Given an arbitrary transfer function, such as

$$H(s) = 100{{s + 1} \over {(s + 10)(s + 100)}} = 100{{s + 1} \over {{s^2} + 110s + 1000}}$$

the question naturally arises: "How can we display this function?" The most useful way to display this function is with two plots, the first showing the magnitude of the transfer function and the second showing its phase. One way to do this is by simply entering many values for the frequency (with s=jω), calculating the magnitude and phase at each frequency and displaying them. This is what a computer would naturally do. For example if you use MATLAB® and enter the commands

>> mySys=tf(100*[1 1],[1 110 1000]) mySys = 100 s + 100 ------------------ s^2 + 110 s + 1000 >> bode(mySys)

you get a plot like the one shown below. The asymptotic solution is given elsewhere.

However, there are reasons to develop a method for drawing Bode diagrams manually. By drawing the plots by hand you develop an understanding about how the locations of poles and zeros effect the shape of the plots. With this knowledge you can predict how a system behaves in the frequency domain by simply examining its transfer function. On the other hand, if you know the shape of transfer function that you want, you can use your knowledge of Bode diagrams to generate the transfer function.

The first task when drawing a Bode diagram by hand is to rewrite the transfer
function so that all the poles and zeros are written in the form (1+s/ω_{0}).
The reasons for this will become apparent when deriving the rules for a real pole. A derivation will be done
using the transfer function from above, but it is also possible to do a more generic derivation. Let's rewrite the
transfer function from above.

$$\eqalign{ H(s) &= 100{{s + 1} \over {(s + 10)(s + 100)}} = 100{{1 + s/1} \over {10 \cdot (1 + s/10) \cdot 100 \cdot (1 + s/100)}} \cr &= 0.1{{1 + s/1} \over {(1 + s/10)(1 + s/100)}} \cr} $$

Now let's examine how we can easily draw the magnitude and phase
of this function when s=*j**ω*.

First note that this expression is made up of four terms, a constant (0.1), a
zero (at s=-1), and two poles (at s=-10 and s=-100). We can rewrite the function
(with s=*jω*) as four individual phasors (i.e., magnitude and phase), each phasor is within a set of square brackets to make them more easily distinguished from each other..

$$\eqalign{ H(j\omega ) &= 0.1{{1 + j\omega /1} \over {(1 + j\omega /10)(1 + j\omega /100)}} \cr &= \left[ {\left| {0.1} \right|\angle \left( {0.1} \right)} \right]{{\left[ {\left| {1 + j\omega /1} \right|\angle \left( {1 + j\omega /1} \right)} \right]} \over {\left[ {\left| {1 + j\omega /10} \right|\angle \left( {1 + j\omega /10} \right)} \right]\left[ {\left| {1 + j\omega /100} \right|\angle \left( {1 + j\omega /100} \right)} \right]}} \cr} $$

We will show (below) that drawing the magnitude and phase of
each individual phasor is fairly straightforward. The difficulty lies in trying
to draw the magnitude and phase of H(*jω*). We can write H(*jω*)
as a single phasor:

$$\eqalign{ H(j\omega ) &= \left( {\left| {0.1} \right|{{\left| {1 + j\omega /1} \right|} \over {\left| {1 + j\omega /10} \right|\left| {1 + j\omega /100} \right|}}} \right)\left( {\angle \left( {0.1} \right) + \angle \left( {1 + j\omega /1} \right) - \angle \left( {1 + j\omega /10} \right) - \angle \left( {1 + j\omega /100} \right)} \right) \cr &= \left| {H(j\omega )} \right|\angle H(j\omega ) \cr \cr \left| {H(j\omega )} \right| &= \left| {0.1} \right|{{\left| {1 + j\omega /1} \right|} \over {\left| {1 + j\omega /10} \right|\left| {1 + j\omega /100} \right|}} \cr \angle H(j\omega ) &= \angle \left( {0.1} \right) + \angle \left( {1 + j\omega /1} \right) - \angle \left( {1 + j\omega /10} \right) - \angle \left( {1 + j\omega /100} \right) \cr} $$

Drawing the phase is fairly simple. We can draw each phase
term separately, and then simply add (or subtract) them. The magnitude term is not so straightforward
because the magnitude terms are *multiplied*, it would be
much easier if they were added - then we could draw each term on a graph and just *add* them. A method for doing this is outlined below.

