Harmonic Response and Bode plots Luigi Palopoli...

Teoria dei sistemi.Harmonic Response and Bode plots

Luigi Palopoli

December 5, 2017

Table of contents

Harmonic Response

Bode Plots

Nyquist Plots

Response to harmonic signals

I We have seen that any LTI system responds to cosωt with

|H(jw)| cos (ωt + ∠H(jw))

I A different story is if the input signal is

u(t) = 1(t) cos (ωt)

Response to harmonic signals (cont)I In this case we can see that the response is given by

|H(jw)| cos (ωt + ∠H(jw)) + f (t)

where f (t) is a terms containing the natural modes stemmingfrom the poles of the system

I If the system is BIBO stable the term f (t) vanishes after atransient phase and we end up, at the steady state, with theharmonic response term

|H(jw)| cos (ωt + ∠H(jw))

Frequency Response

Frequency Response

Proposition

The Laplace transform H(s) evaluated at s = jω for LTIstable systems accounts for the steady state response to acosinusoidal signal. Its modulus multiplies the amplitude ofthe oscillation and the phase corresponds to the phase ofthe oscillation.

Meaning of poles and zeros

I The physical meaning of poles is quite clear: they give rise toexponential signals.

I As concerns the response to harmonic signals, the poles havean impact on the duration of the transient

I The ploes have a direct impact on the system’s open loopstability

I The meaning of the zeros is way less clear.

Meaning of poles and zeros (cont.)

I If a zero is on the imaginary axis (say at jω0), then a signallike cosω0t does not produce any response at the steady state.

I Generally speaking, for a zero zi a signal having Laplacetransform U(s) = A/(s − zi ) is blocked (blocking property ofthe zeros).

I The zeros have important implications on the stability of theclosed loop system and on the system’s performance.

Frequency Response

I The frequency response can be estimated by using differentharmonic signals, and measuring for each, amplitude andphase.

I From the experimental data on the frequency response it ispossible to reconstruct the transfer function.

I The frequency response is the transfer function evaluated fors = jω:

G (jω) = G (s)|s=jω

I It is now important to see how the shape of G (jω) relates tothe position of zeros and poles.

Graphical plots of the frequency response

I Since G (jω) is a complex function of ω it can be representedin two ways:

I Magnitude and phase plot (Bode diagrams)I Real and imaginary parts (Nyquist diagrams)

Bode Diagrams

I Bode plots (or diagrams) are graphical representation of thefrequency response G (jω), which use a logarithmic scale for ω

I They consist ofI Magnitude plots: i.e., |G (jω)| vs logωI Phase plots: i.e., ∠G (jω) vs logω

I The use of logarithmic scale has two advantages:

1. Possibility of spanning with ω across sevral orders of magnitude2. Possibility of creating complex siagrams by the simple

composition of simpler ones.

Example

Example of bode plots

G (s) = s+1s2+s+10

= |G (s)| e j∠G(s)

Magnitude

I The magnitude is defined as

|G (jω)|dB = 20 log10 |G (jω)|

I ExamplesI |G (jω)| = 2→ |G (jω)|dB = 6I |G (jω)| = 10→ |G (jω)|dB = 20I |G (jω)| = 100→ |G (jω)|dB = 40

Bode Form

I In order to draw the Bode diagram, it is convenient to writethe transfer function into the following Bode form

G (s) = K

∏i (1 + τzi s)

∏i (1 + 2 ζi

αnis + s2

α2ni

)

sg∏

i (1 + τpi s)∏

i (1 + 2 ξiωni

s + s2

ω2ni

)

I Example

G (s) =s + 10

s2 + 4s + 2=

=10(1 + s

10)

2(1 + 2s + s2

2 )

= 5(1 + s

10)

1 + 2√2√2s + s2√

22

Basic Building blocks

Based on the discussion above, we have the following terms thatallow us to construct the bode plot:

I K : the gain, which can be positive or negative

I s±1: poles or zeros in the origing

I (1 + τzi s), (1 + τpi s): real zero, real pole

I (1 + 2 ζiαni

s + s2

α2ni

): complex conjugate zero pair

I (1 + 2 ξiωni

s + s2

ω2ni

): complex conjugate pole pair

We will now evaluate the contribution to the bode diagram of eachof these terms, setting s = jω. The different contribution will thenbe composed together.


Positive Constant (K > 0)

Basic Building blocksPositive Constant (K < 0)

Basic Building blocksTerm s = jω (Zero in the origin).

