Section 7 Frequency Response Analysisweb.engr.oregonstate.edu/~webbky/MAE4421_files... ·...

MAE 4421 – Control of Aerospace & Mechanical Systems

SECTION 7: FREQUENCY‐RESPONSE ANALYSIS

K. Webb MAE 4421

Introduction2

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Introduction

We have seen how to design feedback control systems using the root locus

In this section of the course, we’ll learn how to do the same using the open‐loop frequency response

Objectives: Review frequency‐response fundamentals Relate a system’s frequency response to its transient response

Determine static error constants from the open‐loop frequency response

Determine closed‐loop stability from the open‐loop frequency response

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System Response to a Sinusoidal Input

Consider an ‐order system poles: , , … Real or complex Assume all are distinct

Transfer function is:

⋯(1)

Apply a sinusoidal input to the system

sin

Output is given by

⋯∙ (2)

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System Response to a Sinusoidal Input

Partial fraction expansion of (2) gives

⋯ (3)

Inverse transform of (3) gives the time‐domain output

⋯ cos sin (4)

transient steady state

Two portions of the response: Transient Decaying exponentials or sinusoids – goes to zero in steady state Natural response to initial conditions

Steady state Due to the input – sinusoidal in steady state

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Steady‐State Sinusoidal Response

We are interested in the steady‐state response

cos sin (5)

A trig. identity provides insight into :

cos sin sinwhere

tan

Steady‐state response to a sinusoidal input

sin

is a sinusoid of the same frequency, but, in general different amplitude and phase

sinWhere (6)

and tan

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sin → sin

Steady‐state sinusoidal response is a scaled andphase‐shifted sinusoid of the same frequency Equal frequency is a property of linear systems

Note the term in the numerator of (3) will affect the residues Residues determine amplitude and phase of the output Output amplitude and phase are frequency‐dependent

sin

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Gain – the ratio of amplitudes of the output and input of the system

Phase – phase difference between system input and output

Systems will, in general, exhibit frequency‐dependent gain and phase

We’d like to be able to determine these functions of frequency The system’s frequency response

Linear Systemsin sin

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A system’s frequency response, or sinusoidal transfer function, describes its gain and phase shift for sinusoidal inputs as a function of frequency.

Frequency Response9

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Frequency Response

System output in the Laplace domain is⋅

Multiplication in the Laplace domain corresponds to convolution in the time domain

∗

Consider an exponential input of the form

where is the complex Laplace variable:

Now the output is

∗

⋅ (1)

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Frequency Response

⋅ (1)

We’re interested in the steady‐state response, so let the upper limit of integration go to infinity

⋅

⋅ (2)

Time‐domain response to an exponential input is the time‐domain input multiplied by the system transfer function

What is this input?(3)

If we let → 0, i.e. let → , then we have

⋅ (4)

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Euler’s Formula

Recall Euler’s formula:

cos sin (5)

From which it follows that

cos (6)

and

sin (7)

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Frequency Response

We’re interested in the sinusoidal steady‐state system response, so let the input be

cos 2

A sum of complex exponentials in the form of (3) We’ve let → in the first term and → in the second

(8)

According to (4) the output in response to (8) will be

⋅ ⋅ (9)

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Frequency Response

⋅ ⋅ (9)

is a complex function of frequency Evaluates to a complex number at each value of Has both magnitude and phase Can be expressed in polar form as

(10)

whereand ∠

It follows that

(11)

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Frequency Response

Using (11), the output given by (9) becomes

(12)

(13)

where, again

and ∠ (14)

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Frequency response Function –

is the system’s frequency response function Transfer function, where →

| → (15)

A complex‐valued function of frequency

at each is the gain at that frequency Ratio of output amplitude to input amplitude

∠ at each is the phase at that frequency Phase shift between input and output sinusoids

Another representation of system behavior Along with state‐space model, impulse/step responses, transfer

function, etc. Typically represented graphically

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Plotting the Frequency Response Function

is a complex‐valued function of frequency Has both magnitude and phase Plot gain and phase separately

Frequency response plots formatted as Bode plots Two sets of axes: gain on top, phase below Identical, logarithmic frequency axes Gain axis is logarithmic – either explicitly or as units of decibels (dB)

Phase axis is linear with units of degrees

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Bode Plots

Logarithmic frequency axes

Units of magnitude are dB Magnitude

plot on top

Phase plot below

Units of phase are degrees

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Interpreting Bode Plots

Bode plots tell you the gain and phase shift at all frequencies: choose a frequency, read gain and phase values from the plot

For a 10KHz sinusoidal input, the gain is 0dB (1) and the phase shift is 0°.

