side reader for r.h.small “direct-radiator loudspeaker ... · what are thiele/small parameters?...

©2013 Katsuyuki Tsubohara, All rights reserved.

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Last update: 13 May 2013

Side reader for R.H.Small “Direct-Radiator Loudspeaker System Analysis”

© 2013 Katsuyuki Tsubohara, All rights reserved.

Do not re-distribute anything in this document.

Disclaimer

This article may include miscalculations and/or wrong theories. The author accept no responsibility whatsoever for any direct or indirect damages, loss, prejudice or emotional distress caused by use of this document.

Contents

� Introduction � What are Thiele/Small parameters? � Equivalent circuit of loudspeaker driver � Step-by-step derivations of the equations shown in the paper � Further readings


2 / 26

INTRODUCTION

This article is a side reader for the classic paper by R.H.Small “Direct-Radiator Loudspeaker System Analysis”. You

should purchase the original paper at first. It’s available on AES website. Don’t illegal download –I hope you support AES.

The methods and the technical terms described are de-facto standard now and help to design woofers, to read technical

documents and to access updated knowledge.

Supposed readers are around 2 year experienced driver engineers. If you are at a level as trainee, stop reading this and

assemble some drivers with parts available in your lab. It grows your practical skill far more. Because: Any engineer

without practical experience is useless even if he understands basic theories. It’s easy to obtain required LF performance by

following 4 equations. They are enough to calculate linear domain response of woofers and don’t need to understand the

background in usual cases.

MS

MS

MSMS

S M

K

CMf

ππ 2

1

2

1 == MS

MS

MS

TS C

M

DCR

lBR

Q ⋅+

= 22

1

[ ] [ ].25222 1007.4407 3 ltrDMSmDMSDoAS SCSCSCcV

MS××=×≅≡ ρ (MKSA units)

[ ]%

2

2224

2

222

1045.52 MSE

D

MSE

Doo

MR

SlB

MR

SlB

c××≅⋅= −

πρη

In addition, the analysis has an important assumption called “piston range”:

Wave length is enough longer than the diameter of the diaphragm

Thus the basic theories do not contribute to obtain good performance in mid and high frequency range. It’s clearly stated in

the paper, at “ASSUMPTION AND APPROXIMATIONS” paragraph.

In contrast to that the low-frequency response can be simply simulated by a few parameters, outside piston range, the

radiation phenomena is complex and the effects by the shape of diaphragm and the baffle are not ignorable. Also

diaphragms have many resonant modes in high frequencies also contribute to the performance in the range. It’s almost

impossible to be solved by simple calculation. Through the history of loudspeakers, many attempts using parametric method

to analyze behaviors of diaphragms have been done but have never been successful like methods by Thiele and Small.

I.E., a posteriori method, experimental guess is required to design Mid-Woofers, Full-ranges and Tweeters. In this era,

FEA simulation is available but importance of practical experiences has not been reduced at all. FEA calculates the response

but does not show how to improve the response. FEA is useful to save time by reducing number of times engineers create

actual samples, but to use it effectively, the guess by human is still required. That’s what makes loudspeaker design so

interesting.

In conclusion young engineers should experience cut-and-try approach. Basic theories are to be studied later. The author

learned it in the first year as a driver designer, but forgot in the next year and studied again 3 years later from scratch.


3 / 26

Listed below are motivations to learn from the Small’s:

� Enforce your skill with logical thinking

� Give clear technical information, especially in OEM scenes

� Understand de-facto standard technical terms and access updated knowledge

� Understand distortion phenomena by relating it to nonlinearities of the parameters (it’s only way to

understand how to reduce distortions of woofer drivers)

� Design enclosures (in this case, you should read all papers by Thiele and Small)

� Learn how to summarize technical report –the paper has only 13 pages but covers almost everything about

woofer system design.

� Learn how to sophisticate design approach (some kinds of definitions for LF parameters have been

established through history. The set defined by Thiele and Small has gotten the de-facto standard position by

its efficiency in design process)

You want to know the background? Congratulations, you have done 1st step to be a true pro.

The author wishes this article help the world filled with good music and nice sound.

What are Thiele/Small parameters?

The linear-domain response of a woofer driver is determined by 7 parameters, related to the specifications of the parts

(driver parameters):

MSM , MSR , MSC , Bl , eR , DS , ��

The translated parameters in useful definitions for enclosure designers (system parameters):

SF , MSQ , ESQ , ASV , oη , ( )ωjVCZ , ��

Above 2 sets have one-on-one relationship. So, when an enclosure designer shows a driver designer his requirements by the

system parameters, the driver designer can design a driver matches the requirements systematically. To avoid system

engineers order theoretically impossible drivers to driver designers, an equation for checking system parameters’ validity is

shown:

ASES

So V

Q

f

c⋅⋅=

3

3

24πη


4 / 26

Equivalent circuit of loudspeaker driver

The paper starts with an acoustic equivalent circuit of loudspeaker system. If you haven’t majored acoustics or electronics,

you may be fed up. In fact, the problem to be solved is just a simple forced oscillation and the motion equation is familiar to

people who learned dynamics. “Equivalent Circuit” sounds difficult but it’s just a rewritten motion equation.

