CALIFORL\:!IA STATE UNIVEHSI'l'Y! NORTHRIDGE
REDUCTION OF RFI BY SELEC'l'IVE
ZERO SWITCHING OF SCR DII"l1V1F1RS
A graduate project submitted in partial
satisfaction of the requirements for the
deg-ree of Master of Science in
Eng-ineering
by
-
Chiang S. Pien
' -1'-11.\ y ,
1 9 7 .5
•
I
l
1
The graduate project of Chiang S. Pien is approved:
Committee Chairman
California State University, Northridge
MAY., 1975
ii
.J
TABLE OF CONTENTS
Page
LIST OF F:tGURES
iv
v
SYMBOLS AND ABBREVIATIONS
- · ABSTRACT
. . . . . .. .. . .
vi
Section
I.
II.
III.
IV.
V.
INTRODUCTION
. ... . .. .. .
....
THE RF ENERGY IN AM BAND WITH ONE CYCLE
MODEL (FREQUENCY DOMAIN) • • • • •
1
4
POWER SIGNAL TO LOAD WITH ONE CYCLE
MODEL (TIME DOMAIN)
. . • • • . •
19
COMPARISON THE RF ENERGY IN THE AM BAND
USING BO'rH PHASE CON'l'ROL AND SELEC~C'IVE
ZERO SWITCHING • . . • • . • . • • • .
26
The RF Energy With Sine
Half Wave Rectifier
26
The RF Energy With 8 Cycle Packet
32
41
CONCLUSIONS
REFEHENCES
42
APPENDIX.:f.S .
43
I.
The algebra form of JF\
II.
Program 5
III.
Program 6
IV.
Program 7
v.
2
...
. ... . . . . .
. . . . . .. .
Program 8, 9, 10
.
. . .
iii
. . . . .. .
...
.....
43
49
52
55
...
58
LIST OF FIGURES
Page
Figure
.......
4
1.
Chopped Sine Waveform
2.
The Envelope of Sample Function .
3.
RF Power of Phase Control at S
4.
Sine Half Wave Rectifier and its
Differentiation Form • . • • •
5.
The RF Power of Sine Half Wave Rectifier
6.
Selective Zero Switching by 8 Cycle
Packet • . • • • • • • •
7.
Phase
and~
Control Diagram •
iv
8
.....
17
......
27
= ~
30
.
.
...
33
40
-
SYMBOLS AND ABBREVIATIONS
AM
Amplitude Modulation
1J.
Micro
f(t) Differentiation,of f(t)
1
! Fig
Figure
Fn
Fourier Transform
RF
Radio Frequency
RFI
Radio Frequency Interference
s
Chopped Length
SCR
Silicon Control Rectifier
Ts
Switching time
Wo
Fundamental Angular Frequency
K
Ratio
ll-
Much smaller than
Almost equal
v
---~- -·-~-------~--------·--·:
,------------------------------··-------
--------·-----
---------------~--------------·
-----------------·-- -··-------
------·----------
'
ABSTRACT
REDUCTION OF RFI BY SELECTIVE ZERO
SWITCHING OF SCR DIMMER
by
Chiang S. Pien
Master of Science in Engineering
May, 1975
The objective of this project is to investigate the
possibility of reducing Radio Frequency Interference {RFI)
by selective zero switching control instead of phase control of SCR dimmer.
was assigned.
The Radio Frequency (RF) power of phase con-
trol was analyzed.
occur at 90°.
A simple one chopped waveform model
The maximum interference was found to
_The RF power of sine half wave rectifier and
_the RFpower of 8 cycle selective zero switching were also
analyzed.
Both of them present a low level of RF power.
The reduction of RFI was thus predicted by compositing both
burst firing and phase control.
vi
SECTION I
INTRODUCTION
Noise level becomes more and more significant if the
. power of SCR dimmer in light stage is given quite large
i.e., 10 KW.
'!'he cable of an SCR dimmer can act as an
antenna and radiating wide band Radio Frequency Interference (noise) where the E and H field are proportional
to dv/dt and di/dt respectively.
In this study Radio
Frequency Interference (RFI) is analyzed.
There are two
reasons which motivate me to pursue this topic.
First, I
am fascinated by solving problems using the Fourier
Transform.
Second, the new rapidly growing microprocesso.t·s
technology in the solution of real problem just strike us
(Dr. Nelson and me) construct such an mathematics model to
do the switching control randomly by micro-controllers.
There are hvo most conm1on forms of control in P.aC.
circuit:
1.
Burst firing, in which power is supplied for an inte-
gral number of cycles or half cycles and swi"lching is performed at the voltage zero points.
It is more suitable for
the control of resistive loads and since the switching
action taker.; place at zero current and the level of RF
voltage generated is relatively low (V=IR-!t).
For this
reason burst firing control is unsuitable for certain
loads.
It can be the cause of unacceptable flicker in
1
2
lighting loads. [3].
2.
Phase control, in which power is supplied for con-
trolled fractions of half cycles.
In phase control the
thyristor may be switched on at any point in the waveform
but is
switch~d
off at the current zero.
So phase control
circuits are extremely effective in controlling the
brightness of lamps, but the switching action is very
rapid, taking between 2 us and 4 us.
Rapid current and
voltage changes can cause severe interference to sound
reception especially in Broadcast Bands.
In fact, in TV
production, it is common practice to turn off wireless RF
microphones during lighting changes.
The SCR dimmers presently used in theatrical applications use LC filters to suppress RFI.
In principle,
filtering is a satisfactory solution, but inductors which
can carry curren·ts of up to 100 amps are very costly.
The
zero switching approach suggested here appears to have the
potential to reduce RFI by as much as 60 db without
filtering.
By using one microcomputer to control a large
number of dimmers, it appears that system costs could be
significantly reduced.
For a SCR dimmer control circuit, each time a
thyristor energizes a resistive circuit, load current goes
from zero to the load limited current value in a few
micro-seconds.
The frequency analysis of such wave forms
shows an infinite spectrum of energy in '"hich the amplitude
is inversely proportional to frequency.
The paper will be
3
presented here to develop the formula by Fourier transform.
SECTION II
The RF energy in AM Band by switching one cycle model
- Frequency domain.
General formula by a chopped sine waveform f(t)
f(t)
Y
=
AsinWo(S+'I's)
t
T
Figure I.
Assume:
Chopped Sine Waveform
Ts
=
3 x l0- 6 second
m
=
dy is
not a constant
dx
A
=
instantaneous Voltage Amplitude
=
117 xJ2 volts
s = T2N = T26'
N
=
Const.
= 10
Ts <S..:: T/2
Wo
=
2J[fo
For Fourier transform:
f(t)~Pn
4
Ts
=
ts
5
=
Fn
1
T
+ 1
T
JS+TsAsin~7o
(S+Ts)
S
Ts
(t-S)
JS+Ts
T/2 A . W t
s1n o
JT/2+S+Ts
T T/ 2 +S
_ 1
e
-jnWotdt
e
i
-jnWotdt
1
+T
.
AslnWo(S+Ts)
Ts
JT
.
(t-(T/ 2 +S))
.
e-JnWotdt
T/ 2 +~+TsAs1nWot e
'Assume f(t) is voltage V(t)
The E field radiation is due to f(t) = df (t)
dt
.
E o<f (t)
= ~t
~t
£2 (t) =
AsinWo (S+Ts) (t-S)
Ts
(AsinWot)
= ~t
=
f4(t)
Note:
f
1
=
=
AsinWo (S+Ts)
Ts
Wo A Cos\,Yot
AsinWo (S+Ts) (tTs
(T/2+6))
(Asin~~ (S+TsJ
=
, f
d
dt
3
(AsinWot)
=
WoA cosWot
is the skew ra·t.e of rising time.
Now we rewrite the Fourier Transform again:
Fn ==
!.
T
r
S+Ts
6
AsinWo (S+'J:'_s)
Ts
e-jnWotdt
-jntV'ot
dt
6
+ Wo
.T
T/s ACosWot e-jnWotdt
S+Ts
J
_ 1
T
T/2+S+Ts
T/ 2 +S
Asin(S+Ts)
-jnWotdt
Ts
-· e
+·;wo
_-T
; Fn
J
J
T
T/2+S+Ts
A CosWot e-jnWotdt
=
T F
• 1
JS+Ts
la
=
.
