R. W. Ericksonecen5797/course_material/... · 2018-10-19 · 20 log 10 f f 0 n 20n log 10 f f 0 f f...
Transcript of R. W. Ericksonecen5797/course_material/... · 2018-10-19 · 20 log 10 f f 0 n 20n log 10 f f 0 f f...
R. W. EricksonDepartment of Electrical, Computer, and Energy Engineering
University of Colorado, Boulder
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions1
Chapter 8. Converter Transfer Functions
8.1. Review of Bode plots8.1.1. Single pole response8.1.2. Single zero response8.1.3. Right half-plane zero8.1.4. Frequency inversion8.1.5. Combinations8.1.6. Double pole response: resonance8.1.7. The low-Q approximation8.1.8. Approximate roots of an arbitrary-degree polynomial
8.2. Analysis of converter transfer functions8.2.1. Example: transfer functions of the buck-boost converter8.2.2. Transfer functions of some basic CCM converters8.2.3. Physical origins of the right half-plane zero in converters
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions2
Converter Transfer Functions
8.3. Graphical construction of converter transferfunctions
8.3.1. Series impedances: addition of asymptotes8.3.2. Parallel impedances: inverse addition of asymptotes8.3.3. Another example8.3.4. Voltage divider transfer functions: division of asymptotes
8.4. Measurement of ac transfer functions andimpedances
8.5. Summary of key points
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions3
The Engineering Design Process
1. Specifications and other design goals are defined.2. A circuit is proposed. This is a creative process that draws on the
physical insight and experience of the engineer.3. The circuit is modeled. The converter power stage is modeled as
described in Chapter 7. Components and other portions of the systemare modeled as appropriate, often with vendor-supplied data.
4. Design-oriented analysis of the circuit is performed. This involvesdevelopment of equations that allow element values to be chosen suchthat specifications and design goals are met. In addition, it may benecessary for the engineer to gain additional understanding andphysical insight into the circuit behavior, so that the design can beimproved by adding elements to the circuit or by changing circuitconnections.
5. Model verification. Predictions of the model are compared to alaboratory prototype, under nominal operating conditions. The model isrefined as necessary, so that the model predictions agree withlaboratory measurements.
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions4
Design Process
6. Worst-case analysis (or other reliability and production yieldanalysis) of the circuit is performed. This involves quantitativeevaluation of the model performance, to judge whetherspecifications are met under all conditions. Computersimulation is well-suited to this task.
7. Iteration. The above steps are repeated to improve the designuntil the worst-case behavior meets specifications, or until thereliability and production yield are acceptably high.
This Chapter: steps 4, 5, and 6
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions5
Buck-boost converter modelFrom Chapter 7
+–
+–
L
RC
1 : D D' : 1
vg(s) Id(s) Id(s)
i(s) +
v(s)
–
(Vg – V)d(s)Zout(s)Zin(s)
d(s) Control input
Lineinput
Output
Gvg(s) = v(s)vg(s)
d(s) = 0
Gvd(s) = v(s)d (s) vg(s) = 0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions7
Design-oriented analysis
How to approach a real (and hence, complicated) systemProblems:
Complicated derivationsLong equationsAlgebra mistakes
Design objectives:Obtain physical insight which leads engineer to synthesis of a good designObtain simple equations that can be inverted, so that element values can
be chosen to obtain desired behavior. Equations that cannot be invertedare useless for design!
Design-oriented analysis is a structured approach to analysis, which attempts toavoid the above problems
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions8
Some elements of design-oriented analysis,discussed in this chapter
• Writing transfer functions in normalized form, to directly expose salientfeatures
• Obtaining simple analytical expressions for asymptotes, cornerfrequencies, and other salient features, allows element values to beselected such that a given desired behavior is obtained
• Use of inverted poles and zeroes, to refer transfer function gains to themost important asymptote
• Analytical approximation of roots of high-order polynomials• Graphical construction of Bode plots of transfer functions and
polynomials, toavoid algebra mistakesapproximate transfer functionsobtain insight into origins of salient features
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions6
Bode plot of control-to-output transfer functionwith analytical expressions for important features
f
0˚
–90˚
–180˚
–270˚
|| Gvd ||
Gd0 =
|| Gvd || �Gvd
0 dBV
–20 dBV
–40 dBV
20 dBV
40 dBV
60 dBV
80 dBV
Q =
�Gvd
10-1/2Q f0
101/2Q f0
0˚
–20 dB/decade
–40 dB/decade
–270˚
fz /10
10fz
1 MHz10 Hz 100 Hz 1 kHz 10 kHz 100 kHz
f0
VDD' D'R C
L
D'2/ LC
D' 2R2/DL(RHP)
fz
DVg
t(D')3RC
Vg
t2D'LC
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions9
8.1. Review of Bode plots
Decibels
G dB = 20 log10 G
Table 8.1. Expressing magnitudes in decibels
Actual magnitude Magnitude in dB
1/2 – 6dB
1 0 dB
2 6 dB
5 = 10/2 20 dB – 6 dB = 14 dB
10 20dB1000 = 103 3 ! 20dB = 60 dB
Z dB = 20 log10
ZRbase
Decibels of quantities havingunits (impedance example):normalize before taking log
5! is equivalent to 14dB with respect to a base impedance of Rbase =1!, also known as 14dB!.60dBµA is a current 60dB greater than a base current of 1µA, or 1mA.