Why do we have to use the equivalent circuit? It’s that there are far more electronics engineers than acoustics’ and we

should use their accomplishments like Spice.

Necessary basics are shown below.

If you don’t familiar about these, go to a bookstore. There are so many plain textbooks for electronics.

Precalculus of exponent function:

tjtj ejedt

d ωω ω= , .1

constej

dte tjtj +=∫ωω

ω

Forget about the integral constant by assuming initial condition zero. It just complexes.

Ohm’s law expanded to AC circuits:

( ) ( ) ( )tjt IZE ω=

Series impedance calculation:

K+++= 321 ZZZZ

Parallel impedance calculation:

K+++=321

1111

ZZZZ

Impedances of inductor (L), capacitor (C) and resistor (R):

( ) ( )

( ) ( )

( ) ( )

=

=

=

tRtR

tCtC

tLtL

RIE

ICj

E

LIjE

ω

ω1

i.e.

( )

( )

( )

=

=

=

RZ

CjZ

LjZ

jR

jC

jL

ω

ω

ω

ω

ω1

Let’s start. The definitions for the symbols are shown in Small’s. It’s skipped here (this is a side reader). Model of a

loudspeaker driver mounted on an infinite-baffle is shown in Fig.1 and Fig.2

Fig.1 Mechanical model Fig.2 Model of motor and amplifier


5 / 26

Note) Radiation resistance ARℜ is small in direct-radiator systems and ignorable to analyze the movement of the

diaphragm.

The motion equation for Fig.1 is

( ) ( ) ( ) ( )ttDMStDMS

tDMS Blixdt

dRx

Cx

dt

dM +−−= 1

2

2

(I)

The solved Maxwell’s equation for Fig.2 is

( ) ( ) ( ) ( )tgEtDtg iRRxdt

dBle +=− (II)

Second member in left part is counter EMF which disturbs amplifier give energy into driver.

From Eq.(II)

( )( )

( )tDgEgE

tgt x

dt

d

RR

Bl

RR

ei

+−

+= (III)

By substituting Eq.(III) to (I)

( ) ( ) ( )( )

( )

+−

++−−= tD

gEgE

tgtDMStD

MStDMS x

dt

d

RR

Bl

RR

eBlx

dt

dRx

Cx

dt

dM

12

2

(IV)

Simplified Eq.(IV) is

( ) ( ) ( )( )

gE

tgtD

MStD

gEMStDMS RR

Blex

Cx

dt

d

RR

lBRx

dt

dM

+=+

+++ 122

2

2

(V)

By assuming the amplifier gives effective voltage ge and angular velocity ω

( ) ( )tj

jgtg eee ωω2= (VI)

It’s a forced oscillation problem and the solution for Dx is sinusoidal with angular velocity ω

If it’s strange, go to the bookstore again and buy a textbook for dynamics.

For physicians, familiar subject is displacement Dx but to deliver the equivalent circuit, velocity Du is to be used. Why?

–you will find later. Forget about the question. Dx is sinusoidal so Du is sinusoidal too,

( ) ( )tj

jDtD euu ωω2= (VII)

Note) The symbols in the analysis indicate effective values (RMS)

From (VII),

( ) ( ) ( )tDtDtD uj

dtux ∫ ==ω1

(VIII)

( ) ( ) ( )tDtDtD ujudt

dx

dt

d ω==2

2

(IX)

By substituting Eqs. (VII), (VIII) and (IX) into Eq.(V),


6 / 26

( ) ( ) ( ) ( )tggE

tDMS

tDgE

MStDMS eRR

Blu

Cju

RR

lBRuMj

+=+

+++

ωω 122

(X)

In the paper, “acoustic equivalent circuit” is used and “mechanical equivalent circuit” doesn’t appear. Nowadays, the

mechanical one is widely in use because for driver engineers, it’s easier to understand. So here express the mechanical

equivalent circuit first and then deform it to the acoustic one.

A series LCR circuit is shown in Fig.3. Don’t think why. You’ll find later.

Fig.3 Series LCR circuit

By assuming AC source voltage of angular velocity ω. Impedance of series LCR is

(XI)

By applying Ohm’s law

( ) ( ) ( ) ( ) ( )tttttotalt ICj

RILIjIZEω

ω 1++== (XII)

By comparison between Eq.(X) and Eq.(XII), you’ll find relationship of

( ) ( )tDt uI ⇔ , ( ) ( )tggE

t eRR

BlE

+⇔ , MSML ⇔ ,

++⇔

gEMS RR

lBRR

22

, MSCC ⇔

So they are same shape equations. Now mechanical equivalent circuit is delivered

Fig.4 Mechanical equivalent circuit of driver

Through analyzing the relationship between � and �, velocity of diaphragm by input voltage is calculated.

Next subject is the acoustic equivalent circuit.

In acoustics, the volume velocity of air driven by the diaphragm is more important than the velocity of diaphragm �.

Equation should be deformed using relationship:

( )( )

D

tDtD S

Uu = (XIII)

CjRLjZZZZ CRLtotal ω

ω 1++=++=


7 / 26

By substituting this into Eq. (X)

( ) ( ) ( ) ( ) ( )tggE

tDDMS

tDDgED

MStD

D

MS eRR

BlU

SCjU

SRR

lB

S

RU

S

Mj

+=+

+++

ωω 122

(XIV)

To understand acoustic phenomena, “pressure” is better than “force”.

