8.1. review of bode plots - uic engineeringvahe/spring2013/ece412/bode-erickson.pdf · fundamentals...

Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions9

8.1. Review of Bode plots

Decibels

GdB

= 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

ZdB

= 20 log10

ZRbase

Decibels of quantities havingunits (impedance example):normalize before taking log

5Ω is equivalent to 14dB with respect to a base impedance of Rbase =1Ω, also known as 14dBΩ.60dBµA is a current 60dB greater than a base current of 1µA, or 1mA.


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

GdB

= 20 log10

ff0

n

= 20n log10

ff0

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


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

ω0

with ω0 = 1RC


G(jω) and || G(jω) ||

Im(G(jω))

Re(G(jω))

G(jω)

|| G(jω

) ||

∠G(jω)

G( jω) = 11 + j ω

ω0

=1 – j ω

ω0

1 + ωω0

2

G( jω) = Re (G( jω))2

+ Im (G( jω))2

= 1

1 + ωω0

2

Let s = jω:

Magnitude is

Magnitude in dB:

G( jω)dB

= – 20 log10 1 + ωω0

2dB


Asymptotic behavior: low frequency

G( jω) = 1

1 + ωω0

2

ωω0

<< 1

G( jω) ≈ 11

= 1

G( jω)dB≈ 0dB

ff0

– 1

–20dB/decade

ff00.1f0 10f0

0dB

–20dB

–40dB

–60dB

0dB

|| G(jω) ||dB

For small frequency,ω << ω0 and f << f0 :

Then || G(jω) ||becomes

Or, in dB,

This is the low-frequencyasymptote of || G(jω) ||


Asymptotic behavior: high frequency

G( jω) = 1

1 + ωω0

2

ff0

– 1

–20dB/decade

ff00.1f0 10f0

0dB

–20dB

–40dB

–60dB

0dB

|| G(jω) ||dB

For high frequency,ω >> ω0 and f >> f0 :

Then || G(jω) ||becomes

The high-frequency asymptote of || G(jω) || 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 .

ωω0

>> 1

1 + ωω0

2≈ ω

ω0

2

G( jω) ≈ 1ωω0

2=

ff0

– 1


Deviation of exact curve near f = f0

Evaluate exact magnitude:at f = f0:

G( jω0) = 1

1 +ω0

ω0

2= 1

2

G( jω0) dB= – 20 log10 1 +

ω0

ω0

2

≈ – 3 dB

at f = 0.5f0 and 2f0 :

Similar arguments show that the exact curve lies 1dB belowthe asymptotes.


Summary: magnitude

–20dB/decade

f

f0

0dB

–10dB

–20dB

–30dB

|| G(jω) ||dB

3dB1dB

0.5f0 1dB

2f0


Phase of G(jω)

G( jω) = 11 + j ω

ω0

=1 – j ω

ω0

1 + ωω0

2

∠G( jω) = tan– 1Im G( jω)

Re G( jω)

Im(G(jω))

Re(G(jω))

G(jω)||

G(jω) |

|

∠G(jω)∠G( jω) = – tan– 1 ω

ω0


-90˚

-75˚

-60˚

-45˚

-30˚

-15˚

0˚

f

0.01f0 0.1f0 f0 10f0 100f0

∠G(jω)

f0

-45˚

0˚ asymptote

–90˚ asymptote

Phase of G(jω)

∠G( jω) = – tan– 1 ωω0

ω ∠G(jω)

0 0˚

ω0–45˚

∞ –90˚


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– π / 2 ≈ f0 / 4.81fb = f0 eπ / 2 ≈ 4.81 f0


Phase asymptotes

fa = f0 e– π / 2 ≈ f0 / 4.81fb = f0 eπ / 2 ≈ 4.81 f0

-90˚

-75˚

-60˚

-45˚

-30˚

-15˚

0˚

f

0.01f0 0.1f0 f0 100f0

∠G(jω)

f0

-45˚

fa = f0 / 4.81

fb = 4.81 f0


Phase asymptotes: a simpler choice

-90˚

-75˚

-60˚

-45˚

-30˚

-15˚

0˚

f

0.01f0 0.1f0 f0 100f0

∠G(jω)

f0

-45˚

fa = f0 / 10

fb = 10 f0

fa = f0 / 10fb = 10 f0


Summary: Bode plot of real pole

0˚∠G(jω)

f0

-45˚

f0 / 10

10 f0

-90˚

5.7˚

5.7˚

-45˚/decade

–20dB/decade

f0

|| G(jω) ||dB 3dB1dB

0.5f0 1dB

2f0

0dBG(s) = 1

1 + sω0


8.1.2. Single zero response

G(s) = 1 + sω0

Normalized form:

