Complete Response

8/8/2019 Complete Response

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AC circuit analysis I

Consider i = √2R I cosωt

Resistance

v R = Ri =√

2R I cosωt =√

2V R cosωt (1)

orV R ∠0o

I ∠0o = R ∠0o (2)

Inductance

v L = Ld i

dt

=√

2ωLI (

−sinωt ) =

√2ωLI cos(ωt + 90o )

(3)V L∠90o

I ∠0o =

ωL I ∠90o

I ∠0o = j ωL = ωL∠90o (4)

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AC circuit analysis II

Capacitance

i C = C d v

dt =√

2ωC V (− sinωt ) (5)

=√

2ωC V cos(ωt + 90o ) (6)

V ∠0o

I C ∠90o =

V ∠0o

ωC V ∠90o =

1

ωC ∠−90o = − j

ωC (7)

Resistance:Voltage across it is in phase with current

Inductance:Voltage across it leads the current through it by 90o

Capacitance:Voltage across it lags the current through it by 90o

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Example

Voltagev = 120

(2) cos(1000t + 90o ) V

is applied to the circuit whereR = 15Ω, C = 83.3µF , andL = 30mH . Find i(t)

Solution:V = 120∠90o V

I R = 1R

V = 115120∠90o = 8∠90o

= 0 + j 8 A

I C = j ωC V

= (0.0833∠90o )(120∠90o )

= 10∠180o = −10 + j 0 A

I L = 1 j ωL

V = 120∠90o

30∠90o

= 4∠0o = 4 + j 0 A

By Kirchhoff’s current law, I = 0orI = I R +I C +I L = (0−10+4)+ j (8+0+0)= −6 + j 8 = 10∠127o A

i (t ) = 10

(2) cos(1000t + 127o ) A

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AC Impedance

Ohms law of sinusoidal circuit

V = Z I (8)

where Z is complex

Z = V

I = Z ∠φZ = R + j X (9)

where

Z =√

R 2

+X 2

, φZ = arctan

X

R

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Example

Voltage v = 12

(2) cos 5000t Vis applied to the circuit. Find theindividual and combinedimpedances and the current i(t)

Figure: Circuit Elements

Figure: Impedances

Solution:Z R = R = 20 = 20 + j 0 = 20∠0o Ω

Z L = j ωL = j 5000× 0.003 = j 15Ω= 15∠90o Ω

Z C = − j 1ωC

= − j 5000×10−5

= − j 20 = 20∠− 90o Ω

Z RL = Z R +Z L = 20 + j 15 = 25∠37o Ω

Z eq = ZRL.ZC

ZRL+ZC = 25∠37o .20∠−90o

20+ j 15− j 20

= 500∠−53o

20− j 5 = 24.3∠− 39o

= 18.9− j 15.3Ω

I = V Z = 12∠0o

24.3∠−39o = 0.49∠39o A

i (t ) = 0.49

(2) cos(5000t + 39o ) A

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AC Admittance

Y =I

V =

1

Z Siemens (10)

Y R =

1

R ∠0o = G ∠

0o

(11)

Y L =1

j ωL=

1

ωL∠− 90o (12)

Y C = j ωC = ωC ∠90o (13)

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Example

Solving previous problem by usingadmittances

Figure: Phasor diagram of currents

calculated from admittances

Solution:Y C = 1

ZC = 1

20∠−90o =0.05∠90o = 0 + j 0.05 S

Y RL = 1ZRL

= 125∠37o

= 0.04∠− 37o = 0.032− j 0.024 S

Y eq = Y C + Y RL

= 0 + j 0.05 + 0.032− j 0.024= 0.041∠39o S

I C = Y C V

= 0.05∠90o

× 12∠0o

= 0.6∠90o

A

I RL = Y RL V

= 0.04∠− 37o × 12∠0o

= 0.48∠− 37o A

I = I C + I YL

= 0.6 + j 0.6 + 0.384− j 0.288= 0.49∠39o AAlso, I = Y eq V

= 0.041∠39o × 12∠0o = 0.49∠39o A

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General Circuit Analysis

1 Transform time functions to phasors and convert element values toimpedance/ admittances

2 Combine impedance/admittance to simplify circuit

3 Determine the desired response in phasor form

4 Draw phasor diagram to check calculations and display result

5 Transform phasor result to time function

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Example

LT = 3mH . For desired voltagev = 28.3 cos(5000t + 45o )V,determine the current i(t) andbranch currents