To relate right part to pressure, dividing both parts by DS then

( ) ( ) ( ) ( ) ( ) ( )tgDgE

tD

DMS

tD

DgED

MStD

D

MS eSRR

BlU

SCjU

SRR

lB

S

RU

S

Mj

+=+

+++ 22

22

22

1

ωω (XV)

Defining acoustic parameters as following

2DMSAS SRR ≡ ,

2DMSAS SMM ≡ ,

2DMSAS SCC ≡

Then

( ) ( ) ( ) ( ) ( ) ( )tgDgE

tDAS

tD

DgE

AStDAS eSRR

BlU

CjU

SRR

lBRUMj

+=+

+++

ωω 1

2

22

(XVI)

Replacing the electronic symbols to the acoustic symbols as following then the acoustic equivalent circuit is delivered.

( ) ( )tDt UI ⇔ , ( ) ( ) ( )tgDgE

t eSRR

BlE

+⇔ , ASML ⇔ , ( )

++⇔ 2

22

DgE

ASSRR

lBRR , ASCC ⇔

Fig.5 Acoustic equivalent circuit of driver

Note) Descripting L, C and R by derivation and integral as shown below, it’s able to obtain the equivalent circuit

without using complex analytics. But without understanding complex analytics, it’s difficult to understand the

paper.

( ) ( )tLtL Idt

dLE = , ( ) ( )∫= dtI

CE tCtC

1, ( ) ( )tRtR RIE =

Now you’re at the start-line to read the classic.

R U ready?


8 / 26

Step-by-step derivations of the equations shown in the paper

Because the paper is an academic article not a textbook, detailed deformations for the equations are skipped. For most of

us, it’s usual to take hours to obtain only one equation by ourselves. To “battle” against equations is meaningful to be a

physician or a mathematician. Great Scott we audio engineers in 21st century are always pressed for time and it’s too heavy.

But it’s certain that the best way to understand background of a technical paper is to obtain the equations by ourselves. So

here given are step-by-step deformations.

This article is just a typing of author’s hand calculations. The deformations are not optimized completely and you’ll find

his bad habitual manners. The author wishes it to be corrected by somebody smarter.

Notes)

� Numbers of equations are obverse to Small’s.

� Refer to the paper when figure with number appears in texts.

� Refer to the paper for the definitions of symbols.

� By the author’s cup of tea, � = �� is always assumed.

� By the author’s glass of beer, ≡ is used in definition equations.

� Practical equations are indicated by (‡)

� If you are a driver engineer and the mechanical parameters are familiar than the acoustic ones, keeping following 3

equations in mind is suggested. �� would be the only parameter with letter AS used in daily work.

2

2

2

DASMS

DASMS

DASMS

SRR

SMM

SCC

=

=

=

i.e. 2

2

2

DMSAS

DMSAS

DMSAS

SRR

SMM

SCC

≡

≡

≡

(†)

� Some equations are shown in both parameters of acoustic and mechanical using Eqs. (†).

� For the sake of clarifying infinite-baffle cases and in-enclosure cases,

�� and �� are replaced by ��, ��,�� and ��.

First subject is to simulate frequency response of bass-reflex or passive-radiator systems. See Fig.1.

Symbols U are obverse to volume velocity of air. Obviously,

LPDo UUUU ++= (1)

Note) This works as Kirchhoff’s 1st theorem. In case systems have equivalent characteristics of Kirchhoff’s theorems

and Ohm’s Law, it can be written by equivalent circuit. It’s able to descript these kinds of systems simpler using

matrix calculations, but it’s not the subject of this side reader.

L+++= kjin iiii , L+++= rqpn eeee , nnn ize =

Fig.2 is an equivalent circuit for enclosed systems. It seems complex but the analysis starts from simpler case

(Infinite-Baffle) and you’ll understand later. Don’t worry.

By basic theories of acoustics, the radiated sound power is


9 / 26

ARoA UP ℜ= 2 (2)

Note) This equation has the same shape of electric power’s RIP2=

In piston range ka ≪ 1

( )coAR πωρ 22≅ℜ (3)

Note) By Eq.(2) and Eq.(3), flat F-resp. equals to constant acceleration in piston range. That’s why all direct radiator

systems have flat energy response in the range.

Note) It’s so time-consuming to obtain Eq.(3). Shown below is the rigorous calculation taken from L.L.Beranek

“Acoustics” (McGraw-Hill, New York, 1954).

Radiation Impedance for One Side of a Plane Piston in Infinite Baffle

a : Radius of Piston

k: wave number ( λπ2≡k )

( )( )

( )

( )

kaW

where

WWWK

WWWWJ

Kk

cj

ka

JcajXZ

W

W

kaoka

oMMM

2

753533

2

8642642422

21

22

7

2

53

1

222

7

22

5

2

3

1

212

212

≡

⋅⋅+

⋅−=

⋅⋅⋅−

⋅⋅+

⋅−=

+

−=+ℜ=

K

L

π

πρρπ

To be a true tweeter pro, at least the knowledge how above behaves is required. Without the knowledge,

you’ll never find why surface-driven plane drivers have HF cutoff. Google it, you’ll find some graphical

plots of radiation impedance.