-jnWot
AslnWo(S+Ts). e
dt
Ts
Asin~~(S+Ts)
= AsinWo (S+Ts)
--~T-s--
J
~+Ts e-jnWotdt
1
-jnWo e
-jnWot] S+Ts
S
=
AsinWo(S+Ts). --=-=....::;.1_.,.. • (e -jnWo (S+Ts) -e -jnWob)
nWo(-j)
Ts
=
AsinWo (S+Ts). e_-jnWo~e -jnWn~ e-jnWoTs
~. 2j
2
=
=
AsinWo (B+Ts) • e -jnWo (S+Ts/2!~ +jnWoTs/2_e -jn~"7oTsf2)
nWoTs
2j
2
AsinWo (S+Ts). e -jnw·o (S+Ts/2).
Sin?W~Tsc
nWOTs
2
=
. nWoTs
Asinl"lo (S+Ts) • Sln 2
• e -jnWo (S+Ts/2)
nWoTs
~
= AsinWo(S+Ts)
Sin (d).e-jriWo(S+Ts/2)
--d-
7
nWoTs
Where d =
2
The sin x/x envelope depends on Ts, the pulse width.
This sin x/x envelope is characteristic of periodic square
' pulses.
Different pulse shapes result in different envel-
ope shapes.
We find the s.ide lobe will be decrease
following with each zero point. (see Figure
,(T/2
T.F 2 ls+Ts AwoCoswot e
=
AWo
= AWo
J~~;s
J
=
CosWot e -jnvlotdt
e-jWot(n-l)+e-jwot(n+l)dt
AWo ( e -jWot (n-1)] T/2 +e -jwot (n+l)
2
-jWo(n-1)
S+Ts
-jWo(n+l)
A
-2j (n-1).
+ -2j A(n+l)
T.F3
dt
T/2
jWot+ -jwot -jWotdt.
S+Ts e
e
.e
2
= AwoJ~~;s
=
-jnWot
2.)
(
(
e
e
-j\'lo (n-1) T/2
-jWo(n+l)T/2
-e
-e
J
T/2 )
S+Ts
-jWo (n-1) (S+Ts))
-jWo(n+l) (S-t. Ts))
...
J
=
-AsinWo (S-.+-Ts)
Ts
=
-AsinWo(£+Ts)
l
-jnWot
Ts
-jnWo e
=
-
T/2+iii+Ts • e -jnWotdt
T/2+S
J
T/2+s+Ts
T/2+S
) -jnWo (T/2+S+Ts/2)
. W0 (co+T
e
S .e
A S1ll
----~----~~~·
nWoTs
(-2j)
• ( e -jnWots/2_ejnWots/2)
·
8
I
~
ASinWo (S+Ts)
Ts
.
fl (t)
I
Sinx
)-(x
-Sa
X
1
Ts
2
Ts
x=
nNoTs
1
3
Ts
2
I
I
i
l
I
I
Figure 2.
The Envelope of Sample Function
9
=
-AsinWo(£+Ts).e~jnWo(T/2+£+Ts/2).Sin(nWots/2
nWots 2
= -AsinWo {S+Ts)
-J
T.F 4 -
Sin (ntvots/2)
-jn~vo (T/2+S+Ts/2)
nwoW2 .e
T
-jnWot
T/ 2 +S+TsAWoCosWot e
dt
J ~/2+S+Ts
ejWot+e -jnWot ~e -jnWotdt
=
AWo
=
AwoJ T
e -jWot (n-1) + -jWot (n+l) dt
T/2+S+Ts
·
e
2
=
AWo ( e -jn~'Jot (n-l)] T
-jWot (n-1) T
)
-2-jWo(n-1)
T/2+S+Ts +e -jWo (tl+l)
T/ 2 +S+Ts
J
~
=
.
(e-jWo(n-1)T_e-jWo(n-1) (T/2+S+Ts))
-J2 n-1)
+
A
( -jWo(n+1)T -jWo(n+1) (T/2+S+Ts))
-j2 (n-t-1) e
.
-e
= jA
( -jW (n-1) T -jWo (n-1) (T/2+S+Ts))
2{n-l)
e
-e
+ jA _ ( -jWo(nt1)T_ -jWo(n+1) (T/2+S+Ts))
2(n+1) e
e
= AsinWo(S+Ts).
s·
nWots
ln---2-nWots
e
-jnWo (£+Ts/2)
2
.
s~ (nWots)
· W (
/ )
·r,:t •j'))
- AsinWo (S+Ts) • .Ln _2___ . e -Jn o S+Ts 2 • e -J\nwor nWots
-y-
10
+ .
A
{ e -jWo {n-1) T/2_ 6 -jWo {n-1) {S+Ts))
J 2 (n-1)
.
A
( +jWo {n+l) T/2 -jwo {n+l) {S+Ts)\
+ J2(n+l)
e
-e
)
+ .
A
J2(n-l)
.
A
+ J2 (n+l)
=
{ 6 -jWo{n-l)T_ 6 -jWo(n-l) (T/2+S+Ts·))
( 6 -jWo(n+l)T_ 6 -jWo(n+l)(T/2+S+T·s·})
. nWots/2
.
( S+Ts ;·2 )~ _ -JnWot~2.
.
~1 )
AsinWo(S+Ts)sln
-JnWo
1
nWots
•e
•
e
·
2
+ j
A
(·e -jWo (n-1) T/2+e -jlvo (n-1) T/2 • 2
2(n-l) • ·
- (e -jWo (n-1) (S+Ts) +e -jWo (n-1)
+ J'
A . (. -jWo(n+l)T/2+ -jWo(n+l)T/2 )(2
2(nil)• e
e
-(e -jWo (n+l) (S+Ts) +e
We know T
=
(S+Ts+T/2~)
=
27(
Wo
WoT
AsinWo(S+Ts)
+ J.
A
(·
2 ( n-1) • , e
= 2}[
also, e
. nWots/2
8111
nW~ts
~jwo (n+l) (S+Ts+T/2)))
= (-1) n
-in}i
.
(S
)(
.
I )
e-JnWo +Ts. 1-e-JnV~ 2
-jWo(n-l)T/2(l+ -jWo(n-l)T/2))
•
e
-e-jWo (n-1) (S+Ts). (l+e -jWo (n-1)
+
T/2~)
.
A
• ( e -jl.Vo (ntl) T/2. (l+e-jWo (n+l) T/2)\
J 2 (n+ 1)
)
-e ~jwo (nt1) (S+Ts) • {l+e -jWo (n+ 1) T/2.,)
11
T.Fn
. nWot$12
= AsinWo(S+Ts)slnnWots
(
)
.e-jnWo(S+Ts~-1- (-1)n
2
+ ·
A
J 2 (n-1)
(f
1 +e-jV'Jo(n-1)T/_2)·(e-jWo(n-1)T/2
(S+Ts~)
-e-jWo(n-1)
·
A
{ 1( +e -jWo (n+~T/2_).·( c -jWo (n+1) T/2
+ J2 (n+1) \ 1
(S+Ts)))
-e-jWo(n+l)
. nWots
.
.
(
) (
.
= AsinWo(S+Ts).~_2_ .e-JnV'Jo S+Ts 12 • 1-(-1)n)
-nNots
...,.
;/;,
·
A
{ 1 +e-j {n+l)Jr).fe-j (n+l)7C_e-jWo{n+1) (S+Ts))
+ J2{n+1)
\
for
e
j (n-1)7[= Cos (n-l)Jr. +j sin (n-1)7[
=
1
= -1
but
n
=
N:
odd
N:
even
=
T.Fn
even J
0
..
Both' Sine and-Cosine terms are presented by odd nunilier n.
so we take n
T.Fn
=
=
AsinWo(S+Ts)
odd
. nWots
2
• e-jnWo(B+Ts/ 2 ) .2
81 n
~
-2-
+ ·
A
J2(n-l)
•
( 2 ).
(l-e-jWo (n-·1) (£+Ts))
12
·
A
+ J2(n+l)
•
Note
( 2 ). (l-e -jWo (n+l) (.S+Ts)_ ).
. · nWoTs •
for n <<" .. s.1.n . 2
1
t=
nWoTs
2
In considering with radiating field, we only deal with
large n.
=
T.Fn
..
