Asymptotic Bode

Goal:

• Express the frequency response of a system defined by a transfer function as piecewise linear funtions.

Preparations

import numpy as np

from sympy import *
from sympy.physics.control.lti import TransferFunction

from mathprint import *

import matplotlib.pyplot as plt
plt.rcParams.update({
    "text.usetex": True,
    "font.family": "Helvetica",
    "font.size": 10,
})

Define variables that we are going to use repetitively. We also need to define specific prpoperties of the variables.

t        = symbols('t'       , real=True)
s        = symbols('s'       , complex=True)
omega    = symbols('omega'   , positive=True)
a, b     = symbols('a b'     , real=True)
zeta     = symbols('zeta'    , positive=True)
omega_n  = symbols('omega_n' , positive=True)

Asymptotic Bode Plots of Low-Order Transfer Functions

These are the fundamental steps that we must do to plot the frequence response (Bode plot) of a system defined by a transfer function \(H(s)\):

• Do substitution: \(s = j \omega\)

• Compute \(H( j \omega )\)

• The result is a complex number:

  • Extract the magnitude: \( M = \big| H(j \omega) \big| \)

  • Convert the magnitude into dB: \( M_{dB} = 20 \log_{10} M \)

  • Extract the the phase: \( \phi = \angle H(j \omega) \)

It is very easy to execute those steps by using a computer. On the other hand, when we simply want to sketch the Bode plot without using a computer, we can use the asymptotic method to draw the approximated version of the Bode plot.

Case 1: A pole at the origin of the s-plane

Let us start by first defining the transfer function:

H1 = 1/s
mprint('H(s)=', latex(H1))

\[\displaystyle H(s)=\frac{1}{s}\]

After the substitution, we then obtain:

Hjw  = H1.subs(s, I*omega)
M1   = simplify(Abs(Hjw))
Mdb1 = simplify(20*log(M1, 10))
Phi1 = arg(Hjw)
mprintb("M  =", latex(M1))
mprintb("M_{dB} =", latex(Mdb1))
mprintb("\\phi =", latex(Phi1))

\[\displaystyle \boxed{M =\frac{1}{\omega}}\]

\[\displaystyle \boxed{M_{dB} =- \frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}}}\]

\[\displaystyle \boxed{\phi =- \frac{\pi}{2}}\]

Magnitude plot:

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title = '$ M_{dB} $'
p1 = plot(Mdb1, (omega, 0.01 , 100), size=(4, 2), title=title, show=True,  xscale='log')

Phase plot:

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title = '$\\phi $'
p2 = plot(Phi1, (omega, 0.01 , 100), size=(4, 2), title=title, show=True,  xscale='log')

Case 2: A zero at the origin of the s-plane

Let us start by first defining the transfer function:

H2 = s
mprint('H(s)=', latex(H2))

\[\displaystyle H(s)=s\]

After the substitution, we then obtain:

Hjw  = H2.subs(s, I*omega)
M2   = simplify(Abs(Hjw))
Mdb2 = simplify(20 * log(M2, 10))
Phi2 = arg(Hjw)
mprintb("M  =", latex(M2))
mprintb("M_{dB} =", latex(Mdb2))
mprintb("\\phi =", latex(Phi2))

\[\displaystyle \boxed{M =\omega}\]

\[\displaystyle \boxed{M_{dB} =\frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}}}\]

\[\displaystyle \boxed{\phi =\frac{\pi}{2}}\]

Magnitude plot:

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title = '$M_{dB}$'
p1 = plot(Mdb2, (omega, 0.01 , 100), size=(4, 2), title=title, show=True,  xscale='log')

Phase plot:

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title = '$\\phi $'
p2 = plot(Phi2, (omega, 0.01 , 100), size=(4, 2), title=title, show=True,  xscale='log')

Case 3: A single real pole

Let us start by first defining the transfer function:

H3 = 1/(s+a)
mprint('H(s)=', latex(H3))

\[\displaystyle H(s)=\frac{1}{a + s}\]

After the substitution, we then obtain:

Hjw  = H3.subs(s, I*omega)
M3   = simplify(Abs(Hjw))
Mdb3 = simplify(20 * log(M3, 10))
Phi3 = arg(Hjw)
mprintb("M =", latex(M3))
mprintb("M_{dB} =", latex(Mdb3))
mprintb("\\phi =", latex(Phi3))

\[\displaystyle \boxed{M =\frac{1}{\sqrt{a^{2} + \omega^{2}}}}\]

\[\displaystyle \boxed{M_{dB} =- \frac{10 \log{\left(a^{2} + \omega^{2} \right)}}{\log{\left(10 \right)}}}\]

\[\displaystyle \boxed{\phi =\arg{\left(\frac{1}{a + i \omega} \right)}}\]

Now, let us plot the results for both the magnitude and the phase. To do this, we must first set a numerical value for \(a\). Let us set \(a=1\).

a_ = 1

Next, let us write down the approximated version of the plots, for the magnitude and the phase. What we do here is a bit based on observations.

Magnitude plot:

Mhatdb3 = Piecewise((-20 * log(a, 10),     omega <= a),
                    (-20 * log(omega, 10), True))
mprintb("\\hat{ M }_{dB} = ", latex(Mhatdb3))

\[\begin{split}\displaystyle \boxed{\hat{ M }_{dB} = \begin{cases} - \frac{20 \log{\left(a \right)}}{\log{\left(10 \right)}} & \text{for}\: a \geq \omega \\- \frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}}\end{split}\]

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title = '$M_{dB}$'
p1 = plot(Mdb3.subs(a, a_),    (omega, a_/100 , 100*a_), size=(4, 2), title=title, show=False,  xscale='log')
p2 = plot(Mhatdb3.subs(a, a_), (omega, a_/100 , 100*a_), show=False)
p1.append(p2[0])
p1.show()

Phase plot:

Phihat3 = Piecewise((0, omega <= 0.1*a),
                    (-pi/4 * log(omega/(0.1*a), 10) , omega < a*10), 
                    (-pi/2, True))
mprintb("\\hat{ \\phi }  = ", latex(simplify(Phihat3)))

\[\begin{split}\displaystyle \boxed{\hat{ \phi } = \begin{cases} 0 & \text{for}\: 0.1 a \geq 1.0 \omega \\- \frac{\pi \log{\left(\frac{10.0 \omega}{a} \right)}}{4 \log{\left(10 \right)}} & \text{for}\: \omega < 10 a \\- \frac{\pi}{2} & \text{otherwise} \end{cases}}\end{split}\]

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title = '$ \\phi $'
p1 = plot(Phi3.subs(a, a_),    (omega, a_/100 , 100*a_), size=(4, 2), title=title, show=False,  xscale='log')
p2 = plot(Phihat3.subs(a, a_), (omega, a_/100 , 100*a_), show=False)
p1.append(p2[0])
p1.show()

Case 4: A single real zero

Let us start by first defining the transfer function:

H4 = s+a
mprint('H(s)=', latex(H4))

\[\displaystyle H(s)=a + s\]

After the substitution, we then obtain:

Hjw  = H4.subs(s, I*omega)
M4   = simplify(Abs(Hjw))
Mdb4 = simplify(20 * log(M4, 10))
Phi4 = arg(Hjw)
mprint("M  =", latex(M4))
mprint("M_{dB} =", latex(Mdb4))
mprint("\\phi =", latex(Phi4))

\[\displaystyle M =\sqrt{a^{2} + \omega^{2}}\]

\[\displaystyle M_{dB} =\frac{10 \log{\left(a^{2} + \omega^{2} \right)}}{\log{\left(10 \right)}}\]

\[\displaystyle \phi =\arg{\left(a + i \omega \right)}\]

Now, let us plot the results for both the magnitude and the phase. To do this, we must first set a numerical value for \(a\). Let us set \(a=1\).

Next, let us write down the approximated version of the plots, for both the magnitude and the phase. What we do here is a bit based on observations.