One way to transform multiplication into addition is by using the logarithm. Instead of using a simple logarithm, we will use a deciBel (named for Alexander Graham Bell). (Note: Why the deciBel?) The relationship between a quantity, Q, and its deciBel representation, X, is given by:

$$X = 20 \cdot log{_{10}}\left( Q \right)$$

So if Q=100 then X=40; Q=0.01 gives X=-40; X=3 gives Q=1.41; and so on.

If we represent the magnitude of H(s) in deciBels several things happen.

$$\eqalign{ \left| {H(s)} \right| &= \left| {0.1} \right|{{\left| {1 + j\omega /1} \right|} \over {\left| {1 + j\omega /10} \right|\left| {1 + j\omega /100} \right|}} \cr 20 \cdot {\log _{10}}\left( {\left| {H(s)} \right|} \right) &= 20 \cdot {\log _{10}}\left( {\left| {0.1} \right|{{\left| {1 + j\omega /1} \right|} \over {\left| {1 + j\omega /10} \right|\left| {1 + j\omega /100} \right|}}} \right) \cr &= 20 \cdot {\log _{10}}\left( {\left| {0.1} \right|} \right) + 20 \cdot {\log _{10}}\left( {\left| {1 + j\omega /1} \right|} \right) + 20 \cdot {\log _{10}}\left( {{1 \over {\left| {1 + j\omega /10} \right|}}} \right) + 20 \cdot {\log _{10}}\left( {{1 \over {\left| {1 + j\omega /100} \right|}}} \right) \cr &= 20 \cdot {\log _{10}}\left( {\left| {0.1} \right|} \right) + 20 \cdot {\log _{10}}\left( {\left| {1 + j\omega /1} \right|} \right) - 20 \cdot {\log _{10}}\left( {\left| {1 + j\omega /10} \right|} \right) - 20 \cdot {\log _{10}}\left( {\left| {1 + j\omega /100} \right|} \right) \cr} $$

The advantages of using deciBels (and of writing poles and zeros in the form (1+s/ω_{0}))
are now revealed. The fact that the deciBel is a logarithmic term transforms
the multiplications and divisions of the individual terms to additions and subtsractions. Another benefit is
apparent in the last line that reveals just two types of terms, a constant term
and terms of the form 20·log10(|1+jω/ω_{0}|). Plotting the constant
term is trivial, however the other terms are not so straightforward. These
plots will be discussed below. However, once
these plots are drawn for the individual terms, they can simply be added together
to get a plot for H(s).

If we look at the phase of the transfer function, we see much the same thing: The phase plot is easy to draw if we take our lead from the magnitude plot. First note that the transfer function is made up of four terms. If we want

$$\angle H(s) = \angle \left( {0.1} \right) + \angle \left( {1 + j\omega /1} \right) - \angle \left( {1 + j\omega /10} \right) - \angle \left( {1 + j\omega /100} \right)$$

Again there are just two types of terms, a constant term and terms of the form
(1+jω/ω_{0}). Plotting the constant term is trivial; the other
terms are discussed below.

The discussion above dealt with only a single transfer function. Another derivation that is more general, but a little more complicated mathematically is here.

Following the discussion above, the way to make a Bode Diagram is to split the function up into its constituent parts, plot the magnitude and phase of each part, and then add them up. The following gives a derivation of the plots for each type of constituent part. Examples, including rules for making the plots follow in the next document, which is more of a "How to" description of Bode diagrams.

Consider a constant term,

$$H(s) = H(j\omega) = K$$

Clearly the magnitude is constant

$$\left| {H(j\omega )} \right| = |K|$$

The phase is also constant. If K is positive, the phase is 0° (or any even multiple of 180°). If K is negative the phase is -180°, or any odd multiple of 180°. We will use -180° because that is what MATLAB® uses. Expressed in radians we can say that if K is positive the phase is 0 radians, if K is negative the phase is -π radians.

$$\eqalign{
The magnitude (in dB is calculated as $$20 \cdot {\log _{10}}\left( {15} \right) = 23.5$$. |

- For a constant term, the magnitude plot is a straight line.
- The phase plot is also a straight line, either at 0° (for a positive constant) or ±180° (for a negative constant).