G( jω) = 1 + ωω0

2

∠G( jω) = tan– 1 ωω0

Magnitude:

Use arguments similar to those used for the simple pole, to deriveasymptotes:

0dB at low frequency, ω << ω0

+20dB/decade slope at high frequency, ω >> ω0

Phase:

—with the exception of a missing minus sign, same as simple pole


Summary: Bode plot, real zero

0˚∠G(jω)

f045˚

f0 / 10

10 f0 +90˚5.7˚

5.7˚

+45˚/decade

+20dB/decade

f0

|| G(jω) ||dB3dB1dB

0.5f01dB

2f0

0dB

G(s) = 1 + sω0


8.1.3. Right half-plane zero

Normalized form:

G( jω) = 1 + ωω0

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 – sω0

∠G( jω) = – tan– 1 ωω0


+20dB/decade

f0

|| G(jω) ||dB3dB1dB

0.5f01dB

2f0

0dB

0˚∠G(jω)

f0

-45˚

f0 / 10

10 f0

-90˚

5.7˚

5.7˚

-45˚/decade

Summary: Bode plot, RHP zero

G(s) = 1 – sω0


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) = 1

1 +ω0s

G(s) =

sω0

1 + sω0


Asymptotes, inverted pole

0˚

∠G(jω)

f0

+45˚

f0 / 10

10 f0

+90˚5.7˚

5.7˚

-45˚/decade

0dB

+20dB/decade

f0

|| G(jω) ||dB

3dB

1dB

0.5f0

1dB2f0

G(s) = 1

1 +ω0s


Inverted zero

Normalized form, inverted zero:

An algebraically equivalent form:

Again, the inverted-zero format emphasizes the high-frequency gain.

G(s) = 1 +ω0s

G(s) =1 + s

ω0

sω0


Asymptotes, inverted zero

0˚

∠G(jω)

f0

–45˚

f0 / 10

10 f0

–90˚

5.7˚

5.7˚

+45˚/decade

–20dB/decade

f0

|| G(jω) ||dB

3dB

1dB

0.5f0

1dB

2f0

0dB

G(s) = 1 +ω0s


8.1.5. Combinations

Suppose that we have constructed the Bode diagrams of twocomplex-values functions of frequency, G1(ω) and G2(ω). It is desiredto construct the Bode diagram of the product, G3(ω) = G1(ω) G2(ω).

Express the complex-valued functions in polar form:

G1(ω) = R1(ω) e jθ1(ω)

G2(ω) = R2(ω) e jθ2(ω)

G3(ω) = R3(ω) e jθ3(ω)

The product G3(ω) can then be written

G3(ω) = G1(ω) G2(ω) = R1(ω) e jθ1(ω) R2(ω) e jθ2(ω)

G3(ω) = R1(ω) R2(ω) e j(θ1(ω) + θ2(ω))


Combinations

G3(ω) = R1(ω) R2(ω) e j(θ1(ω) + θ2(ω))

The composite phase isθ3(ω) = θ1(ω) + θ2(ω)

The composite magnitude is

R3(ω) = R1(ω) R2(ω)

R3(ω)dB

= R1(ω)dB

+ R2(ω)dB

Composite phase is sum of individual phases.

Composite magnitude, when expressed in dB, is sum of individualmagnitudes.


Example 1: G(s) =G0

1 + sω1

1 + sω2

–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

f1

100 Hz

f2

2 kHz

G0 = 40 ⇒ 32 dB–20 dB/decade

0 dB

f1/1010 Hz

f2/10200 Hz

10f1

1 kHz

10f2

20 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 = ω1/2π = 100 Hz, f2 = ω2/2π = 2 kHz


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:


Example 2, continued

One solution:

A(s) = A0

1 + sω1

1 + sω2

Analytical expressions for asymptotes:For f < f1

A0

1 + sω1

1 + sω2

s = jω

= A011

= A0

For f1 < f < f2

A0

1 + sω1

1 + sω2

s = jω

= A0

sω1 s = jω

1= A0

ωω1

= A0ff1

8.1. review of bode plots - uic engineeringvahe/spring2013/ece412/bode-erickson.pdf · fundamentals...

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