1. The terms tends to zero (−∞ in dB) for ω → 0

2. Modulus = 1 (0 in dB) for ω = 0)

3. Phase = π/2

4. Slope of magnitude plot is 20dB every decade.


Term s = jω (Zero in the origin).

1. The terms tends to zero (−∞ in dB) for ω → 0


3. Phase = π/2

4. Slope of magnitude plot is 20dB every decade.

5. If the zero has multiplicity h then the slope will be h · 20dB/decade

Basic Building blocksTerm s−1 = (jω)−1 (Pole in the origin).

1. The modules tends to ∞ (∞ in dB) for ω → 0


3. Phase = −π/2

4. Slope of magnitude plot is −20dB every decade.


Term s−1 = (jω)−1 (Pole in the origin).

1. The modulus tends to ∞ (∞ in dB) for ω → 0


3. Phase = −π/2

4. Slope of magnitude plot is −20dB every decade.

5. If the multiplicity of the pole in the origin is h, the slope is−h · 20dB/decade


Term (1 + sτ) = (1 + jωτ): real zero.

1. For ω → 0, the modules is 1 (0 dB), and the phase is 0

2. For ωτ � 1

I the magnitude tends to infinity with a slope of 20db/decadeI for ω = 1/τ the modulus (

√1 + ω2τ 2 is

√2 (3dB).

I the phase tends to −π/2 if τ is negative and to π/2 if τ ispositive.

I the phase is ±π/4 for ω = 1/τ


Term (1 + sτ) = (1 + jωτ): real zero.

τ > 0 τ < 0


Term 1(1+sτ) = 1

(1+jωτ) : real pole

1. For ω → 0, the modules is 1 (0 dB), and the phase is 0

2. For ωτ � 1

I the magnitude tends to infinity with a slope of −20db/decadeI for ω = 1/τ the modulus ( 1√

1+ω2τ 2is 1√

2(−3dB).

I the phase tends to −π/2 if τ is positive and to π/2 if τ isnegative.

I the phase is ±π/4 for ω = 1/τ


Term 1(1+sτ) = 1


τ > 0 τ < 0

Asymptotic Diagrams

I In order to simplify the drawing of bode plots, it is possible toplot an approximate diagram using the asympotic behaviour

I The asymptotic diagram is constructed by drawing linesegments corresponding to small values of ω and to highvalues of ω

I The break point in the magnitude diagram corresponds to topoint where the asymptotic phase plot becomes tangent tothe actual phase plot.

I Let us go back to the example of a real pole


Term 1(1+sτ) = 1


I |G (jω)| = 1√1+(ωτ)2

I ωτ � 1→ |G (jω)||dB ≈ −20 log10 1 = 0

I ωτ � 1→ |G (jω)||dB ≈ 20 log101|ωτ | =

20 log101|τ | − 20 log10(ω), which is 0 for ω = 1

τ


Term 1(1+sτ) = 1


I In the breakpoint the value of the actual bode plot is −3dB.


Term 1(1+sτ) = 1


For the phase plot remember that the value is −π/4 for ω = 1/τ .

Basic Building blocksTerm 1

(1+sτ) = 1(1+jωτ) : real pole

I How to find ωa and ωb?I In the point 1/τ the tangent is:

d∠G (jω)

d log10 ω

∣∣∣∣ω=1/τ

=d − arctanωτ

dω

dω

d log10 ω

∣∣∣∣ω=1/τ

=

=−τ

1 + (ωτ)2de loge ω

d log10 ω

∣∣∣∣ω=1/τ

=

=−τ

1 + (ωτ)2de

log10 ωlog10 e

d log10 ω

∣∣∣∣∣ω=1/τ

=

=−τ

1 + (ωτ)21

log10 ee

log10 ωlog10 e

∣∣∣∣ω=1/τ

=

=−ωτ

1 + (ωτ)21

log10 e

∣∣∣∣ω=1/τ

=

= −1

2

1

log10 e


Term 1(1+sτ) = 1


I How to find ωa and ωb?I We need to find the intesection between the phase asymptotes

and the tangent in ω = 1τ

π/4

log101τ − log10 ωa

=1

2 ∗ log10 e→ 1

τωa= log10 (π/2 log10 e)

π/4

log10 ωb − log101τ

=1

2 ∗ log10 e→ τωb = log10 (π/2 log10 e)

Break Points

ωa = 1/τ4.81 , ωb = 4.81 · 1τ

Asymptotic Diagram – Summary

I The summary of what we just said is that it is possible tosimply draw a simplified version of the diagram by theasymptotes.