The volume of air pushed out from enclosure equals to the volume vacuumed out from enclosure, thus

Bo UU −= (4)

Obviously, when an enclosure has separated chambers

,321 LBBBo UUUU ++= (5)

Defining input power by “Consumed power at the resistor when the driver in the system is replaced by a resistor having the

same resistance equals to the DCR of voice coil”,

( ) EEg

gE

gE

gg

gE

E

EE R

RR

eR

RR

ee

RR

R

RP

2

2

22

1

+=⋅

+=

⋅

+≡ (6)

Note) Today, input power is usually defined by nominal impedance. It gives approx. 10% to 20% less value.

By defining efficiency by ratio of radiated acoustic power and above input power,


10 / 26

( )Eg

EgARo

EEg

g

ARo

E

A

Re

RRU

RRR

e

U

P

P2

22

2

2 +⋅ℜ=

+

ℜ=≡η (7)

Note) Modern amplifiers have very small output impedance. If cable is short enough, $ is ignorable.

To start from simple case, infinite-baffle condition is now assumed in Fig.2 and

No air leakage,

0≅LU i.e. ∞≅ALR

No port or passive radiator

0=PU i.e. ∞=APR

Infinite volume of chamber

∞≅ABC

No acoustic damping

0≅ABR

To simplify, counter EMF and mechanical loss are summed

( ) 2

22

DEg

ASATSRR

lBRR

++≡ (8)

Then Fig.3 is delivered. Calculating %&,

Note) The form of equation would be strange for engineers who have not studied about active filters. To separate

efficiency and shape of frequency response curve, following is required (normalization).

12

)(lim =∞→

ωω

jIBG i.e. ( ) 12

≅⇒>> ωωω jIBs G

( )

( )

( ) ( )

( )( )

( )

( )( )

( )

( ) )(

2

2

2

2

2

1

1

1

11

ω

ω

ω

ωωω

ω

ωωω

ω

ωωω

ωω

jIBASDEg

g

ASASASAT

ASAS

ASDEg

g

ASASASAT

AS

DEg

g

ASASASAT

AS

DEg

g

ASAS

ATDEg

g

jo

GMSRRj

Ble

CMjCRj

MCj

MSRRj

Ble

CMjCRj

Cj

SRRj

Ble

CMjCRj

Cj

SRR

Ble

MjCj

RSRR

Ble

U

⋅+

≡

++⋅

+=

++⋅

+=

++⋅

+=

++⋅

+=

(9)(10)

From Eq.(9)


11 / 26

( ) ( )( ) ( ) ( )

( ) ( )2

2222

222

2

ω

ωω

ω

ωω

jIB

ASDEg

g

jIBjIBASDEg

g

ASDEg

g

oo

o

GMSRR

lBe

GGMSRRj

Ble

MSRRj

Ble

UU

U

⋅+

=

⋅+−

⋅+

=

=

∗

∗

From Eqs. (2), (3) and (7)

( )

( ) ( )( )

( )2

22

22

2

222

2222

222

2

2

ω

ω

ω

πρ

πωρ

ω

η

jIB

ASDE

o

Eg

EgojIB

ASDEg

g

j

GMSR

lB

c

Re

RR

cG

MSRR

lBe

⋅⋅=

+⋅⋅⋅

+=

(11)

Note) For driver engineers, calculations by the mechanical parameters are shown below. It is enough to calculate SPL

curve of a driver on an infinite-baffle in piston range.

'& =(&2*+

∙-./.

$01�.2��

. =(&2*+

∙-./.1�

.

$023�.

( ) ( ) ( )

( )( )

( )( ) ( )

( )( ) ( ) ( )( )

( ) ( )( )

( ) 22222

224

22

224

22

224

2

2

2

2

2

1

11

11

11

ATASASAS

ASAS

ATASASASATASASAS

ASAS

ATASASASATASASAS

ASAS

ATASASAS

ASAS

ATASASAS

ASAS

jIBjIBjIB

RCMC

MC

RCjMCRCjMC

MC

RCjMCjRCjMCj

MC

RCjMCj

MCj

RCjMCj

MCj

GGG

ωωω

ωωωωω

ωωωωω

ωωω

ωωω

ωωω

+−=

−+−+−=

+−+−++=

+−+−−⋅

++=

= ∗

Thus


12 / 26

( )

( )

( )

( )

21

222

2222

224

21

2

2

22

2422

2

222

4

2424

21

22222

224

1

1

1

+++−

=

+++

−

⋅=

+−=

EgMSMSMSMS

MSMS

DEgD

MSDMS

D

MSDMS

D

MSDMS

ATASASAS

ASAS

jIB

RR

lBRCMC

MC

SRR

lB

S

RSC

S

MSC

S

MSC

RCMC

MC

G

ωω

ω

ωω

ω

ωωω

ω

To calculate impedance curve, deforming Fig.3 by basic theories for electric circuit are shown below.

By separating $�4 to $�� and counter EMF,

By applying dual transformation

By applying Thieving’s theorem to the current source and the left resistor

�-/5$ 6 $071�

-./.

