10
The
abril
-
2023
-
remin
Escirut
he en
agine
What
would
be the
equivalentcircuit
left
to
the
R
before)
From
Reg
or
L
Rell(Ry Ry3 ReCR,
+
=
=
+ly)
x4]
k2 (23
+
+
thevenin:
for
mo
Ot
I
Equivalents.
The
renin
Ra
S
R4
=
Circuit
of
-Da
I
given
circuit
-o
d
+
+
1
+
Re
-g.
-
b
-
Steps
a)
Compute VT
we
obtain
1
+
the
voltage
between
a,
b
nodes
R,
-I open circuit)
Quinte
V
UTH
+
H
est
*
abierto,
Ob
-
atantior
por
van
d)
&T#
la
1
Law
I
sources
Usources
are
shorts
are
open
si
no
linea.
VR,
=
V,(x)
=
Voltage
Law
ke
in
an
por
Ve
=
ahie
corriente,
WTH
r
V,
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+
Dividen
vat-> I
p
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mini
->
6
q
A
Riz
h,
=
11RuR,z
R3
dog
a
-
Lo-
b
isits
Draw
Thevenin
the
R
=
s
+
+
+
1
⑰inSre=e
-
Once
Van
the
V
=
+
circuit
is
=
y
+
Fey
Pry
I
=
&4
simplified,
we
Ir
(0Hn)
Power)
could
find:
Ample The
12
venin
-
abril
2023
-
-
Given
Circuit:
R3
Ri
diningto
t=
,
e.g
Il
UTH:
RTH 1
+Rs
>
questE
=
Ho
-
+
at
UTH
-
-
-
No
*
Source
transformation
hay
voltage
un
dropin
t
F
-
I
*
z
UT=
1
-
&
2
(Active
(4)
-
samatoria de
voltios individuales
*
Unknown
There fore,
m1i
sign)
V,
Ia,2,
-
+
-
Iq4
Ia
#
Is, Ir
=
Be
into
(Fx F1)
-
R, Ikz
-
,Ix
-
v,
I(1,
=
~,
IR
+
=
0
Fe
kzIrn
+
-
R2)
-
, It
(k, R)
1.F
+
v
,
2,
B
+
+
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+
FA
kz
+
A
-
A
=
=
-
0
=
=
INEOUT
=
Vi
Vi
Far
+
er
Substituting
UTH
Therefore,
n
Alternative
way
I:
->
UTH
0
Vn't Ibe
=
bat
obtain
kr L
->
1
Van F,lz
I, analyzethe mesh
-
To
+
=
m,:
(Active sign)
V,
-
>
*
Consume
IR
W
Inz F,R,
=
I,R
+
#(Re)
-
-
I
=
Unknown
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=
UTH
#
1,R Ixl2
F+
2
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+
=
(+Ia
a Fa
=
+
-
↓
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+Iaa El
-
(2, R2)
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[V,
Vi =
2].
I1
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(2,
(2)
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'
P
anyo
-
I
VTH
·
↳
↳-
T
UTH
Vi
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beg
I
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=
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v,
=
->
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·
UTH
=
)x]
I
+
Vi
R,
~
Ng
↓+Y
⑧
a
V
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+
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ab
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gumin
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Vz
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I
Green
Norton
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ge
Ry
Ir
mple)
&
bu,
k
=
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H
=
Obtain
directlyIn,
R
R,
m
⑧
↓
-
LAW
-
>
shortat
tetput
te
⑧
Now
consider
the
,
with
circuit
is
↳
isresi
·
*
C
it-ite
x
1
=
1
samethe
MaxPower
Imagine
Transfer
complicated
a
converted
was
I7-abail-2023
circuit
nations
PIW
=
UTH
no
RroA=
ROAD I."
-
P
der=
VTHF,
or
I, U+H
I
would
like
minimize
transfer
maximize
power
obtain
Ron
to
max
to
the
load
the
transfer.
loss
and
to
With
-
-Gas
=
Example,
61
organo
4
a)
find
or
the
source
power
associatedwith
the
6) state
whether the 60
absorbing or
delivering
is
source
power.
P Ir
Procedure
=
e
·on
the
-r
61
FR4
39
y
3
se
Suman
resistores.