Consider a simple real pole

$$H\left( s \right) = {1 \over {1 + {s \over {{\omega _0}}}}},\quad \quad H\left( {j\omega } \right) = {1 \over {1 + j{\omega \over {{\omega _0}}}}}$$

The frequency ω_{0} is called the break frequency, the corner frequency
or the 3 dB frequency (more on this last name later).

The magnitude is given by

$$\eqalign{ & \left| {H\left( {j\omega } \right)} \right| = \left| {{1 \over {1 + j{\omega \over {{\omega _0}}}}}} \right| = {1 \over {\sqrt {{1^2} + {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} }} \cr & {\left| {H\left( {j\omega } \right)} \right|_{dB}} = 20 \cdot {\log _{10}}\left( {{1 \over {\sqrt {1 + {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} }}} \right) \cr} $$

Let's consider three cases for the value of the frequency:

**Case 1)** ω<<ω_{0}. This is the
low frequency case. If ω<<ω_{0}

$\sqrt {1 + {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} \approx 1$, and ${\left| {H\left( {j\omega } \right)} \right|_{dB}} \approx 20 \cdot {\log _{10}}\left( {{1 \over 1}} \right) = 0$

This low frequency approximation is shown in blue on the diagram below.

**Case 2)** ω>>ω_{0}. This is the high frequency case.
If ω>>ω_{0}

We can write an approximation for the magnitude of the transfer function

$\sqrt {1 + {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} \approx \sqrt {{{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} \approx {\omega \over {{\omega _0}}}$, and

$${\left| {H\left( {j\omega } \right)} \right|_{dB}} \approx 20 \cdot {\log _{10}}\left( {{{{\omega _0}} \over \omega }} \right)$$

The high frequency approximation is at shown in green on the diagram
below. It is a straight line with a slope of -20 dB/decade going through
the break frequency at 0 dB (if ω=ω_{0} the approximation simplifies to 0 dB; ω=10·ω_{0} gives an approximate gain of 0.1, or -20 dB and so on). That is, the approximation goes through 0 dB at ω=ω_{0}, and for every factor of 10 increase
in frequency, the magnitude drops by 20 dB..

**Case 3)** ω=ω_{0}. The break frequency. At this
frequency

$${\left| {H\left( {j{\omega _0}} \right)} \right|_{dB}} = 20 \cdot {\log _{10}}\left( {{1 \over {\sqrt {1 + {{\left( {{{{\omega _0}} \over {{\omega _0}}}} \right)}^2}} }}} \right) = 20 \cdot {\log _{10}}\left( {{1 \over {\sqrt 2 }}} \right) \approx - 3\;dB$$

This point is shown as a red circle on the diagram.

To draw a piecewise linear approximation, use the low frequency asymptote up to the break frequency, and the high frequency asymptote thereafter.

Interactive Demo |

The resulting asymptotic approximation is shown highlighted in transparent magenta. The maximum error between the asymptotic approximation and the exact magnitude function occurs at the break frequency and is approximately -3 dB.

The rule for drawing the piecewise linear approximation for a real pole can be stated thus:

*For a simple real pole the piecewise linear asymptotic Bode plot for
magnitude is at 0 dB until the break frequency and then drops at 20 dB per
decade as frequency increases (i.e., the slope is -20 dB/decade).*

The phase of a single real pole is given by is given by

$$\angle H\left( {j\omega } \right) = \angle \left( {{1 \over {1 + j{\omega \over {{\omega _0}}}}}} \right) = - \angle \left( {1 + j{\omega \over {{\omega _0}}}} \right) = - \arctan \left( {{\omega \over {{\omega _0}}}} \right)$$

Let us again consider three cases for the value of the frequency:

**Case 1)** ω<<ω_{0}. This is the low frequency case with ω/ω_{0}→0.
At these frequencies We can write an approximation for the phase of the
transfer function

$$\angle H\left( {j\omega } \right) \approx -\arctan \left( 0 \right) = 0^\circ = 0\;rad$$

The low frequency approximation is shown in blue on the diagram below.