5$ 6 $071�. $�� 8�� 2��

5$ 6 $071��-/

5$ 6 $071�

.

-./.

2�� 8�� 1

$��


13 / 26

By replacing the voltage source using transformer

Now the source voltage is � , the output of amplifier. Focused on the electric impedance, it’s equivalent to

Defining electric parameters as following, Fig.4 is delivered

830� ≡�9

:

�:;:2��, <=0� ≡

�:;:

�9: 8��, $0� ≡

�:;:

�9: ∙

>

?@A

Note) Using the mechanical parameters,

830� 3BA

�:;:, <=0� -./.83�, $0�

�:;:

?BA

Because of the ignored radiation impedance, 1� does not contribute to the motion of diaphragm thus does not

appears in the electric impedance.

By defining C� as resonant frequency of Fig.4 then

� ∙1�

-/

1�.

-./.∙ $

1�.

-./.∙ $0

8�� 2��

1

$��

�

1:1�

-/

1�.

-./.∙ $

1�.

-./.∙ $0

2�� 8��

1

$��

�

$ $0

1�.

-./.2��

-./.

1�. 8��

-./.

1�. ∙

1

$��


14 / 26

( )

MSMS

D

MSDMS

ASAS

D

ASDAS

CESMES

sss

MC

S

MSC

MC

S

lBC

lB

SM

LC

FT

=

⋅=

=

⋅=

===

2

2

2

22

22

2

222 211 πω

(12)

The Q factors are

2

22

22

2

2

22

22

21

DAS

DAS

ASAS

ASAS

DAS

DAS

ASAS

ESMESS

MS

SR

lB

lB

SM

MC

MC

SR

lB

lB

SM

MC

RC

Q

⋅⋅=

⋅⋅=

≡ ω

ASASS

ASASASAS

RC

RCMC

ω1

1

=

⋅=

MSMSS

D

MSDMSS

RC

S

RSC

ω

ω

1

1

22

=

⋅=

(13)

EMS

S

EDAS

S

EMESS

ES

RlB

M

RlB

SM

RC

Q

⋅⋅=

⋅⋅=

≡

22

22

2

ω

ω

ω

(14)

For driver designers, forms without ω� works better because they are directly related to the specifications of the parts

MS

MSEES

MSMS

MSMS

ClB

MRQ

CR

MQ

22=

=

(‡)


15 / 26

Note) Q factor is ratio of energy stored and dissipated. To get physical image, considering via the mechanical equivalent

circuit would be helpful. Shown below is the mechanical acoustic circuit ignoring output impedance of amplifier;

$.

Q factor in series LCR circuit is defined by (inverse of parallel circuit case):

CRR

CLQ

oω11

=≡−

E0� and E3� are delivered directly through above.

Note that counter EMF acts like resistance but no energy is dissipated. It just disturbs energy input into driver.

Too big Magnet reduces the force around C�.

The definition of �� is

�� ≡ (&+.8�� (15)

If “Volume of air having same acoustic compliance…” sounds too academic, “Volume of closed enclosure which

gives 41% higher resonance” would be nice.

From the electric equivalent circuit,

( )( ) ( )ESCES

CESESESMESCES

ESCESMESjRLC RLj

LjRRCLj

RLjCjZ

ESCESMES ωωω

ωωω

++=++=−2

1 11

( )

( )

( ) 12

2

+⋅+

⋅⋅+=

++⋅+=

ES

CESMESCES

ES

CES

ESE

CESESESMESCES

CESESE

jVC

RL

jCLj

R

Lj

RR

LjRRCLj

LjRR

Z

ωω

ω

ωωω

ω

�-/

$0

-./.

$0

$3� 83� 23�


16 / 26

( )

( )

( )

( )

( )

+++=

+⋅+

⋅⋅+=

+⋅+

⋅⋅+=

+⋅+

⋅⋅+=

+⋅+

⋅⋅+=

1

11

1

1

1

1

22

22

22

22

2

MSSs

MSSESE

MESESSSs

MESESSS

ESE

MESES

MESCESSs

MESES

MESCESS

ESE

MESES

CESSs

MESES

CESS

ESE

MESES

CESCESMESMESCES

MESES

CESCESMES

ESE

QTjTj

QTjRR

CRTjTj

CRTj

RR

CR

CLTjTj

CR

CLTj

RR

CR

LTjTj

CR

LTj

RR

CR

LLCjCLj

CR

LLCj

RR

ωωω

ωωω

ωω

ωω

ω

ωω

ω

ωω

ω

(16)

Today, measurement tools using laser are used in loudspeaker manufactures. They can measure displacement of diaphragm

directly and the method described in “Measurement of Driver Parameters” is not widely in use now. Modern amplifiers

have ignorable output impedance and “Measurement of Amplifier Source Impedance” has lost its importance.

So, let’s skip.

E0� definition is “the Q $0 acting alone i.e. with $ = 0”. In condition cable resistance and output impedance of

amplifier are not ignorable, it’s obvious $0 should be replaced by $0 6 $ by according to the electric equivalent circuit.