For a 10MHz sinusoidal input, the gain is ‐32dB (0.025), and the phase shift is ‐176°.

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Interpreting Bode Plots

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Decibels ‐ dB

Frequency response gain most often expressed and plotted with units of decibels (dB) A logarithmic scale Provides detail of very large and very small values on the same plot

Commonly used for ratios of powers or amplitudes

Conversion from a linear scale to dB:

20 ⋅ log

Conversion from dB to a linear scale:

10

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Decibels – dB

Multiplying two gain values corresponds to adding their values in dB E.g., the overall gain of cascaded systems

⋅

Negative dB values corresponds to sub‐unity gain Positive dB values are gains greater than one

dB Linear

60 1000

40 100

20 10

0 1

dB Linear

6 2

‐3 1/√2 0.707‐6 0.5

‐20 0.1

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Value of Logarithmic Axes ‐ dB

Gain axis is linear in dB A logarithmic scale Allows for displaying detail at very large and very small levels on the same plot

Gain plotted in dB Two resonant peaks

clearly visible

Linear gain scale Smaller peak has

disappeared

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Value of Logarithmic Axes ‐ dB

Frequency axis is logarithmic Allows for displaying detail at very low and very high frequencies on the

same plot

Log frequency axis Can resolve

frequency of both resonant peaks

Linear frequency axis Lower resonant

frequency is unclear

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Gain Response – Terminology

Corner frequency, cut off frequency, ‐3dB frequency: Frequency at which

gain is 3dB below its low‐frequency value

2

This is the bandwidthof the system

Peaking Any increase in gain

above the low frequency gain

1.45

2 0.23

~5 of peaking

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Frequency‐Response Factors26

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Transfer Function Factors

Numerator and denominator of a transfer function can be factored into first‐ and second‐order terms

⋯ 2 2 ⋯⋯ 2 2 ⋯

Can think of the transfer function as a product of the individual factors

For example, consider the following system

2

Can rewrite as

⋅1

⋅1

2

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⋅1

⋅1

2

Think of this as three cascaded transfer functions, ,

1 12

or

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In the Laplace domain, transfer function of a cascade of systems is the product of the individual transfer functions In the time domain, overall impulse response is the convolution of the individual impulse responses

Same holds true in the frequency domain Frequency response of a cascade is the product of the individual frequency responses

Or, the product of individual factors

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Frequency Response Components ‐ Example

Consider the following system

20 201 100

The system’s frequency response function is

20 201 100

As we’ve seen we can consider this a product of individual frequency response factors

20 ⋅ 20 ⋅11 ⋅

1100

Overall response is the composite of the individual responses Product of individual gain responses – sum in dB Sum of individual phase responses

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Gain response

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Phase response

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In this section, we’ll look at a method for sketching, by hand, a straight‐line, asymptotic approximation for a Bode plot.

Bode Plot Construction33

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Bode Plot Construction

We’ve just seen that a system’s frequency response function can be factored into first‐ and second‐order terms Each factor contributes a component to the overall gain and phase responses

Now, we’ll look at a technique for manually sketching a system’s Bode plot In practice, you’ll almost always plot with a computer But, learning to do it by hand provides valuable insight

We’ll look at how to approximate Bode plots for each of the different factors

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Bode Form of the Transfer function

Consider the general transfer function form:⋯ 2 ⋯⋯ 2 ⋯

We first want to put this into Bode form:

1 1 ⋯ 2 1 ⋯

1 1 ⋯ 2 1 ⋯

The corresponding frequency response function, in Bode form, is

1 1 ⋯ 2 1 ⋯

1 1 ⋯ 2 1 ⋯

Putting into Bode form requires putting each of the first‐ and second‐order factors into Bode form