I Example: G (jω) = (1 + jωτ)

Asymptotic Diagram – Summary

I Example: G (jω) = 11+jωτ

Asymptotic Diagram – Example

I Example: G (s) = 0.5 s+2(s+0.1)(s+10)

I Bode form:G (s) = 1+0.5s(1+10s)(1+0.1s)

G (s) = G1(s)G2(s)G3(s)

G1(s) = 1 + 0.5s

G2(s) =1

1 + 10s

G3(s) =1

1 + 0.1s


G1(s) = 1 + 0.5s


G2(s) =1

1 + 10s


G3(s) =1

1 + 0.1s


G (s) = G1(s)G2(s)G3(s)

Basic Building Blocks

I Let us now consider a couple of complex conjugate poles:

G (s) =1(

1 + s2

ω2n

+ 2δ sωn

)mG (jω) =

1(1− ω2

ω2n

+ 2δ jωωn

)m

I Let us consider the case m = 1 (the case m > 1 is treatedsimilarly using the additive property of magnitude and phase)

Basic Building Blocks - Complex pair

I Magnitude

20 log10 |G (jω)| = 20 log101√(

1− ω2

ω2n

)2+ 4δ2 ω

2

ω2n

I Phase

∠G (jω) = − arctan2δ ωωn

1− ω2

ω2n

Complex pole pair - Magnitude

I For ω/ωn � 1

20 log10 |G (jω)| = 20 log101√(

1− ω2

ω2n

)2+ 4δ2 ω

2

ω2n

≈ 0

I For ω/ωn � 1

20 log10 |G (jω)| = 20 log101√(

1− ω2

ω2n

)2+ 4δ2 ω

2

ω2n

≈ 40 log10 ωn − 40 log10 ω

The amplitude decreases with a rate of −40 dB/decade

I We can still build and asymptotic diagram, but it can be veryfar from the actual one (the deviation could be infinite)

Complex pole Pair - Magnitude

I Cases:

0 ≤ δ ≤ 1/√

2→ The amplitude plot has a maximum

0 ≤ δ ≤ 1/2→ The curve lies above the asymptotic plot

(intesection with x axis greater than ωn)

1/2 ≤ δ ≤ 1/√

2→ Intersection with x axis smaller than ωn

1/√

2 ≤ δ ≤ 1→ The curve is below the aymptotic plot

(no intersection with x axis)

Complex Poles Pair - Example

Complex Poles Pair - Resonance PeakI The maximum value (when exists) is called resonance peakI The ω for which it is attained is called resonance pulsationI If we set u = ω/ωn, the resonance pulsation can be found by

minimising (1− u2)2 + 4δ2u2

I By computing the derivative and setting to zero, one finds theresonance pulsation ωR as ωR = ωn

√1− 2δ2 and he

correspondign peak: 12δ√1−δ2

Complex Poles Pair - Phase Plot

The asymptotic diagram is quite easy:

1. The phase changes between 0 and −π2. The phase is −π/2 on the breakpoint

3. The connection on the asymptote can be found computing thetanget of ∠G (jω) = φ(u) = − arctan 2δu

1−u2 with u = ω/ωn:

4. The tangent is as follows:

dφ

d log10 ω

∣∣∣∣ω=ωn

=dφ

du

du

d log10 ω

∣∣∣∣u=1

= − 1

δ log10 e


The intesection between the asymptotes and the tangent is asfollows:

1

δ log10 e=

π/2

log10 ωn − log10 ωa=

π/2

log10 ωb − log10 ωn

log10ωn

ωa= log10

ωb

ωn=πδ log10 e

2ωn

ωa=ωb

ωn= 4.81δ

Which leads us to:

Break Points

ωa = 14.81δ

ωn, ωb =(4.81δ

)· ωn

Complex Pole Pair - Summary

Complex Zero Pair - Summary

Reasoning in the same way, we can work with a complex conjugatepair of zeros:

Complex Zero Pair - Summary

Important Observation

For poles (and zeros) the sign of the pole (of the zero) hasan impact only on the phase plot, but it does not changethe magnitude plot. In particular, if δ < 0 the phase willbe positive for poles and negative of zeros (the opposite forδ > 0).

Summary

For poles (zeros) with negative real part:

I Every pole(zero) determines a break point from which theslope of the magnitude decreases (increases) by 20dB/decade.

I The phase decreases (increases) asymptotically of π/2

I In the break point the phase decreases (increases) of π/4

I For a complex conjugate pair of poles we can have aresonance peak.