. .. nWots/2
2A sinWo(S+Ts)
nnWots/ 2
51
.e
-jnWp(S+Ts/2}
ZA • -j Wo (n-1} (S+Ts} ( +j Wo (n-1} (S+Ts}
- ~2 ~j~(-n--~l} e
2
e
2
. Wo (n-1) (S+Ts)'
-e-J
2
) ·
2A
. Wo (n+l) (e+Ts >t +. ~.Vo (n+l) (S+Ts)
2J(n+l} e-J--------2----r J
2
-
_ .Wo (n+l) (S+Ts)'
-e J
2
:}
=
2A sinWo(S+Ts}
. nWoTs/2
.nwo (S+Ts/2)
n· W T
• e-J
n o.s
51
2
2A
-jWo(n-1} (S+Ts)
(n-l) . e .
2
. Wo(n-1) (B+Ts}
s1.n
2
-jWo(n+1) (S+Ts}
2A
- (n+l). e
2
. ~vo(n+l} (S+Ts)
s1.n
2
-
Let b
=
vJo(n+l) {S+Ts)
2
c
=
Wo(n-1) {S+Ts)
2
d
=
nWots
2
13
T.Fn
=
-
.
sind -jnWo(S+Ts/2)
2AslnWo(S+Ts)---d- e
2A
n-1
e
-]c
sine
2A
-'b
- n+l e J sinb
The algebra form of RF energy as follows:
I 12 = (
F
-
2Af) 2
(
2
.
sind)
(slnWo(S+Ts)~
.
·
sind . ( Cos ,nwo (S+Ts/2)
(2Af) 2 n2 s1nWo
(S+Ts)
--d-
•
(~
sin2b+sinc) +si~
2
+-( 2Af)
n
b. sin
nWo (S+Ts/2))
2
2
2
2
• (sin b+sin c+2 sinb.sinc.Cos (b-e))
See Appendix I
From Paraval's theothem, the total energy
frequency band f 1<f<f 2 is E
=J~i ~~ 2 Jf.
any harmonic frequency f is just JFJ
Now
2
in the
The energy at
.
I would like to find the energy contained in AM
Band width from 600 KHZ to 1200 KHZ.
If power signal is
60 cycle, thEWI start the No. of n from 10000 to 20000.
Also, this function is zero when n
=
even.
no. of n from 10001 to 2000l.by increment 2.
So I take the
14
T
for S =
Note:
if
sinWo (S+Ts)
= sin(~
+ WoTs)
= Cos (WoTs)
=
1
·;cwoTs ...<t)
I
o (Cos(nWo(S+Ts/2V o
(<~
sin2b+sinc)
+ sin2bosin nWo(S+Ts/2))
+ (2Af)
-2
2
2
o (Sin2b+sin c+2 sinb.sin c·Cos
n
Program 1:
S
=
!'
~Double
I
=
switching at
=
0
(19852o56)**2
Ts ·= 0. 3E-5
R
=
3.1415926536
S
=
60*R*Ts
J.
Do
B
=
I *
=
I
10001,
s
D = I* (R/2.+8)
E
=
( I+ 1) * ( R/ 4 • + s )
F
=
(I+l) * (R/4.+s)
90°
precision ANS, SUM
SUM = O.
c
a =
20001,~
2
(b-e~
15
=
ANS
C*(SIN(B)/B)**2- 2.*C/I *SIN (B)/B *
(COS(D}*(O.S*SIN(2.*E}+SIN(F}) +
(SIN(E)}**2*SIN(D}) +
C/FLOAT(I*I)*((SIN(E})**2*(SIN(F)}**2-
:
2. *SIN(E)*SIN(F}*SIN(2.*S))
SUM
=
SUM+ANS
' 1 CONTINUE
Y = 10. * D LOG LO (SUM)
WRITE (61,4} Y
4 FORMAT (1X, F11.4}
STOP
END
* * * * * *
•
2
. . Y = 10 1og10 ( F i)
Program 2:
I
=1
=0
c =
=
101.1867 db
Dimension IA (5001)
KT
Ts
=
(19857. 56) **2
0.3E-5
.R
=
3.1415926536
S
=
60. *R*TS
Do 1
I = 100011 20001, 2
s
B
=
I *
D
=
I
E
=
(I+1) * (R/4+s)
R
=
(I -1 } * (R/ 4 . + s)
* (R/2.+s)
1
<!===-
ANSA(5001}
1
YA(5001)
..
'·
16
ANS = C * (sin(B)/B)**2-2.*C/I*sin(B)/B*
(Cos(D)*(0.5*sin(2.*E)+sin(F)) + (Sin(E)**2
* sin(D))
+ C/FLOAT (I*I) *
((sin(E))**2 + (sin(F))**2-2.*sin(E)*sin(F)
*sin(2.*s))
. Y = 10. *ALOG 10 (ANS)
IA(KT) = I
ANSA(KT) = ANS
. YA(KT) = Y
KT = KT+l
1 Continue
KT=KT-1
WRITE (61,5) (IA(K) ,ANSA(K) ,YA(K) ,K=l,KT)
5 FORMAT ( ( 1 X , 3 ( I 7 , 4 X I E 11 . 5 i
2XI F 9 . 3 ) )
WRITE (6l,loo) SUM
100 FORMAT (lX, Fll.4)
STOP
END
"*
*
..)'€
JOB, 70090 87, PIEN,5,6000,PIEN, John
SCHED, CORE
=
80
FTNU (L,S,X)
Th e m1n1mum
· ·
KHZ
f
IF1 2
(Third zero) .
1
= Ts
occurs at 666 KHZ
(Second zero) and 999
The first zero of envelope is at
333 KHZ, and RFI become small whenever it is
approach to zero.
(See
Fig~
3) .
17
db
70
lC·x
x-~ -x~
I
I
i
I
I
l
50
I
'
\
X
'X
J
I
I
t
~
I
"'
I
I
I
I I
l
\I
\ )\
oJ
10
Figure 3.
..·
I I
)(
600 KHZ
X....
I
1
i.
\
~- 1<.--~><-
\
I
'/.
\
30
·X-- ... x .. 'X
I
~
800
KHZ
1000
KHZ
f
1200
KHZ
RF Power of Phase Control at S=~
18
·If I start the switching position from
e =
1~8°
to
16.2° by increment 1.8° each time, we can find the RF
power goes up from 70 db to 90 db (See Appendix V) .
I start to switch the SCR dimmer from
e =
18° to 162°
by increment 18° each time, we can find the RF power goes
up until 90° then decreases (See Appendix V).
SECTION III
:Power Signal to Load With One Cycle Mode (Time Domain)
:p
=
1 im
:If f(t)
p
=
~a
J~a r
f
ct >1
2
dt
is also periodic
lj T
o
T
f
2
(t)dt
The period is T/2 for one power cycle.
1 IT
If we assume the function f ( t) is a voltage, then T
o
.
2
f(t) dt represents the average power delivered by f(t) to
a one ohm resistor.
F(f(t)}
=
lJT f(t) 2 dt
Also T
0
=II
Fl
2
/
If
F(f}.The mean-square value of periodic function
f(t) is the sum of the mean-square values of its harmonics.
It should be noted that the power content of a periodic
function depends only on the amplitude of its harmonics
and independent of their phases.
signal.
2
= A J~1 2
(l-Cos2Wot)dt
19
e = o s = o
20
~
=
72.62
=
228.14
sin 2 Wot/2)
,..._
{/1)
s =
20
I ~~2
=
2
=
A
=
72.62 (3.121)
2
J!1
-· 226.64
2
( l-Cos2Wot) dt
21
--- - -- ----
K
=
T.P 1
e=
36° 1
T.P
=
2
2
= A
= A
K =
e=
99.34%
(t}dt
A2 s1.n
. 2 Wotdt
((;
( ;
7\2 (.4
Wo
5
dt
-{;~ 2
Cos 2 Wot d,
-
io) - ~Wo
T
+
._.+
~~
~Wo
sin 2 Wot
);~ 2)
sin 2 Wo. 2s)
. 27C)•
21 . sl.ns-
(2. 988)
217
T.P
2
T.Pt
54°,
T~P,2
--
s
2.
217
278.14
=
T.Pt
=
0.95 == 95%
-- 3 T
20
T.P 3 - 2 JT/2
3s
=
2
(J ;~ 2
2
---
T/2 (1-Cos 2 Wot)dt
2s
= 72.62
=
f
2
2
----·
10
T/2
2s
A
= A
=
I
J
=
---
T
s =
2 JT/2
2s
=. 2
=
226.64
228.14
=
T.Pt
-- ----·--
fV2
20r
f
2
(t)dt
2
2
.
A sin Wot dt
.
