PROCESS CONTROL SYSTEM DESIGN Process Control System Design LECTURE 2: FREQUENCY DOMAIN ANALYSIS

(1)

Frequency Domain Analysis PROCESS CONTROL SYSTEM DESIGN - (c) Daniel R. Lewin

1 -2

054414 Process Control System Design

LECTURE 2:

FREQUENCY DOMAIN ANALYSIS

Daniel R. Lewin

Department of Chemical Engineering

Technion, Haifa, Israel

Objectives

Draw Bode and Nyquist plots of an arbitrary SISO

linear system

Both by hand and using MATLAB

Sketch the temporal response of a SISO linear

system, given its Bode plot

Both by hand and using MATLAB/SIMULINK

Determine an appropriate transfer function given

the Bode plot of a SISO linear system.

Both by hand and using MATLAB

(2)

3 -2

Frequency Response

P(s)

( )

Usin tω Ysin t

(

ω + φ

)

( )

s i

( )

p s →= ωp iω

( )

Amplitude ratio, AR = Y U p i= ω

( )

{

}

1Im p i

_{

{

_{( )}

( )

_}

}

Phase shift, arg p i tan

Re p i

− ω

φ = ω =

ω φ ω

Frequency Response Example

Response of dead sea pond to cyclic perturbation:

( )

dc _{0.05c 0.02E,c 0} _{0;E t} _{3sin t ,} _rad/h

dt 12 π = − + = = ω ω = Solution:

( )

(

)

(

)

(

)

− = + ω − ω = π = = φ ω = π = − 0.05t c t 0.0225e 0.0229 sin t 1.383 p 12 0.0229 3 0.0076

12 1.383 rad (phase lag) φ

(3)

5 -2

Frequency Response – 1

o

_System

( )

{

( )

}

_{

{

( )

_{( )}

}

_}

(

)

p 2 2 1 1 K AR p i 1 Im p i

arg p i tan tan Re p i − − = ω = + τ ω ω φ ω = ω = = −τω ω

( )

Kp p s s 1 = τ +

The ultimate response has the following characteristics:

(

p

)

(

)

K 1 i 1 i 1 i − τω ⇒ + τω − τω

( )

Kp p i i 1 ⇒ ω = τω +

(

)

(

p 2 2

)

K 1 i 1 − τω ⇒ + τ ω

Example (Dead Sea Pond):

( )

0.002 0.04 p s s 0.05 20s 1 = = + +

( )

(

)

2 1 0.04 AR p i . For 12,AR 0.0076 1 400 tan− 20 . For 12, 1.383 = ω = ω = π = + ω φ ω = − ω ω = π φ = −

Freq. Response Plots – 1

o

_System

Bode Plot

L og ( A R) Pha se

In the Bode magnitude plot, log (AR) is plotted against log (ω)

In the Bode phase plot, Phase (in rad or degrees) is plotted against log (ω)

( )

_tan−1

( )

φ ω = −ω

( )

1 ₂ p i 1 ω = + ω AR(0.01) = 1.000 AR(0. 10) = 0.995 AR(1.00) = 0.707 AR(10.0) = 0.099 AR(100) = 0.010 φ(0.01) = -0.57o φ(0.10) = -5.71o φ(1.00) = -45o φ(10.0) = -84.2o φ(100) = -89.4o

( )

1 p s s 1 = +

(4)

7 -2

Freq. Response Plots – 1

o

_System

L og ( A R) Pha se p p 2 2 0 K lim K 1 ω→ _{+ τ ω} = p p 2 2 K lim K 1 ω→∞ _{+ τ ω} = τω

(

)

1 o lim tan− 90 ω→∞ −τω = −

(

)

1 o 0 lim tan− 0 ω→ −τω = Asymptotes

Asymptotes join at breakpoint” located at ω = 1/τ rad/min

Asymptotes join at “break point”

Bode Plot

( )

