Réponses des exercices du chapitre 8 (partie 1)w3.gel.ulaval.ca/~lehuy/gel16132/sol_ch8_a.pdf · 2...

1

Réponses des exercices du chapitre 8 (partie 1)

8.1 a)

On écrit:

Alors:

b)

On écrit:

Alors:

c)

On écrit:

On a: /{ u(t)-u(t-T)} =

Alors:

d)

On a: /{ e-5tu(t)} =

Alors: /{ t2e-5tu(t)} =

e)

On a: /{ u(t-2)} =

Alors: /{ e-0.5tu(t-2)} =

f)

On écrit:

f t( ) A 1 eαt–

–( )u t( )=

f t( ) Au t( ) eαt–

Au t( )–=

F s( ) As----

As α+------------–=

f t( ) A pour 0 t T≤ ≤0 … ailleurs

=

f t( ) A u t( ) u t T–( )–{ } Au t( ) Au t T–( )–= =

F s( ) As---- e

Ts– As----×–

As---- 1 e

Ts––[ ]= =

f t( ) Aeαt–

pour 0 t T≤ ≤0 … ailleurs

=

f t( ) Aeαt–

u t( ) u t T–( )–{ }=

1s---

eTs–

s----------–

F s( ) A1

s α+------------

eT s α+( )–

s α+-----------------------–

=

f t( ) t2e

5t–u t( )=

1s 5+-----------

d2

ds2

--------1

s 5+-----------

2

s 5+( )3-------------------=

f t( ) e0.5t–

u t 2–( )=

e2s–

s----------

e2 s 0.5+( )–

s 0.5+-------------------------

f t( ) 12e1.5t–

500t 0.785–( )u t( )cos=

12e1.5t–

500t 0.785–( )u t( )cos 6e1.5t–

ej 500t 0.785–( )

ej 500t 0.785–( )–

+{ }u t( )=

2

Alors:

8.2

a)

b)

c)

d)

e)

f)

8.4 Circuit transformé (domaine de Laplace):

f t( ) 6ej0.785–

e1.5– j500+( )t

6ej0.785

e1.5– j500–( )t

+

u t( )=

F s( ) 6ej0.785–

s 1.5 j500–+( )--------------------------------------6e

j0.785

s 1.5 j500+ +( )--------------------------------------+=

F s( ) 2s 6+

5s3

15s2

30s 20+ + +-----------------------------------------------------

0.267s 1+-------------

0.176 2.428∠s 1 j1.732–+---------------------------------

0.176 2.428–∠s 1 j1.732+ +-----------------------------------+ += =

f t( ) 0.267et–

0.352et–

1.732t 2.428+( )cos+{ }u t( )=

F s( ) 0.5s2

s– 0.5+

3s3

9s2

12s 6+ + +-----------------------------------------------

0.667s 1+-------------

0.416 2.215∠s 1 j–+

-------------------------------0.416 2.215–∠

s 1 j+ +-----------------------------------+ += =

f t( ) 0.667et–

0.832et–

t 2.215+( )cos+{ }u t( )=

F s( ) 5s 1–

0.2s3

0.6s– 0.4–------------------------------------------

10

s 1+( )2-------------------

5s 1+-----------–

5s 2–-----------+= =

f t( ) 10tet–

5et–

– 5e2t

+{ }u t( )=

F s( ) 4

0.5s3

1.5s2

2s 1+ + +-----------------------------------------------------

8s 1+-----------

4s 1 j–+-------------------–

4s 1 j+ +-------------------–= =

f t( ) 8et–

8et–

tcos–{ }u t( )=

F s( ) 2s 6+

5 s 1+( ) 0.5s2

s 2.5+ +( )------------------------------------------------------------

0.4s 1+-----------

0.282 2.356–∠s 1 j2–+

-----------------------------------0.282 2.356∠

s 1 j2+ +-------------------------------+ += =

f t( ) 0.4et–

0.564et–

2t 2.356–( )cos+{ }u t( )=

F s( ) s 3+

7s 2s2

6s 4+ +( )----------------------------------------

0.036s 2+-------------

0.143s 1+-------------–

0.107s

-------------+= =

f t( ) 0.036e2t–

0.143et–

– 0.107+{ }u t( )=

+

-

Vs

I1

v2

Vs = 100/s

50

+

-

50

10020000/s20000/s

V1 V2

3

Méthode des noeuds:

On calcule V1 et V2 par la méthode Cramer.

