Yaser Zhian Fanafzar Game Studio IGDI, Workshop 07, January 2 nd, 2013.
06 Yaser Rahmati
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Transcript of 06 Yaser Rahmati
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NATURAL AND STEP RESPONSES OF
RLC CIRCUI TS
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NATURAL RESPONSE OF A
PARALLEL RALC CIRCUIT
V
R LVd I C
dV
dt
R
dV
dt
V
L Cd V
dt
d V
dt RC
dV
dt
V
LC
t
1
0
10
10
00
2
2
2
2
Ordinary, second-order
differential equation with
constant coefficients.Therefore, this circuit is
called a second-order
circuit.
CL R V
++
V0
I0iciL iR
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GENERAL SOLUTION OF SECOND-
ORDER DIFFERENTIAL EQUATIONS
Assume that the solution of the differential equation is of
exponential form V=Aestwhere A and s are unknown constants.
As e
As
RCe
Ae
LC
Ae ss
RC LC
st st
st
st
2
2
0
10
( )
This equation can besatisfied for all values of t
only if A=0 or the term in
parentheses is zero.
A=0 cannot be used as a general solution because to do so
implies that the voltage is zero for all time-a physical
impossibility if energy is stored in either inductor or
capacitor.
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THE CHARACTERISTIC
EQUATION
ss
RC LC2 1 0 is called the characteristic equation
The two roots of the characteristic equation are
sRC RC LC
sRC RC LC
1
2
2
2
1
2
1
2
1
1
2
1
2
1
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If either s1or s2is substituted into Aest, the assumed solution
satisfies the differential equation, regardless of the value of A
V Ae V A es t s t1 1 2 21 2 ,
These two solutions as well as their summation V=V1+V2satisfy
the differential equation
V A e A es t s t 1 21 2
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FREQUENCIES
The behavior of V(t) depends on the values of s1and s2.
Writing the roots in a notation as widely used in the literature
s s
LC
1
2
0
2
2
2
0
2
0
2 1
,
=1
2RC
The exponent of e must be dimensionless, so both s1and s2(and
hence and 0) must have the dimension of the reciprocal oftime (frequency). s1and s2are referred to as complex
frequencies, is called the neper frequency, and 0 is called the
resonant radian frequency. They have the unit radians per second
(rad/s)
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The nature of the roots s1and s2depends on the values of and
0. There are three possible cases:
1) If 022, both roots will be complex and, in addition, will be
conjugates of each other. The voltage response is said to be
underdamped.
3) If 02=2, roots will be real and equal. The voltage response
is said to be critically damped.
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EXAMPLE
CL R V
+ Find the roots of the characteristic
equation if R=200 , L=50 mH, and
C=0.2 F.
1
2
10
400 0 2125 10
1 10 10
50 0 2
10
125 10 15625 10 10 5000
125 10 15625 10 10 20000
64
3 6
8 2
1
4 8 8
2
4 8 8
RC
LC
s
s
s
( )( . ).
( )
( )( . )
/
. .
. .
rad / s
rad
rad / s
rad / s
0
2 2
Overdamped
+
V0
I0
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Repeat the problem for R=312.5
1
2
10
625)(02
8000
1 10 10
50 0 210
8000 0 64 10 10 8000 6000
8000 0 64 10 10 8000 6000
6
3 6
8 2
1
8 8
2
8 8
RC
LCs
s j
s j
( . )( )
( )( . )/
.
.
rad / s
rad
rad / s
rad / s
0
2 2
Underdamped
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Find the value of R for a critically damped circuit.
For critical damping, 2=02
1
2
110
1 10 102 10 0 2
250
2
8
4
6
4
RC LC
RCR
( )( . )
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THE OVERDAMPED RESPONSE
When the roots of the characteristic equation are real and
distinct, the response is said to be overdamped in the form
V t A e A es t s t
( ) 1 21 2
The constant A1and A2are to be determined by the initial
conditions V(0+) and dV(0+)/dt.