**Case 2)** ω>>ω_{0}. This is the high frequency case with ω/ω_{0}→∞.
We can write an approximation for the phase of the transfer function

$$\angle H\left( {j\omega } \right) \approx - \arctan \left( \infty \right) = - 90^\circ = - {\pi \over 2}\;rad$$

The high frequency approximation is at shown in green on the diagram below. It is a horizontal line at -90°.

**Case 3)** ω=ω_{0}. The break frequency. At this
frequency

$$\angle H\left( {j\omega } \right) = - \arctan \left( 1 \right) = - 45^\circ = - {\pi \over 4}\;rad$$

This point is shown as a red circle on the diagram.

Interactive Demo |

A piecewise linear approximation is not as easy in this case because the high and low frequency asymptotes don't intersect. Instead we use a rule that follows the exact function fairly closely, but is also somewhat arbitrary. Its main advantage is that it is easy to remember. The rule can be stated as

*Follow the low frequency asymptote at 0° until one tenth the break frequency
(0.1·ω _{0}) then decrease linearly to meet the high frequency asymptote
at ten times the break frequency (10·ω_{0}). *This line is shown above. Note that there is no error at the break
frequency and about 5.7° of error at

$$H(s)=\frac{1}{1+\frac{s}{5}}$$

The second example shows a double pole at 30 radians per second. Note that the slope of the asymptote is -40 dB/decade and the phase goes from 0 to -180°.

$$H(s)=\frac{1}{\left(1+\frac{s}{30}\right)^2}$$

- For a simple real pole the piecewise linear asymptotic Bode plot for
magnitude is at 0 dB until the break frequency and then drops at 20 dB per
decade (i.e., the slope is -20 dB/decade). An n
^{th}order pole has a slope of -20·n dB/decade. - The phase plot is at 0° until one tenth the break frequency and
then drops linearly to -90° at ten times the break frequency.
An n
^{th}order pole drops to -90°·n.

There is another approximation for phase that is occasionally used. The approximation is developed by matching the slope of the actual phase term to that of the approximation at ω=ω_{0}. Using math similar to that given here (for the underdamped case) it can be shown that by drawing a line starting at 0° at ω=ω_{0}/e^{π/2}=ω_{0}/4.81 (or ω_{0}·e^{-π/2}) to -90° at ω=ω_{0}·4.81 we get a line with the same slope as the actual function at ω=ω_{0}. This approximation is slightly easier to remember as a line drawn from 0° at ω_{0}/5 to -90° at ω_{0}·5. The latter is shown on the diagram below.

Although this method is more accurate in the region around ω=ω_{0} there is a larger maximum error (more than 10°) near ω_{0}/5 and ω_{0}·5 when compared to the method described previously.

The piecewise linear approximation for a zero is much like that for a pole Consider a simple zero:

$$H(s)=1+\frac{s}{\omega_0},\quad H(j\omega)=1+j\frac{\omega}{\omega_0}$$

The development of the magnitude plot for a zero follows that for a pole. Refer to the previous section for details. The magnitude of the zero is given by

$$\left| {H\left( {j\omega } \right)} \right| = \left| {1 + j{\omega \over {{\omega _0}}}} \right|$$

Again, as with the case of the real pole, there are three cases:

- At low frequencies, ω<<ω
_{0}, the gain is approximately 1 (or 0 dB). - At high frequencies, ω>>ω
_{0}, the gain increases at 20 dB/decade and goes through the break frequency at 0 dB. - At the break frequency, ω=ω
_{0}, the gain is about 3 dB.