For poles (zeros) in the origin with multiplicity n

I The initial slope is -n · 20 dB/decade (+n · 20 db/decade).

I The initial phase is -n · · ·π/2 (+n · π/2)

I The contibution of the poles (zeros) must take vale K forω = 1

Bode diagram – Example 1

Consider

G (s) =100(1 + 10s)

s(1 + 2s)(1 + 0.4s + s2)

Basic data:

K = 100(40dB)

z1 = −0.1

p1 = 0

p2 = −0.5

p3,4 = −0.2± j√

0.96(ωn = 1, δ = 0.2)

Example – Magnitude

Example – Phase plot

For the Phase plot we have to compute the break points:

I z1: ω1a = 0.1/4.81 = 0.0208, ω1b = 0.1 · 4.81 = 0.481

I p2: ω2a = 0.5/4.81, ω2b = 0.5 · 4.81 = 2.405

I p3,4: ω3a = 1/(4.810.2) = 0.73, ω3b = 1 · 4.810.2 = 1.369

Attention: The pole in the origin introduces a phase shift of π/2

Example – Phase


Consider

G (s) =100(1 + 10s)

s(1 + s)2

Basic data:

K = 100(40dB)

z1 = −0.1

p1 = 0

p2 = p3 = −1

Example 2 – Magnitude

Example 2 – Phase plot


I z1: ω1a = 0.1/4.81 = 0.0208, ω1b = 0.1 · 4.81 = 0.481

I p2,3: ω2a = 1/4.81, ω2b = 1 · 4.81 = 4.81

Attention: The pole in the origin introduces a phase shift of π/2.The pole with multiplicity 2 determines a phase change equal to π

Example 2 – Phase


Consider

G (s) =0.1s(1 + s)

(1 + 5s)2(1 + 0.2s)(1− 0.1s)

Basic data:

K = 0.1(−20dB)

z1 = 0

z2 = −1

p1 = p2 = −0.2

p3 = −5

p4 = 10

Example 3 – Magnitude

Example 3 – Phase plot


I z2: ω2a = 1/4.81 = 0.208, ω2b = 1 · 4.81 = 0.481

I p1,2: ω1a = 0.2/4.81 = 0.0416, ω2b = 0.2 · 4.81 = 0.962

I p3: ω3a = 5/4.81 = 1.0395, ω3b = 5 · 4.81 = 24.05

I p4: ω4a = 10/4.81, ω4b = 10 · 4.81 = 48.1

Example 3 – Phase

Nyquist Plots

A different and useful way to represent the tansfer function is by apolar diaggram.

I GIven our transfer function G (s), we follow a path along theimaginary axis leaving to the left (by circlets of infinitesimalradius) the poles on the imaginary axis

Nyquist Plots

A different and useful way to represent the tansfer function is by apolar diaggram.

I By noting that G (−jω) = G ?(jω),

I We can therefore construct the plot for positive frequenciesand complete it with the symmetric plot correpsonign to thenegative path.

I The two branches join on the origin for strictly properfunctions

Nyquist PlotsThe Easiest way to construct the Nyquist diagram is to draw teh bodediagram first (moving to a linear scale for the magnitude) and then usethe plot for the Nyquist.Example:

G (s) =1

s + 1.

Bode Plots.

Nyquist Plot

Nyquist Plots

Example 1:

G (s) =100

(s + 10)2=

1

(1 + s/10)2.

Bode Plots.

Nyquist Plot

Nyquist Plots

Example 2:

G (s) =10(1− s)

(s + 1)2(s + 10)=

1− s

(s + 1)2(1 + s/10).

Bode Plots.

Nyquist Plot

Nyquist Plots

Example 2:

G (s) =10(1− s)

(s + 1)2(s + 10)=

1− s

(s + 1)2(1 + s/10).

Nyquist Plot (close up):

Nyquist Plot (qualitative):

Nyquist Plots

Example 3:

G (s) =100

s(s + 10)2=

1

s(1 + s/10)2.

In this case we have a pole in the origin with multiplicity 1Bode Plots.

The Bode plots reveal that the Nyquist plots makes half a rotation atinfinity clockwise when moving from ω = 0− to ω = 0+

Nyquist Plots

Example 3:

G (s) =100

s(s + 10)2=

1

s(1 + s/10)2.

Nyquist Plot (qualitative)

Nyquist Plot (Exact)

Harmonic Response and Bode plots Luigi Palopoli...

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