V A AdV
dts A s A
V VdV
dt
i
C
iV
R
I
c
c
( ) ,( )
( )( ) ( )
( )
00
00 0
0
1 2 1 1 2 2
0
0
0
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EXAMPLE
CL R V
+ Fin V(t), if R=200 , L=50 mH, and
C=0.2 F. V0=12V, I0=30 mA.
+
V0
I0iciL iR
i I mA
iV
R
V
RmA
i i i mA
dV
dt
i
CkV s
L
R
c
c L R
c
( )
( )( )
( ) ( ) ( )
( ) ( )
./
0 30
00 12
20060
0 0 0 90
0 0 90 10
0 2 10450
0
0
3
6
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From the previous example, we determined s1=-5000 rad/s and
s2=-20000 rad/s. Then
V t Ae A e
V A A
dV t
dtAe A e
dVdt
A A
A V A V
V t e e V t
t t
t t
t t
( )
( )
( )
( )
( ) ( )
1
5000
2
20000
1 2
1
5000
2
20000
1 23
1 2
5000 20000
0 12
5000 20000
0 5000 20000 450 10
14 26
14 26 0
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0 0.5 1 1.5 2 2.5
x 10-4
-6
-4
-2
0
2
4
6
8
10
12
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i tV t
e e mA t
i t CdV tdt
e e
e e mA t
i t i t i t e e mA t
R
t t
c
t t
t t
L R c
t t
( )( )
( )
( ) ( ) . ( )
( )
( ) ( ) ( )( )
20070 130 0
0 2 10 70000 520000
14 104 0
56 26 0
5000 20000
6 5000 20000
5000 20000
5000 20000
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THE UNDERDAMPED RESPONSE
When 02>2, the roots of the characteristic equation are complex,
and the response is underdamped.
s j
s j
d
d d
1 0
2 2
2 0
2 2
( )
dis called the damped radian frequency.
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V t Ae A e
Ae e A e e
e A t jA t A t jA t
e A A t j A A t
j t j t
t j t t j t
t d d d d
t
d d
d d
d d
( )
( cos sin cos sin )
[( )cos ( )sin ]
( ) ( )
1 2
1 2
1 1 2 2
1 2 1 2
A1and A2are complex conjugates. Therefore, their sum is a
real number and their difference is imaginary. Then j(A1-A2)is also a real number. Denoting B1=A1+A2, and B2=j(A1-A2)
V t e B t B t t d d( ) ( cos sin )
1 2
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DAMPING FACTOR
The trigonometric functions indicate that the response is
oscillatory; that is, the voltage alternates between positive and
negative values. The rate at which the voltage oscillates is fixed
with d. The rate at which the amplitude decreases is determinedby . Because determines how quickly the amplitude
decreases, it is called as the damping factor. If there is no
damping, =0 and the frequency of oscillations is 0. When
there is a dissipative element, R, in the circuit, is not zero and
the frequency of oscillations is, d, less than 0. Thus, when isnot zero, the frequency of oscillation is said to be damped.
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EXAMPLE
CL R V
++
V0
I0iciL iR Find V(t) if R=20 k, L=8H, C=0.125F,
V0=0, and I0=-12.25 mA
12
102 20 10 0125)
200
1 10
8 0125)10
9798
200 979 8
200 979 8
6
3
6
3
2
0
2 2
RC
LC
j j
j j
d
d
d
( ) ( .
( .
.
.
.
rad / s
rad / s
rad / s
s rad / s
s rad / s
0
0
2
1
2
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V(0+)=V0=0, then iR(0+)=V(0+)/R=0.
ic(0+)=-iL(0
+)=12.25 mA
dV
dt
B B V
V t e t V t
d
t
( ) .
.
,
( ) sin .
0 12 25 10
0125 1098000
098000
100
100 979 8 0
3
6
1 2
200
V / s
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THE CRITICALLY DAMPED
RESPONSE
The second-order circuit is critically damped when 2=02. The
two roots of the characteristic equation are equal
s s RC1 21
2
For a critically damped circuit, the solution takes the following
formV t D te D et t( ) 1 2
D1and D2are constant which must be determined using the
initial conditions V(0+) and dV(0+)/dt
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EXAMPLE
CL R V
++
V0
I0iciL iR Determine the value of R for a critically
damped response when, L=8H,
C=0.125F, V0=0, and I0=-12.25 mA
From the previous example 02=106. Then
101
240003
RC R
Again, from the previous example V(0+)=0 and dV(0+)/dt=98000 V/sThen D1=0 and D2=98000 V/s.