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions10
Bode plot of fn
G = ff0
n
Bode plots are effectively log-log plots, which cause functions whichvary as fn to become linear plots. Given:
Magnitude in dB is
G dB = 20 log10ff0
n
= 20n log10ff0
ff0
– 2
ff0
2
0dB
–20dB
–40dB
–60dB
20dB
40dB
60dB
flog scale
f00.1f0 10f0
ff0
ff0
– 1
n = 1n =
2
n = –2
n = –120 dB/decade
40dB/decade
–20dB/decade
–40dB/decade
• Slope is 20n dB/decade
• Magnitude is 1, or 0dB, atfrequency f = f0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions11
8.1.1. Single pole response
+–
R
Cv1(s)
+
v2(s)
–
Simple R-C example Transfer function is
G(s) = v2(s)v1(s)
=1
sC1
sC + R
G(s) = 11 + sRC
Express as rational fraction:
This coincides with the normalizedform
G(s) = 11 + s
t0
with t0 = 1RC
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions12
G(j!) and || G(j!) ||
Im(G(jt))
Re(G(jt))
G(jt)
|| G(jt
) ||
�G(jt)
G( jt) = 11 + j t
t0
=1 – j t
t0
1 + tt0
2
G( jt) = Re (G( jt)) 2 + Im (G( jt)) 2
= 11 + t
t0
2
Let s = j!:
Magnitude is
Magnitude in dB:
G( jt)dB
= – 20 log10 1 + tt0
2 dB
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions13
Asymptotic behavior: low frequency
G( jt) = 11 + t
t0
2
tt0
<< 1
G( jt) 5 11
= 1
G( jt)dB5 0dB
ff0
– 1
–20dB/decade
ff00.1f0 10f0
0dB
–20dB
–40dB
–60dB
0dB
|| G(jt) ||dB
For small frequency,! << !0 and f << f0 :
Then || G(j!) ||becomes
Or, in dB,
This is the low-frequencyasymptote of || G(j!) ||
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions14
Asymptotic behavior: high frequency
G( jt) = 11 + t
t0
2
ff0
– 1
–20dB/decade
ff00.1f0 10f0
0dB
–20dB
–40dB
–60dB
0dB
|| G(jt) ||dB
For high frequency,! >> !0 and f >> f0 :
Then || G(j!) ||becomes
The high-frequency asymptote of || G(j!) || varies as f-1.Hence, n = -1, and a straight-line asymptote having aslope of -20dB/decade is obtained. The asymptote hasa value of 1 at f = f0 .
tt0
>> 1
1 + tt0
25 t
t0
2
G( jt) 5 1tt0
2= f
f0
– 1
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions15
Deviation of exact curve near f = f0
Evaluate exact magnitude:at f = f0:
G( jt0) = 1
1 + t0t0
2= 1
2
G( jt0) dB= – 20 log10 1 + t0
t0
25 – 3 dB
at f = 0.5f0 and 2f0 :
Similar arguments show that the exact curve lies 1dB belowthe asymptotes.
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions16
Summary: magnitude
–20dB/decade
f
f0
0dB
–10dB
–20dB
–30dB
|| G(jt) ||dB
3dB1dB0.5f0 1dB
2f0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions17
Phase of G(j!)