The compensated Q, E0 is defined by

( )

E

EGES

EgMESS

E

R

RRQ

RRC

Q

+⋅=

+≡ ω

(21)

In the same way when mechanical loss is added by absorbing materials,

ATASS

T RCQ

ω1≡

(27)

Note) The most familiar Q factor for driver engineers, E4� does not appear in the paper. It’s because of “System

Analysis”.


17 / 26

ESMS

ESMS

AS

AS

ASAS

ASE

AS

ASE

AS

AS

AS

AS

AS

ASEAS

AS

ASEAS

AS

DE

AS

AS

AS

DE

AS

TS

QQ

QQ

C

M

RC

M

lB

R

C

M

lB

R

C

M

R

C

M

lBRRC

M

lBRRC

M

SR

lBR

C

M

SR

lBR

Q

+=

+

⋅=

⋅+

=

+≡

1

1

1

1

22

22

22

22

2

22

2

22

(‡)

See Fig.6. It’s a simplified acoustic equivalent circuit for “system” described in Fig.2.

Note) In this era, Many Spice simulators are available for free and we can simulate any complex system. Fig.6 works as

basic in the approach.

Obviously,

Eg

gg RR

Blep

+=

(23)

( )AS

ASATjAS CjMjRZ

ωωω

1++= (24)

ASABAB Cj

RZω

1+= (25)

AAABAS

g

D

ZZZ

p

U

+=

Thus


18 / 26

( )

( ) gjsystemAS

g

AAABASABAS

g

AA

AA

AAABAAASABAS

AAAB

AAAB

AA

g

AAAB

AAABAS

AAAB

AA

AAABAS

g

AAAB

AA

DAAAB

AA

o

pGMj

pZZZZZ

pZ

Z

ZZZZZZ

ZZ

ZZ

Z

p

ZZ

ZZZZZ

Z

ZZZ

p

ZZ

Z

UZZ

Z

U

⋅⋅=

⋅++

=

⋅⋅++

+⋅+

=

⋅

++

⋅+

=

+⋅

+=

⋅+

=

−

−

ωω1

1

1

1

1

( )

( )AAABASABAS

AS

g

oAS

jsystem

ZZZZZ

Mj

p

UMj

G

++=

≡

ω

ω

ω

(26)

Note) To understand intuitively why �� has the shape of ω2��GH

IJ , discussing about Fig.2 would be helpful.

Ignoring the leakage (when enclosure has leakage, it’s defect) and assuming ω ≫ ��,

>>

>>

CjMj

RMj

ωω

ω

1

i.e.

≅

≅

01

0

Cj

R

ω

Thus the only impedance to consider in the case ω ≫ �� is jω2�� so

( )AS

gjo Mj

PU S

ωωω

ω → >>

The air in an enclosure is compressing and expanding, not moving. So M�� does not include any M. So

�� is always normalized. Describing mathematically,


19 / 26

( )

( )

( )

1

0101

1

1

1

1

1

lim

lim

lim

lim

=

⋅++=

⋅++=

++=

=

−

−

∞→

−

∞→

−

∞→

∞→

ASAP

AS

AP

AS

AB

AS

AS

AS

AB

AS

AS

AS

APABASABAS

jsystem

jsystem

MM

Mj

ZMj

Z

Mj

Z

Mj

Z

Mj

Z

Mj

ZZZZZ

G

G

ω

ωωωω

ω

ω

ω

ωω

ωω

i.e. ( ) 1≅⇒>> ωωω jsystems G

Note) When leakage is not ignorable

( )ALAB

ALjsystem RR

RG

+=

∞→ ωωlim

This shows leakage causes low SPL in piston range. By understanding radiation phenomena outside piston range,

you’ll find SPL does not decrease by leakage in case wave length is smaller than diameter of diaphragm and

enclosure is large enough. This knowledge is useful to find out the cause of low SPL defect. The most important

thing to analyze defects is accumulated past trouble records of course, but sometimes backing to basic theories

solves the problem faster.

Next subject is to deliver frequency response of driver on infinite-baffle by method of filter design.

Eqs.(10), (12) and (27) are shown here again

The author thinks it’s waste of time academic textbooks always require flipping pages again and again.

( )( )

( ) ASASASAT

ASASjIB

CMjCRj

MCjG

2

2

1 ωωω

ω ++≡

(10)

ASAS

S

S MCT ==2

2 1

ω (12)

ATASST RC

Qω

1≡ (27)

In summary,


20 / 26

( )

( )( )

( )( ) 1

1

22

22

2

2

++=

++=

TSS

S

ASASSS

ASAT

ASAS

jIB

QTjTj

Tj

CMjCR

j

MCj

G

ωωω

ωωω

ω

ωω

(28)

Eq.(28) is 2nd-order HPF function widely known by electronics or DSP engineers as

( )( )

( ) ( ) 1122

22

++=

OO

OjIB

TjaTj

TjG

ωωω

ω

(29)

Much knowledge about designing 2nd-order HPF has been obtained by electronics engineers, that’s why we use “equivalent

circuit” to design loudspeakers.