V t te V t t( ) 98000 01000
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THE STEP RESPONSE OF A
PARALLEL RLC CIRCUIT
IC
L R V
+t=0ic iL iR
i i i I
iV
RC
dV
dtI
L R c
L
V Ldi
dt
dV
dtL
d i
dt
i
L
R
di
dt LC
d i
dt I
d i
dt RC
di
dt
i
LC
I
LC
L L
L
L L
L L L
2
2
2
2
2
2
1
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The equation describing the step response of a second-order
circuit is a second-order differential equation with constant
coefficients and with a constant forcing function. The solution of
this differential equation equals the forced response which is inthe same form of the forcing function (constant for a step input)
plus a response function identical in form to the natural response.
Thus, the solution for the inductor current is in the form
i IL f
function of the same form
as the natural response
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EXAMPLE
IC
L R V
+t=0ic iL iR C=25nF, L=25mH, R=400
The initial energy in the
circuit is zero. I=24 mA.
Find iL(t)Since there is no initial energy in the circuit, iL(0
+)=0 and V(0+)=0
V(0+)=L[diL(0+)/dt]=0, then diL(0
+)/dt=0
0
2
12
8
9
4
8
0
2
1 10
25)(25)16 10
1
2
10
2 400 25)5 10
25 10
LC
RC
(
( )(rad / s
2
Overdamped
circuit
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s
s
1
4 4
2
4 4
5 10 3 10 20000
5 10 3 10 80000
rad / s
rad / s
As t , circuit reaches dc steady state where inductor is short
and capacitor is open. All of the input current flows through theinductor. Then If=24 mA.
i t A e A e A
i A A
di
dtA e A e
di
dtA A
L
t t
L
L t t
L
( )
( )
( )
24 10
0 24 10
20000 80000
020000 80000 0
3
1
20000
2
80000
3
1 2
1
20000
2
80000
1 2
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A mA A mA
i t e e mA t Lt t
1 2
20000 80000
32 8
24 32 8 0
( ) ( )
EXAMPLE:If the resistor in the circuit is increased to 625,
find iL(t) in the circuit.
Since L and C remain fixed, resonant frequency has the same
value. But neper frequency decreases to 3.2x104rad/s. Withthese values circuit is underdamped with complex conjugate
roots.
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s j
s j
i t I B e t B e t
I A
i B
di
dt
e B t B t
e B t B t
di
L f
t
d
t
d
f
L
L t
d d
t
d d
L
14 4
2
4 4
1 2
3
3
1
1 2
1 2
32 10 2 4 10
32 10 2 4 10
24 10
0 24 10 0
0
. .
. .
( ) cos sin
( )
( cos sin )
( sin cos )
(
rad / s
rad / s
+ d
)
dtB Bd 1 2 0
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B mA B mA
i t e t t mA t Lt
1 2
32000
24 32
24 24 24000 32 24000 0
( ) ( ( cos sin ))
EXAMPLE: Find iL(t) if R=500
The resonant frequency remains the same, but neper frequency
becomes 4x104rad/s. These values correspond to critical damping.
Roots of the characteristic equation are real and equal at s=-40000
i t I D te D e AL ft t( ) 1
40000
2
40000
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I A
i D
didt
e D t D D e
di
dtD D
D mA s D mA
i t te e mA t
f
L
L t t
L
L
t t
24 10
0 24 10 0
00
960000 24
24 960000 24 0
3
3
2
1 2 1
2 1
1 2
40000 40000
( )
( )
( )
/
( ) ( )
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0 0.5 1 1.5 2 2.5
-4
0
0.005
0.01
0.015
0.02
0.025
Underdamped
Critically dampedOverdamped
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V V Ldi
dt
di
dtA s
i t I D te D e AI A
i D D A
didt
D D D A s
i t te
L c
L L
L f
t t
f
L
L
L
t
( ) ( )( ) ( )
/
( )
( )
( ) /
( ) ( .