Magnitude plot:

Mhatdb4 = Piecewise((20 * log(a, 10), omega <= a),
                    (20 * log(omega, 10), True))
mprintb("\\hat{ M }_{dB} = ", latex(Mhatdb4))

\[\begin{split}\displaystyle \boxed{\hat{ M }_{dB} = \begin{cases} \frac{20 \log{\left(a \right)}}{\log{\left(10 \right)}} & \text{for}\: a \geq \omega \\\frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}}\end{split}\]

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title = '$M_{dB}$'
p1 = plot(Mdb4.subs(a, a_),    (omega, a_/100 , 100*a_), size=(4, 2), title=title, show=False,  xscale='log')
p2 = plot(Mhatdb4.subs(a, a_), (omega, a_/100 , 100*a_), show=False)
p1.append(p2[0])
p1.show()

Phase plot:

Phihat4 = Piecewise((0, omega <= 0.1*a),
                    (pi/4 * log(omega/(0.1*a), 10) , omega < a*10), 
                    (pi/2, True))
mprintb("\\hat{ \\phi } = ", latex(simplify(Phihat4)))

\[\begin{split}\displaystyle \boxed{\hat{ \phi } = \begin{cases} 0 & \text{for}\: 0.1 a \geq 1.0 \omega \\\frac{\pi \log{\left(\frac{10.0 \omega}{a} \right)}}{4 \log{\left(10 \right)}} & \text{for}\: \omega < 10 a \\\frac{\pi}{2} & \text{otherwise} \end{cases}}\end{split}\]

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title = '$ \\phi $'
p1 = plot(Phi4.subs(a, a_),    (omega, a_/100 , 100*a_), size=(4, 2), title=title, show=False,  xscale='log')
p2 = plot(Phihat4.subs(a, a_), (omega, a_/100 , 100*a_), show=False)
p1.append(p2[0])
p1.show()

Case 5: Complex Conjugate Pole Pair

Let us start by first defining the transfer function:

H5 = 1 / (s**2 + 2*zeta*omega_n*s + omega_n**2)
mprint('H(s)=', latex(H5))

\[\displaystyle H(s)=\frac{1}{\omega_{n}^{2} + 2 \omega_{n} s \zeta + s^{2}}\]

By substituting \(s\) with \(j \omega\), we can obtain:

Hjw  = H5.subs(s, I*omega)
M5   = simplify(Abs(Hjw))
Mdb5 = expand_log(20 * log(M5, 10))
Phi5 = arg(Hjw)
mprint("M =", latex(M5))
mprint("M_{dB} =", latex(Mdb5))
mprint("\\phi =", latex(Phi5))

\[\displaystyle M =\frac{1}{\sqrt{\omega^{4} + 4 \omega^{2} \omega_{n}^{2} \zeta^{2} - 2 \omega^{2} \omega_{n}^{2} + \omega_{n}^{4}}}\]

\[\displaystyle M_{dB} =- \frac{10 \log{\left(\omega^{4} + 4 \omega^{2} \omega_{n}^{2} \zeta^{2} - 2 \omega^{2} \omega_{n}^{2} + \omega_{n}^{4} \right)}}{\log{\left(10 \right)}}\]

\[\displaystyle \phi =\arg{\left(\frac{1}{- \omega^{2} + 2 i \omega \omega_{n} \zeta + \omega_{n}^{2}} \right)}\]

Resonance peak:

• Complex conjugate pole pair occurs when \(\zeta < 1\).

• When \(\zeta < 1\), the system becomes underdamped and resonace peak appears in the magnitude plot.

• Resonance peak appears at \(\omega_p\) and its peak magnitude is \(M_p\).

omega_p5 = solve(Eq(diff(M5, omega), 0), omega)
mprintb("\\omega_p=", latex(omega_p5[0]))

\[\displaystyle \boxed{\omega_p=\omega_{n} \sqrt{1 - 2 \zeta^{2}}}\]

Mp5 = simplify(M5.subs(omega, omega_p5[0]))
Mdbp5 =simplify(Mdb5.subs(omega, omega_p5[0]))
mprintb("M_p = ", latex(Mp5))
mprintb("M_{p(dB)} = ", latex(Mdbp5))

\[\displaystyle \boxed{M_p = \frac{1}{2 \omega_{n}^{2} \zeta \sqrt{1 - \zeta^{2}}}}\]

\[\displaystyle \boxed{M_{p(dB)} = - \frac{10 \log{\left(- 4 \omega_{n}^{4} \zeta^{2} \left(\zeta^{2} - 1\right) \right)}}{\log{\left(10 \right)}}}\]

Now, let’s express the asymptotic Bode plot as piecewise functions. At the moment, we will do this by intuition.