The rule for drawing the piecewise linear approximation for a real zero can be stated thus:

*For a simple real zero the piecewise linear asymptotic Bode plot for
magnitude is at 0 dB until the break frequency and then increases at 20
dB per decade (i.e., the slope is +20 dB/decade).*

The phase of a simple zero is given by:

$$\angle H\left( {j\omega } \right) = \angle \left( {1 + j{\omega \over {{\omega _0}}}} \right) = \arctan \left( {{\omega \over {{\omega _0}}}} \right)$$

The phase of a single real zero also has three cases (which can be derived similarly to those for the real pole, given above):

- At low frequencies, ω<<ω
_{0}, the phase is approximately zero. - At high frequencies, ω>>ω
_{0}, the phase is +90°. - At the break frequency, ω=ω
_{0}, the phase is +45°.

The rule for drawing the phase plot can be stated thus:

*Follow the low frequency asymptote at 0° until one tenth the break frequency
(0.1 ω _{0}) then increase linearly to meet the high frequency asymptote
at ten times the break frequency (10 ω_{0}).*

This example shows a simple zero at 30 radians per second. The asymptotic approximation is magenta, the exact function is the dotted black line.

$$H(s)=1+\frac{s}{30}$$- The plots for a real zero are like those for the real pole but mirrored about 0dB or 0°.
- For a simple real
*zero*the piecewise linear asymptotic Bode plot for magnitude is at 0 dB until the break frequency and then*rises*at +20 dB per decade (i.e., the slope is +20 dB/decade). An n^{th}order*zero*has a slope of +20·n dB/decade. - The phase plot is at 0° until one tenth the break frequency and
then
*rises*linearly to +90° at ten times the break frequency. An n^{th}order*zero**rises*to +90°·n.

A pole at the origin is easily drawn exactly. Consider

$$H\left( s \right) = {1 \over s},\quad H\left( {j\omega } \right) = {1 \over {j\omega }} = - {j \over \omega }$$

The magnitude is given by

$$\eqalign{

\left| {H\left( {j\omega } \right)} \right| &= \left| { - {j \over \omega }} \right| = {1 \over \omega } \cr

{\left| {H\left( {j\omega } \right)} \right|_{dB}} &= 20\cdot{\log _{10}}\left( {{1 \over \omega }} \right) = - 20\cdot{\log _{10}}\left( \omega \right) \cr} $$

In this case there is no need for approximate functions and asymptotes, we can plot the exact funtion. The function is represented by a straight line on a Bode plot with a slope
of -20 dB per decade and going through 0 dB at 1 rad/ sec. It also goes
through 20 dB at 0.1 rad/sec, -20 dB at 10 rad/sec... Since there are no parameters (i.e., ω_{0}) associated with this function, it is always drawn in exactly the same manner.

The rule for drawing the magnitude for a pole at the origin can be thus:

*For a pole at the origin draw a line with a slope of -20 dB/decade
that goes through 0 dB at 1 rad/sec.*

The phase of a simple zero is given by (H(jω) is a negative imaginary number for all values of ω so the phase is always -90°):

$$\angle H\left( {j\omega } \right) = \angle \left( { - {j \over \omega }} \right) = - 90^\circ $$

The rule for drawing the phase plot for a pole at the origin an be stated thus:

*The phase for a pole at the origin is -90°.*

This example shows a simple pole at the origin. The exact (dotted black line) is the same as the approximation (magenta).

No interactive demo is provided because the plots are always drawn in the same way.

- For a simple pole at the origin draw a straight line with a slope of -20 dB per decade and going through 0 dB at 1 rad/ sec.
- The phase plot is at -90°.
- The magnitude of an n
^{th}order pole has a slope of -20·n dB/decade and a constant phase of -90°·n.

A zero at the origin is just like a pole at the origin but the magnitude increases with increasing ω, and the phase is positive 90° (i.e. simply mirror the graphs for the pole around the origin around 0dB or 0°).

This example shows a simple zero at the origin. The exact (dotted black line) is the same as the approximation (magenta).

- The plots for a zero at the origin are like those for the pole but mirrored about 0dB or 0°.
- For a simple zero at the origin draw a straight line with a slope of +20 dB per decade and going through 0 dB at 1 rad/ sec.
- The phase plot is at +90°.
- The magnitude of an n
^{th}order zero has a slope of +20·n dB/decade and a constant phase of +90°·n.