G( jt) = 11 + j t
t0
=1 – j t
t0
1 + tt0
2
�G( jt) = tan– 1Im G( jt)Re G( jt)
Im(G(jt))
Re(G(jt))
G(jt)
|| G(jt
) ||
�G(jt)�G( jt) = – tan– 1 t
t0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions18
-90˚
-75˚
-60˚
-45˚
-30˚
-15˚
0˚
f
0.01f0 0.1f0 f0 10f0 100f0
�G(jt)
f0
-45˚
0˚ asymptote
–90˚ asymptote
Phase of G(j!)
�G( jt) = – tan– 1 tt0
! "G(j!)
0 0˚
!0–45˚
# –90˚
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions19
Phase asymptotes
Low frequency: 0˚High frequency: –90˚Low- and high-frequency asymptotes do not intersect
Hence, need a midfrequency asymptote
Try a midfrequency asymptote having slope identical to actual slope atthe corner frequency f0. One can show that the asymptotes thenintersect at the break frequencies
fa = f0 e– / / 2 5 f0 / 4.81fb = f0 e/ / 2 5 4.81 f0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions20
Phase asymptotes
fa = f0 e– / / 2 5 f0 / 4.81fb = f0 e/ / 2 5 4.81 f0
-90˚
-75˚
-60˚
-45˚
-30˚
-15˚
0˚
f
0.01f0 0.1f0 f0 100f0
�G(jt)
f0
-45˚
fa = f0 / 4.81
fb = 4.81 f0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions21
Phase asymptotes: a simpler choice
-90˚
-75˚
-60˚
-45˚
-30˚
-15˚
0˚
f
0.01f0 0.1f0 f0 100f0
�G(jt)
f0
-45˚
fa = f0 / 10
fb = 10 f0
fa = f0 / 10fb = 10 f0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions22
Summary: Bode plot of real pole
0˚�G(jt)
f0-45˚
f0 / 10
10 f0
-90˚
5.7˚
5.7˚
-45˚/decade
–20dB/decade
f0
|| G(jt) ||dB 3dB1dB0.5f0 1dB
2f0
0dB G(s) = 11 + s
t0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions23
8.1.2. Single zero response
G(s) = 1 + st0
Normalized form:
G( jt) = 1 + tt0
2
�G( jt) = tan– 1 tt0
Magnitude:
Use arguments similar to those used for the simple pole, to deriveasymptotes:
0dB at low frequency, ! << !0
+20dB/decade slope at high frequency, ! >> !0
Phase:
—with the exception of a missing minus sign, same as simple pole
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions24
Summary: Bode plot, real zero
0˚�G(jt)
f045˚
f0 / 10
10 f0 +90˚5.7˚
5.7˚
+45˚/decade
+20dB/decade
f0
|| G(jt) ||dB3dB1dB
0.5f01dB
2f0
0dB
G(s) = 1 + st0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions25
8.1.3. Right half-plane zero
Normalized form:
G( jt) = 1 + tt0
2
Magnitude:
—same as conventional (left half-plane) zero. Hence, magnitudeasymptotes are identical to those of LHP zero.Phase:
—same as real pole.The RHP zero exhibits the magnitude asymptotes of the LHP zero,and the phase asymptotes of the pole
G(s) = 1 – st0
�G( jt) = – tan– 1 tt0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions26
+20dB/decade
f0
|| G(jt) ||dB3dB1dB
0.5f01dB
2f0
0dB
0˚�G(jt)
f0-45˚
f0 / 10
10 f0
-90˚
5.7˚
5.7˚
-45˚/decade
Summary: Bode plot, RHP zero
G(s) = 1 – st0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions27
8.1.4. Frequency inversion
Reversal of frequency axis. A useful form when describing mid- orhigh-frequency flat asymptotes. Normalized form, inverted pole:
An algebraically equivalent form:
The inverted-pole format emphasizes the high-frequency gain.
G(s) = 11 + t0
s
G(s) =st0
1 + st0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions28
Asymptotes, inverted pole
0˚
�G(jt)
f0+45˚
f0 / 10
10 f0
+90˚5.7˚
5.7˚
-45˚/decade
0dB
+20dB/decade
f0
|| G(jt) ||dB
3dB
1dB
0.5f0
1dB2f0
G(s) = 11 + t0
s
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions29
Inverted zero
Normalized form, inverted zero:
An algebraically equivalent form:
Again, the inverted-zero format emphasizes the high-frequency gain.