Mhatdb5 = Piecewise((-40 * log(omega_n, 10), omega <= omega_n),
                    (-40 * log(omega, 10), omega > omega_n))
mprintb("\\hat{ M }_{dB} ( \\omega ) = ", latex(Mhatdb5))

\[\begin{split}\displaystyle \boxed{\hat{ M }_{dB} ( \omega ) = \begin{cases} - \frac{40 \log{\left(\omega_{n} \right)}}{\log{\left(10 \right)}} & \text{for}\: \omega \leq \omega_{n} \\- \frac{40 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}}\end{split}\]

Magnitude plot:

For plotting, we need to define some numerical values for:

omega_n_ = 1
zeta_ = [0.1, 0.2, 0.5] # plot for several values

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title = '$ M_{dB} $'
p1 = plot(Mdb5.subs(([omega_n, omega_n_],[zeta, zeta_[0]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='r', size=(5, 3), title=title, show=False,  xscale='log', legend=True)
p2 = plot(Mdb5.subs(([omega_n, omega_n_],[zeta, zeta_[1]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='b', show=False)
p3 = plot(Mdb5.subs(([omega_n, omega_n_],[zeta, zeta_[2]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='g', show=False)
p4 = plot(Mhatdb5.subs(([omega_n, omega_n_],[zeta, zeta_[0]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)

p1[0].label = "$\\zeta=" + str(zeta_[0]) + "$"
p2[0].label = "$\\zeta=" + str(zeta_[1]) + "$"
p3[0].label = "$\\zeta=" + str(zeta_[2]) + "$"
p4[0].label = ""

p1.append(p2[0])
p1.append(p3[0])
p1.append(p4[0])

p1.show()

The general shape of the asymptotic magnitude plot is not affected by \(\zeta\).

We are not going to include the resonance peak for the asymptotic plots. However, we can still compute this peak value by using the equation expressed by \(M_{p(dB)}\):

print(Mdbp5.subs(([omega_n, omega_n_],[zeta, zeta_[0]])).evalf()) # for zeta[0]
print(Mdbp5.subs(([omega_n, omega_n_],[zeta, zeta_[1]])).evalf()) # for zeta[1]
print(Mdbp5.subs(([omega_n, omega_n_],[zeta, zeta_[2]])).evalf()) # for zeta[2]

14.0230481407449
8.13608784304507
1.24938736608300

Resonance peak happens at \(\omega = \omega_{p}\):

print(omega_p5[0].subs(([omega_n, omega_n_],[zeta, zeta_[0]])).evalf())

0.989949493661167

Phase plot:

Phihat5 = Piecewise((0, omega <= (omega_n / 10**zeta)),
                    (-pi/(2*zeta) * log(omega/(10**(-zeta)*omega_n), 10) , omega < (omega_n*10**zeta)), 
                    (-pi, True))
mprintb("\\hat{ \\phi } ( \\omega ) = ", latex(Phihat5))

\[\begin{split}\displaystyle \boxed{\hat{ \phi } ( \omega ) = \begin{cases} 0 & \text{for}\: \omega \leq 10^{- \zeta} \omega_{n} \\- \frac{\pi \log{\left(\frac{10^{\zeta} \omega}{\omega_{n}} \right)}}{2 \zeta \log{\left(10 \right)}} & \text{for}\: \omega < 10^{\zeta} \omega_{n} \\- \pi & \text{otherwise} \end{cases}}\end{split}\]

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title = '$ \\phi $'
p1 = plot(Phi5.subs(([omega_n, omega_n_],[zeta, zeta_[0]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='r', size=(5, 3), title=title, show=False,  xscale='log', legend=True)
p2 = plot(Phi5.subs(([omega_n, omega_n_],[zeta, zeta_[1]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='b', show=False)
p3 = plot(Phi5.subs(([omega_n, omega_n_],[zeta, zeta_[2]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='g', show=False)

p4 = plot(Phihat5.subs(([omega_n, omega_n_],[zeta, zeta_[0]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)
p5 = plot(Phihat5.subs(([omega_n, omega_n_],[zeta, zeta_[1]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)
p6 = plot(Phihat5.subs(([omega_n, omega_n_],[zeta, zeta_[2]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)


p1[0].label = "$\\zeta=" + str(zeta_[0]) + "$"
p2[0].label = "$\\zeta=" + str(zeta_[1]) + "$"
p3[0].label = "$\\zeta=" + str(zeta_[2]) + "$"
p4[0].label = ""
p5[0].label = ""
p6[0].label = ""

p1.append(p2[0])
p1.append(p3[0])
p1.append(p4[0])
p1.append(p5[0])
p1.append(p6[0])

p1.show()