The magnitude and phase plots of a complex conjugate (underdamped) pair of poles is more complicated than those for a simple pole. Consider the transfer function (with 0<ζ<1):

$$H(s) = {{\omega _0^2} \over {{s^2} + 2\zeta {\omega _0}s + \omega _0^2}} = {1 \over {{{\left( {{s \over {{\omega _0}}}} \right)}^2} + 2\zeta \left( {{s \over {{\omega _0}}}} \right) + 1}}$$

The magnitude is given by

$$\eqalign{

\left| {H(j\omega )} \right| &= \left| {{1 \over {{{\left( {{{j\omega } \over {{\omega _0}}}} \right)}^2} + 2\zeta \left( {{{j\omega } \over {{\omega _0}}}} \right) + 1}}} \right| = \left| {{1 \over { - {{\left( {{\omega \over {{\omega _0}}}} \right)}^2} + j2\zeta \left( {{\omega \over {{\omega _0}}}} \right) + 1}}} \right| = \left| {{1 \over {\left( {1 - {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} \right) + j\left( {2\zeta \left( {{\omega \over {{\omega _0}}}} \right)} \right)}}} \right| \cr

&= {1 \over {\sqrt {{{\left( {1 - {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} \right)}^2} + {{\left( {2\zeta {\omega \over {{\omega _0}}}} \right)}^2}} }} \cr

{\left| {H(j\omega )} \right|_{dB}} &= - 20 \cdot {\log _{10}}\left( {\sqrt {{{\left( {1 - {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} \right)}^2} + {{\left( {2\zeta {\omega \over {{\omega _0}}}} \right)}^2}} } \right) \cr} $$

As before, let's consider three cases for the value of the frequency:

**Case 1)** ω<<ω_{0}. This is the low frequency case.
We can write an approximation for the magnitude of the transfer function

$${\left| {H(j\omega )} \right|_{dB}} = - 20 \cdot {\log _{10}}\left( 1 \right) = 0$$

The low frequency approximation is shown in red on the diagram below.

**Case 2)** ω>>ω_{0}. This is the high frequency case.
We can write an approximation for the magnitude of the transfer function

$${\left| {H(j\omega )} \right|_{dB}} = - 20 \cdot {\log _{10}}\left( {{{\left( {{\omega \over {{\omega _0}}}} \right)}^2}} \right) = - 40 \cdot {\log _{10}}\left( {{\omega \over {{\omega _0}}}} \right)$$

The high frequency approximation is at shown in green on the diagram below. It is a straight line with a slope of -40 dB/decade going through the break frequency at 0 dB. That is, for every factor of 10 increase in frequency, the magnitude drops by 40 dB.

**Case 3)** ω≈ω_{0}. It can
be shown that a peak occurs in the magnitude plot near the break frequency.
The derivation of the approximate amplitude and location of the peak are
given here. We make the approximation that a peak exists only when

0<ζ<0.5

and that the peak occurs at ω_{0} with height 1/(2·ζ).

To draw a piecewise linear approximation, use the low frequency asymptote up to the break frequency, and the high frequency asymptote thereafter. If ζ<0.5, then draw a peak of amplitude 1/(2·ζ) Draw a smooth curve between the low and high frequency asymptote that goes through the peak value.

As an example for the curve shown below ω_{0}=10, ζ=0.1,

$$H(s) = {1 \over {{{{s^2}} \over {100}} + 0.02\zeta s + 1}} = {1 \over {{{\left( {{s \over {10}}} \right)}^2} + 0.2\left( {{s \over {10}}} \right) + 1}} = {1 \over {{{\left( {{s \over {{\omega _0}}}} \right)}^2} + 2\zeta \left( {{s \over {{\omega _0}}}} \right) + 1}}$$

The peak will have an amplitude of 1/(2·ζ)=5.00 or 14 dB.

Interactive Demo |

The resulting asymptotic approximation is shown as a black dotted line, the exact response is a black solid line.