Solution:Z T = j ωLT = j 5000× 3× 10−3 = j 15ΩY C = j ωC = j 5000

×10−5 = j 0.05 S

Z C = 1Y C

= − j 20Ω

Z L = j ωL = j 5000× 4× 10−3 = j 20ΩY RL = 1

ZRL= 1

20+ j 20 = 0.025√

2∠− 45o

= 0.025− j 0.025 SY = Y C + Y RL = j 0.05 + (0.025

− j 0.025)

= 0.025 (2)∠45o SI = Y V = (0.025

(2)∠45o )(20∠45o )

= 0.7∠90o Ai (t ) = 0.7

√2 cos(5000t + 90o ) A

Phasor diagram is drawn on basis of following

calculated valuesI C = Y C V = (0.05∠90o )(20∠45o ) = 1∠135o

I RL = Y RLV = (0.035∠− 45o )(20∠45o )= 0.7∠0o AV R = R I RL = (20∠0o )(0.7∠0o ) = 14∠0o V

V L = Z LI RL = (20∠

90o

)(0.7∠

0o

) = 14∠

90o

V9/34

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Natural Response

Natural Response: Behavior due to internal energy storage

First order system

Capacitance with initial voltageV 0(or initial charge Q 0 = C V 0) issuddenly shorted across a resistanceR at time t = 0

At t > 0 , using KVLv = 0 = v C − v R

= V 0 − 1C

t

0i dt − R i

Differentiating to eliminate theintegral

R d i dt + 1C i = 0

Solving,Let i = A e st , then d i

dt = sA e st

This impliesR s A e st + 1

C A e st = 0

or

(Rs +1

C )A e st

= 0A=0 is a trivial solution, the other beingRs + 1

C = 0

i.e. s = − 1RC

and hence

i (t ) = A e (1/RC )t

To evaluate A: Apply initial condition tooriginal eqn.V 0 − 0− R i (0+) = 0or i (0+) = V 0/R

= A e s (0) = A

This implies,

i = V 0R e −t /R C

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Example

At time t = 0,short circuit thecoil of 2H inductance and 10Ω

resistance. If the initial current inthe coil, I o , is 20 A, predict thecurrent i after 0.2 s has elapsed.

Solution:1. By Kirchhoff’s voltage law,

v = 0 = −v L − v R = −L

d i

dt − R i 2. The homogenous equation isL d i

dt + R i = 0

3. Assuming an exponential solution,we writei = A e st

where s and A are to be determined.

4. Substituting into the homogenous equation,LsA e st + RA e st = (sL + R )A e st = 0If sL + R = 0, s = −R

Land i = A e −(R /L)t

5. At t = 0+

i = I o = A e 0 = A or A = I o = 20

Hence the solution isi = I o e −(R /L)t = 20 e −(10/2)t = 20 e −5t AAfter 0.2 s the current isi = 20 e −5×0.2 = 20× 0.368 = 7.36 AThe current decreases continuously, but never

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Steps to determine natural response

1 Write the governing eqn. using Kirchoff’s laws

2 Reduce this to homogeneous differential eqn.

3 Assume an exponential solution with undetermined constants4 Determine the exponents from the homogenous eqn.

5 Evaluate the coeff. from the given conditions

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Natural Response

Second order system

Assume initial voltage V 0 exists on

capacitance C

Using KVL,v = 0

= −L d i dt − R i + V 0 − 1

C

t

0i dt

By differentiation and rearrangement

Ld 2i dt 2 + R

d i dt + 1C i = 0

Assuming exponential solution, i = A e st

s 2L A e st + s R A e st + 1C A e st = 0or (s 2L + R s + 1

C )A e st = 0

Satisfied when,s 2L + R s + 1C

= 0Roots are

s 1, s 2

=

−R 2L

±

( R 2L

)2

−1

LC

General soln:i = A1 e s 1t + A1 e s 2

A1 and A2 are determined by initialcondition

s 1 and s 2 are determined by circuit

constants

Soln to characteristic eqn leads to threecases

Roots are real & distinct

Roots are complexRoots are real & e ual 13/34

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Example(Roots Real & distinct)