10
-
-
los
tion
·
or
0
GV0
-
I
u
161
+At
60
19.28
-
a)
I(6U);
--+X
onex
P Iv
=
-
=
- - -
161
-
-
out⑧
-
19.28
Ilo
⑱
E
r
-
THEN I NB
R
=
I
↓
=
ENs
=
↓
-
=82A
Therefore,
3 746V ( 0.821](6r)
-
=
=
Pr
-
=
4.92
W
-
b)Asorbing
or
-
Delivering
(andaber
Activge iveon
conventi
sign
Position
Super
&,
In
-
get
↳3 ar
I
sources
are
open!
↳sources
are
closed!
deni._rize
L
Ine
I
In,
Ic
=
Un Vn,
=
Practice
I
R,
Circuit namet
+
Vna
sources
+
-
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sources)
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P FU
I I,(u,)
me
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t
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2nd)
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usgene
tumer
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e
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Inductors
("Bobina)
Inductor
In
Capacitors
and
-
one
Store
o
dependenciat
del
-
!
tiempo
Vilt)
L
=
Energy
(H)
2:Henry
I
103
- voltage
the
at
equal
is
inductor
the
to
value
variation
times
L
of
i(t)
orgrties
a)
If
in
constant
THEN
*
i
= >
u
t
constant:
is
x
0
=
0
=
0
=
"
- short
&
b) If
is
changes
0
=
Wi
(t)
has
zuhen in
NOT
> @
charse
be
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more
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value
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will
c) i, (t)
a
ic
Higher
changes
jumps!
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#
17 -abril-2023
Crando
ke
libera
T
circuito
se
corriente,
2
E
=
opposition
el
aprimos
to
L 1 Henry
5
=
=
motion
too
-
much
resistance
~ Inertia,
&
Momentum
00
Prest
Opposite
to
motion
⑱
ete
o
or
ascending
to
descending
Carset or -
↳
V IR
=
-ir(t)
t,
I t
i(t) i)
=
Entaneous
-
---)
=
⑨
R
(0)
Stability
ex
Initial Conditions
↑
o
to
=>
for
Analysis
=
f
t
Time
1
=
:
Inductor:
with
circuit
a
*,+(t)d
i(0)
e
*
constant:time
exponential
factor
become
ti, t,
Initial Condition
5 1
in
↳
+
=
-
⑧ t
which
ve
(I.C.)
t
oomt
O
Given
in10
in
Amplitude"
these
4
107
=
I. C
Currentsource
a
I
plot:
,
x
=
x
IfDecay
and
THEN
is
1
L notcommentireintheI
(Initial
Condition
S
?
=
modeled
is
source
our
conditions:
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as
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ett)
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5
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in(0t) i,(07)
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[10.(-5). t]
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·
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1100)(10-3)
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50
0.1H_
OSest
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12
!
-
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⑱
i
sa
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10(2
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st
=
10
it=
H
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0
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Le
-
t
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(0)]
1goet-1
I
↓
--
1
102
St
source i
-
st
Capacitors
ie(t)
C:Farad
->
I
-
ic(t)
V.
(
(t)
c
=
V(t) i)!i,(t)dt v(0)
+
=
e
I.
Properties:
wiLA)
a) If
constant,
is
THEN
i,
(t)
-0
0
=
=
b)
If
c)
viCA)
fast
will changes
THE i,(t)4q
can
4 RC
=
NOT
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[1.c seconds]
=
open!
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As
. . . .
bi
Inductors
time
passes
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3
t
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Data
or
i(t)
0
=
100mH
W
=
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to
in(t) ?
=
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Procedure
i(0)
n(t) di
=20.55 (102)
i,(t)
5
est]
10
-
+
d
10
+
10
10(e
-
↳
10(e
=
↳
+
0
=
-
styt
sit)
-
10
=
-Sest
=
1))t
=)(-se]
=
I. C.
=
u
10
=
i(0)
=
(0.1)(6)1-
=
10.i(t)
find:(t) 4
=
10-
I.C
St
gs(0)]
!x e
=
10(e
-
()
=
+0
=10
ic(t)
2]
10
+
-
10
=
-
st
e
-
i,(0)
-
*
s0)
10
=
=
e
I
in(t)
=10
one t
OlkitGoose
atural
of
R
Analysis
A)
- ist
X
m
-
Un
V is.