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First‐Order Factors in Bode Form

First‐order frequency‐response factors include:

, ,

For the first factor, , is a positive or negative integer Already in Bode form

For the second two, divide through by , giving

1 and

Here, , the corner frequency associated with that zero or pole, so

1 and

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Second‐Order Factors in Bode Form

Second‐order frequency‐response factors include:

2 and

Again, normalize the coefficient, giving

1 and /

Putting each factor into its Bode form involves factoring out any DC gain component

Lump all of DC gains together into a single gain constant,

⋯ ⋯

⋯ ⋯

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Frequency response function in Bode form ⋯ ⋯

⋯ ⋯

Product of a constant DC gain factor, , and first‐and second‐order factors

Plot the frequency response of each factor individually, then combine graphically Overall response is the product of individual factors Product of gain responses – sum on a dB scale Sum of phase responses

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Bode plot construction procedure:1. Put the sinusoidal transfer function into Bode form2. Draw a straight‐line asymptotic approximation for the

gain and phase response of each individual factor3. Graphically add all individual response components

and sketch the result

Next, we’ll look at the straight‐line asymptotic approximations for the Bode plots for each of the transfer function factors

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Bode Plot – Constant Gain Factor40

Constant gain

Constant Phase

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Bode Plot – Poles/Zeros at the Origin41

0: zeros at the origin

0: poles at the origin

Gain: Straight line Slope ⋅ 20 ⋅ 6

0 at 1

Phase: ∠ ⋅ 90°

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Bode Plot – First‐Order Zero42

Single real zero at

Gain: 0 for

20 6 for

Straight‐line asymptotes intersect at , 0

Phase: 0° for 0.1 45° for 90° for 10

° for 0.1 10

1

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Bode Plot – First‐Order Pole43

Single real pole at

Gain: 0 for

20 6 for

Straight‐line asymptotes intersect at , 0

Phase: 0° for 0.1 45° for 90° for 10

° for 0.1 10

1

1

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Bode Plot – Second‐Order Zero44

Complex‐conjugate zeros:

,

Gain: 0 for ≪ 40 12 for ≫ Straight‐line asymptotes intersect at

, 0

‐dependent peaking around

Phase: 0° for ≪ 90° for

180° for ≫ ‐dependent slope through

Sketch as step‐change at for low , 90°/ for high , or in between

21

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Bode Plot – Second‐Order Pole45

12 1

Complex‐conjugate poles:

,

Gain: 0 for ≪ 40 12 for ≫ Straight‐line asymptotes intersect at

, 0

‐dependent peaking around

Phase: 0° for ≪ 90° for

180° for ≫ ‐dependent slope through

Sketch as step‐change at for low , 90°/ for high , or in between

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Bode Plot Construction – Example46

Consider a system with the following transfer function

10 20400

The sinusoidal transfer function:

10 20400

Put it into Bode form

10 ⋅ 20 20 1

⋅ 400 400 1

0.5 20 1

⋅ 400 1

Represent as a product of factors

0.5 ⋅ 20 1 ⋅1⋅

1

400 1

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Bode Plot Construction – Example

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Bode Plot Construction – Example

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Polar Frequency Response Plots49

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Polar Frequency Response Plots

is a complex function of frequency Typically plot as Bode plots Magnitude and phase plotted separately Aids visualization of system behavior

A real and an imaginary part at each value of A point in the complex plane at each frequency Defines a curve in the complex plane A polar plot Parametrized by frequency – not as easy to distinguish frequency as on a Bode plot

Polar plots are not terribly useful as a means of displaying a frequency response However, an important concept later, when we introduce the Nyquist stability criterion

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Polar Frequency Response Plots

Identical frequency responses plotted two ways: Bode plot and polar plot

Note uneven frequency spacing along polar plot curve Dependent on frequency rates of change of gain and phase

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Relationship between Frequency Response and Transient Response

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Transient/Frequency Response Relationship

We have seen relationships – some exact, some approximate – between closed‐loop pole locations and closed‐loop transient response