The rule for drawing the piecewise linear approximation for a complex conjugate pair of poles can be stated thus:

*For the magnitude plot of complex conjugate poles draw a 0 dB at low
frequencies until the break frequency, ω _{0}, and then drops with a slope of -40 dB/decade. If ζ<0.5 we draw a peak of height at ω_{0}, otherwise no peak is drawn.*

$$\left| {H(j{\omega _0})} \right| \approx {1 \over {2\zeta }},\quad {\left| {H(j{\omega _0})} \right|_{dB}} \approx - 20 \cdot {\log _{10}}\left( {2\zeta } \right)$$

Note: The actual height of the peak and its frequency are both slightly less than the approximations given above. An in depth discussion of the magnitude and phase approximations (along with some alternate approximations) are given here.

The phase of a complex conjugate pole is given by is given by

$$\eqalign{

\angle H(j\omega ) &= \angle \left( {{1 \over {{{\left( {{{j\omega } \over {{\omega _0}}}} \right)}^2} + 2\zeta \left( {{{j\omega } \over {{\omega _0}}}} \right) + 1}}} \right) = - \angle \left( {{{\left( {{{j\omega } \over {{\omega _0}}}} \right)}^2} + 2\zeta \left( {{{j\omega } \over {{\omega _0}}}} \right) + 1} \right) = - \angle \left( {1 - {{\left( {{\omega \over {{\omega _0}}}} \right)}^2} + 2\zeta \left( {{{j\omega } \over {{\omega _0}}}} \right)} \right) \cr

&= - \arctan \left( {{{2\zeta {\omega \over {{\omega _0}}}} \over {1 - {{\left( {{\omega \over {{\omega _0}}}} \right)}^2}}}} \right) \cr} $$

Let us again consider three cases for the value of the frequency:

**Case 1)** ω<<ω_{0}. This is the low frequency case.
At these frequencies We can write an approximation for the phase of the
transfer function

$$\angle H\left( {j\omega } \right) \approx -\arctan \left( {{\omega \over {{\omega _0}}}} \right) = 0^\circ = 0\;rad$$

The low frequency approximation is shown in red on the diagram below.

**Case 2)** ω>>ω_{0}. This is the high frequency case.
We can write an approximation for the phase of the transfer function

$$\angle H(j\omega ) = - 180^\circ $$

The high frequency approximation is at shown in green on the diagram below. It is a straight line at -180°.

**Case 3)** ω=ω_{0}. The break frequency. At this
frequency

$$\angle H(j\omega ) = - 90^\circ $$

The asymptotic approximation is shown below for ω_{0}=10, ζ=0.1, followed by an
explanation

$$H(s) = {1 \over {{{{s^2}} \over {100}} + 0.02\zeta s + 1}} = {1 \over {{{\left( {{s \over {10}}} \right)}^2} + 0.2\left( {{s \over {10}}} \right) + 1}} = {1 \over {{{\left( {{s \over {{\omega _0}}}} \right)}^2} + 2\zeta \left( {{s \over {{\omega _0}}}} \right) + 1}}$$

Interactive Demo |

A piecewise linear approximation is a bit more complicated in this case, and there are no hard and fast rules for drawing it. The most common way is to look up a graph in a textbook with a chart that shows phase plots for many values of ζ. Three asymptotic approximations are given here. We will use the approximation that connects the the low frequency asymptote to the high frequency asymptote starting at

$$\omega = {{{\omega _0}} \over {{{10}^\zeta }}} = {\omega _0} \cdot {10^{ - \zeta }}$$

and ending at

$$\omega = {\omega _0} \cdot {10^\zeta }$$

Since ζ=0.2 in this case this means that the phase starts at 0° and then breaks downward at ω=ω_{0}/10^{ζ}=7.9 rad/sec. The phase reaches -180° at ω=ω_{0}·10^{ζ}=12.6 rad/sec.

As a practical matter If ζ<0.02, the approximation can be simply a vertical
line at the break frequency. One advantage of this approximation is that it is very easy to plot on semilog paper. Since the number 10·ω_{0} moves up by a full decade from ω_{0}, the number 10^{ζ}·ω_{0} will be a fraction ζ of a decade above ω_{0}. For the example above the corner frequencies for ζ=0.1 fall near ω_{0} one tenth of the way between ω_{0} and ω_{0}/10 (at the lower break frequency) to one tenth of the way between ω_{0} and ω_{0}·10 (at the higher frequency).