22
(1 - Cos 2 Wot) dt
= A
2
2
2
+
=
((T- - . -3
A
r iii
. .
e
= 12°
194.2
228.14
= o..-85 =
A2 (·(T2 _ 5T) + 12Wo
=
3iT 1 .
A2
(
-5- + 2
Wo
=
72.62 (1.8849 +
=
158.2
Sln
T.P4
T.Pt
=
85%
s = 5T
1
=
a = 90°
3iT)
S
194.2
T.P 3
=- =
T.Pt
T.p5
.
72.62 (2. 6745)
K
=
1
Wo \ 10 + 2 • s1n
=
K
1
,
sin 2 Wo 3 T)
20 T) + -2 Wo
20
=
I
2.
158.2
228.16
=
• sin 2 Wo. T5 \}
4/i)
-5
21
0.693
X
0.587)
=
s = T/4
J
T/2
5s
T)+ 2WO
1
6
sin
69.3%
23
7T2
= A2 (
Wo
T.P5
K
ll4.07)
'
1
= T.Pt
e =
=
= 2
s =
108°,
3
10
1
sin 2 wo
+ 2Wo
sin
;
11
+
~
~/I
)
• <-o. s87)
= 69.943
K
T.P6
69.9403
126°,
b
= T.Pt- 228.16
e =
= A2
,..
=
wo
30.65%
7T
= -20
~~.
4
A2
=
0.3065
(n _
= A2 (
=
=
(
~
33.909
+
L
Wo
+
1
2Wo
.
Sln
1
10
fo T)
24
K = T.P7 _ 3~.909 =
0.1486 = 14.86%
T.Pt - 228.14
·e
= 144 o,
=
72.62 (0.1527)
=
11.094
11.094
= 0.048
228.14
= 4.8%'
K
=
e
= 162° 1
= 2.
9T
s = 20
J
~/2
. 2Wot d t
A2 Sln
2oT
= A2
(n - ~0
T) +
-
1 Sln
.
5911 )
2Wo
=(72.62 *0.02)
=
K
=
e=
1. 4 72
P9
pt
-=
180°,
1. 472
228.76
s = 2T
=
0.00645
=
0.6%
25
2
£ (t)dt
=
0
I can drat.v the diagram using the results of this
, section and the results of Program 6 (See Appendix lii).
Thus, the phase control diagram is achieved by 9 Sample
Data (See Fig. 7).
SECTION IV
Comparison the RF Energy in the AM Band Using Both Phase
Control and Selective Zero Switching
The RF Energy With Sine Half Wave Rectifier -
Fn
=
2
T
JO
T/2
A Wo Cos Wot e
-jnWotdt
+
(See Fig. 4)
2J T/ 2
T
T
- .nWotdt
O.e J
2AWof T/2 •C
-jltWotdt
= --T- O
. OS Wot e
.
·tt·
ejWot+e-jWot
2AWo
=
A~o ( -j (n:l)Wo
T
+
=
T
=
T/2
0
2
e
-j (n-l)WotJ T/2
0
1
e -j (n+l) Wot] T /2)
-j (n+l) Wo
0
AWo( .
+
T.Fn
J
=
1
( -j (n-1) Wo· T/2_ 1 )
·
- J ( n-1) Wo e
1
-j (n+l)·Wo
( -j (n+l) Wo T/2 ))
e
-1
. A
( e -j (n-1) _ 1)+ - . - A
_ _ (e -j (n+l)Jz;
-J(n-1)
. -J(n+1l
= -j(n~l)
-i)
(cos(n-l)it-1) + --j(n!l) (cos(n+1)it-1)
26
27
f(t)
'
-I
f(t)
=
ASinWot
0< t
=
0
-T/2 <. t 4
< T/2
0
f(t)
df{t)
d t -- A Wo Cos Wot
=
Figure 4.
0
0..::::: t .c. T/2
-T/2 ..:::::t.c:::. 0
Sine Wave Half Rectifier
and its Differentiation Form
28
------
Cos (n-l)it = 1
= -1
{
n:
odd
n:
even
T.Fn = 0
for n = odd
T.Fn :\: 0
for n = even
A
.
A
T.Fn = j (n-1) + j-(n+l) ·
-- -j ( 2A)( n=l + n!l )
n
= -J"2 A·-2n -1
=
-j4An
2
n -1
. -j4A
n
.
(for large n)
T2. IF 12= T 2 .Fn*Fn *
16A 2
= -2-.
n
'
.
2 2
16A f
IF 12=
2
n
for f = 60 HZ
A= 117 x 1.414 volts
------
, ..
29
Program 3:
SUM 1 = 0.
I
=
0
=
Do 11 I
Z
=
•
0.15769 E 10
' ANSl =
=
Yl
10000, 20000,2
z I
(I*I)
lO.*Alo~
lO(ANSl)
SUM 1
=
SUM 1+ Yl
WRITE
=
(61, 6) I, ANSl, Yl
6 FORMAT (lX, I7, 4X, Ell.4, 2X, F9.3)
11 CONTINUE
WRITE (61.10) SUMl
10 FORMAT (lX, FlO .. 4)
STOP
END
(see
.X.
'"*
Fig.
s) *'
30
Fh
2
d
10
f... ...
5
f
600 KHZ 800 1000 1200
KHZ KHZ KHZ
Figure 5.
The RF Power of Sine HAlf Wave Rectifier
(See Appendix IV)
31
Program 4:
=
SUM 1
I
= o
=
DO 11 I
Z
I
0.
=
10000, 20000, 2
0.15769 E 10
ANS 1
= z I
SUM 1
=
(I*I)
SUl-1 1 + ANS 1
11 CONTINUE
Yl = 10
*
A LOb; 10 (SUM 1)
WRITE (61,6) Yl
.
6 FOID1AT (lX, Fll.4)
STOP
END
* *
NOTE:
.·, Yl
FTNU(L,X)
=
* * *
*
No single integral S requiredo
10 log 10 ( 1Fhl
2
)
=
45.9585db
Now I just recall the former result at g
e=
90, Yl
=
10 log 10 <IFI
2
>
=
=
T/4
101.1867db
Obviously Y 7Yl
It means we will get a lot of RF Power by phase control,
but we can get small RF Power by doing zero out selected
1/2 cycles.
32
The RF Energy ·with 8 Cycle Packet
Let's take the window as 8 cycles, from the Fourier
Transform (See Fig. 6) •
T.Fn
=j~T
. . ·n'2Jt
2
dt
A sin - t e, J
MT- t
T
=J-~T
. n'27Ct
e-J~dt
3.....t
A sin
T
(M = 8)
Now if we chop the first one half cycle out:
...
=
f = df
dt
d(ASinWot) = WoA Cos Wot
dt
:. 8T. F 15/16
J
J
ST
-jnlli'ot
T/ 2 AWo Cos Wot e
8 dt
.nwot
8T
T/ Cos Wot e -J-8- dt
2
= AWo
J
= AWo
f T/2
8T
2
= ~f8T
2
T/2
.nwot
( ejWot+e-jWot) e - ] -.8dt
e
jWo(l-n)t +e-j~~o ( 1+~) t dt
8dt
.
n
<
eJWo(l-B)t 8T
AWo
= ·-2-
jWo
.
(1-~)
AWo
T/2
e-jWo(l+~)tfT
-jWo(l+~)
= -2-
J
+ AWo
-2-
e
T/2
jWo(l-£)8T jWo(l-~)T/2
8
-e
8
jWo (1-£)
8
+
(e-jWo(l+~)BT_e-jWo(l+~)T/2)
AWo
-jwo (l+n)
8
33
r· -------- ------------------------------------------------------- ---------I
8T------
The numbers indicated in the picture
above presents the chopped sequence.
I
Figure 6. Selective Zero Switching by 8 Cycle Packet
I
i
I
i
:I
1•.
•••••·•~··•-r-"'--•••"--"<'"
·~.r ••-·~·-•••••••••
34
=
A •
1·
t
. e
·n
8
2J (1--)
+ A •
jWo (l-n ) 8T
8
-e jWo (1-~) T/2 )
1: . -.(e-jWq (1+~) 8T_e -jWo ( 1·+~) T/2)
-2j(l+n)
·8
·
A ·
( j (1-n) 167C j (1·-n)7f.)
8
-e
8
:. 8 T.F15/16 = - - - ·- . e
2j (1-!!)
8
, ( -j (1+!!.) 167L
A
----. e
2j (1+~)
8
-e
-j (1+n)7[.)