Case 6: Complex Conjugate Zero Pair

Let us start by first defining the transfer function:

H6 = (s**2 + 2*zeta*omega_n*s + omega_n**2) 
mprint('H(s)=', latex(H6))

\[\displaystyle H(s)=\omega_{n}^{2} + 2 \omega_{n} s \zeta + s^{2}\]

Hjw  = H6.subs(s, I*omega)
M6   = simplify(Abs(Hjw))
Mdb6 = simplify(20 * log(M6, 10))
Phi6 = arg(Hjw)
mprint("M  =", latex(M6))
mprint("M_{dB}  =", latex(Mdb6))
mprint("\\phi  =", latex(Phi6))

\[\displaystyle M =\sqrt{\omega^{4} + 4 \omega^{2} \omega_{n}^{2} \zeta^{2} - 2 \omega^{2} \omega_{n}^{2} + \omega_{n}^{4}}\]

\[\displaystyle M_{dB} =\frac{10 \log{\left(\omega^{4} + 4 \omega^{2} \omega_{n}^{2} \zeta^{2} - 2 \omega^{2} \omega_{n}^{2} + \omega_{n}^{4} \right)}}{\log{\left(10 \right)}}\]

\[\displaystyle \phi =\arg{\left(- \omega^{2} + 2 i \omega \omega_{n} \zeta + \omega_{n}^{2} \right)}\]

Resonance peak:

Complex conjugate zero pair occurs when \(\zeta < 1\). Resonance peak appears in the Bode magnitude plot at \(\omega = \omega_n\). The magnitude at this resonance peak can be computed as:

omega_p6 = solve(Eq(diff(M6, omega), 0), omega)
mprintb("\\omega_p=", latex(omega_p6[0]))

\[\displaystyle \boxed{\omega_p=\omega_{n} \sqrt{1 - 2 \zeta^{2}}}\]

Mp6 = simplify(M6.subs(omega, omega_p6[0]))
Mdbp6 =simplify(Mdb6.subs(omega, omega_p6[0]))
mprintb("M_p = ", latex(Mp6))
mprintb("M_{p(dB)} = ", latex(Mdbp6))

\[\displaystyle \boxed{M_p = 2 \omega_{n}^{2} \zeta \sqrt{1 - \zeta^{2}}}\]

\[\displaystyle \boxed{M_{p(dB)} = \frac{10 \log{\left(4 \omega_{n}^{4} \zeta^{2} \left(1 - \zeta^{2}\right) \right)}}{\log{\left(10 \right)}}}\]

Magnitude plot:

We first present the piecewise logarithmic representation of the asymptotic Bode plot. After that, we present the plot for several diffrent values for \(\zeta\).

Mhatdb6 = Piecewise((40 * log(omega_n, 10), omega < omega_n),
                    (40 * log(omega, 10), omega >= omega_n))
mprintb("\\hat{ M } ( \\omega ) = ", latex(Mhatdb6))

\[\begin{split}\displaystyle \boxed{\hat{ M } ( \omega ) = \begin{cases} \frac{40 \log{\left(\omega_{n} \right)}}{\log{\left(10 \right)}} & \text{for}\: \omega < \omega_{n} \\\frac{40 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}}\end{split}\]

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title = '$ M_{dB} $'
p1 = plot(Mdb6.subs(([omega_n, omega_n_],[zeta, zeta_[0]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='r', size=(5, 3), title=title, show=False,  xscale='log', legend=True)
p2 = plot(Mdb6.subs(([omega_n, omega_n_],[zeta, zeta_[1]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='b', show=False)
p3 = plot(Mdb6.subs(([omega_n, omega_n_],[zeta, zeta_[2]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='g', show=False)
p4 = plot(Mhatdb6.subs(([omega_n, omega_n_],[zeta, zeta_[0]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)

p1[0].label = "$\\zeta=" + str(zeta_[0]) + "$"
p2[0].label = "$\\zeta=" + str(zeta_[1]) + "$"
p3[0].label = "$\\zeta=" + str(zeta_[2]) + "$"
p4[0].label = ""

p1.append(p2[0])
p1.append(p3[0])
p1.append(p4[0])

p1.show()

The general shape of the asymptotic magnitude plot is not affected by \(\zeta\).