By Eq.(11), the efficiency of the driver where ω ≫ �� is

22

22

2

2

2

222

22

22

2

2

2

MSDE

o

D

MSDE

Do

ASDE

o

o

MSR

lB

c

S

MSR

SlB

c

MSR

lB

c

⋅=

⋅=

⋅=

πρ

πρ

πρ

η

(30)(31)

From Eqs.(12) ,(14), (15), (30), the efficiency calculation by the system parameters is

22

22

2

22

22

2

1

2

1

ASDEAS

AS

ASDE

o

o

MSR

lB

cCc

V

MSR

lB

c

⋅⋅=

⋅⋅=

π

πρ

η

ASASES

ASS

ASDEAS

AS

ES

DASES

ASDEAS

AS

MCQc

Vf

MSRCc

V

Q

SMRf

MSRCc

VlB

3

223

2

223

22

2

2

2

=

⋅=

⋅=

ππ

π


21 / 26

ASES

S VQ

f

c⋅⋅=

3

3

24π

(32)

Substituting velocity of sound; 345m/s, where �� is expressed in litters

ASES

So V

Q

f ⋅⋅×= −3

10106.9η (33)

Where �� is expressed in NOP

ASES

So V

Q

f⋅⋅×= −

310107.2η

(34)

Note) Enclosure designers MUST check performance requests for driver designers via above.

Scientifically impossible requests do not work at all.

Discussion about linear domain is finished now. Next subject is large signal performances; nonlinearity and power

handling capacity. The description given by Thiele and Small is the first quantitative approach and updated knowledge is

based on it. One of the examples is “Loudspeaker Nonlinearities –Causes and Symptoms” by Wolfgang Klippel (available

on Klippel GmbH website).

First, assume driver has displacement limit: ��

Note) How to determine �� is not shown in the paper and is still subject to debate (refer to the Klippel’s).

Small suggested to consider:

1) Damage to suspensions

2) FM distortion

3) Harmonic distortion or AM distortion

Using Google, many sites will be found explaining it as the length of Voice Coil out of Gap of motor, but it’s out

of date. Suspensions give more distortion in many cases. Klippel’s approach is performance-based (10%THD or

IMD) and works nicely for subwoofers, but because it uses near-field measurement, IMD in midrange or higher

caused by surround resonance or <�� is not taken account. Further discussions are still required.

Defining system parameter �� as moving air volume at �� thus

maxxSV DD = (35)

The velocity of the diaphragm by assumption that the driver is linear and works precisely as the equivalent circuit shown in

Fig.6 is


22 / 26

gAAASABABAS

AAAB

gABAAABASAAAS

ABAA

g

ABAA

ABAAAS

ABAAAS

g

D

PZZZZZ

ZZ

PZZZZZZ

ZZ

P

ZZ

ZZZ

ZZZ

P

U

⋅++

+=

⋅++

+=

⋅

++

=

+=

1

1

%� is volume velocity of air i.e. velocity of the diaphragm is

D

DD S

Uu =

Displacement of diaphragm �� is calculated by the integral (note that it’s effective value),

( )

( )

( )

AAASABABAS

AAAB

ED

E

AAASABABAS

AAAB

EDEG

Eg

AAASABABAS

AAAB

DEG

g

D

D

jD

ZZZZZ

ZZ

RSj

BlP

ZZZZZ

ZZ

RSj

Bl

RR

Re

ZZZZZ

ZZ

SRR

Ble

Sj

uj

x

+++⋅⋅=

+++⋅⋅

+=

+++⋅

+⋅=

=

1

1

11

1

22

1

2

ω

ω

ω

ω

ω

( )

( ) ( )ω

ω

σ

ωω

ω

jsystemxPxE

jsystemx

E

MSE

AAASABABAS

AAAB

ASAS

ED

E

XkP

XkR

BlCP

ZZZZZ

ZZ

CjCj

RSj

BlP

2

1

2

1

22

1 11

≡

⋅⋅≡

+++⋅⋅⋅⋅=

(36)

Q0R:S��T� = UQ0 ∙ -/83� U$0V equals to the displacement of the diaphragm of a driver mounted on an infinite-baffle by DC

voltage equivalent to Q0.

Note) Peak value equals to effective value in DC case.


23 / 26

EMS

MSEG

g

PBlCx

xCRR

eBl

=

=−+

⋅ 01

To express S��T� by the system parameters, E0� and �� are to be deformed to

ES

MSS

EMSESES Q

M

R

lBlBMRQ

ωω =⇔=22

22

222

DoASMSASoAS ScVCCcV ρρ =⇔=

Then

( )2

1

22

2

1

2

1

22

1222

=

=

⋅=

=

ESDoS

AS

ESS

MSMS

ES

MSS

E

MSPx

QSc

V

Q

CC

Q

M

R

lBC

ρωωωσ

(37)

( )ωjsystemX : normalized system displacement function

xk : system displacement constant of unity or less

( )AAASABABAS

AAAB

ASjsystemx ZZZZZ

ZZ

CjXk

+++⋅= 11

ωω (38)

Note) To understand why Eq.(36) has such shape, let’s check the behavior. Discussions are similar to ��.