On calcule le courant I1:

Le courant i1(t) est la transformée inverse de I1:

→

150------

s20000---------------

150------+ +

150------–

150------–

150------

s20000---------------

1100---------+ +

V1

V2

Vs

50------

0

=

0.04 55–×10 s+ 0.02–

0.02– 0.03 55–×10 s+

V1

V2

0.02Vs

0=

V1

0.02Vs 0.02–

0 0.03 55–×10 s+

0.04 55–×10 s+ 0.02–

0.02– 0.03 55–×10 s+

------------------------------------------------------------------------------------0.02Vs 0.03 5

5–×10 s+( )×

0.04 55–×10 s+( ) 0.03 5

5–×10 s+( )× 0.02( )2–

------------------------------------------------------------------------------------------------------------------= =

V16 0.01s+

0.254–×10 s

20.035s 8+ +

-------------------------------------------------------------- Vs× 6 0.01s+

0.254–×10 s

20.035s 8+ +

--------------------------------------------------------------100

s---------×= =

V1s 600+

0.254–×10 s

20.035s 8+ +( )s

----------------------------------------------------------------------40000 s 600+( )

s s2

1400s 320000+ +( )----------------------------------------------------------= =

V2

0.04 55–×10 s+( ) 0.02Vs

0.02– 0

0.04 55–×10 s+ 0.02–

0.02– 0.03 55–×10 s+

------------------------------------------------------------------------------------0.02Vs 0.02×

0.04 55–×10 s+( ) 0.03 5

5–×10 s+( )× 0.02( )2–

------------------------------------------------------------------------------------------------------------------= =

V24

0.254–×10 s

20.035s 8+ +

-------------------------------------------------------------- Vs× 4

0.254–×10 s

20.035s 8+ +

--------------------------------------------------------------100

s---------×= =

V2400

0.254–×10 s

20.035s 8+ +( )s

----------------------------------------------------------------------0.16

8×10

s s2

1400s 320000+ +( )----------------------------------------------------------= =

I1

Vs V1–

50-------------------

100s

---------s 600+

0.254–×10 s

20.035s 8+ +( )s

----------------------------------------------------------------------–

50-------------------------------------------------------------------------------------

2 s2

1000s 80000+ +( )

s s2

1400s 320000+ +( )----------------------------------------------------------= = =

I12 s

21000s 80000+ +( )

s s2

1400s 320000+ +( )----------------------------------------------------------

0.5s

-------1.053

s 287.7+----------------------

0.447s 1112.3+-------------------------+ += =

i1 t( ) 0.5 1.053e287.7t–

0.447e1112.3t–

+ +{ }u t( )=

4

La tension v2(t) est la transformée inverse de V2:

→

b) La durée du régime transitoire est:

En régime permanent, on a: i1 = 0.5 A et v2 = 50 V

c) Cas où vs=100cos(400ππt)u(t)

On a:

V20.16

8×10

s s2

1400s 320000+ +( )----------------------------------------------------------

50s

------17.44

s 1112.3+-------------------------

67.44s 287.7+----------------------–+= =

v2 t( ) 50 17.44e1112.3t–

67.44e287.7t–

–+{ }u t( )=

0 0.005 0.01 0.0150

0.5

1

1.5

2

Cou

rant

i 1(t) ,

A

Exercice 8.4 − Excitation echelon

0 0.005 0.01 0.0150

10

20

30

40

50

Tens

ion

v 2(t) ,

V

Temps , s

5max1

287.7-------------

11112.3----------------( , ) 17.4ms=

Vs100s

s2

400π( )2+

-------------------------------=

V16 0.01s+

0.254–×10 s

20.035s 8+ +

-------------------------------------------------------------- Vs× 6 0.01s+

0.254–×10 s

20.035s 8+ +

--------------------------------------------------------------100s

s2

400π( )2+

-------------------------------×= =

V140000s s 600+( )

s2

400π( )2+[ ] s

21400s 320000+ +( )