=
(0) I6) Initiality)
v(t)
L
=
I.C.
krL
S
n,
vx(t)
Y
+
it)R
C
-
k
=
0
Consume
-
0Eorder)
=
is
R
=
Supply
0
=
1)
+
vi
F
=
Mattal
⑭
i(t) Io
=
e
ft
R
u 63.89
i
,
general
In
I
-
=
-
ait)a i)t)s
-
*
at
xi(t) te
A, io obtained
=
I.
11
at
C
i(0)
a(0)
i(0) A=
=
-
11
in
(1)
I=A
A
-
Step
o
--
o-
response
t-I.
o
=
a)
Asis
M1
=
Consume
(mp:
=
V
vx
=
v,(t)
+
i(t)k
=
b)
MATLABsolution
5
v
-
i(t)k
=
=
L
-
⑭
[v-(
i(t)
:(t)
C)
V
=
-
-
[r
10
-
2]
+
-
1
ie
+
exp())/
+
I0R] Et
R
I (u ok)
=
=
(
+
-
t
V IR
Draw
=
I
nee
............
-
A
-
et
t
intere
I
e
↳mpte
L and
associations
aero
eee:
↳
·
Log
Series
o
0
-
=
cer
-
cert. Ial
4
ring
a
-
--
Log
-
·
Li
I
0
a
5
=
-
-
Leg
in
-
4
=
C
+
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mewor
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response
t
-
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of
e
is
Many
hir
#
=
I
H
C
~(0) e,0
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i.
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Vo
=
2,(t)
E
L
=
⑧
-
T
ey-w
W
-
--
iR ic
KCL
=
=
KUL
a)
b)
-----
Analysis
solution
c) Drau
·
Step
Wi(t)
=?
Response of
in
(MATLABL
RC
ic
It
·r=
a)
Analysis
b)
solution
c) Drau
We(t)
=
?
->
(MATLAB)
Basic
Es
I
-
for
Math
Imaginario
Al
26-abril
Circuits
TE,
Or)
LG,
a)
(.0,
E
=
a
z
zY
-ke(z)
Polar
3
1)
Z,*
RE
0
2 Rx0
-
ImaginaryNumbers
Dj
500
j
=
form
Reo
=
Rk(0 180)
+
=
=
=
Ejemplo:
-
360° 1460
=
F
=j!.j
=
Ij=
j
-100
=
-
6)
j
Complex
a
=
zx
=x
z
=
-
z
j
=
2
Polar
form
-----j
j
+
-2-
Find
z
1
=
a
to
Change
12
=
=>
jb
Practice:
F
Conjugate
jd
+
-
-
(j4)2
-4x25
=
=
=
1
1
b
=
25
z
j
-
-
circuits:
i
jb
Rk0 Rk(0 360)
3)
In
2022
+
a
=
-
Rej8 RKO
=
=
i
Re(z)
Find
z
Rej8 RKO
=
=
1:R
Step
Step
2:O
(+
b*
=
Segundo
(-)
=
cuadrante
=
=
Step
1:
180-tan"
L
=180
(i)
153.43.
26.57
-
32.23
=
=
2.23k 153.438
·z
2)
v5
=
+
(I)
tan"
=
⑦=tan"
x
(
=
=
-
243.430
S
G
R N2
=
R x()
x, 74.47
=
=
-
Step
tan'(a)
2:8:
0
fercer
tan(t)
=
cuadrante
2
=
-
180
tan(E)
+
--116.56°
=
(W
+300w
243.43°Cy
=
4.47
or
③
2k
120° 2ej(120)
=
=
z
-
or
2cos(-120) +j2sin(z)
z( z)
-
z
e
-
=
1
jz) )
+
-
-
j1.73
4) V(t)
750
=
(1600t
cos
find:
a)
Radians
frequency
in
c) frequency
in
b)
d) period
of
2) Phase
f) Phase
angle
voltage
of
amplitude
Max
signal
in
angle
in
signal
second
per
Hertz
the
+30) volts
seconds
is
degrees
radians
Procedure:
a)
(cost
f(x)
If
=
750cos-
v(t)
=
·
Umax 750(Volts)
-
350-
=
b) v(t)
7SOcos(t 30t
+
=
p
COS
Then:w
cos(wt)
=
Then
2) I
cos(wt0)
1600 radIs
F
=
=
se
30°
=
=
=
I Tf
=
O
=)
I
e
est
d)
Other
examples:
the
following
Consider
elements
resistor
b) 32 mH
inductor
as
9001
-)
SMAcapacitor
If
Al
find
frequency
Procedure:
R:zc R
=
zn
:zc
5000
Sec
Imperance
the
each.