G(s) = 1 + t0s
G(s) =1 + s
t0
st0
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions30
Asymptotes, inverted zero
0˚
�G(jt)
f0–45˚
f0 / 10
10 f0
–90˚
5.7˚
5.7˚
+45˚/decade
–20dB/decade
f0
|| G(jt) ||dB
3dB
1dB
0.5f0
1dB
2f0
0dB
G(s) = 1 + t0s
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions31
8.1.5. Combinations
Suppose that we have constructed the Bode diagrams of twocomplex-values functions of frequency, G1(!) and G2(!). It is desiredto construct the Bode diagram of the product, G3(!) = G1(!) G2(!).
Express the complex-valued functions in polar form:G1(t) = R1(t) e je1(t)
G2(t) = R2(t) e je2(t)
G3(t) = R3(t) e je3(t)
The product G3(!) can then be writtenG3(t) = G1(t) G2(t) = R1(t) e je1(t) R2(t) e je2(t)
G3(t) = R1(t) R2(t) e j(e1(t) + e2(t))
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions32
Combinations
G3(t) = R1(t) R2(t) e j(e1(t) + e2(t))
The composite phase ise3(t) = e1(t) + e2(t)
The composite magnitude is
R3(t) = R1(t) R2(t)
R3(t)dB
= R1(t)dB
+ R2(t)dB
Composite phase is sum of individual phases.Composite magnitude, when expressed in dB, is sum of individual
magnitudes.
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions33
Example 1: G(s) = G0
1 + st1
1 + st2
–40 dB/decade
f
|| G ||
� G
� G|| G ||
0˚
–45˚
–90˚
–135˚
–180˚
–60 dB
0 dB
–20 dB
–40 dB
20 dB
40 dB
f1100 Hz
f22 kHz
G0 = 40 � 32 dB–20 dB/decade
0 dB
f1/1010 Hz
f2/10200 Hz
10f11 kHz
10f220 kHz
0˚
–45˚/decade
–90˚/decade
–45˚/decade
1 Hz 10 Hz 100 Hz 1 kHz 10 kHz 100 kHz
with G0 = 40 ! 32 dB, f1 = "1/2! = 100 Hz, f2 = "2/2! = 2 kHz
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions34
Example 2
|| A ||
� A
f1
f2
|| A0 ||dB +20 dB/dec
f1 /10
10f1 f2 /10
10f2
–45˚/dec+45˚/dec
0˚
|| A' ||dB
0˚
–90˚
Determine the transfer function A(s) corresponding to the followingasymptotes:
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions35
Example 2, continued
One solution:
A(s) = A0
1 + st1
1 + st2
Analytical expressions for asymptotes:For f < f1
A0
1 + st1➚
1 + st2➚
s = jt
= A011 = A0
For f1 < f < f2
A0
1➚ + st1
1 + st2➚
s = jt
= A0
st1 s = jt
1 = A0tt1
= A0ff1
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions36
Example 2, continued
For f > f2
A0
1➚ + st1
1➚ + st2
s = jt
= A0
st1 s = jt
st2 s = jt
= A0t2t1
= A0f2f1
So the high-frequency asymptote is
A' = A0f2f1
Another way to express A(s): use inverted poles and zeroes, andexpress A(s) directly in terms of A!
A(s) = A'
1 + t1s
1 + t2s
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions37
8.1.6 Quadratic pole response: resonance
+–
L
C Rv1(s)
+
v2(s)
–
Two-pole low-pass filter example
Example
G(s) = v2(s)v1(s)
= 11 + s L
R + s2LC
Second-order denominator, ofthe form
G(s) = 11 + a1s + a2s2
with a1 = L/R and a2 = LC
How should we construct the Bode diagram?
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions38
Approach 1: factor denominator
G(s) = 11 + a1s + a2s2
We might factor the denominator using the quadratic formula, thenconstruct Bode diagram as the combination of two real poles:
G(s) = 11 – s
s11 – s
s2
with s1 = – a12a2
1 – 1 – 4a2
a12
s2 = – a12a2
1 + 1 – 4a2
a12
• If 4a2 ! a12, then the roots s1 and s2 are real. We can construct Bode
diagram as the combination of two real poles.• If 4a2 > a1
2, then the roots are complex. In Section 8.1.1, theassumption was made that !0 is real; hence, the results of thatsection cannot be applied and we need to do some additional work.
