This page demonstrates the techniques described previously to take a transfer function defined by the user,
identify the constituent terms, draw the individual Bode plots, and then combine them to obtain the overall
transfer function of the transfer function. Finally, an example is given of how to use the Bode plot to find the
time domain response of the system to a sinusoidal input.
This page displays best on a large monitor.
Be patient it can take several seconds to load, and several seconds each time a change is made.
Define transfer function
Define `H(s)=C\frac{N(s)}{D(s)}`
This tool is designed as an introduction to drawing Bode plots by hand. As such, it only works for
systems with the following restrictions:
Proper Systems (order of den >= order of num)
Maximum of 14 terms
Monic Polynomials (it works for most polynomials, but may fail for some non-monic polynomials).
A monic
polynomial is one whose highest power has a coefficient of 1 ("s2+3s+2" is
monic,
"2s2+6s+4" is not). If you enter a non-monic polynomial, the code will
attempt to
rewrite it as a monic polynomial.
Minimum phase systems; poles and zeros have negative real parts (no poles or zeros in right half
of
s-plane)
There must be no poles and zeros at the origin that cancel each other out (for example
\[\frac{s}{{{s^2} + s}} = \frac{s}{{s\left( {s + 1} \right)}} = \frac{1}{{s + 1}}\]
would display incorrect results.
This tool is designed as an introduction to drawing Bode plots by hand. As such, it only works for
minimum phase systems (i.e., systems whose poles and zeros are not positive; they are no poles and zeros
in the right half of the s-plane). Your system has poles and/or zeros with positive real parts, so the
Bode plot will not be correct. You can refer to the section below to determine the locations of all
poles and zeros.
Note that the magnitude plot will still be correct,
but the phase plot will not be. If a zero is in the right half of the s-plane, its phase decreases
(rather than increasing) around the break frequency. If a pole is in the right half of the s-plane, its
phase increases (rather than decreasing) around the break frequency.
Your system has poles and/and or zeros on the jω axis. Normal rules for Bode plots cannont
be applied so the asymptotic approximations will be incorrect (the rules don't apply because the
transfer function is either going to infinity (at pole), or going to zero (at zero). The zero is a
problem because the dB scale is logarithmic and can't be evaluated at zero.
Your system has multiple poles and/or zeros at the origin. If a system has multiple poles or zeros at the
origin, components of the phase plot may appear to disagree with plots produced by other tools (e.g.,
MATLAB, Mathematica, …). This is because the phase of a complex number is not unique; you can add
any integer multiple of 360° (2π radians) and the angles are equivalent. As an example the phase
of the number -1 could be +180°, or -180°, or ±540°…
Likewise the phase of s2 (where s=j·ω) could be ±180°,
±540°…
Identify individual terms and convert transfer function to standard form
We start with the transfer function (note: the process here follows that on the
second
page
of the file BodeRules.pdf):
Show numeric transfer function; this gets replaced.
We rewrite it by factoring into real poles & zeros, complex poles & zeros and poles & zeros
at the origin.
Show original symbolic transfer function; this gets replaced.
Define all symbols; this gets replaced.
Next we write all the poles and zeros is our standard form.
Show modified symbolic transfer function; this gets replaced.
Rewrite the constant:
Define K; this gets replaced.
So
Transfer function in standard form; this gets replaced.
Now the transfer function is in the form we need to apply our rules to draw the Bode plot.
Draw individual terms
In this section we draw the Bode plots of each of the indivuidual termas enumerated above. Select one of the
terms by selecting the corresponding radio button. The selected term will be highlighted on the graphs with a
thicker line. An explanation that describes how to draw the asymptotic magnitude and phase plots for the
selected term is written out below.
Combine individual terms to get final asymptotic plot
The graphs below show the individual plots (with the same color scheme as above), along with the combined plot (a
thicker gray line) which is the sum of the individual plots for both magnitude and phase; in other words the
asymptotic plot of the complete transfer function. This addition of terms can be complicated, but is
conceptually quite easy.
For magnitude, start at the left side of the graph. All of terms will be equal to 0 with the exception of
the constant term and any poles and zeros at the origin. This gives us a place to start drawing the sum.
From here on we simply pay attention to the slope. If there are no poles and zeros at the origin, the slope
is zero, else it is determined by those poles and zeros at the origin. We continue with this slope until we
get to the first finite pole or zero at which point the slope increases or decreases by a multiple of 20
dB/decade. For example a 1st order pole decreases the slope by 20 dB/dec, a second order pole by 40.... We
keep moving to the right, increasing or decreasing the slope as we come to zeros or poles.
We do the same for phase. At the left side only poles and zeros at the origin, and the sign of the constant
contribute to the initial phase. As we move to the right the slope will increase (for a zero), or decrease
(for a pole).
Show Exact Bode Plot (and a time domain example)
Exact plot
On the Bode plot, the gray lines represent the asymptotic plot, adn the black line is the exact solution. The
pink dots show the magnitude and phase of the Bode plot at a frequency chosen by the user (see below).
Time domain (sinusoidal steady state) response from Bode plot
Input sinusoid: (Note: it can take
several seconds to
refresh after changing variables.) The input to the system is
A·cos(ω·t + φ°)
Input = 1.0 · cos(1.0 · t + 0°); replace this.
Output sinusoid The output is M·A·cos(ω·t +(φ +
θ)°). We get M and θ from the Bode plot. cos(ω · t + (φ + θ)°); replace this.