8
8
=
E
(1)
+ j
G. (1)
See Appendix II
. If we throw away first one cycle, using the same formula
we get:
8T.F 14116
( -j(l+g)l6~ -j(l+n)2~
e
8
-e
8
)
=A
2 j ( 1+!!.)
8
A
(
e
-j {l+g) 167r.
8
-e
-j (1+£)
8
27[)
2j(l+n)
8
=E
(2) + j G(2) (See Appendix II)
35
If we throw away three half cycles:
· A ·
8T.Fl3/16 =
( j (1·-n) 167C
J. (1-n) 97L
8
-e
8
·
. e-
2j (1-E.)
8
+e
j (1-n) 8Jl j (1-E.)
8
-e
8
2Tt.)
·(e-j (l+~)f67t_ej (1+~)9il.
A
2j(l+£)
8
=
+e
.:. j ( 1 +.!!) 8/[ -j ( 1 +£)
8
-e
8
E
(3}
+ j G (3)
9JC)
(See Appendix II}
Throw away four half cycles:
8T.F 12 /lG = A
( j (1-!!.) 167i j (1-n) 87l:.
----· e
8
+e
8
2j ( 1-n)
8
A
( -j (l+!l) 16li -j (1+£) 8Jt -j (l+n) 107[
---- . e
8
+e
8
-e
8
2'(l+n)
J
8
.
=
E (4) + j G(4)
(See Appendix II)
36
Throw away five one half cycles:
·A
( j (1·-£) 1671: j (1·-n) 107[ j (l-n)
8T.F11116 = -------· e
8
-e
8
+e
8
2j (1-~)
87[
-A
-e -j
(1+~) 37[J
= E (5) + j
G(5)
(See Appendix II)
Throw away six one.half cycles:
8T.Fl0/16 = A ____ ·(ej (1-~) 167[+ej (1-~) 8Jt_ej (1-~) 117[
2j (1-n)
8
-e
j (1-!!)
8
A
37[)
(e-j(l+~)l6n_e-j(l+~)lln+e-j(l+~)sn
2j (1+!!)
8
-e -j '(1+~)
3TC)
= E (6) + j G (6) (See Appendix
II)
37
Throw away seven one half cycles;
A
·( -j(l+n )16n _ -j(l+n )lln+ -j(l+n )
8
8
8
2j (l+n
e
e
e
a>
-e -j
=
(1+~) 47[.)
E (7) + j G (7)
(See Appendix II)
Throw away eight one half cycles:
8T.f8/16
=
A
A
• (e
(e
+j (1-n) 16
-j
8
-e
j(l-n)l2
8
(1+~) 16
2j (l+n)
8
-e -j(l+£)4)
8
= E ( 8) + j
G ( 8)
(See Appendix II)
BTL
38
Program 5: (See Appendix II)
'The total average power of 8 cycle for the time domain is:
, Pav
=
!_J
8T
BT f 2 (t) dt
0
J ~T
= 2A2
(l-Co~2Wot) dt
(J ~T dt
=
A2
=
A
2
=
2
A . 8T
-J~TCos
2 Wot dt:)
~
2Wo sin 2 Wot] 08T)
.
BT -
(
For one half cycle:
8T.Pav
Kl
15/16
=
=
A
2
BT.Pl5/16
8T.Pav
(8T-T/2)
=
15
16
Using the same method we get:
K2
14
= 16
K4
=
12
16
11
K5 = .16
K6
=
10
16
K7
=
9
16
1
8
Ka = 16 - 2
39
Thus,
the~-control
diagram is achieved by using the
results of Program 5 (See Appendix II) , and different
numbers of power ratio K (See Fig. 7).
\
40
Prf
db
100
...,. t
Phase Control
I
I
I
I
I
;u Control
10
10%
Figure 7.
'
I
l., >-·•·-··••·~·v•-··-···
....... ,._.,,.,. ...
~-~--~···~·
·"
100%
PL%
Phase and;u-control Diagram
SECTION V
CONCJ...,USIONS
We can use modified burst control to zero out selected
1/2 cycles.
We analyzed simple example of 8 full cycle
packet which given 15/16, 7/B, 3/4, etc. power control
levels at significants less RFI levels then comparable
power control levels using phase control (see Fig. 7).
Hence, if we increase the packet size to 32 one half
cycles, we can get enough power levels for TV or theatrical lighting applications.
Thus packet size would
require that the lamp response to greater than 32/60 sec.
so that no flicker would result.
For power levels from
1/2 power to full power, we can zero out integer numbers
of 1/2 cycles.
For power levels from
o~l/2
cycle we feel
that phase control can be carefully applied by switching
at near but non-zero levels such that the required power
levels can be obtained.
---------
41
REFERENCES
1.
ERA,· Radio Interference From Thyris·t·or· co·nt·ro1
circuit, 1969.
2.
Froehlich, John P. Information Tran·s·mittaT and
Conununicatiori. Systems, 1969.
3.
General Electrpc, SCR Manual, 1972.
4.
Hse, Hwei P.
5.
Lynn, Paul A.
1973.
6.
Mix, Dwight F.
7.
NASA.
Fourier Analysis, 1970.
The Analysis and PYocess·ing Signals,
Random Signal Analysis, 1968.
Radio Frequency Interference Handbook, 1971.
42
APPENDIX I
43
APPENDIX I
= (X+Y+Z)•(X*+Y&+Z*)
= XX*+YY*+ZZ*+XY*+XZ*
+ YX*+YZ*+ZX*+ZY*
Let:
.nwo (S+Ts/2)
X= 2A sinWo(S+Ts) .si~d e-J
-2A
-jc sine
n-1 e
Y
=
Z
= - 2A e -jbsinb
n+l
Also,
(XY* 1 YX*)
(XZ* 1 :zX*) Pairs
'
(YZ *, ZY*)
2
2
Rearrange: T
{F[
= XX*+YY*+ZZ*+(XY*+YX*)
.
+ (XZ*+ZX*)+(YZ*+ZY*)
2
For
T FnFn*
=
(2AsinWo (S+Ts) sidnd e -jnWo(S+T~) -~A
n-1
.
Sind +jnWo(S+Ts/2)
2A
.(2AslnWo(S+Ts)·-d- e
- n-1
e
2A
jb
)
s1nc - n-+:-l e sinb
JC .
44
45
=
4A 2 sin 2 Wo(S+Ts)si~ d
2
d
2
+ 4A
(n-1) 2
. 2
. 2 . 4A 2
s1n c+
s1n b
(n+l) 2
.
sind -jnWo (S+Ts/2) · 2A
jc .
2As1nWo(S+Ts)---d- e
• n-l e s1nc
2A e -jcsinc ( 2A . (S+T )sind
jnWo (S+Ts/2)
- n-l
s1n
s ---d- e
. b)
. Wo (c+T
- 2As1n
o
s )sind
--d- e -jnWo(S+Ts/2) · (2A
n+ 1 e jb s1n
2A
-·b
- n+l e J sinb
2Asin(S+Ts)sind ejnWo(S+Ts/2)
d
2A
-jc
2A
'b
+ ~l e
sine• n+l eJ sinb
On large n over 10000,
~
n+l
------
n
~
T•FnFn*
2
2
. 2w0 (co+Ts )sin
= 4A2 Sln
- - --d + -4A22
d
n
2
8A
- -n-
2
. 2 C + -4A- Sl.n
. 2b
S1n
2
n
. d
.
(S+Ts ) s1nc---d. s1n
s1nWo
.
e
jnWo(S+Ts/2);-e -jnWo(S+Ts/2)
2
2
8A
- -n-
(c
) . bsind
.
s1.nWo
~+Ts s1n ---d-
e
j(b-nWo(S+Ts/2)+ -j(b-nWo(S+Ts/2)
e
2
46
+ BA
2
sinb sine.· e.j,(b-.e)+e-j(b.-.e)
n
2
'
BA 2
~.
f.
' d
- I1 sinWo (S+Ts)· sin·~s· 1 ~ .cos \n Wo (S+Ts/21
8A2
. . ' d
- I1 sinWo(S+Ts). sinb·s 1 ~ .cos (b-nWo(S+Rs/2~
2
8A
+ --2 sinb.sine·Cos(b-e)
n
BA 2 sinb sine Cos(b-e)
+ --n2
-
-----~
8 ~ 2 sinWo (S+Ts) • si~d • ~inc • Cos ( nWo ( S+Ts/2))
+ sinb • Cos (b-nWo (S+Ts/2.~
= 4A 2 sin 2Wo(£+Ts)sin
2
g
+
d
2
2
4 A~ . s1n
. 2b + --. 2 e+--4A s1n
8A
n
n2
n2
sinb sine . Cos (b-e)
2
8A
--n-
--a-·
. T.o7 (C'
) sind(Slne•COS
.