(:
is
admittance
and
of
90
=
jw)
=
j(5000)(0.032)
=j(00(2)
=
j
=
=
t
jwa
=
=
10.000005)
40(c)
15000
=
Admittances
2:y
=
=
0.0111s
1
=
11.11ms
=>
zR
2:y
C:
Y
=
==
=
E
=
=
I
j0.006255
=
-
j(40)
=
-j40
7
=
j6.25ms
-
j0.025 7;asms
=
=
Perez
Cruz
James
3>
Prof. Edgar
123507
#
Homework#2
⑪
·
Natural
t
->
-
pum
-
formulas
~(0) e,0
=
I.C
Vo
=
L
=
↓t
V(0) Y
=
=
-----
Wi(t)
solution
c) Drau
ab
current
on
=
Analysis
b)
D. E
Base
v (t)
=
KUL
a)
w
iR ic
KCL
unknown
-
T
=
try
voltage
I
is
m,
14
2023
Collado
V(t) ?
↓
X
on
-
=
=
C
-
Based
is
W
I
Analysis:
RL
of
response
0-abril
=?
(MATLABL
V IR
=
Us is.R
=
Vc(0) Vc,
=
ic(t)
kUL
->
KCL: is in
Y
=
o
c
&
Va Vc(0)
+
R
=
=
m,:
itt)
In OUT
=
c
+
=
Supply
0
=
dVcCt)
dt
Vi 0
0
=
D.E.
=
Consume
-
i(t)R
=
- it
MATLAB
d
As
bymatLAB
i(t) I.
=
-
e
2t
A
->
inos.,
·Responsesofe
ic
mo-
I. C.
Fri
->
a)
b)
Analysis
solution
c) Drau
ab
kUL
Wc(t)
3
=
=
(MATLAB)
?
m,iV Va
It ict)
=
=
consume
Y
+
+
bL MATLAB
& =
Di
(F= (F1
-
-
i(t) Ia
=
10
-
a)
x
[I1
-
exp(*))]/
x
IoR] Et
R
i(t)
)
E
=
-
(F yF)c
=
DRAW
-
Et
V IR
=
-------
T
=
A
Assymptote horizontal
Graph
provided
by
MATLAB
RC
Circuit
1
tresponsi
-
mayo
-2023
KCL:IN OUT
It
->
ic
=
IA FR Ic
=
+
to
-
N
KUL
1
&a
Natural
=0
-
W(t)
-
+
=
11
first order) Homogeneous
=
i(t)
&cult:
-
=
qi(t)
Akt
=
me
A
2
Fawatt
11
Ir Ic
0
( (i
e
0
=
Respons_ +
=
=
Vi
+
=
0
N
THE
m1:2
ALL: IN=OUT
Is
Ir Ic
wr
rate
If
=>
Trial
or
function
b
t
=
et
de
t
k
=
Cardinate
If
W,
not par t
for
Then
U
natural
E v(t)
Step
+
Wc,
=
tot
v
=
response
rom
ct
Response (A
type
Responsel
forced
of
risteristic
KCL IN 0LT
ic
->
=
=
I Ie+In
=
18
↓
11
M
x
=
),
cdi
e
+
-
y
(t)=k) E
+
-
at=
1
itNor
(1storder)
D. E
2, ToT
-
*
>
Vc,
Vc,
+
-
=
Non-Homogeneous
-
&
=
and
t
=
0
homogeneous
H
for
Homogeneous
/Step Response S
↑om
A
Exists
Non
t
L
P.
t., Evaluate
at
aka Segndo
*
-
If
ist)
xame
fant, THEN upp
is
constant
is
sinusoidal, THEN M, p
simusoidal
is
Also,
Ie
If
Y, TOT Wa
V,
=
AND
STEP
t
As
-
it
LANDIDATE
Response
so
t
As
+
p
future
e
->
11.