We are not going to include the resonance peak for the asymptotic plots. However, we can still compute this peak value by using the equation expressed by \(M_{p(dB)}\):

print(Mdbp6.subs(([omega_n, omega_n_],[zeta, zeta_[0]])).evalf()) # for zeta[0]
print(Mdbp6.subs(([omega_n, omega_n_],[zeta, zeta_[1]])).evalf()) # for zeta[1]
print(Mdbp6.subs(([omega_n, omega_n_],[zeta, zeta_[2]])).evalf()) # for zeta[2]

-14.0230481407449
-8.13608784304507
-1.24938736608300

Resonance peak happens at \(\omega = \omega_{p}\):

print(omega_p6[0].subs(([omega_n, omega_n_],[zeta, zeta_[0]])).evalf())

0.989949493661167

Phase plot:

Similar to the prevoius section, we first present the piecewise logarithmic representation of the asymptotic Bode plot. After that, we present the plot for several diffrent values for \(\zeta\).

Phihat6 = Piecewise((0, omega <= (omega_n / 10**zeta)),
                    (pi/(2*zeta) * log(omega/(10**(-zeta)*omega_n), 10) , omega < (omega_n*10**zeta)), 
                    (pi, True))
mprintb("\\hat{ \\phi } ( \\omega ) = ", latex(Phihat6))

\[\begin{split}\displaystyle \boxed{\hat{ \phi } ( \omega ) = \begin{cases} 0 & \text{for}\: \omega \leq 10^{- \zeta} \omega_{n} \\\frac{\pi \log{\left(\frac{10^{\zeta} \omega}{\omega_{n}} \right)}}{2 \zeta \log{\left(10 \right)}} & \text{for}\: \omega < 10^{\zeta} \omega_{n} \\\pi & \text{otherwise} \end{cases}}\end{split}\]

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title = '$ \\phi $'
p1 = plot(Phi6.subs(([omega_n, omega_n_],[zeta, zeta_[0]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='r', size=(5, 3), title=title, show=False,  xscale='log', legend=True)
p2 = plot(Phi6.subs(([omega_n, omega_n_],[zeta, zeta_[1]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='b', show=False)
p3 = plot(Phi6.subs(([omega_n, omega_n_],[zeta, zeta_[2]])),    (omega, omega_n_/10 , 10*omega_n_), line_color='g', show=False)

p4 = plot(Phihat6.subs(([omega_n, omega_n_],[zeta, zeta_[0]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)
p5 = plot(Phihat6.subs(([omega_n, omega_n_],[zeta, zeta_[1]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)
p6 = plot(Phihat6.subs(([omega_n, omega_n_],[zeta, zeta_[2]])), (omega, omega_n_/10 , 10*omega_n_), line_color='k', show=False)

p1.append(p2[0])
p1.append(p3[0])
p1.append(p4[0])
p1.append(p5[0])
p1.append(p6[0])

p1[0].label = "$\\zeta=" + str(zeta_[0]) + "$"
p2[0].label = "$\\zeta=" + str(zeta_[1]) + "$"
p3[0].label = "$\\zeta=" + str(zeta_[2]) + "$"
p4[0].label = ""
p5[0].label = ""
p6[0].label = ""


p1.show()

Summary

Finally, let us summarize all boxed formulas.

Asymptotic Bode as piecewise functions

The following table shows piecewise approximations of a Bode plot.

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from pandas import DataFrame
from IPython.display import Markdown

def makelatex(args):
    return ["${}$".format(latex(a)) for a in args]

Hs   = [H1, H2, H3, H4, H5, H6]
Mdbs = [Mdb1, Mdb2, Mhatdb3, Mhatdb4, Mhatdb5, Mhatdb6]
Phis = [Phi1, Phi2, Phihat3, Phihat4, Phihat5, Phihat6]

dic = {'$ H(s) $': makelatex(Hs), 
       '$ M_{dB} $': makelatex(Mdbs), 
       '$ \\phi $': makelatex(Phis)}

df = DataFrame(dic)
Markdown(df.to_markdown(index=False))