Determine and plot the currentresponse as a function of timeafter the switch is closed

Solution:Roots of the characteristic eq. are

s 1, s 2 =−

4

2×1 ± (4

2×1)2

−1

1×13

=−

2±

1

The general solution for current isi = A1 e −t + A2 e −3t

At t = 0+

i = i 0 = 0 = A1 e 0 + A2 e 0 = A1 + A2

Therefore, A2 =−

A1

Also, voltage iR across the resistance is zeroTherefore,v L = −v C = V 0 and L d i

dt = V o or

d i dt

= V o

L= −A1 e 0 − 3A2 e 0

= −A1 − 3A2 = −A1 + 3A1 = +2A1

Solving,

A1 = + V o 2L = + V o

2 and A2 = −A1 = −V o 2

i = V o

2e −t − V o

2e −3t

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Example(Roots complex)

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Example(Roots complex)

Let L=1H, C = 117F and R = 2Ω.

Derive an expression for thenatural response.

Solution:

s = − R 2L ± ( R 2L )2 − 1LC

= − 22×1 ±

( 22×1 )2 − 1

1× 117

= −1±√−16

s 1 =−

1 + j 4 and s 2 =−

1−

j 4

The natural response is

i = A1 e (−1+ j 4)t + A2 e (−1− j 4)t

can also be written as using euler’s eqn.i = e −αt [(A1 + A2)cosωt + j (A1

−A2)sinωt ]

= e −αt [B 1 cosωt + B 2 sinωt ]= A e −αt

damping function

sin(ωt + θ) Oscillating function

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Example

Rewrite previous example in theform of a damped sinusoid

Solution:

α = R 2L

= 1 and ω = 1

LC − R 2

4L2 = 4

Hence, i = A e −t sin(4t + θ)At t = 0+,v L = d i

dt = V 0 = −v C since i L = 0

andi = 0 = A e 0 sin(0 + θ) = A sin θHence, θ = 0d i dt

= V 0 = A e 0 4 cos(0)− A e 0 sin(0) = 4AorA = V 0/4

Therefore,i = V 0

4e −t sin4t

T = 2πω = 2π

4 = 1.57 sec

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Example(Roots real & equal)

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Example(Roots real & equal)

In example for real & distinct roots,substitute

R = 2Ω, L = 0.5H ,C = 0.5F

On calculation, s 1 = s 2 = −2

Hence,

i = (A1 + A2 t )e st

By applying initial condition A can be determined

A = 2 V 0t e −2t

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Real & distinct rootsOverdamped Response(No oscillation)

Complex rootsUnderdamped Response(Dying oscillation)

Real & equal rootsCritically damped (No physical significance,border line betweenabove two oscillating output)

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Complete Response of a circuit

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Complete Response of a circuit

To determine the transient behaviour of a circuit, complete response of circuit has to be found.It includes

Natural ResponseDue to energy stored in inductance and/or capacitances.

Determined by assuming that energy has been stored by an externalsource and then that source is removed

Forced ResponseProduced by external energy sources such as batteries or generator

Determined by assuming that a sufficient time has elapsed so thatall natural response components have died away or at least havebecome negligibly small

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Example I

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Example I

Consider an R-L circuit as shown.Herei (0−) = I 0 = V 0

R = i (0)

Find the complete responseresponse i(t) for t ≥ 0 when asinusoidal voltage sourceV m cosωt is connected at t = 0

Solution:Applying KVL for t ≥ 0, we getL d i

dt + Ri = V m cosωt

ord i dt

+ R L

i = V mL

cosωt

Particular Solution:i p (t ) = A cos(ωt − θ)

= B 1 cosωt + B 2 sinωt On substituting i p (t ) in above eqn,we get

B 1 = R V mR 2+ω2L2 & B 2 = ωLV m

R 2+ω2L2

Thus,

i p (t ) =V m

√R 2+ω2L2 [(R

√R 2+ω2L2 )cosωt +( ω√

R 2+ω2L2)sinωt ]

=> i p (t ) = V m√R 2+ω2L2

cos(ωt − tan−1(ωLR

))