PAS T
-
L
·
D
W,
=
t -
·
=
As
t
I
AR
-
0
=
*
If
W(0, Y,
THEN
~(0)
E, i
↓
+
↓
io
=
+
A
Be
t
it
+
11
-
()
ta
-
~
Fa
=
Tor
=
=
I
V(t)
Be,(0)
+
-
④
V IR
Fa
=
IAk
0
B
+
=
.B Vo
I
-
=
·
Patting
⑪
v(t)
Vc,H
2,1
=
+
(vo
(() IAR
=
If
Together *
All
it
current
0
=
Capacitor
in
②
desired
is
ca
If
i(t)
THEr
1r]e kt;A
-
+
=
c)F
()
(k I1]e)
=
W
⑧
(0
=
(v
+
-
+
=
↑
-
)( s) t
F
-
1
YRCt
F(y)e
it
S
v)
-
-
=
-
-
+
i(t) (Ir
-
=
RL
e
g)et
R
↓
I
in
Circuit
to
were
Circuit
KVL
I
L
m
onsume
m1:
=
i
m1:v, Vs Vv,(t) vr(t) n(t)
=
+
=
+
↳libk+
use
=
-is
ratil
Wi
"
x
=
ve
V
V
+
A
=ji +
& &in
0
D.
E
=
Homogeneous
If
=
THE
-
in
·
--
ui(t)
Al
i(t) Ac t
=
Ae -t t
t
te
-
()t
⑯
-
)
=
ie
hesponse
a
If
i
L
THEN
(A)
ToT
Vc,
We,
=
in (A)
evaluate
t
and
H
+
t
Tor
as
-
0
0
=
Past
-
Centerenecom
Tratag
la
↑wtere
-
amette
-
-
If
i(A)+T
in i,
+
=
THEN
)it) or=A Be it
+
=
+
e
Candidate
i. (A)
Evaluating
in(t)
A 0
at
=
(see future)
( g,
=
in
=
⑪
Ast
int)
in,+in,H
to t
=
A
=
in
1)
0
=
Be
Be/10)
+
+
=
un
in
4
B i
=
·
B
+
=
-
E
-
There fore,
]e*; =
+
Si
①
Entice.
Ous
LI
#
②
⑳
Examen
#2
A
-
1) Analysis
2) ic
?
3 Draw
=
1)
I Hand)
Analysis
2) Ur ?
3) Draw
=
4)
Find
(paid)
is
Thevenin, Nodal, KVL, KLL, Mesh current, AL
&R Basic Math (complexconjugate, imperance
tedioso (
E. D. (10
mas
Practice:
⑲
1) Analysis
2) ic
?
3 DRAW
=
X
(Hand)
①
Analysis
KCL:n1+is ic
IN OUT
=
=
o
=
0
kVL
=
m1
t
I Ic
V,
-
+
+
Ic
Ei
Ac
Vs
bt
=
-
qi(t)
Akt
=
=
t 0
=
-
Ae
Lu
()(0)
-
=
11
n(0)
i(t)
0
=
=
+
=
un(0)
hallsensor
- gder)
In
X(t)
encoder
⑪
v
=
A
=
X
b
=
c
=
p
v
v,
B
=
B V
+
=
Vc(t)
-
V,R
v2, 2,H
=
+
[Wo -V, R)
V. (A) v, 1
+
=
lo
i
(r
2,
-
y
V,k]et]
zeE
(0
=
+
ert;A
v
(wo
+
Y,R]( )
-
-
-V(- )
(Vs1
=+
vo)
+
Re
is(t) (V,
=
-
e
e
etct
) et
t
⑦
2) Analysis
3) Dr"
(and
4)
I
-
·
Find
is
a) Analisis
XC1:n1
=
I=
I
n2
KWL:m1
=
Ir In
Ic
1
0
+
FIN 0uT
+
t
It
-
+
=
+
It
0
=
-
0
=
w
r
+
[int)r +
=
+1in
-I
->
i(t)
i,(0)
-
=
xe
=
i
:- kit)
(A)
i(t) te t
-
Ae(0)
=
i,(0)
is
0
=
A
=
A
=
-eioet
=
Ae
-
t
Ac-Et
*
serve:
Lets
↓
kVL-m:
(active)
Vi
V
v
0
=
x
+
=
0
=
Ve
Here
⑤
t
c
KCL
-
=
v
b
tion
!