(n Wo (S+T~)) + sinb
S1nvv0
~+Ts
• .cosb" cos(nwo (S+Ts/2~
2
\I
+ sinb"
sin (n· Wo. (S+Ts/2~)
2
2
4A s1n
4A s1n
. 2 c + --. 2b
+ --n2
8A 2
+ --- sinb sine Cos (b-e)
n2
n2
47
'
sin2b •
(,
·~
Cos f•Wo(S+Ts/2~
2
. 2b
. (,
+ s1n
•Sln \n
Wo(S+Ts/2~
'
2
2
= 4A sin wo ( S+Ts.) ·s·in·~ d + .4.A ~ (sin 2 b+sin 2 c
n
d
+ 2 sinh· sine• Cos (b-e))
-
8 ~ 2 sinWo (S+Ts) si~d. (sine • Cos n Wo (E+Ts/2)
+
T·~-~
~
2
sin2b.Cos nWo(S+Ts/2) + sin b.sin nWo(S+Ts/2?)
[F( 2 = en~rgy/unit
'
=
BW in
HZ
energy density spectrum
2 2
+ 4A f
(sin 2 b + sin 2 c + 2 sinh sine Cos (b-cU
n2
2
8A f 2 sinWo (S+Ts)
2
+ sin b· sin (n Wo
fFJ 2 =
(2Af
.
+
2
< :~)
si~d. ( CosnWo (S+Ts)-(~
sin2b + sine)
(S+Ts/2~)
sinWo(S+Ts)si~~)
2
2
2
2
• (sin b + sin c + 2 sinh sine Cos
(b-e.~
48
(2Af)
2 .2
.
·s·ind ( Cos n Wo(S+Ts/2)
SlnWo(S+Ts)---dn
•(~
=
2
sin2b + sine)·+ ·sin b.sin ( n
Wo(S+Ts/2~
'
· ·sind)
(2Af) 2 ·, ( sinr7o (S+Ts)
---d- 2
(2Af) 2
•(
~
.~
SinWo (S+Ts.)
si~d • (~os
2
sin2b+sinc) + sin b. sin
2
.
n• Wo (S+Ts/2)
(n · Wo · ( S+Ts/~ l))
+ ( 2 A:~ . (sin b+sin e +2sinb sine Cos (b-e~
2
2
---~---
I:
APPENDIX II
'
PROGRAM 5
~.
...
49
F"· .....
1
~··
tr;r---~-
_______ ANSI.FORT~.O.N<2.3)/~AS"fER .......... !~HEGER WORD SIZE : : .. 1 , ..• CPTICN .IS .. OFF_:, . 0.. OPTION._IS. _____ Of.f...______ Q~/
r
~ j ~ ~ ....--·---~---··--·
g6~~t~~-q~F ~ ~ ~ ~ ~ ~- ( ~ h §~t-1 ~ ~.h ~.t~ L.. .......... . . ...............
0~03
0~04
C~GS
00 15
15
'
ooo6
K=1~8
SU~lKl=J~ .
R=3.1~1S~26536
---·-·-· ------ · · · n=1
n.
~
1 • z. ·1 4 • 3 • 1 s
· .... ···- · · · - ........ ---- ....... ·· .. · · - ·. . ··-· .. -·. · .. . . ·· . . . -- · .. ·-- ..
00 1 I=1CCu1,20001
02C7
Z=I
0~08
OJC9
8=~·(1.-ZIS.)
0 iJ 10 - - - . - - · .. C ==¥- ( 1 t Z
~
0~11
IF\~)
.
....
_...... __________ . . . . . . . . . . . ______
---------·
··------------~------
--
r
,
-··----··------...........
:·· .. ..............
... .. . ...
. .... ........
.. .
f(11=n/(1.-ZI8.l•(S!N(16~•Bl-SIN(8))+0/(1.+Z/A.)•(SINC16.•C)•
1SIN(C>>
Q~12
J~13
.
.
.
.
.
GC1l=D/(1.t71~.)•(C0S(16.•C)-C0StCl)-0/(1.•718.)•(COS(16.•B,•
D•; 14 -------·-· ·1 CCS ( 8 l ) . .
. .
. ------------------0115
EC2l=DIC1.-ZIB.,•<SIN(16.•B)-3!Nt2.•Bll+CIC1.+Z/8.)•(SIN<16.•Cl0116
. 1SINC2.~C)l
,
!) iJ 17
G( 2 ) 0 I ( 1 • + ZI '3 • ) ~ (COS ( 16 • "C)-COS ( 2. "'C) ) -0 I ( 1 , - Z I 8 • ) .., ( C0 S ( 16 ~ "6) ..
0, 18D:: 1 g
CC20
=
=
. 1 c c s (2 • "" ~ ) l
.
.
. - . ----~--- · - - E ( 3) ll I ( 1 • - 7. 18. ) >~- <SIN ( 16 • ..,. 8) - 3 IN ( 9. "'8 l + SIN ( 8 • • 3 ) - S I~: ( 2. • 8) )
1+01(1.+ZI8.l~tS!N<R.~Cl+SIN(16.¥C)-SIN(S.~Cl-SIN<2.•C))
0021
, G()l=rJ/(1.+71t3.l • (C0S(1S. 4 CitCO'S(.'\ ... Cl•COSt'3 • ..,.C)-GOSt2."'Cll
olJ 2 2 -------1- o1 t 1 • - z1 '3 • • ... <co s <1 6 • "' !3 , + co s < e• ~ e) - c o s <g • -'H~ l - co s c 2 • .>$. s > > .. - - - - - - - - - - - - - - · - - 0 G2 3
E ( 4 l 1J I ( 1 •- Z I 8. l .¥ { S TN ( :\. • ~) +SIN ( 16. • n)- SIt; ( 1 C• ,.I\ l -SIN ( 2 .... 8) lQJ 2 4
1 + 0 I ( 1 • + Z I '1. l '~' (SIN U~ .... Cl + S IN ( 1 G• "C l - S I~~ ( 1 G• "'C ) - S HI ( 2 • .,. C ) l
00 2S
G ( 4 l =DI (1 • + Z I 8 • ) "' ( C0 S { 1 S • ""C l + C0 S <8. + C ) - CC S ( 1 C • "'C ) - C :1 S ( 2 • • C ) ),
J ~ 2 6 -------···1- CI ( 1 • - Z I 3 • ) • ( C:) S ( 1 6 • "' ~ ) + C() 3 ( ~ • + R l - C0 3 ( 1 0 • • :3 ) - C0 S ( 2 • "" 3 ) )
... --·
=
EC~l=DI<1.-718.)•CSIN(8.•Pl+SIN(16.•Bl-S1N(10.~ll-SIN(3.•a))
0027
00 28
1 t CI ( 1 • t Z I '\ • ) "' ( S I N ( .~ ..... C ) + S IN (1 6 • + C l - S I N ( 1C • • C ) - S I ~J ( 3 • -"' C ) )
0 "l 2 9 .