Also have relationships between closed‐loop frequency response and closed‐loop transient responses

Applicable to second‐order systems:

2

Also applicable to higher‐order systems that are reasonably approximated as second‐order Systems with a pair of dominant second‐order poles

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Damping ratio vs. overshoot vs. peaking

Natural frequency vs. risetime vs. bandwidth

Qualitative 2nd‐order time/freq. response/pole relationships

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Frequency Response Peaking

For systems with 0.707, the gain response will exhibit peaking

Can relate peak magnitudeto the damping ratio

12 1

Relative to low‐frequency gain

And the peak frequency to the damping ratio and natural frequency

1 2

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Can also relate a system’s bandwidth (i.e., ‐3dB frequency, ) to the speed of its step response

Bandwidth as a function of and

1 2 4 4 2

Bandwidth as a function of 1% settling time and

4.61 2 4 4 2

Bandwidth as a function of peak time and

11 2 4 4 2

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Steady‐State Error from Bode Plots57

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Static Error Constants

For unity‐feedback systems, open‐loop transfer function gives static error constants Use static error constants to calculate steady‐state error

lim→

lim→

lim→

We can also determine static error constants from a system’s open‐loop Bode plot

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Static Error Constant – Type 0

For a type 0 system

→

At low frequency, i.e. below any open‐loop poles or zeros

Read directly from the open‐loop Bode plot Low‐frequency gain

100 303 200

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For a type 1 systemlim→

At low frequencies, i.e. below any other open‐loop poles or zeros

and

A straight line with a slope of / Evaluating this low‐frequency asymptote at yields the velocity constant,

On the Bode plot, extend the low‐frequency asymptote to Gain of this line at 1 is

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85 0.1 5010 125

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For a type 2 systemlim→

At low frequencies, i.e. below any other open‐loop poles or zeros

and

A straight line with a slope of / Evaluating this low‐frequency asymptote at yields the acceleration constant,

On the Bode plot, extend the low‐frequency asymptote to Gain of this line at 1 is

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1600 0.1 5100

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Nyquist Stability Criterion64

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Stability

Consider the following system

We already have a couple of tools for assessing stability as a function of loop gain, Routh Hurwitz Root locus

Root locus: Stable for some values of Unstable for others

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Stability

In this case gain is stable below some value

Other systems may be stable for gain abovesome value

Marginal stability point: Closed‐loop poles on the imaginary axis at

For gain

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Open‐Loop Frequency Response & Stability

Marginal stability point occurs when closed‐loop poles are on the imaginary axis Angle criterion satisfied at

1 and ∠ 180°

Note that

is the open‐loop frequency response Marginal stability occurs when:

Open‐loop gain is: Open‐loop phase is:

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Stability from Bode Plots

Here, stable for smaller gain values 0 when∠ 180°

Often, stable for larger gain values 0 when∠ 180°

Root locus provides this information Bode plot does not

Varying simply shifts gain response up or down

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Open‐Loop Frequency Response & Stability

A method does exist for determining stability from the open‐loop frequency response:

Nyquist stability criterion Graphical technique Uses open‐loop frequency response Determine system stability Determine gain ranges for stability

Before introducing the Nyquist criterion, we must first introduce the concept of complex functional mapping

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Complex Functional Mapping

Consider a complex function⋯⋯

Takes one complex value, , and yields a second complex value, In other words, it maps to

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Mapping of Contours

provides a mapping of individual points in the s‐plane to corresponding points in the F‐plane

Can also map all points around a contour in the s‐plane to another contour in the F‐plane

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Mapping of Contours

Recall how we approached the application of the angle criterion Vector approach to the evaluation of a transfer function at a particular point in the s‐plane

∏ ∏

∠ Σ∠ Σ∠

Can take the same approach to evaluating complex functions around contours in the s‐plane

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Mapping Contours – Example 1

Map contour by in a clockwise direction Contour does not enclose the zero

Here, , so and ∠ ∠

As is evaluated around , ∠ never exceeds 0° or 180° does the same:

Does not rotate through a full 360° Contour does not encircle the origin

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Map contour by in a clockwise direction Contour does not enclose the pole

