely far to the left of the bode plot (because
a logarithmic scale never reaches zero), so finding the value of the
gain at ω = 0 essentially sets that value to be the gain for the Bode
plot from all the way on the left of the graph up till the first break
point. The value of the slope of the line at ω = 0 is 0 dB/Decade.

From each pole break point, the slope of the line decreases by
20 dB/Decade. The line is straight until it reaches the next break
point. From each zero break point the slope of the line increases by
20 dB/Decade. Double, triple, or higher amounts of repeat poles and
zeros affect the gain by multiplicative amounts. Here are some examples:

-   2 poles: -40 dB/Decade
-   10 poles: -200 dB/Decade
-   5 zeros: +100 dB/Decade

## Bode Phase Plots

**Bode phase plots** are plots of the phase shift to an input waveform
dependent on the frequency characteristics of the system input. Again,
the Laplace transform does not account for the phase shift
characteristics of the system, but the Fourier Transform can. The phase
of a complex function, in \"real+imaginary\" form is given as:

$$\angle T(j\omega) = \tan^{-1}\left(\frac{X}{R}\right)$$

## Bode Procedure

Given a frequency response in pole-zero form:

$$T(j\omega) = A\frac{\prod_n|j\omega + z_n|}{\prod_m|j\omega + p_m|}$$

Where A is a non-zero constant (can be negative or positive).

Here are the steps involved in sketching the approximate Bode magnitude
plots:

Here are the steps to drawing the Bode phase plots:

## Examples

### Example: Constant Gain

### Example: Integrator

### Example: Differentiator

### Example: 1st Order, Low-pass Filter (1 Break Point)

## Further reading

-   Circuit Theory/Bode Plots
-   Bode Plots on
    ControlTheoryPro.com