0 00 0 50
25 102000
24 10
0 24 10 29 10 5 10
0 2000 2200
24 2 2 10 5
3
1
40000
2
40000
3
3
2
3
2
3
1 2 1
6 40000
e mA t t 40000 0)
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V t Ldi
dt
te
e e
te e V
L
t
t t
t t
( )
( )[( . )( )
. ( ) ]
.
25 10 2 2 10 40000
2 2 10 5 40000 10
2 2 10 50
3 6 40000
6 40000 40000 3
6 40000 40000
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THE NATURAL AND STEP RESPONSE
OF A SERIES RLC CIRCUIT
R L
C+
V0
I0
i
Ri Ldi
dt Cid V
Rdi
dt
Ld i
dt
i
C
t
1
0
0
00
2
2
d i
dt
R
L
di
dt
i
LC
s RL
sLC
2
2
2
0
1 0
sR
L
R
L LC
R
L LC
1 2
2
2
0
2
0
2 2
1
2
1
,
rad / s rad / s
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These equations are in the same form that of the equations for
the parallel RLC circuit. Therefore, the solution will be
overdamped, critically damped, or underdamped depending
on relative magnitudes of the resonant frequency and neperfrequency.
i t Ae A e
i t B e t B e t i t D te D e
s t s t
t
d
t
dt t
( )
( ) cos sin( )
1 2
1 2
1 2
1 2 (overdamped)
(underdamped) (critically damped)
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R L
C+
Vci
+ VR + VL
+V
t=0V Ri L
di
dtV
i CdV
dt
di
dtC
d V
dt
c
c c
2
2
d V
dt
R
L
dV
dt
V
LC
V
LC
V t V Ae A eV t V B e t B e t
V t V D te D e
c c c
c f
s t s t
c f
t
d
t
d
c f
t t
2
2
1 2
1 2
1 2
1 2
( )( ) cos sin
( )
(overdamped) (underdamped)
(critically damped)
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t=0
+
100V
100mH
+
Vc 560i
The capacitor is initially charged
to 100V. At t=0, switch closes.
Find i(t) and Vc(t) for t0
0
2
3 6
8
3
6
0
2
0
2 2
1 10 10
100 0110
2
560
2 100
10 2800
784 10
9600
LC
R
L
( )( )
( . )
( ).
rad / s
Underdamped
rad / s
2
d
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i t B e t B e t
i B
V Ri L didt
di
dt
V
LA s
didt
B e t B e t
di
dtB B A
t
d
t
d
c
c
t t
( ) cos sin
( )
( ) ( ) ( )
( ) ( )/
sin cos
( )
1 2
1
3
22800
22800
2 2
0 0
0 0 0 0
0 0 100
10010 1000
2800 9600 9600 9600
09600 1000
1000
9600
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i t e t t
V t iR Ldi
dt
V t t t e V t
t
c
c
t
( )
.
sin
( )
( ) ( cos . sin )
1
9 6
9600 0
100 9600 2917 9600 0
2800
2800
A
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EXAMPLE
0.4
F
+
Vc+48V
t=0
280 0.1HThere is no initial stored
energy in the circuit at t=0.
Find Vc(t) for t0.