Here, 1/ , so 1/ and ∠ ∠

∠ oscillates over some range well within 0° and 180° rotates through the negative of the same range Contour does not encircle the origin

1

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Now, contour encloses a single zero

, so and ∠ ∠

rotates through a full 360° in a clockwise direction does the same:

Contour encircles the origin in a clockwise direction

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Now, contour encloses a single pole

1/ , so 1/ and ∠ ∠

rotates through a full 360° in a clockwise direction rotates in the opposite direction Contour encircles the origin in a CCW direction

1

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Now, contour encloses two poles

, so and ∠ ∠ ∠

and each rotate through a full 360° in a clockwise direction rotates in the opposite direction Contour encircles the origin twice in a CCW direction

1

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∠ and∠ rotate through 360° in a CW direction Their contributions rotate in opposite directions ∠ does not rotate through a full 360° Contour does not encircle the origin

Now, contour encloses one pole and one zero

, so and ∠ ∠ ∠

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Complex Functional Mapping of Contours

Some observations regarding complex mapping of contour in a CW direction to contour : If does not enclose any poles or zeros, does not encircle the origin

If encloses a single pole, will encircle the origin once in a CCW direction

If encloses two poles, will make two CCW encirclements of the origin

If encloses a pole and a zero, will not encircle the origin

Next, we’ll use these observations to help derive the Nyquist stability criterion

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Nyquist Stability Criterion

Our goal is to assess closed‐loop stability Determine if there are any closed‐loop poles in the RHP

Consider a generic feedback system:

Closed‐loop transfer function

1

Closed‐loop poles are roots (zeros) of the closed‐loop characteristic polynomial:

1

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Can represent the individual transfer functions as

and

The closed‐loop polynomial becomes

1 1

From this, we can see that: The poles of 1 are the poles of , the open‐loop poles

The zeros of 1 are the poles of , the closed‐loop poles

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To determine stability, look for RHP closed‐loop poles Evaluate 1 CW around a contour that encircles the entire right half‐plane Evaluate 1 along entire

‐axis Encircle the entire RHP with an infinite‐radius arc

If 1 has one RHP pole, resulting contour will encircle the origin once CCW

If 1 has one RHP zero, resulting contour will encircle the origin once CW

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Total number of CW encirclements of the origin, , by the resulting contour will be

# of RHP poles of 1 # of RHP zeros of 1

Want to detect RHP poles of , zeros of 1 , so

# of closed‐loop RHP poles

# of open‐loop RHP poles

# of CW encirclements of the origin

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Basis for detecting closed‐loop RHP polesMap contour encircling the entire RHP through closed‐loop characteristic polynomial

Count number of CW encirclements of the origin by resulting contour

Calculate the number of closed‐loop RHP poles:

Need to know: Closed‐loop characteristic polynomial Number of RHP poles of closed‐loop characteristic polynomial

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Open‐loop transfer function Easy to use for mapping – we know poles and zeros

Resulting contour shifts left by 1 – that’s all

Now, count encirclements of the point

Instead, map through

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Nyquist stability criterion If a contour that encloses the entire RHP is mapped through the open‐loop transfer function, , then the number of closed‐loop RHP poles, , is given by

where

# of CW encirclements of # of open‐loop RHP poles

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Want to detect net clockwise encirclements# CW encirclements ‐ # CCW encirclements

Draw a line from in any

direction Count number of times contour crosses the line in each direction

2

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Nyquist Diagram

The contour that results from mapping the perimeter of the entire RHP is a Nyquist diagram

Consider four segments of the contour:

①②

③

④

1) Along positive ‐axis, we’re evaluating ) Open‐loop frequency response

2) Here, → Maps to zero for any physical system

3) Here, evaluating ) Complex conjugate of segment ① Mirror ① about the real axis

4) The origin Sometimes a special case – more later

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Nyquist Criterion – Example 1

Apply the Nyquist criterion to determine stability for the following system

First evaluate along segment ①, ‐axis This is the frequency response

Read values off of the Bode plot

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Segment ① is a polar plot of the frequency response

All of segment ②, arc at , maps to the origin

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Segment ③ is the complex conjugate of segment ①Mirror about the real axis