--------------------------------------------------------------------------------------------=

V24

0.254–×10 s

20.035s 8+ +

-------------------------------------------------------------- Vs× 4

0.254–×10 s

20.035s 8+ +

--------------------------------------------------------------100s

s2

400π( )2+

-------------------------------×= =

V21.6

7×10 s

s2

400π( )2+[ ] s

21400s 320000+ +( )

--------------------------------------------------------------------------------------------=

5

Le courant i1(t) est la transformée inverse de I1:

→

La tension v2(t) est la transformée inverse de V2:

→

I1

Vs V1–

50-------------------

100s

s2

400π( )2+

-------------------------------s 600+

0.254–×10 s

20.035s 8+ +( )s

----------------------------------------------------------------------–

50------------------------------------------------------------------------------------------------------------

2s s2

1000s 80000+ +( )

s2

400π( )2+[ ] s

21400s 320000+ +( )

--------------------------------------------------------------------------------------------= = =

I12s s

21000s 80000+ +( )

s2

400π( )2+[ ] s

21400s 320000+ +( )

--------------------------------------------------------------------------------------------0.196

s 1112.3+-------------------------

0.052s 287.7+----------------------

0.904 0.252∠s j1256.6–

-------------------------------0.904 0.252–∠

s j1256.6+-----------------------------------+ + += =

i1 t( ) 1.808 1256.6t 0.252+( )cos 0.052e287.7t–

0.447e1112.3t–

+ +{ }u t( )=

V21.6

7×10 s

s2

400π( )2+[ ] s

21400s 320000+ +( )

--------------------------------------------------------------------------------------------7.663

s 1112.3+-------------------------

3.359s 287.7+----------------------–

3.698 2.192–∠s j1256.6–

-----------------------------------3.698 2.192∠s j1256.6+

-------------------------------+ += =

v2 t( ) 7.396 1256.6t 2.192–( )cos 7.663e1112.3t–

3.359e287.7t–

–+{ }u t( )=

0 0.005 0.01 0.015−2

−1

0

1

2

3

Cou

rant

i 1(t) ,

A

Exercice 8.4 − Excitation sinusoidale

0 0.005 0.01 0.015−10

−5

0

5

10

Tens

ion

v 2(t) ,

V

Temps , s

6



On calcule VA et VB par la méthode Cramer:

Les courants I1, I2 et la tension V3 sont calculés à partir de VA et VB:

+

-Vs

I1

V3

Vs = 100/s

200+

-

0.5s

50

150

I20.1s VA VB

10.1s----------

10.5s----------

1200---------+ +

1–0.5s----------

1–0.5s---------- 1

150--------- 1

50------ 1

0.5s----------+ +

VA

VB

Vs

0.1s----------

Vs

150---------

=

12s

------1

200---------+

2–s

------

2–s

------ 4150---------

2s---+

VA

VB

10Vs

s------------

Vs

150---------

=

VA

10Vs

s------------

2–s

------

Vs

150---------

4150---------

2s---+

12s

------1

200---------+

2–s

------

2–s

------ 4150---------

2s---+

-------------------------------------------------------------

10Vs

s------------

4150---------

2s---+

×Vs

150---------

2s---

×+

12s

------1

200---------+

4150---------

2s---+

× 2s---

2–

---------------------------------------------------------------------------300 7s 500+( )

s2

2475s 150000+ +( )------------------------------------------------------- Vs×= = =

VB

12s

------1

200---------+

10Vs

s------------

2–s

------Vs

150---------

12s

------1

200---------+

2–s

------

2–s

------ 4150---------

2s---+

-------------------------------------------------------------

12s

------1

200---------+

Vs

150---------×

10Vs

s------------

2s---

×+

12s

------1

200---------+

4150---------

2s---+

× 2–s

------ 2

–

------------------------------------------------------------------------------s2

2400s 600000+ +

4 s2

2475s 150000+ +( )---------------------------------------------------------- Vs×= = =

7

Les courants i1(t), i2(t) et la tension v3(t) sont obtenus en effectuant la transformation inverse sur I1, I2 et V3.