Considering Fig.2 in condition of no leakage and ω ≫ ��,

0lim =∞→

AA

AB

Z

Zω

( )

0

0101

0001

111

11

lim

lim

lim

=⋅++

⋅+⋅=

⋅++

⋅+⋅=

+++⋅=

∞→

∞→

∞→

AS

AP

AB

AS

AS

AS

AB

AS

AS

AP

AB

ASAS

AS

APASABABAS

APAB

AS

jsystemx

Cj

ZZ

MjZ

MjZ

MjZ

Z

Z

MjMj

Cj

ZZZZZ

ZZ

Cj

Xk

ω

ωωω

ωωω

ω

ω

ω

ωω

So W�� is always LPF function. Next, considering ω ≪ �� case then


24 / 26

>>

>>

MjCj

RCj

ωω

ω

1

1

i.e.

≅≅

0

0

Mj

R

ω

So the impedances to consider are 1 �ω8��⁄ , 1 �ω8��⁄ and 1 �ω8�T⁄ thus

( )

ASAPAB

APAB

AB

AS

AP

AS

AB

AS

AP

AS

AB

AS

AP

AS

ASASABASAPASABASAPASASAS

ABASAPAS

ASABAPABAPAS

ABAP

AS

APASABABAS

APAB

AS

jsystemx

CCC

CC

CC

CC

CC

CC

CC

CC

ZCjZCjZCjZCjZCjZCj

ZCjZCj

ZZZZZZ

ZZ

Cj

ZZZZZ

ZZ

Cj

Xk

+++=

+⋅+

+=

⋅+⋅+⋅+=

+++⋅=

+++⋅=

→

→

→

→

ωωωωωωωω

ω

ω

ω

ω

ω

ωω

0

0

0

0

lim

1lim

11lim

lim

�� is to be normalized, ( ) 1lim0

=→ ωω jX thus

ASAPAB

APABx CCC

CCk

+++=

In infinite-baffle cases, 8�� = ∞ and 8�T = ∞. In bass-reflex systems, 8�T = ∞. Both of these have unity W�.

When a bass-reflex system inputted LF signal below port resonant frequency, not only the sound is not

represented but also it causes harmonic distortion and IMD in higher frequency range. If it’s available, such signal

should be cut by an active HPF. Especially in 2-way or single driver systems, such IMD harms harmony and

pianists will never like the sound.

In infinite-baffle case, W� = 1, M�� = ∞ and M�� = 0 thus

( )ASAS

jIB ZCjX

11 ⋅=ωω

Applying the conditions to Eq.(26),

( )AS

ASjIB ZMjG

1⋅= ωω

From Eq.(12)

( ) ( ) ( ) ( ) ( )ωωω ωω jIB

S

jIB

ASAS

jIB GTj

GMCj

X222

11 ==

From Eq.(28)


25 / 26

( ) ( ) ( ) 1

122 ++

=TSS

jIBQTjTj

Xωωω

(39)

Note) This is 2nd order LPF function

Then the condition that peak displacement reaches �� is delivered from Eq.(36)

( ) ( ) maxmax2

1

2 xXP jsystemPxE =⋅ ωσ

Q0? defined by the condition is

( ) ( )

2

max

max

≡

ωσ jsystemxPx

ERXk

xP

(40)

Using system parameters, by substituting Eq.(35) and (37) then

( ) ( )( )( )

( )

( )2

max

2

22

2

max

22

2

2

2

max

22max

12

2

1

1

2

1

ω

ω

ω

πρ

ρπ

σ

jsystemxAS

DESSo

jsystemxAS

ESDoS

D

D

jsystemx

Px

ER

XkV

VQfc

XkV

QScf

S

V

Xkx

P

⋅=

⋅⋅⋅=

⋅⋅= −

(41)

Q�? defined by Eq.(32) and (41) is

( )

( )2

max

2

243

2

max

2

22

3

32

4

4

ω

ω

ρπ

πρπη

jsystemx

DSo

jsystemxAS

DESSo

ES

ASS

ERo

AR

Xk

Vf

c

XkV

VQfc

Qc

Vf

P

P

⋅=

⋅⋅=

≡

In infinite-baffle cases, W� = 1 thus

( )( )

2

max

22

ω

πρjIBAS

DESSoIBER

XV

VQfcP ⋅=

(44)

( )( )

2

max

2434

ω

ρπ

jIB

DSoIBAR

X

Vf

cP ⋅=

(43)

As described above, modern measurement tools deliver parameters by displacement. So “APPENDIX” is also skipped here.

So that’s all.


26 / 26

Further readings

� Articles on J. Audio Eng. Soc.:

R.H.Small

“Closed-Box Loudspeaker Systems Part I: Analysis”

“Closed-Box Loudspeaker Systems Part II: Synthesis”

“Vented-Box Loudspeaker Systems Part I: Small-Signal Analysis”

“Vented-Box Loudspeaker Systems Part II: Large-Signal Analysis”

A.N.Thiele

“Loudspeakers in Vented Boxes: Part I”

“Loudspeakers in Vented Boxes: Part II”

� L.L.Beranek “Acoustics” (McGraw-Hill, New York, 1954)

� W. Klippel “Loudspeaker Nonlinearities –Causes and Symptoms” on Klippel GmbH website

Thank you.

K.Tsubohara

side reader for r.h.small “direct-radiator loudspeaker ... · what are thiele/small parameters?...

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