1 p s s 1 = + -1

Gradient of high frequency asymptote is -1 on a log-log scale

Freq. Response Plots – 1

o

_System

w=logspace(-2,2,200);s=i*w; p=1./(s+1); AR=abs(p); ph=180*phase(p)/pi; subplot(2,1,1) loglog(w,AR,'-r’) subplot(2,1,2) semilogx(w,ph,'-r')

(5)

9 -2

Freq. Response Plots – 1

o

_System

Nyquist Plot

In the Nyquist plot, p(iω), which is a complex number, is plotted directly, as a locus from ω = 0, to ω = ∞. -0.57 1.000 0.01 -89.4 0.010 100 -84.2 0.099 10.0 -45.0 0.707 1.00 -5.71 0.995 0.10 φ(o₎ AR ω w=logspace(-2,2,200);s=i*w; p=1./(s+1); plot(p)

Frequency Response – 1

o

_System

Nyquist Plot Bode Plot

(6)

11 -2

Frequency Response–Integrator

( )

{

( )

}

_{

{

_{( )}

( )

_}

}

( )

p 1 1 K AR p i Im p i

arg p i tan tan

Re p i 2 − − = ω = ω ω π φ ω = ω = = −∞ = − ω

( )

Kp p s s =

(

)

p K i i i − ω ⇒ ω − ω

( )

Kp p i i 1 ⇒ ω = ω + p iK − ⇒ ω

Frequency Response–Integrator

(7)

13 -2

Frequency Response Interpretation

Consider two first order systems:p s1

( )

=_{5s 1}1₊ and p s2

( )

=_{s 1}1₊

These two systems are excited by a series of steps (a square wave, approximately) as shown below.

Note that p1responds more sluggishly than p2, because its time constant is 5 times larger.

Note also that the exciting signal is only approximately a square wave...

Frequency Response Interpretation

This is because it is actually made up of a series of harmonic terms. In fact, it can be shown that a true square wave can be expressed as the infinite series:

( )

(

₍

(

₎

)

( )

i 1

1 1

3 5

sin 2 i 1 1 t

u t sin t sin 3t sin 5t 2 i 1 1 ∞ ∞ = − + ⎡ ⎤ ⎣ ⎦ = = + + + − +

∑

…

More accurate approximations of the true square wave are obtained by using more terms in the above expansion.

Any signal can be expressed as a Fourier expansion of the form:

( )

u t sin t

ω

=

∑

α ω ω

where α(ω) is the amplitude of the signal at ω.

(8)

15 -2

Frequency Response Interpretation

( )

(

)

7 ₃1 ₅1 ₁₃1

u t =sin t + sin 3t + sin 5t +…+ sin 13t Let’s display the magnitudes of the terms in the expansion to seven terms:

on the Bode magnitude plot. This is equivalent to a step (integrator)…

Frequency Response Interpretation

Equivalent to the square wave (step)

(9)

17 -2

Frequency Response Interpretation

Frequency Response – Lead/Lag

( )

{

( )

}

_{

{

_{( )}

( )

_}

}

( )

{

( )

}

_{

{

_{( )}

( )

_}

}

(

)

2 2 1 2 p 1 1 1 1 1 1 2 1 1 2 2 2 AR p i p i K 1 Im p i

arg p i tan tan Re p i

Im p i

arg p i tan tan Re p i − − − − = ω = ω = + τ ω ω φ ω = ω = = τω ω ω φ ω = ω = = −τω ω

( )

(

)

1 p p s =K s 1τ +

( )

(

)

1 p p i K i 1 ⇒ ω = τω +

( )

(

)

2 p p s =K −τ +s 1 ⇒p i₂

( )

ω =K_p

(

− τω +i 1

)

(10)

19 -2

Frequency Response – Lead/Lag

+1

Gradient of high frequency asymptote is +1 on a log-log scale

Frequency Response – 2

o

_System

( )

(

)

(

)

( )

{

( )

}

_{

{

_{( )}

( )

_}

}

p 2 2 2 2 1 1 2 2 K AR p i 1 2 Im p i 2

arg p i tan tan Re p i 1 − − = ω = − τ ω + ξτω ω ⎛ − ξτω ⎞ φ ω = ω = = _⎜ _⎟ ω ⎝ − τ ω ⎠

