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Control and Guidance

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EETAC

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J. Carlos Aguado

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ESAII-UPC

Unit 3. FREQUENCY RESPONSE 3.1 Concept of frequency response. Gain and phase 3.2 Frequency response diagrams 3.2.1 Bode diagrams 3.2.1.1 Building process 3.2.1.2 Basic factors 3.2.1.3 Non-minimum phase zeros 3.2.2 Nyquist representation 3.2.3 Gain margin and phase margin 3.2.4 Effects of compensators

3.1 CONCEPT OF FREQUENCY RESPONSE. GAIN AND PHASE The frequency response is defined as the stationary state time response from a linear stable system to a periodic input. (It can be extended to systems that possess integrators or a single pair of imaginary poles but only in an approximate way). We will see that this response, for a linear system, will always be periodic with the same frequency, but the amplitude and phase can be modified. Therefore, it is enough to study separately the modification of gain and phase to completely characterize the frequency response: sin wt

ytrans(t)+yss(t)

G(s)

Obviously a sinusoidal input will provoke a time response formed by transient components (that will tend to zero as time tends to infinite) and stationary ones. Thanks to the Euler's formula, we can write down a sinusoidal input as an exponential. Let us see know how a linear system can change its amplitude and phase but never its frequency:

sin wt

G(s) t→¥

ejwt

G(s) t→¥

L {e j ωt u(t )}=

1 s−j ω

yss(t) = K sin(wt+f)

yss(t) = K ej(wt+f)

1 1 L {sin (ω t )u(t )}= 2 ω 2 =ω s+ j ω s−j ω s +ω

Just for simplicity reasons let us write down the proof with (negative) real simple poles. The only necessary condition, remember, is stability. But let us suppose: Control and Guidance

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G(s)=

N (s) n

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EETAC

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J. Carlos Aguado

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ESAII-UPC

where ℜ(− si )< 0 ∀ i . Just to simplify visually let si ∈ℝ ∀ i and s i ≠s j ∀ i , j

(s+ si ) ∏ i=1 Therefore, a stable system, negative real part in their poles and its response will tend to zero because it will be multiplied by negative exponential functions. The supposition of only real and simple poles is not necessary at all, only makes easier to write down the response. ωt

j x (t)=e u(t )=(cos ω t +sin ω t )u (t ) B1 Bn N ( s) 1 A Y (s)= =( Heaviside) = + + ...+ s− j ω s+ s1 s− j ω (s+ s1 )( s+s2 )... (s+s n ) s+s n −s t −s j t y (t )=( Ae ω +B 1 e +...+ B n e )u (t ) y (t) = Ae j ω t + 0 + ... + 0= Ae j ω t = Ax(t ) 1

t →∞

nt

t→∞

As we can see, all the negative exponential functions that multiply the Bi coefficients will tend to zero and in stationary state only the A coefficient will remain. How can we compute it? According to the Heaviside partial fraction expansion method, we can see that it will be complex, with amplitude and phase:

A=(s− j ω )Y ( s)|s= j ω =G( s)|s= j ω =G( j ω )

As a consequence, the phase and amplitude for the frequency response are just the ones corresponding to the evaluation of the transfer function with the argument s=jw , where w is the frequency of the sinusoidal input. We have used an exponential input, but a sine can be expressed as:

Example Let us study the frequency response of a system with transfer function:

Kω n K = 1+τ s s+ ω n We know that we have to replace s with s=j w : Kω n K' K = G( j ω )= = 1+τ j ω j ω +ω n j ω + ω n G(s)=

and find its module and phase:

|G ( j ω )|=

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|K | |1+ τ j ω |

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=

K K' = 2 2 2 2 √ 1 +τ ω √ ω +ω n 2

Frequency Response

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EETAC

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J. Carlos Aguado

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ESAII-UPC

−1 −1 ω −1 ϕ (G ( jω ))=0 −ϕ (1+τ j ω )=−tg τ ω =−tg τ ω =−tg ω n 1

Habitually we will study it in extreme cases of frequency for sinusoidal inputs:

K −1 sin(ω t−tg τ ω )→K sin ω t 2 2 √1 + τ ω K { G ( jω ) } 2 2 2 →0 √ 1 +τ ω ω →∞ ⇒ y(t )= ϕ (G ( j ω))=−tg−1 τ ω → −π 2

ω →0⇒ y(t)=

2

{

What can we observe? When the input is a slow periodic signal, the output will be more or less the signal amplified by the gain factor K. In fact this will only happen for a constant input. But for fast inputs the output will tend to zero (and it will have a phase of 90 degrees with respect to the input). Therefore, any first order systems allows the pass of low frequency signals (multiplying them by a gain), and reduces the output to virtually zero for high frequency signals. This is called in electronics a low-pass filter, and can be constructed with just a capacitor and a resistor: for low frequency sinusoidal input voltages, the output voltage will be the same, and for high frequency ones the output will be virtually zero:

In fact, any physical system with moving parts will have a similar behavior, it can be able to follow slow signals but after some threshold it will have a limiting speed and will not be able to follow very fast inputs.

3.2. FREQUENCY RESPONSE DIAGRAMS Taking into account that the stationary state response from a linear stable system to a sinusoidal input is another sinusoidal output with equal frequency but probably different amplitude and phase, we only need to record how these two quantities vary with the input frequency. A very convenient way is to draw graphs, and we will see two possibilities for that: the logarithmic double Bode representation of Bode and the single Nyquist graph.

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EETAC

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J. Carlos Aguado

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ESAII-UPC

3.2.1 Bode Diagrams Bode diagrams have been very popular in the field of automatic control, because they can be sketched by hand, new factors are just graphically summed and they allow to predict the behavior in response to slow and fast periodic inputs and even to get the main time response characteristics (speed, precision, presence or absence of overshoot). So closed-loop Bode diagrams give us the same information as time response and more, and, on the other hand, the openloop Bode can be used to study the stability of the closed-loops system similarly to the root locus. Bode representation consists of two Cartesian graphs, representing the gain and phase that a linear stable system would introduce to a periodic input depending on its frequency. That frequency is represented in decimal logarithmic scale (each separation multiplies the frequency by a factor of 10), thus allowing the collection of a very wide range of input frequencies, selecting which of them we want to depict with more or less detail (habitually the closest ones to zero). The graph of the gain is also logarithmic and the phase has also this behavior without the need for more logarithms. Therefore the contribution of factors that are multiplied can be easily summed up. According to that, the phase will be represented either in degrees or radians, but the gain needs some logarithm units, the decibels:

1 decibel=1 db=1 dB=20 log10 |G ( j ω)| Nevertheless usually the frequencies are represented in their real values, not their logarithm, but spaced according to their logarithm. Of course, the values shown for both the magnitudes and the frequencies are always positive: the logarithmic positive infinite corresponds to the real infinite, but the negative logarithmic infinite means a zero real value, and the logarithmic zero stand for the unit real value. Location of the gain and phase axis on the frequency one is arbitrary, chosen to leave to its immediate right the interesting cutoff frequencies (to collect important frequency response changes). 3.2.1.1 Building process  Rewriting of G(jw) as the product of basic factors, from four possible types:

{

K gain factor ±1 ( j ω) integrators and derivators ±1 jω +1 first order factors (real zeros and poles) K

(

(( )



)

2

jω jω +1 +K 2 K1 K1

)

±1

second order factors (complex zeros and poles)

}

Asymptotic representation of each factor's contribution for gain and phase

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J. Carlos Aguado

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If necessary, include the modified contribution by non-minimum phase zeros

 Combine (add up) all the contributions, a table of factors-frequencies is very advisable for that