→

→

→

La durée du régime transitoire est:

En régime permanent, on a: i1 = 0 , i2 = 2 A , v3 = 100 V

I1

Vs VB–

150--------------------

3s s 2500+( )

4 s2

2475s 150000+ +( )----------------------------------------------------------

Vs= =

I2

VA VB–

0.5s---------------------

s– 6000+

2 s2

2475s 150000+ +( )----------------------------------------------------------

Vs= =

V3 VBs2

2400s 600000+ +

4 s2

2475s 150000+ +( )---------------------------------------------------------- Vs×= =

I1s– 6000+

2 s2

2475s 150000+ +( )----------------------------------------------------------

100

s---------× 0.5

s 2500+

s2

2475s 150000+ +--------------------------------------------------×= =

I10.518

s 62.2+-------------------

0.018s 2412.8+-------------------------–= i1 t( ) 0.518e

62.2t–0.018e

2412.8t––{ }u t( )=

I2s– 6000+

2 s2

2475s 150000+ +( )----------------------------------------------------------

100

s---------× 50

s– 6000+

s s2

2475s 150000+ +( )----------------------------------------------------------×= =

I22s---

2.074s 62.2+-------------------–

0.074s 2412.8+-------------------------+= i2 t( ) 2 2.074e

62.2t–– 0.074e

2412.8t–+{ }u t( )=

V3s2

2400s 600000+ +

4 s2

2475s 150000+ +( )----------------------------------------------------------

100

s---------× 25

s2

2400s 600000+ +

s s2

2475s 150000+ +( )----------------------------------------------------------×= =

V3100

s---------

77.8s 62.2+-------------------–

2.8s 2412.8+-------------------------+= v3 t( ) 100 77.8e

62.2t–– 2.8e

2412.8t–+{ }u t( )=

5 max1

62.2----------

12412------------{ , }× 80ms=

8

Cas où vs=100cos(500ππt)u(t)

On a:

Le courant I1 et la tension V3 sont calculés à partir de VA et VB:

Le courant i1(t) et la tension v3(t) sont obtenus en effectuant la transformation inverse sur I1 et V3.

→

→


0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.080

0.2

0.4

0.6

i 1(t)


0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.080

0.5

1

1.5

2

i 2(t)

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.080

50

100

v 3(t)

Temps , s

Vs 100s

s2

500π( )2+

-------------------------------×=

I1

Vs VB–

150--------------------

3s s 2500+( )

4 s2

2475s 150000+ +( )----------------------------------------------------------

Vs3s s 2500+( )

4 s2

2475s 150000+ +( )----------------------------------------------------------

100s

s2

500π( )2+

-------------------------------××= = =

V3 VBs2

2400s 600000+ +

4 s2

2475s 150000+ +( )---------------------------------------------------------- Vs× s

22400s 600000+ +

4 s2

2475s 150000+ +( )---------------------------------------------------------- 100

s

s2

500π( )2+

-------------------------------××= = =

I10.5s

2s 2500+( )

s2

500π( )2+( ) s

22475s 150000+ +( )

--------------------------------------------------------------------------------------------0.0008s 62.2+-------------------

0.013s 2412.8+-------------------------–

0.256 0.02∠s j1570.8–----------------------------

0.256 0.02–∠s j1570.8+

--------------------------------+ += =

i1 t( ) 0.0008e62.2t–

0.013e2412.8t–

– 0.512 1570.8t 0.02+( )cos+{ }u t( )=

V325s s

22400s 600000+ +( )

s2

500π( )2+( ) s

22475s 150000+ +( )

--------------------------------------------------------------------------------------------1.95

s 2412.8+-------------------------

0.12s 62.2+-------------------–

11.62 0.08–∠s j1570.8–

--------------------------------11.62 0.08∠s j1570.8+----------------------------+ += =

v3 t( ) 23.24 1570.8t 0.08–( )cos 0.12e62.2t–

– 1.95e2412.8t–

+{ }u t( )=

5 max1

62.2----------

12412------------{ , }× 80ms=

9

En régime permanent, on a: i1 = , v3 =


Méthode de mailles:

On calcule J1 et J2 par la méthode de Cramer:

0.512 1570.8t 0.02+( )cos 23.24 1570.8t 0.08–( )cos

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

−0.5

0

0.5i 1(t)


0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

−20

−10

0

10

20

v 3(t)

Temps , s

+

-

Vs

i1

V2

Vs = 100/s

50

+

-20020000/s

0.2s

J1 J2

0.2s 5020000

s---------------+ +

20000–s

------------------

20000–s

------------------ 20020000

s---------------+

J1

J2

Vs

0=

10

Le courant I1 et la tension V2 sont calculés à partir de J1 et J2.




