Unformatted text preview: Page 1 of 2 CHE 361: Frequency Response Analysis
Chap. 14 SEMD3 = “Frequency Response Analysis” = 3 “terms” involved
Review Chap. 5.2.3 = 1st  order Sinusoidal Response, 5.4.2 = 2nd  order
Sinusoidal Response, For physical systems = Examples 5.2 & 5.6
1 = "frequency"  physical interpretation is that the input to the process is
changing with time as u '(t ) = A sin(ωt ) with ω in radians/time
ω radians/time
f = frequency in cycles/time =
2π radians/cycle
P = period of oscillation, time per cycle in time units = 1/f
2 = "response"  for linear systems, if the input is a sine wave, the output
(response) after a “long” time is also a sine wave with a shifted
“argument” but the same frequency and in general a different
amplitude. The shift and amplitude depend on the frequency of the
input oscillation, thus the "frequency response" also implies a
"frequency dependance". The term “long time” is long compared to
the time constants of the system.
3 = "analysis" looking at a range of frequencies all at once and drawing
conclusions, extending the "physical" input/output ideas of
oscillatory behavior into math concepts based on complex
functions of complex variables.
14.1  Sinusoidal Forcing of a Firstorder Process
K
=
AR
, φ = Phase Angle = − tan −1 (ωτ )
22
ω τ +1
14.2  Sinusoidal Forcing of an nthorder Process ( n negative real poles )
and "Shortcut" Method for Finding the Frequency Response of G(s) Model
Substitute jω for s. Put resulting complex number into the form:
=
G ( jω ) Re (ω ) + Im (ω ) j Then amplitude ratio (AR) of G at ω = (Re2 + Im2)1/2 and the phase angle (PA)
of G at ω = tan1 (Im/Re) ... caution: pay attention to the quadrant of the
complex number by examining the signs of Re and Im !
Two examples:14.1, 14.2 pgs 253254 SEMD3 Page 2 of 2 14.3 Bode Diagrams = plots of AR and PA as functions of ( ω ) frequency
Firstorder System (stable) = frequency response of a "pole" in LHP
5 steps include ....
AR: low ω asymptote, AR = K with slope = 0
at ω = 1/τ, AR = 0.707 K (asymptotes meet at "corner" frequency)
high ω asymptote, slope of AR( ω ) =  1
PA: low ω asymptote, PA = 0 deg.
at ω = 1/τ, PA =  45 degrees (at "corner" or "break" frequency)
high ω asymptote = 90 deg.
Secondorder System: given K , τ and ζ : AR = K (1 − ω τ ) + ( 2ζωτ )
222 2 −2ζωτ and PA = φ = tan 1 22
1 − ω τ 2ndorder 14.3.3 = note two 1storder systems in series = K x “pole1” x “pole2”
AR overall (total) = product of ARi ( individual factors )
PA overall (total) = sum of PAi ( individual factors )
Process Zero 14.3.4 : AR & PA
Since LHP zero can cancel LHP pole, that product ratio = 1, AR of zero at 1/τ
= AR1 of pole at 1/τ so that combined AR = 1 for all ω. Likewise, PA of zero
at 1/τ = PA of pole at 1/τ so that combined PA = 0 for all ω.
For a zero in the RHP at 1/τz, where τz < 0, the AR is the same as if τz had the
same size but different sign, while the PA has opposite sign for the two cases.
Thus a RHP zero has a PA that starts at 0 and goes to 90 deg, similar to the PA
for a LHP pole.
Time delay 14.3.5: AR = 1 for all ω and PA = (ωθ) radians (see Figure 13.6 or
14.4).
Example 14.3: Bode plot for K = 5, time delay = 0.5, zero at 2.0,
poles at 0.05 and 0.25.
Table 14.2 Examples of G(s) and Bode Plots  Also see CHE 361 Handout
with plots from MATLAB
Skip 14.4 = Frequency Response Characteristics of Feedback Controllers
CHE 461 PID controller analysis
14.5 Nyquist Diagram of Gp(s)  plot of Re(ω) vs. Im(ω) on Cartesian
coordinates for ω from 0 to ∞ and the "reflection" for ω from 0 to ∞. ...
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This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.
 Winter '08
 Staff
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