( )

p 2 2 K p s s 2 s 1 = τ + ξτ +

( )

₍

₎

p 2 2 K p i 1 i2 ⇒ ω = − τ ω + ξτω

( )

(

)

(

)

(

)

(

)

2 2 p 2 2 2 2 K 1 i2 p i 1 i2 1 i2 − τ ω − ξτω ⇒ ω = − τ ω + ξτω − τ ω − ξτω

( )

(

)

(

)

(

)

(

)

(

)

(

)

2 2 p p 2 2 2 2 2 2 2 2 K 1 K 2 p i i 1 2 1 2 − τ ω ξτω ⇒ ω = − − τ ω + ξτω − τ ω + ξτω

(11)

21 -2

Frequency Response – 2

o

_System

(

)

p 2 2 p 2 2 2 K lim K 1 ω→∞ = τ ω + τ ω Bode Plot 1 o 2 2 2 lim tan 180 1 − ω→∞ − ξτω ⎛ _{⎞ = −} ⎜ _{− τ ω} ⎟ ⎝ ⎠ -2

Gradient of high frequency asymptote is -2 on a log-log scale

Frequency Response – Complex p(s)

On the Bode log AR plot, the amplitudes of each component

add up

add up:

To construct asymptotes in Bode plots for:

( )

(

₍

1

)(

₎₍

2

)

_{) (}

(

m

₎

)

s p 1 2 n a s 1 a s 1 a s 1 p s K e b s 1 b s 1 b s 1 −θ + + + = + + + … …

( )

p 1 2 1 2

log p s logK log a s 1 log a s 1 log b s 1 log b s 1

= + + + + +

− + − + −

… …

On the Bode linear phase plot scale, the phase of each component add upadd up:

_{

_{( )}

_}

_{

_}

_{

_}

{

}

{

}

φ = = + + + + − + − + − − θ … … 1 2 1 1

arg p s arg a s 1 arg a s 1 arg b s 1 arg b s 1 s For large ω, log AR(ω) vs. log ω has an asymptotic slope of −(n − m)

For large ω, for MP p(s) [no positive zeros or delay terms], φapproaches asymptotic value of −(π/2)×(n − m)

(12)

23 -2

Frequency Response – MP/NMP

MP = Minimum phase, NMP = Non-minimum phase.

RHP zeros: It is convenient to factor the RHP zero with its LHP mirror image: . The AR of this component is AR = 1, which has a phase lag of zero. In contrast, for large ω, p_zhas a phase lag of:

( )

(

) (

)

z

p s = −zs 1 zs 1+ +

( )

{

z

}

{

}

{

}

lim arg p s lim arg zs 1 lim arg zs 1 2 2

ω→∞ =ω→∞ − + −ω→∞ +

= − π − π = −π

NMP systems are those that feature phase lags greater than anticipated based on the system’s AR alone. Those whose phase lag corresponds to the systems AR are MP systems. NMP components are either:

(a) Right-half plane (RHP) zeros, or (b) Dead time

Dead time: The phase lag of is −θω. The AR of this component is AR = 1, which has a phase lag of zero.

( )

s d

p s ₌e−θ

Class Exercise 1 - Sketch p(iω)

Generate a Bode plot for the following transfer function:

( )

(

)

2 0.5s s 1 p s e 10s 1 − + = + Solution.

(

)

2 0.5s 1 s 1, and e 10s 1 − + +

The Bode plot is plotted by combining the contributions of the components:

(13)

25 -2

Class Exercise 1 - Sketch p(iω)

Solution (Cont’d).

( )

(

s 1

)

2 0.5s p s e 10s 1 − + = + s 1+ 0.5s e− (_{10s 1}1₊)2 s 1+ (_{10s 1}1₊)2 -0.287 0.01 -287 10.0 -28.7 1.00 -2.87 0.10 φ(e-0.5s₎ ω

Class Exercise 2 - Determine p(s)

Determine the transfer function, p(s) for the process whose Bode diagram is given above.