3.2.1.2 Basic factors  Gain factors

G (s)=K ⇒|G ( j ω)|=20 log 10 |K |db =constant 0 G (s)=K ⇒ ϕ(G ( j ω))=tan−1 = 0rad if K≥0 K π rad if K1

radians

 Integral and derivative factors (remember that physically these factors do not make much sense. isolated derivators do not exist and integrators are unstable, but they are useful as parts of a system)

1 G (s)= ⇒|G ( j ω)|=0 db−20 log10 |ω|db=straight line with slope -20db/dec s waypoint for the straight line |G( j ω)|ω=1=0 db−20 log10 1 db=0 db 1 0 G(s)= ⇒ϕ (G ( j ω))=tan−1 −tan−1 ω =− π rad 2 1 0 s

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J. Carlos Aguado

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G (s)=s ⇒|G ( j ω)|=20 log10 ω db=straight line with slope 20 db/dec waypoint for the straight line |G( j ω)|ω=1=20 log 10 1 db =0 db G (s)=s ⇒ ϕ(G ( j ω))=tan−1 ω = π rad 0 2

radians

 First order factors, simple real zeros and poles First order factors are characterized uniquely by the value of the zero or pole, which is called, from the frequency point of view, cutoff frequency. Again we build an asymptotic representation and only compute the exact value for that particular point. For simple real poles:

G(s)=

|

|

√

2 ωn ωc 1 ω db ⇒|G( j ω)|=20 log10 =0 db−20 log10 12 +( ω = ω ) c s+ω n s+ω c 1+ j ω c lim |G ( j ω)|=0 db−20 log10 1 db=0 db ω→ 0

lim |G( j ω)|=(ω ≫ωc , 1)=straight line with slope -20db/dec ω→ ∞

|G( j ω)|ω=ω =0 db−20 log10 √ 2 db=−3 db c

G (s)=

ωc ω ⇒ ϕ( G ( j ω))=0−tan−1 ω c s +ω c lim ϕ( G ( j ω))=0 rad ω→ 0 lim ϕ( G ( j ω))=− π rad ω→∞ 2 lim ϕ(G ( j ω))=− π rad ω→ω 4 c

Repeating the same process for the zeros, the only difference will be the sign of logarithmic gain and phase, and therefore the graphs will be exact mirror reflections:

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radians 20 dB/dec

Simple zero

Simple pole

-20 dB/dec

20dB/ dec

Simple zero

Simple zero

Simple pole

Simple pole

-20dB/ dec

 Second order factors, complex zeros and poles Just as before, the second order factors must first also be written in the Bode form, making explicit the cutoff frequency. Then we will show the asymptotic behavior and the exact one only for the cutoff frequency, again only for the poles remembering the zeros will be their mirror reflection with respect to the frequency axis.

((

)

2

ω +2 ζ j ω +1 G ( j ω)= j ω ωc c

)

−1

⇒|G ( j ω)|=0 db−20 log10

√(

)

2

2

2 1−ω2 + 2 ζ ωω db c ωc

(

)

lim |G ( j ω)|=0 db−20 log10 1 db=0 db ω→ 0

lim |G( j ω)|=(ω≫ωc , 1)=straight line with slope -40db/dec ω→ ∞ |G( j ω)|ω=ω =0 db−20 log 10 2 ζ db , if ζ=1 we have a double real pole and the gain is −6 db c

((

−1

) ⇒ ϕ(G ( j ω))=0−tan

ω 2 +2 ζ j ω +1 G ( j ω)= j ω ωc c

)

−1

2ζ ω ωc 2

1−ω2 ωc

lim ϕ( G ( j ω))=0 rad ω→ 0

lim ϕ(G ( j ω))=−π rad =π rad ω→ ∞ lim ϕ( G ( j ω))=−π rad ω→ ω 2 c

As we can see, the gain behavior depends on the damping factor and will show overshoot for underdamped systems. Again the pole swill be shown in the next graphs and the zeros' graphs will be symmetrical with respect to the frequency axis:

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radians

Complex pole

3.2.1.1 Non-minimum phase zeros Once that we know that the frequency response of zeros and poles is symmetric with respect to the frequency axis (0 db in the gain representation, 0 rad in the phase graph), we have to take into account the so-called non-minimum phase zeros. Theoretically, we could also define non-minimum phase poles, but they would be UNSTABLE poles, and therefore the frequency response cannot be defined (the response would be infinite) for them and the Bode diagrams would not reflect reality (BEWARE, the usual Computer Aided Automatic Control programs as MatLab and CC are wrong about it). If we repeat our study for positive real part zeros (non-minimum phase systems), we can see that the gain graph does not change and only the phase graph is reflected, not around 0 rad but around the furthest part from it (the asymptotic behavior as w tended to infinity). Therefore, all the graphs for first order factors will be the following (and equivalent ones are obtained for second order factors):

ω +1 simple real mimimum phase zero j ω c

simple real mimimum phase pole

1 ω +1 jω c

ω −1 simple real non-mimimum phase zero j ω c

(

simple real non-mimimum phase pole

radians

Simple zero Minimum phase or not

Non-minimum p. zero Minimum phase zero Minimum phase pole Non-minimum p. pole WRONG GRAPH

Simple pole Minimum phase or not

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1 ω j ω −1 c

)

Control and Guidance

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Bode diagrams fast building The most habitual form of manually drawing Bode diagrams is by building tables. Those auxiliary tables have as many column as cutoff frequencies plus one, and as many rows as basic factors, plus one for the total. Each cell will represent a contribution in units that, for minimum-phase systems, will represent both a gain asymptote of 20 db/dc and a phase asymptote of p/2. So, for minimumphase systems it will look like:

w=0

w=w c1

w =w ci

w =w c(i+1)

w =w cn

...

... 0

... ...

... m

... ...

... m

... Total

... S

... S

... S

... S

... S

where S is the sum of every column, and m is the multiplicity of every zero and pole, therefore can be reduced to {-2,-1,1,2} (positive values for zeros and negatives for poles). Nevertheless there are a pair of cases where the multiplicity disagrees with the visible exponent, for complex zeros and poles. Therefore a complex zero row will be:

0

...

2

...

2

-2

...

-2

And of course a complex pole will give: 0

...

Meanwhile a derivative part would produce: jw

1

...

1

Without any cutoff frequency, just the same as an integrator: -1

...

-1

This operation has to be refined for non-minimum phase systems, where a slope of 20 db/dc is not always linked to a p/2 phase. So we need to units in every cell:

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w=0

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w =w ci

w=w c1

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w =w c(i+1)

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w =w cn

...

...

...

...

...

...

...

...

...

...

...

...

Total

All of the rest remains the same: non-apparent multiplicity for complex zeros and poles, no cutoff frequency for derivative factors and integrators… The only difference is that a part of the sum (“numerator”) shows the gain slope and the other (“denominator”) the phase asymptote. Finally, we have to observe that the table does not collect the gain factors, because their asymptotic contribution is zero. Obviously, when drawing the diagram, those factors will move up or down the gain asymptotes and can sum to or subtract from the phase p rads.

3.2.2 Nyquist and Nichols representations Nyquist representation is a polar one, again representing the stationary state response from a linear stable system to a sinusoidal input. Its advantage is to represent simultaneously gain and phase, but the price to pay is that the frequency they depend upon cannot be explicitly represented. Consequently, the phase will be represented as an angle and the gain as a module for every positive (physical) frequency. But in the Nyquist graph also the (non-physical, mathematical) negative frequencies should be represented if we want to use the open-loop transfer function to study the closed-loop stability. It is easy because G(-jw) follows a trajectory that constitutes a mirror image of G(jw) with respect to the real axis. But Nyquist representation is linear, not logarithmic. Therefore sometimes the size of the graph can be intractable, the basic factors are summed for the phase but not for the gain, and the input frequency must be added if we need to study a particular point. Of course we can draw them either directly or via a previous Bode double diagram.

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conjugate pole s

simple pole