s
js j
V t V B e t B e t t c cft t
1
2 6
2
1
1400
2
1400
280
2 01
280
0 2
10
01 0 4
1400 48001400 4800
4800 4800 0
( . ) . ( . )( . )
( ) cos sin ,
rad / srad / s
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dV
dt
B B
B V
V t e t e t V t
c
c
t t
( )
( ) ( cos sin ) ,
04800 1400 0
14
48 48 4800 14 4800 0
2 1
2
1400 1400
A CIRCUIT WITH TWO
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A CIRCUIT WITH TWO
INTEGRATING AMPLIFIERS
++
Vo1
R1
++
Vo
R2
C2C1
+
Vg
Assuming ideal opamps, find the relation between Voand Vg
00 0
1
1
1 1
1
1 1
V
RC
d
dtV
dV
dt R C V
g
o
o
g( )
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00 0
1
1
1 1
1
2
2
2 2
1
2
0
2
2 2
1
2
0
21 1 2 2
V
RC
d
dtV
dV
dt R C V
d V
dt R C
dV
dt
d V
dt R C R C V
o
o
o
o
o
g
( )
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EXAMPLE
+
+
Vo1
250k
+ +Vo
0.1F
+
Vg
500k
1F
-5V-9V
9V5V
No energy is stored in the circuit when the input voltage Vg
jumps instantaneously from 0 to 25 mV.
Derive the expression for Vo(t) for 0 ttsat.
How long is it before the circuit saturates?
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1 1000
250 0140
1 1000
500 1
2
40 2 25 10 2
2 2
2
1 1
2 2
2
2
3
0
2
0
R C
R Cd V
dt
dV
dt dx t
V xdx t
o
o t
o
t
( )( . )
( )( )
( )( )
0 t tsat
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The second integrating amplifier saturates when Voreaches 9V
or t=3s. But it is possible that the first opamp saturates before
t=3s. To explore this possibility use the following equation
dVdt R C
V
V t
o
g
o
1
1 1
3
1
1 40 25 10 1
( )
Thus, at t=3s, Vo1=-3V. The first opamp does not reachsaturation at t=3s. The circuit reaches saturation when the
second amplifier saturates.
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TWO INTEGRATING AMPLIFIERS
WITH FEEDBACK RESISTORS
++
Vo1
Ra
++
Vo
Rb
C2C1
+Vg
R1 R2
The reason that the op amp saturates in the integratingamplifier is the feedback capacitors accumulation of charge.
To overcome this problem, a resistor is placed in parallel
with each feedback capacitor.
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0 00 0
1
0 0 0 0
1
1
1 1
1
1 11
1
1 1 1
1 1
1 1
1
2
2
2
1
2
2 2 2
V
R
V
RC
d
dtV
dV
dt R C V
V
R C
R CdV
dt
V V
R C
VR
VR
C ddt
V
dV
dt
V V
R CR C
g
a
o
o
o
o
g
a
o o g
a
o
b
oo
o o o
b
( )
( )
Let
,
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d V
dt
dV
dt R C
dV
dt
dV
dt
V V
R C
V R CdV
dt
R CV
d V
dt
dV
dtV
V
R C R C
o
b
o
o o g
a
o b
o b
o
o o
o
g
a b
2
0
2
2 2
1
1 1
1 1
1 2
2
2
2
2
1 2 1 2 1 2
1 1
1 1 1
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8/10/2019 06 Yaser Rahmati
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The characteristic equation is
s s2
1 2 1 2
1 1 10
The roots of the characteristic equation are real
s s11
2
2
1 1
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8/10/2019 06 Yaser Rahmati
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EXAMPLE
The parameters of the circuit are Ra=100 k, Rb=25 k,
R1=500 k, R2=100 k, C1=0.1F, and C2=1F. The power
supply of each op amp is 6V. The input voltage jumps from 0
to 250 mV at t=0. No energy is stored in the feedback capacitorsat the instant the input is applied. Find Vo(t) for t0.
1 1 1 21 2 2
1 2
2
2
0 05 01
30 200 1000
R C s R C s
V
R C R C
d V
dt
dV
dtV
g
a b
o o
o
. .
= 1000 V / s
+
2
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8/10/2019 06 Yaser Rahmati
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The characteristic equation is s2+30s+200=0. Then
s1=-20 rad/s and s2=-10 rad/s. The final values of the output
is the input voltage times the gain of each stage, because
capacitors behave as open circuits as t
V V
V t Ae A e
VdV
dt
A V A V
V t e e V t
o
o
t t
oo
o
t t
( ) ( )
( )
( )( )
( ) ( )
250 10500
100
100
255
5
0 00
0
10 5
5 10 5 0
3
1
10
2
20
1 2
10 20
and