V
*
-
W FR
=
I
in
=ic
Wait!
stop!
I
are
=
11
11
D. t
=
-
Step Response
T
-
-
U,
TOT
U,
=
p
Yc,H
+
-
Existing
Non
t
s
Homogeneous
Series
A
Natural
R
it
I
Response
Step
and
r(t)
o
Note!
We
this
In
I
need
case,
(A)
Stative)
IR
I I
R
v
=
circuit
L
i,(t)
c
=
=
to
select
I
choose
differential
a) obtain
KUL 0 m,:
=
RLC
formulas
Guide
-
of
variable!
E
consume
↳natic
t
2, V(t)
=
11
=
i(t)a
V(t) x(t)
+
+
t
Gindt
+
+i
L
+
ict)
indt+i
r
Le
+
+
11
It
I.C
V
ix(t)k
=
If
W=
, (0)
1)i(t) Le
0
=
are
+
+
C;L-
Imtegration
;
Differentication
constant
Pitti
and
Eel
&
THEN
-
↑
as
I
Rearranging
=
↳+
int)
2nd
Let
us
follow
i
D
an
order
D. E
association
for
procedure
- "s" a
=
la
b.
potencia
S
There
fore,
x +int)
+
0
=
de
the
Is
Sh
ww
&madratic
formula
In
are
S
-
12
general,
3
C
b
a
=
b
IA,
ac
2a
there
possibilities
5, S2
=
Real
Solutions(b =Yan)
↳fre
Example:
SE Reaoutiopations
1
(b
>
4ac)
(I h
R 1kR
=
1 1 0(
+
+
1mIt
=
=
c
find:
2uf
=
?i? 8i
S"
* if
+5
+
"
*
Sie 10"
10
I
+
n
u
S
=
-
-
0
=
=
0
e
Atac
b
tz.10
100
=-
Ix10")
u
Y
5,
xx10")
-
=
=
-o
5
*
Another
,
Si
example
1
+
=
-
10
I
+
k
=
52
0
=
-
=
=
-
100
-
5,
0")
999,499.75
(continued)
RLC
formatfor
Alternate
i
+
int)
R5
52
Solution
0
+
I
0
=
=
+
f
Therefore
=
for
-
problem
the
Soh
R
I
S
I
ver
a
=
-
series
0
=
L
bac
w
(E)
=-
(RLC
↳
b
es',
Hand
at
+
f
e
(eral)
IES
2
4(i)(i)
=
all
) E2 Neper frequency; (*))
-
where:
=
-
=
Wo
te(Resonantfrequency;(w])
wor=
3
cases:
2 wo
=
2
s,2t
=
> w s,,,
=
I
2
-
=
=
-
Double
Root
S
critically
Damped
2
I
C"w" s,,
Real
S
Real
Over
d
DistinctD
ampe
Roots
jawan
-
2I
wa
/Complex
conjugated) med
TextBook
From
Model
RLC
Solutions
i(t) A,
=
series
(as
assumed
variable
et
Azet (over damped
+
i(t) B,e tcos,
=
1
Brettsinn,
+
(Underdamped)
i) =D, Ae-2A D, et
critically
Damped
L
+
Note:
Each
I
case
a) A., Az
is
!
obtained
from
t
initial conditions
S
Emple:
The
1000.
0.1ufcapacitor
through
inductor
find:
At
a
t
0
=
series
AND
a
a)i(t)
m (t)
b
the
charged
capacition is
is
combination
360 resistor
of
a
A =0
=
t
0
-
Circuit!
t
he
as
-Notetoat
t
-
-
lour-
"(0) )
-
I
discharged
100mH
Procewr
Draw
to