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Example II

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Example II

Homogenous solution:

i h(t ) = e st

Substituting i h(t ) in the homogenous differential eqn, we get

(s + R L

)e st = 0 => s = −R L

& i h(t ) = Ke −R L

t

Complete soln for t ≥ 0

i (t ) = i h(t ) + i p (t ) = K e −R L

t + V m√R 2+ω2L2

cos(ωt − tan−1(ωLR

))

i (0) = I 0 => I 0 = K + V m√R 2+ω2L2

cos(tan−1(ωLR

))

= K + V m√R 2+ω2L2

R √R 2+ω2L2

Implies, K = I 0 − V m R R 2+ω2L2

i (t ) = (I 0 − V mR

R 2 + ω2L2)e −

R L

t

Natural Response

+V m√

R 2 + ω2L2cos(ωt − tan−1(

ωL

R ))

Forced response

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Example III

When t →∞ , e −R

L t → 0 ⇒ i h(t ) → 0Therefore,

i (t ) → i p (t )

Impedance of series R-L combination is

Z = R + j ωL

i p (t ) = V m|Z| cos(ωt − ∠Z )

In phasor form, I p = V m

√2|Z|e − j ∠Z

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Procedure of finding the complete response(typically for

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Procedure of finding the complete response(typically fort ≥ 0) I

1 Write the governing eqn using Kirchoff’s laws2 Reduce this to a linear differential eqn

3 Find the initial conditions(if they are not given) from the conditionof the current prior to t = 0. The number of initial conditions isequal to the order of the differential eqn

4 Find the particular solution by assuming a form of the particularsolution, and substituting it in the differential eqn. Unknownconstants are to be found by substitution in the differential eqn andnot from initial condition

Forcing function Particular solution

constant another constantA e at , ’a’ real or complex B e at

A cosωt B cosωt or B 1 cosωt + B 2 sinωt

A cos(ωt + θ) B cosωt or B 1 cosωt + B 2 sinωt

The particular soln is the forced response

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g p p ( yp yt ≥ 0) II

5 Find the homogenous soln by assuming e st as a soln, substituting inthe homogenous differential eqn and solving for s

First order: s = s 0 , homogeneous soln K e st

Second order:s = s 1, s 2 with s 1

= s 2

⇒ homogeneous solution : K 1 e s 1t + K 2 e s 2t

s = s 0, s 0⇒ homogeneous solution : (K 1 + K 2t )e s 0t

The unknown constants in the homogeneous soln are to be foundlater from the initial conditions.The homogeneous soln is the natural

response6 Add the forced and natural response to get the complete response.

Find the undetermined constants which arose from the homogeneoussoln by applying the initial conditions on the complete response

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First order circuit I

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Step response of an RL circuit

For t ≥ 0

Ld i

dt + R i = V ⇒ d i dt

+ R L

i = V L

Particular soln:i p (t ) = B

⇒ R L

B = V L

⇒ B = V R

i p (t ) =V

R

Homogenous soln:i h(t ) = e st

(s + R L

)e st = 0 ⇒ s = −R L

i h(t ) = K e −R L

t

Complete soln:i (t ) = i h(t ) + i p (t ) = K e −

R L

t + V

R i (0) = 0 ⇒ K + V R

= 0 or K = −V R

i (t ) = V R

(1− e −R L

t ) , t ≥ 0

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First order circuit II

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AC switching transients

v c (0) = 0⇒

i (0) = V m

R For t ≥ 0R i + 1

C

t

− inf i (τ )d τ = V m cosωt

ord i dt

+ i RC

= −ω V mR

sinωt

Particular soln:i p (t ) = A cos(ωt + θ)

I p = Vm/√2

R + 1 j ωC

= V mωC √2√1+ω2R 2C 2

∠π2 − tan−1(ωRC )

i p (t ) = V mωC √1+ω2R 2C 2

cos(ωt + π2 − tan−1(ωRC ))

Homogenous soln:

i h(t ) = K e −t /

RC

Complete soln:i (t ) = i p (t ) + i h(t )i (0) = 0⇒ V m

R = K + V mωC √

1+ω2R 2C 2sin(tan−1(ωRC ))