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Count CW encirclements of Draw a line from 1 in any direction

Here, Closed‐loop RHP poles given by:

No open‐loop RHP poles, so

Two RHP poles, so system is unstable

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This system is open‐loop stable Stable for low enough Nyquist plot will not encircle

Three poles and no zeros Unstable for above some value Nyquist plot will encircle

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For , , and the system is stable Modifying simply scales the magnitude of the Nyquist plot

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Here, the Nyquist plot crosses the negative real axis at

As gain increases real‐axis crossing moves to the left

Increasing by 2x or more results in two encirclements of

Unstable for 60 More later …

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Nyquist Diagram – Poles at the Origin

We evaluate the open‐loop transfer function along a contour including the ‐axis

is undefined at the pole Must detour around the pole

Consider the common case of a pole at the origin

④

This is the special case for segment ④

12

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④

97


Segment ④ contour: for 0° 90° Evaluate around segment ④ as

12

Magnitude:1

212

As → 0lim→

∞

Maps to an arc at

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④

98


Segment ④ traversed in a CCW direction varies from

Phase of the resulting contour:

∠

Negative because it is angle from a pole

Extra phase from additional pole

maps segment ④ to: An arc at Rotating CW from 0° to 90°

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Apply the Nyquist criterion to determine stability for the following system

Use Bode plot to map segment ① Infinite DC gain Starts at at for

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Segment ① starts at at Heads to the origin at All of segment ②, arc at , maps to the origin

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Segment ③ is the complex conjugate of segment ①Mirror about the real axis

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Segment 4 maps to a CW arc at CW, so it does not encircle Can’t draw to scale

Here, No open‐loop RHP poles, so

No RHP poles, so system is stable

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Stability Margins103

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Stability Margins

Recall a previous example

According to the Nyquist plot, the system is stable How stable?

Two stability metrics Both are measures of how close the Nyquist plot is to encircling the point 1

Gain margin and phase margin

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Crossover Frequencies

Two important frequencies when assessing stability:

Gain crossover frequency The frequency at which the open‐loop gain crosses

Phase crossover frequency The frequency at which the open‐loop phase crosses

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Gain Margin

An open‐loop‐stable system will be closed‐loop stable as long as its gain is less than unity at the phase crossover frequency

Gain margin, GM The change in open‐loop gain at the phase crossover frequency required to make the closed‐loop system unstable

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Phase Margin

An open‐loop‐stable system will be closed‐loop stable as long as its phase has not fallen below at the gain crossover frequency

Phase margin, PM The change in open‐loop phase at the gain crossover frequency required to make the closed‐loop system unstable

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Gain and Phase Margins from Bode Plots

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Phase Margin and Damping Ratio,

PM can be expressed as a function of damping ratio, , as

tan

For 65° or so, we can approximate:

100 or

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Frequency Response Analysis in MATLAB110

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bode.m

[mag,phase] = bode(sys,w)

sys: system model – state‐space, transfer function, or other w: optional frequency vector – in rad/sec mag: system gain response vector phase: system phase response vector – in degrees

If no outputs are specified, bode response is automatically plotted – preferable to plot yourself

Frequency vector input is optional If not specified, MATLAB will generate automatically

May need to do: squeeze(mag) and squeeze(phase)to eliminate singleton dimensions of output matrices

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nyquist.m

nyquist(sys,w) sys: system model – state‐space, transfer function, or other w: optional frequency vector – in rad/sec

MATLAB generates a Nyquist plot automatically Can also specify outputs, if desired:

[Re,Im] = nyquist(sys,w)

Plot is not be generated in this case

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margin.m

[GM,PM,wgm,wpm] = margin(sys)

sys: system model – state‐space, transfer function, or other GM: gain margin PM: phase margin – in degrees wgm: frequency at which GM is measured, the phase crossover frequency – in rad/sec

wpm: frequency at which PM is measured, the gain crossover frequency

If no outputs are specified, a Bode plot with GM and PM indicated is automatically generated

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Section 7 Frequency Response Analysisweb.engr.oregonstate.edu/~webbky/MAE4421_files... ·...

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