G ( 5 l = 0 I ( 1 • t 7 I 8. l • ( C 0 S ( 16. :tt- C)+ COS ( /'l, '~'-C) - C0 ~ ( 1 0 • "'C > - C J S ( 3 • • C) )
0 J 3 0 .- .. -------··1 - C I ( 1 • - Z I q • ) + ( C0 S ( 1 S • + '3 ) + C0 S ( '~ • + 8 ) - C0 S ( 1 C • ll- ~ ) - C0 S ( 3 • "'- ~ ) )
- - - - - - - - - - ----'2 J 3 1
E ( F:, l =I) I < 1 • - 7 I 8 • > "' { S I t~ ( ~ • "· ::J ) +S I N < 1 6 • "' 8 > - S I r, (11 • "' 8 > - S I ~~ ( .3 • • l1 ) l
0 :l 3 2
1 + C I ( 1 • t 7 I 11 , l ... ( S I t\ ( "> • "' C ) + S It l ( 1 6 • . ,. C ) - S I :~J ( 11 • + Cl - S I N ( 3 • . ,. C l l
·0n33
G ( 6 l 0 I < 1 • + Z I .o, • l • ( C 0 S {1 S .... C l + C 0 S ( P. • -~"- C ) - C0 5 ( 11 • "'C ) - C•J S C3 • • C ) l
0 j 3 4 · - - - - - · 1- 0 I ( 1 • - Z I ~ • ) ... ( ~ J S <1 6 • •· 0 l + C0 S ( 8 .... t~ ) - C0 S ( 11 • "' i_i ) - C0 S ( 3 • + l ) )
Q0 ~ 5
E: ( 7 ) =;')I C1 • - Z I '0 • ) "' ( S I N C~ • '~'- r3 > +S I N (16 • • i1 l - S I t~ ( i 1 • .,. ::q - S I N ( 4 , ._ g ) )
Q'! 3 6
1 t 0 I { 1 • + Z I 8. l • (SIN <8 • "'C l +SIN ( 16 • ..,. C.)- SIN ( 11 • • C:) - S I' I ( 4. "'C) l
=
GC7l=01(1.+ZI~.)•(COS<16.•Cl+COS(8.+C)-C0~(11.~:l-COS(4.•C))
0037
or: 3 8 .. ·------·--1- o1 < 1 • - z1 s • > ~ <co s c 1 6 ... 81 + cos ( a.... B> - co s < 11 • "' rn -cos
00 3S
Q G4 C
'J041
0 : :.. 2
0043
0044
0}45
C ;: 4 6 . -
0147
OQ48
4g
00 C
J5
c
c ~ ... 11 > >
-------·--·--
E ( ~ ) =Cl/ ( 1 • -?.I 11 • ) + (SIN t 16. -'~- ~\' - S I ~J ( 12. • 1 l +S 1 N( 8 , •.J) -SIN ( 4. • 8) )
1 + 0 I ( 1 • t- ? I ~. l "" ( 3 IN ( 1 6 • -'~- ·~ l - 5 I :J ( 1 2 • .,._ C l + .3 IN ( 8 • • C l -5 It J ( 4. + C ) l
G (8l-=D/ (1.+ZI8.) ¥-{COS (16. •C)-COS l i2. •Cl +CUS (8,•C) -COSt4, ""C))
------- 1- 0 I ( 1 • - ZI o, • l >~- (CDS (16 • • 3 l - C0 S ( 12, ~ 8) +C0 S ( 8 • "'8 ) - C0 S ( 4 • • r:n ) .---------DC5K=1,8
FF=CMr..1 LXC::CKl,G(K))
A~S=FF
·---- --- S U~~ { K l
5
--··1
=CONJG(FF)
SU "-1 { K l +A t\J S
4
~(KJ=tC.•ALOGlOtSUM(K.l
CO~HT~~IJE
C
r:: C:cNT2 PllJ
5 K=E 1 '8 .. .. . . ...................................
""""C - - - · - · - - · - · - - · - - - - - ....... ---------------·---···""""'' __________________ ..
0851
WPTT€Cn1,1ClK,SUM(~),X(K)
QGS2
10 FCD~ATt1X,I2,4X,F11.~t4X,F11.4)
0 'J S 3 ·------· 2 5 C CNT I NU?. .. _ .......... _____ .. _________ .... -------------..---------- ...... __ ............-·---·-·----·--------------·
0054
STOP
.
.
GQ55
END
Ul
0
·-·~·---·--~---··-· -~---··
r-1
---------
1
9852.5926
39.9355
2
ges3.2S46
3~.s3s~
------- .. 3 .. -------- 1 g 7 0 5 !j3 0 ------- ------ 4 2. g 4 s 9
4
19706.2205
42.946C
5
19705.6912
42.·9459
6 ------19700_•.3105 ·------------42. 9447
1
1o?os.6gts
42.9459
8
197G6.2212
42.9460
~.
--------------··---------·--------- ..... ---- ·--------------·--
STOP
--------------------------
----------·--------------------------,-------·---------- ............................ ------ ·--,.---·-·-·------
"'
-------------------..-·-----------------·--· -----------------
__________ _________ __
,_
.;
.
..--·---------·-----:----·-:--------------------------'------·-----.-.-..;-~---·-------~----------:-
-----------------·--·----··---------··-·-·---···----~-------··---·------
. ·------···-·--------·"'-------------- ___ .. _____ --·---· .. ··-·-· ·------·..
c
-------·--·· ··---·
····---·--·--~~....._--
---------·--
_.,_
_____
,
________ ------
-
§
~
..
~
:~
"0
---···-··------------·
"
~
--------
-·-·-- ---·-•·------""e·-·· ----·--•"''"_. ,_,___ ·-• _, -··-- --· ·-·--··---·--:-----·-----·----·-·-r-•--·---·-- -·••\,
------ ----------·-----·-----------·--------- -------· --· ·-·-----·-------..--..._...__ ..____________
,_,
Ul
APPENDIX III
\
PROGRAM 6
52
-~
'
'
$
··:'
--------- l N 0!: 0 1 ________________________ D 0 Uc3l:: P ~::"CIS I 0 N
LN 0002
LN O~C3
LN OJJ4
$
ANS____________ .. ______ ..... ..
-------------------- -----
J=O
002J=1,9
SU~=l:.
--------------- L N OJC5 ____________ CO
e
SLJ_~,
.
LN 0006
LN 0007
LN 0~08
1I-=1UCJ1,200'~1,2
_____________ _
..........
0=0.4~g
--------- l N 0 0 cg ·-------------------LN ~S10
LN 0011
R=3.141592t536
s=6 J.
s
/(.D. T
E= 0 /2G.~J
8=(I+11•(~+3)
LN
J~12
C=<I-ll•(f:+S)
U-!
3~14
0~15
F=2.~c+2,•S
G=I~<z.•~+Sl
0019
012u
----------------------------
SUM=SUM+ANS
1 ,CONTINUE
0 C21 ________________ W); IT E ( 61, 11 l S U'-1 . __________________ ... ------------------- _________________ . __________ -----------------------------------------0022
1~23
OJ2~
11
FC~~~T(1Xt
~2G.6)
Y=10.•DLOG10CSUM)
W~ITE(61,~)J,Y
0 0 2 5 .... ___ . ____ .? F C'\ f'i AT ( 1 X , I 2 , 4 X, F.11 • 4 L .. ·------------0226
2 CONTINUE
cnz7
0028
·-------------------
-------------·-
ST0°
END
"-
··,·
----- h
--
-.---
•••••
·-.
•
-
·--
~
----------
· · - - - · - · - - - - - - ••
--------------
-
~
-
-
·
-
~
'
-
USASI FORTRAN DIAGNOSTIC RESULTS FOR FTN.HAIN
- ____________ .._________
·····---·-··--··
-----NO ····r:.·R RO ~ s··
~
G
----h--------~-
ANS=O-(CSTNfF))•~2•<SINC0)/0) ~•2-2.ti•S!N(Fl•SIN<OliD~tSIN(Cl
·
e
·····-···
iJ"' 17 ______________ 1 .. COS c.,r;) +SIN ( 8) "'COS ( B- G)))_ +0_/fLOAJ ti•!) • (lSI ~l( Cl) ••2 + ( S! N (in) .._.._2+_ _ __
0018- 12 .... SHH8l.,SIN(C)•COS(Fl)
- - . -···---------- --------
·
~
0~16
·~·-·
·
~
LN
LN
______ LN
LN
LN
LN
---~--------- l N
LN
LN
LN
lN
LN
LN
LN
-------·----------·-----------------------
TS=O.JE-5
--- ------ L N 0 0 1 3 . ------------ __ C= I "~- '3
A
w
------·-- ---------------
OBJ,LGO
G
---------·- ------ ___________________________ ·----· --·-------..------·---·-· ----------··--'·-...
,_
----~--"·-·-··--
.....
- ... ·-·-··--··---
--···· .... .,,_____ . ····-··· ·······-
__________. ___ _
0
0
0
-----------·
G
------- ----·
-
(.1'1
----·- ------------ .... --·· ·----- --------------------------- _________w____ _
&
'!ll!'
:J
J.12~834E+10
91.1~D3
1
- - - - - - - - - -f" •
2
---·3
-~
4
--------.---,
6 15 7 5 E +1 {) ---------·····S6e6612
~
----------···-----···-----------··-------- -----
~----~--·-··.
-· .•......