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions12
G(j!) and || G(j!) ||
Im(G(jt))
Re(G(jt))
G(jt)
|| G(jt
) ||
�G(jt)
G( jt) = 11 + j t
t0
=1 – j t
t0
1 + tt0
2
G( jt) = Re (G( jt)) 2 + Im (G( jt)) 2
= 11 + t
t0
2
Let s = j!:
Magnitude is
Magnitude in dB:
G( jt)dB
= – 20 log10 1 + tt0
2 dB
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions39
Approach 2: Define a standard normalized formfor the quadratic case
G(s) = 11 + 2c s
t0+ s
t0
2 G(s) = 11 + s
Qt0+ s
t0
2or
• When the coefficients of s are real and positive, then the parameters !,"0, and Q are also real and positive
• The parameters !, "0, and Q are found by equating the coefficients of s• The parameter "0 is the angular corner frequency, and we can define f0
= "0/2#
• The parameter ! is called the damping factor. ! controls the shape of theexact curve in the vicinity of f = f0. The roots are complex when ! < 1.
• In the alternative form, the parameter Q is called the quality factor. Qalso controls the shape of the exact curve in the vicinity of f = f0. Theroots are complex when Q > 0.5.
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions40
The Q-factor
Q = 12c
In a second-order system, ! and Q are related according to
Q is a measure of the dissipation in the system. A more generaldefinition of Q, for sinusoidal excitation of a passive element or systemis
Q = 2/ (peak stored energy)(energy dissipated per cycle)
For a second-order passive system, the two equations above areequivalent. We will see that Q has a simple interpretation in the Bodediagrams of second-order transfer functions.
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions41
Analytical expressions for f0 and Q
G(s) = v2(s)v1(s)
= 11 + s L
R + s2LC
Two-pole low-pass filterexample: we found that
G(s) = 11 + s
Qt0+ s
t0
2Equate coefficients of likepowers of s with thestandard form
Result: f0 = t02/ = 1
2/ LCQ = R C
L
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions42
Magnitude asymptotes, quadratic form
G( jt) = 1
1 – tt0
2 2+ 1
Q2tt0
2
G(s) = 11 + s
Qt0+ s
t0
2In the form
let s = j! and find magnitude:
Asymptotes are
G A 1 for t << t0
G Aff0
– 2
for t >> t0
ff0
– 2
–40 dB/decade
ff00.1f0 10f0
0 dB
|| G(jt) ||dB
0 dB
–20 dB
–40 dB
–60 dB
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions43
Deviation of exact curve from magnitude asymptotes
G( jt) = 1
1 – tt0
2 2+ 1
Q2tt0
2
At ! = !0, the exact magnitude isG( jt0) = Q G( jt0) dB
= Q dBor, in dB:
The exact curve has magnitudeQ at f = f0. The deviation of theexact curve from theasymptotes is | Q |dB
|| G ||
f0
| Q |dB0 dB
–40 dB/decade
Fundamentals of Power Electronics! Chapter 8: Converter Transfer Functions!45!
Phase asymptotes!
f / f0
�G
0.1 1 10
Increasing Q
–180°
–90°
0°
284 Converter Transfer Functions
At high frequencies where (ω/ω0) > 1, the (ω/ω0)4 term
dominates the expression inside the radical of Eq. (8.62).Hence, the high-frequency asymptote is
(8.64)
This expression coincides with Eq. (8.5), with n = – 2.Therefore, the high-frequency asymptote has slope –40 dB/decade. The asymptotes intersect at f = f0, and are indepen-dent of Q.
The parameter Q affects the deviation of the actualcurve from the asymptotes, in the neighborhood of the cor-ner frequency f0. The exact magnitude at f = f0 is found by substitution of ω = ω0 into Eq. (8.62):
(8.65)
So the exact transfer function has magnitude Q at the corner frequency f0. In decibels, Eq. (8.65) is
(8.66)
So if, for example, Q = 2 ⇒ 6 dB, then the actual curve deviates from the asymptotes by 6 dB at the cor-ner frequency f = f0. Salient features of the magnitude Bode plot of the second-order transfer function aresummarized in Fig. 8.20.
The phase of G is
(8.67)
The phase tends to 0˚ at low frequency, and to –180˚ at high frequency. At f = f0, the phase is –90˚. Asillustrated in Fig. 8.21, increasing the value of Q causes a sharper phase change between the 0˚ and–180˚ asymptotes. We again need a midfrequency asymptote, to approximate the phase transition in the
|| G ||
f0
| Q |dB0 dB
–40 dB/decade
Fig. 8.20 Important features of the magni-tude Bode plot, for the two-pole transferfunction.