On solving,K = V mR (1+ω2R 2C 2)

i (t ) = V mR (1+ω2R 2C 2)e −t /RC +

V mωC √1+ω2R 2C 2

cos(ωt + π2 − tan−1(ωRC )) , t ≥ 0

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Second order circuits I

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Step response of an RLC circuit

v (t ) = V 1 + V 2 u (t )Initial conditions:v C (0) = V 1 , i (0) = 0

v = R i + Ldi

dt + v C v (0) = R i (0) + Li (0) + v C (0)⇒ i (0) = V 2

L

For t ≥ 0,d 2 i dt 2

+ R L

d i dt

+ 1LC

i = 0Homogenous soln is complete soln. Let e st

be a solns 2 + R

Ls + 1

LC = 0

s = − R 2L±

R 2

4L2 − 1LC

Case 1:R < 2 LC (Underdamped)

s = − R 2L± j

1

LC − R 2

4L2 = −α + j β

On applying initial condition:

⇒i (t ) = V 2

βLe αt sin β t

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Case 2: R = 2

LC

(Critically damped)

s =−

R

2L

,−

R

2L

=−α,−α

i (t ) = e −αt [A1 + A2t ]


⇒ i (t ) = V 2L

te −αt

Case 3: R > 2

LC

(Overdamped)

s = − R 2L± j

R 2

4L2 − 1LC

= −α + γ

i (t ) = e −αt [A1e γt + A2e −γt ]


⇒ i (t ) = V 2γL

e −αt [e γt −e −γt ]2 = V 2

γLe −αt sinh(γ t )

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Two port networks I

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v 1 = f 1(i 1, i 2) and v 2 = f 2(i 1, i 2)or

i 1 = φ1(v 1, v 2) and v 2 = φ2(v 1, v 2)

For ideal elements,the functions are linear so equation becomes

v 1 = k 1 i 1 + k 2 i 2 and v 2 = k 3 i 1 + k 4 i 2or

i 1 = k 5 v 1 + k 6 v 2 and i 2 = k 7 v 1 + k 8 v 2

For sinusoidal excitation, the voltage and currents are represented byphasors, the constants are complex impedances or admittances

V 1 = Z 11 I 1 + Z 12 I 2 and V 2 = Z 21 I 1 + Z 22 I 2orI 1 = Y 11 V 1 + Y 12 V 2 and i 2 = Y 21 V 1 + Y 22 V 2

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For linear elements, Z 12 = Z 21 and Y 12 = Y 21 by reciprocity theorem

Figure: T network

V 1 = (Z 11 − Z 12 + Z 12)I 1 + Z 12 I 2V 2 = Z 21I 1 + (Z 22 − Z 21 + Z 21)I 2

Figure: π network

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T and π as three terminal networks called as Y and ∆ networks

Figure: T or Y network

Z 1 = Z 12Z 31Z 12+Z 23+Z 31

Z 2 = Z 12Z 23Z 12+Z 23+Z 31

Z 3 = Z 23Z 31Z 12+Z 23+Z 31

Figure: π or ∆ network

Z 12 = Z 1Z 2+Z 2Z 3+Z 3Z 1Z 3

Z 23 = Z 2Z 3+Z 1Z 2+Z 1Z 3Z 1

Z 31 = Z 3Z 1+Z 1Z 2+Z 3Z 2Z 2

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Example

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Given:

Z a = Z b = Z c = 10∠60o .Determine the impedance of theequivalent ∆

Figure: Y to ∆ conversion

Z ∆ = Z ab = Z a Z b +Z b Z c +Z c Z aZ c

= 3(Z Y )2Z Y

= 3 Z Y = 30∠60o Ω

Impedance of equivalent ∆load = 3× Impedance of balanced Y load

Phase angle of theimpedances and the powerfactor of the load areunchanged

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Example

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Transform T network into π form

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Example

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Determine equivalent seriesresistance of the bridge n/w

Dotted ∆ replaced by equivalent Y

Z 1 = 10×610+6+4 = 3Ω

Z 2 = 4×1010+6+4 = 2Ω

Z 3 = 6×410+6+4 = 1.2Ω

Resistance in series and parallel can

be combined to giveZ ab = (3+3)(1+2)

(3+3)+(1+2) + 1.2

= 2 + 1.2 = 3.2Ω

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Complete Response

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