The rule for drawing phase of an underdamped pair of poles can be stated as

*Follow the low frequency asymptote at 0*°* until *

$$\omega = {{{\omega _0}} \over {{{10}^\zeta }}}$$

*then decrease linearly to meet the high frequency asymptote at -180*°
at

$$\omega = {\omega _0} \cdot {10^\zeta }$$

Other magnitude and phase approximations (along with exact expressions) are given here.

- For the magnitude plot of complex conjugate poles draw a 0 dB at low frequencies,
go through a peak of height,

$$\left| {H(j{\omega _0})} \right| \approx {1 \over {2\zeta }},\quad {\left| {H(j{\omega _0})} \right|_{dB}} \approx - 20 \cdot {\log _{10}}\left( {2\zeta } \right)$$

at the break frequency and then drop at 40 dB per decade (i.e., the slope is -40 dB/decade). The high frequency asymptote goes through the break frequency. Note that in this approximation the peak only exists for

0 < ζ < 0.5

- To draw the phase plot simply follow low frequency asymptote at 0° until
$$\omega = {{{\omega _0}} \over {{{10}^\zeta }}} = {\omega _0} \cdot {10^{ - \zeta }}$$

then decrease linearly to meet the high frequency asymptote at -180° at$$\omega = {\omega _0} \cdot {10^\zeta }$$

If ζ<0.02, the approximation can be simply a vertical line at the break frequency. - Note that the
*shape*of the graphs (magnitude peak height, steepness of phase transition) are determined solely by ζ, and the frequency at which the magnitude peak and phase transition occur are determined solely by ω_{0}.

Not surprisingly a complex pair of zeros yields results similar to that for a complex pair of poles. The magnitude and phase plots for the complex zero are the mirror image (around 0dB or 0°) of those for the complex pole. Sot hat the magnitude has a dip instead of a peak, the magnitude increases above the break frequency and the phase increases rather than decreasing.

The graph below corresponds to a complex conjugate zero
with ω_{0}=3, ζ=0.25

$$H\left( s \right) = {\left( {{s \over {{\omega _0}}}} \right)^2} + 2\zeta \left( {{s \over {{\omega _0}}}} \right) + 1$$

The dip in the magnitude plot will have a magnitude of 0.5 or -6 dB. The break frequencies for the phase are at ω=ω_{0}/10^{ζ}=1.7 rad/sec and ω=ω_{0}·10^{ζ}=5.3 rad/sec.

- The plots for a complex conjugate pair of zeros are like those for the poles but mirrored about 0dB or 0°.
- For the magnitude plot of complex conjugate zeros draw a 0 dB at low frequencies,
go through a dip of magnitude:
$$\left| {H(j{\omega _0})} \right| \approx {2\zeta},\quad {\left| {H(j{\omega _0})} \right|_{dB}} \approx 20 \cdot {\log _{10}}\left( {2\zeta } \right)$$

at the break frequency and then rise at +40 dB per decade (i.e., the slope is +40 dB/decade). The high frequency asymptote goes through the break frequency. Note that the peak only exists for0 < ζ < 0.5

- To draw the phase plot simply follow low frequency asymptote at 0° until
$$\omega = {{{\omega _0}} \over {{{10}^\zeta }}} = {\omega _0} \cdot {10^{ - \zeta }}$$

then increase linearly to meet the high frequency asymptote at 180° at$$\omega = {\omega _0} \cdot {10^\zeta }$$

- Note that the
*shape*of the graphs (magnitude peak height, steepness of phase transition) are determined solely by ζ, and the frequency at which the magnitude peak and phase transition occur are determined solely by ω_{0}.

**Brief review of page: ** This
document derived piecewise linear approximations that can be used to draw different
elements of a Bode diagram. A synopsis of these rules can be found in a separate document.

© Copyright 2005 to 2015 Erik Cheever This page may be freely used for educational purposes.

Erik Cheever Department of Engineering Swarthmore College