----·---·-------------
·-----------~-
0.875369f:"+1C
99,4219
0.1207S2E+11
1GIJ.81gQ
~ ~
13 3 3 ~ 2 E +11··-------------~-------,--------·-------···-·---------------- ____________ .,_____ -----------------------------------
101.2510
0.120636!:"+11
5
---6-------- 1 0 I] • 811f 8 ... --- --- ·-----------------------.
'
0.'373107E+10
\
gg,4107
7
·------- ---------------------
-----!) • !; 6 0 6 6 5!? + 1 0 --------------------------------------
96.633g
,8
~.1?.6918<::+10
:--:-··----g--------------- g 1 • f) 3 52
,,-.,
,_
.s l 0 p
--------·--·- ---- --·--·-·----------------------- ------------------------------------- ----------------~------------- ·-···--- --------·-----·-·--------------------------------
-------
..;
--------------------------------------·-·-·"-··-····-----------------·---------~-~--.-.----
.::
...-......... ________________
,
_,~--
. -·----·. ··-- ....... ---·····
·-····
-------------
£;
0
"-
'-'
s
:;
.;-;
"-'
0
::<
l11
. --- -- -------- -- ------------------------·-·- ----------;jS.-
-------- --· -·· -··------------------...
APPENDIX IV
PROGRAM 7
55
----ANSI
FOFi.T~AN<2•3)/MAST::R-
........
HHEG~R
viOP-0 SIZE:
= 1,"' OPTIOI\· IS -·OFF-· , ..-.Q .. OPTION ·IS--.:;
rq)
~
c
. 0 I ME. NSI 0 N XA F.t-U~ Y ( 51J 2) ,
CALL PLCTS<~,~ 1 0)
----·--lN 0 ~ 1 -
LN DJC2
LN ODC3
LN OCC4
oa~ 5
LN OuC6
---LN
}
---- ..-·
i., _,:
LN 0007
LN J~v8
LN (.)..; \j g
LN 0010
LN 0011
LN OJ12
- - -..··-----lN D:') 1 J
·.. )
LN 0014
LN OJ16
UJ 0115
CALL FACTORCG.ll
CALL PLOT<~.s,c.G,-3)
. ..·~···-··-·· --- -·-· -~- . I = CJ
J=O
OC1I=120~0,2G~J0,20
J=J+1
------------------ z=c • 1 57 6 9::: 111
ANS=Z/(FLOAT(J)¥FLOAT(l))
Y=1~.~~LOG1L{~N5)
XA~RAY(J)=I
.
- ---·- ---- ---- Y M:-\ A Y ( ,J) =1;:; • '"A1.. 0 G10 (A NS )
1 CONTINUE
.
CALL 3CAL~(XARRAY,60.,30C,1)
CALL SCALE<YA~R~Y,34. ,5QG,1)
----·----· LN D0 l 7 - ........___ ----· . C~ l L AXI S t .;.. • C , 0 •
1{5L2ll
LN OC18
·~,_)
YARRA Y ( 5 C2) . ......... .
LN 0019
CALL
LN 0023
END
:.J ,
11 HF Ri~ 0 I N N0 • , 11 , 6 0 • , C• , X A ~ R. AY ( 5 Q1 ) , XARRAY ------·---· .
~XISCC.,0.,6HLOGAN3,5,34.,9t.,YARRAY(5G1)~1ARRAY(5Q2))
CALL LINECXAkFAY,YA~RAY,Su~,l,J,C)
LN GC2J
-------LN D 0 21 ···---------------- CALL ?L 0 T ( C • , 0 • , 9 9 9 l ..
STOP
LN 0022
,_)
~-·"'·
--~·-····~--··-···· ·-·--·-~--~--
C•
.......... ·- ··-·-·"·
.::
E._,:
,,
~
<'
c
;,
'•
~ _)
..
,~..
:.:
__.
__.·
V1
"---...)'
-~----
.. ··- ..
---~-·-···
...
-······
..- ....... -- ..<?.'~....
j
57
(')
c>
<B
;:u
2
>
(')
0
3:
"D
c
-i
I'll
;:u
"tJ
;:u
0
c
c:
~
~VI
2
p
3:
>
0
"'z
c
"'
1-
Vr
i;
~~~~~~~·~·~~··~X~·~-~--~~·~·~"~,.~·~·~··~·~X~w=v~·~·~
·--·· . ------- --·-
---~~-· ----·---~-- ~---- ----~----·---
·-·-. -~-·----·"" --···--·--·- -----·· • ----·-· -
APPENDIX V
PROGRAM 8, 9, 10
58
·- ----·---·· --------·-··- --- ----.---------------- ···-·---- -- _____ .:..__""!
59
PROGRAM 8
•";,
J
ANSI
FORT?QN(2.3)/MAST~~
USASI
FO~TRA~
NO·
INTEG~~
DIAGNOSTIC
ERRO~s---·
WORS
R~SULTS
SIZ~
FJR
-----------------------------
OBJ,LGO
- - - --------·--··--·-- --···-·---·-·--·-----·-
--------·-----------·--·----·--·-·-·--·-···-·-· - ------·--
....
=1
, •
FTN.~AIN
OPTIC~
IS
OF~
~
60
C<
-----------------'--
71.2868
77.2216
8J.7f9S
1
2
3
z.
(
\
(t
·e 3. 1
~sa·----------~-----------
5
~5.1C24
6
86.6649
----1-------- e? • g e14_______________
8
q
--------
89.1169
c:JJ.1133
STOP
n
......
--~--!.·----------------~-------------------------
c
/
(0
rc
\
61
A~SI
FDRTRAN{2.3)/MASTER
INTEGER WORD SIZE
----------------------------------------------l~
00{)1
l~
01G8
lN
LN
ll\
LN
0011
=1
, • CPTICN IS . OFF , C
------------
LN GOG2
------lt\ 0 !J c 3
LN OOC4
U\ (jQ{j5
LN
OOC6
--------- l N
0007
LN OG!J9
lt\ 0010
OGle
OQ13
0014
lt\ OG15
Lt-. n~ 16
Lt-. 0•)17
lt\ 0018
LN OJ 1 g
Lt-. (020
LN 0021
lt\ (j ~ 2 2
-- l N U023
lN 0024
LN 0025
Ll\ QJ26
LN f;j 27
LN GQ28
LN 00 29
LSASI
FORTRA~
DIAGNOSTIC RESULTS FOR FTN.MAIN
-- --------------------------------No ERRCRS- ------------------------------ - -------------------- -------OBJ,U;o
--------------------------- ------------------ -- ---
dOJ.S
6
***********
***********
9
10*7£•oo. _ -··----·- ***A:*******--··· _l
OL-t,L•COl
***********
9
S~Ql•TOT
***********
S
2:l~t·rot _ ···-·-·-· ***********· .... ..,
2L*7£•o6
***********
£
tsts•96
***********
2
£6 6&_~_o&__________ 1r'!'** * * ***** ________ .!....:..
T&96·c&
Tzgs•gs
(
1::9
63
PROGRAM 10
·.- ........
~NSI
INT~G~R
FORT~AN12.3l/M~5T~R
--------------------------------------·
-------~
~---
-__..-- -·
~---
WO~O
____ ..., ---
~-
SIZ~
- -·---<--·-·"-•
=1
-------------------- -- --·· --- ----- -- -----------
-
·-- - -· ... _._
~--
,
----
.
~
-----
LN
lN
LN
LN
LN
LN
UJ
Lt-:
UJ
LN
LN
LN
LN
LN
LN
LN
LN
LN
LN
LN
LN
_______ LN
LN
LN
LN
LN
LN
LN
----------------·usA S r
!=Q;:;_IP il f\- CTAGNOSTIC R::: SUL TS- FOR FTN • .._,AIN
ERRCRS
- 09J, LGC_______________________NO ----- - --------------------------- ----------------
i
- I
-~
---· -·.
~- ------~-
OPTIO~
.... ____ ._..__ ___ _,
IS
-- ---·-··-·-- ------- --
..
~
....... --- •. - ......
-~-
OFF , 0
64
----91·---------
----·-ga cftg:;_____
~
ag.n7~6
92
93
87.9373
-·----·-·-··-~·---
9:4
95
96
97 ______ _
86.613~
tE>.0403
83.1C73
. -- 8 0 • 6 5 T- ----------------------- ··------- --- ______ __,_ -· ---- .. -- -- ----. -~
77.D656
70.974'3
98
99
STOF
/
-----------· -------------------- ·----·····-·· ---· ---·
---·····
-·-···
.. ./.