G →ff0
– 2for ω> ω0
G( jω0) = Q
G( jω0) dB = Q dB
∠G( jω) = – tan– 11Q
ωω0
1 – ωω0
2
Fig. 8.21 Phase plot, second-order poles.Increasing Q causes a sharper phase change.
f / f0
∠G
0.1 1 10
Increasing Q
–180°
–90°
0°• Low frequency asymptote of 0˚"• High frequency asymptote of
–180˚"• Change from 0˚ to –180˚
becomes more sharp as Q is increased"
Fundamentals of Power Electronics! Chapter 8: Converter Transfer Functions!46!
Mid-frequency phase asymptote!
f / f0
�G
0.1 1 10–180°
–90°
0°
f0
–90°
fb
fa0°
–180°
f / f0
�G
0.1 1 10–180°
–90°
0°
f0
–90°
fb
fa0°
–180°
8.1 Review of Bode Plots 285
vicinity of the corner frequency f0, as illustrated in Fig. 8.22. As in the case of the real single pole, wecould choose the slope of this asymptote to be identical to the slope of the actual curve at f = f0. It can beshown that this choice leads to the following asymptote break frequencies:
(8.68)
A better choice, which is consistent with the approximation (8.28) used for the real single pole, is
(8.69)
With this choice, the midfrequency asymptote has slope –180Q degrees per decade. The phase asymp-totes are summarized in Fig. 8.23. With Q = 0.5, the phase changes from 0˚ to –180˚ over a frequencyspan of approximately two decades, centered at the corner frequency f0. Increasing the Q causes this fre-quency span to decrease rapidly.
Second-order response magnitude and phase curves are plotted in Figs. 8.24 and 8.25.
fa = eπ/ 2 – 12Q f0
fb = eπ/ 212Q f0
fa = 10– 1/2Q f0fb = 101/2Q f0
Fig. 8.22 One choice for the midfrequencyphase asymptote of the two-pole response,which correctly predicts the actual slope atf = f0.
f / f0
∠G
0.1 1 10–180°
–90°
0°
f0
–90°
fb
fa0°
–180°
Fig. 8.23 A simpler choice for themidfrequency phase asymptote, whichbetter approximates the curve over theentire frequency range and is consistentwith the asymptote used for real poles.
f / f0
∠G
0.1 1 10–180°
–90°
0°
f0
–90°
fb
fa0°
–180°
8.1 Review of Bode Plots 285
vicinity of the corner frequency f0, as illustrated in Fig. 8.22. As in the case of the real single pole, wecould choose the slope of this asymptote to be identical to the slope of the actual curve at f = f0. It can beshown that this choice leads to the following asymptote break frequencies:
(8.68)
A better choice, which is consistent with the approximation (8.28) used for the real single pole, is
(8.69)
With this choice, the midfrequency asymptote has slope –180Q degrees per decade. The phase asymp-totes are summarized in Fig. 8.23. With Q = 0.5, the phase changes from 0˚ to –180˚ over a frequencyspan of approximately two decades, centered at the corner frequency f0. Increasing the Q causes this fre-quency span to decrease rapidly.
Second-order response magnitude and phase curves are plotted in Figs. 8.24 and 8.25.
fa = eπ/ 2 – 12Q f0
fb = eπ/ 212Q f0
fa = 10– 1/2Q f0fb = 101/2Q f0
Fig. 8.22 One choice for the midfrequencyphase asymptote of the two-pole response,which correctly predicts the actual slope atf = f0.
f / f0
∠G
0.1 1 10–180°
–90°
0°
f0
–90°
fb
fa0°
–180°
Fig. 8.23 A simpler choice for themidfrequency phase asymptote, whichbetter approximates the curve over theentire frequency range and is consistentwith the asymptote used for real poles.
f / f0
∠G
0.1 1 10–180°
–90°
0°
f0
–90°
fb
fa0°
–180°Match slope at f = f0:" Choose same approximation as in real pole case:"
or"
Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions44
Two-pole response: exact curves
Q = '
Q = 5
Q = 2
Q = 1Q = 0.7
Q = 0.5
Q = 0.2
Q = 0.1
-20dB
-10dB
0dB
10dB
10.3 0.5 2 30.7
f / f0
|| G ||dB
Q = 0.1
Q = 0.5Q = 0.7Q = 1Q = 2Q =5Q = 10Q = '
-180°
-135°
-90°
-45°
0°
0.1 1 10
f / f0
�G
Q = 0.2