\( H(s) \)	\( M_{dB} \)	\( \phi \)
\(\frac{1}{s}\)	\(- \frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}}\)	\(- \frac{\pi}{2}\)
\(s\)	\(\frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}}\)	\(\frac{\pi}{2}\)
\(\frac{1}{a + s}\)	\(\begin{cases} - \frac{20 \log{\left(a \right)}}{\log{\left(10 \right)}} & \text{for}\: a \geq \omega \\- \frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}\)	\(\begin{cases} 0 & \text{for}\: \omega \leq 0.1 a \\- \frac{\pi \log{\left(\frac{10.0 \omega}{a} \right)}}{4 \log{\left(10 \right)}} & \text{for}\: \omega < 10 a \\- \frac{\pi}{2} & \text{otherwise} \end{cases}\)
\(a + s\)	\(\begin{cases} \frac{20 \log{\left(a \right)}}{\log{\left(10 \right)}} & \text{for}\: a \geq \omega \\\frac{20 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}\)	\(\begin{cases} 0 & \text{for}\: \omega \leq 0.1 a \\\frac{\pi \log{\left(\frac{10.0 \omega}{a} \right)}}{4 \log{\left(10 \right)}} & \text{for}\: \omega < 10 a \\\frac{\pi}{2} & \text{otherwise} \end{cases}\)
\(\frac{1}{\omega_{n}^{2} + 2 \omega_{n} s \zeta + s^{2}}\)	\(\begin{cases} - \frac{40 \log{\left(\omega_{n} \right)}}{\log{\left(10 \right)}} & \text{for}\: \omega \leq \omega_{n} \\- \frac{40 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}\)	\(\begin{cases} 0 & \text{for}\: \omega \leq 10^{- \zeta} \omega_{n} \\- \frac{\pi \log{\left(\frac{10^{\zeta} \omega}{\omega_{n}} \right)}}{2 \zeta \log{\left(10 \right)}} & \text{for}\: \omega < 10^{\zeta} \omega_{n} \\- \pi & \text{otherwise} \end{cases}\)
\(\omega_{n}^{2} + 2 \omega_{n} s \zeta + s^{2}\)	\(\begin{cases} \frac{40 \log{\left(\omega_{n} \right)}}{\log{\left(10 \right)}} & \text{for}\: \omega < \omega_{n} \\\frac{40 \log{\left(\omega \right)}}{\log{\left(10 \right)}} & \text{otherwise} \end{cases}\)	\(\begin{cases} 0 & \text{for}\: \omega \leq 10^{- \zeta} \omega_{n} \\\frac{\pi \log{\left(\frac{10^{\zeta} \omega}{\omega_{n}} \right)}}{2 \zeta \log{\left(10 \right)}} & \text{for}\: \omega < 10^{\zeta} \omega_{n} \\\pi & \text{otherwise} \end{cases}\)

Resonance peaks

The following table summarizes when a resonance occurs and its peak value.

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Hs   = [H5, H6]
wp   = [omega_p5[0], omega_p6[0]]
Mp   = [Mp5, Mp6]
Mpdb = [Mdbp5, Mdbp6]

dic = {'$H(s)$': makelatex(Hs), 
       '$ \\omega_p $': makelatex(wp), 
       '$ M_p $': makelatex(Mp),
       '$ M_{p(dB)} $': makelatex(Mpdb)}

df = DataFrame(dic)

Markdown(df.to_markdown(index=False))

\(H(s)\)	\( \omega_p \)	\( M_p \)	\( M_{p(dB)} \)
\(\frac{1}{\omega_{n}^{2} + 2 \omega_{n} s \zeta + s^{2}}\)	\(\omega_{n} \sqrt{1 - 2 \zeta^{2}}\)	\(\frac{1}{2 \omega_{n}^{2} \zeta \sqrt{1 - \zeta^{2}}}\)	\(- \frac{10 \log{\left(- 4 \omega_{n}^{4} \zeta^{2} \left(\zeta^{2} - 1\right) \right)}}{\log{\left(10 \right)}}\)
\(\omega_{n}^{2} + 2 \omega_{n} s \zeta + s^{2}\)	\(\omega_{n} \sqrt{1 - 2 \zeta^{2}}\)	\(2 \omega_{n}^{2} \zeta \sqrt{1 - \zeta^{2}}\)	\(\frac{10 \log{\left(4 \omega_{n}^{4} \zeta^{2} \left(1 - \zeta^{2}\right) \right)}}{\log{\left(10 \right)}}\)