J1

Vs20000–

s------------------

0 20020000

s---------------+

0.2s 5020000

s---------------+ +

20000–s

------------------

20000–s

------------------ 20020000

s---------------+

---------------------------------------------------------------------------------------------5 s 100+( )

s2

350s 125000+ +----------------------------------------------- Vs×= =

J2

0.2s 5020000

s---------------+ +

Vs

20000–s

------------------ 0

0.2s 5020000

s---------------+ +

20000–s

------------------

20000–s

------------------ 20020000

s---------------+

---------------------------------------------------------------------------------------------500

s2

350s 125000+ +----------------------------------------------- Vs×= =

I1 J15 s 100+( )

s2

350s 125000+ +----------------------------------------------- Vs× 5 s 100+( )

s2

350s 125000+ +-----------------------------------------------

100s

---------× 500 s 100+( )

s s2

350s 125000+ +( )-------------------------------------------------------= = = =

V2 200J2200 500×

s2

350s 125000+ +----------------------------------------------- Vs× 200 500×

s2

350s 125000+ +-----------------------------------------------

100s

---------× 17×10

s s2

350s 125000+ +( )-------------------------------------------------------= = = =

I1500 s 100+( )

s s2

350s 125000+ +( )-------------------------------------------------------

0.4s

-------0.728 1.85–∠

s 175 j307.2–+---------------------------------------

0.728 1.85∠s 175 j307.2+ +---------------------------------------+ += =

i1 t( ) 0.4 1.456e175t–

307.2t 1.85–( )cos+{ }u t( )=

V21

7×10

s s2

350s 125000+ +( )-------------------------------------------------------

80s

------46.03 2.62∠

s 175 j307.2–+---------------------------------------

46.03 2.62–∠s 175 j307.2+ +---------------------------------------+ += =

v2 t( ) 80 92.06e175t–

307.2t 2.62+( )cos+{ }u t( )=

51

175---------× 29ms=

11


On a:



→

→

0 0.005 0.01 0.015 0.02 0.025 0.03 0.0350

0.2

0.4

0.6

0.8

1

Cou

rant

i 1(t) ,

A

Exercice 8.6 − Excitation échelon

0 0.005 0.01 0.015 0.02 0.025 0.03 0.0350

20

40

60

80

100

Tens

ion

v 2(t) ,

V

Temps , s

Vs 100s

s2

500π( )2+

-------------------------------×=

I15 s 100+( )

s2

350s 125000+ +----------------------------------------------- Vs× 5 s 100+( )

s2

350s 125000+ +-----------------------------------------------

100s

s2

500π( )2+

-------------------------------× 500s s 100+( )

s2

500π( )2+( ) s

2350s 125000+ +( )

-----------------------------------------------------------------------------------------= = =

V2 200J2200 500×

s2

350s 125000+ +----------------------------------------------- Vs× 200 500×

s2

350s 125000+ +-----------------------------------------------

100s

s2

500π( )2+

-------------------------------×= = =

V21

7×10 s

s2

500π( )2+( ) s

2350s 125000+ +( )

-----------------------------------------------------------------------------------------=

I1500s s 100+( )

s2

500π( )2+( ) s

2350s 125000+ +( )

-----------------------------------------------------------------------------------------0.038 2.37∠

s 175 j307.2–+---------------------------------------

0.038 2.37–∠s 175 j307.2+ +---------------------------------------

0.163 1.4–∠s j1570.8–-----------------------------

0.163 1.4∠s j1570.8+---------------------------+ + += =

i1 t( ) 0.076e175t–

307.2t 2.37+( )cos 0.326 1570.8t 1.4–( )cos+{ }u t( )=

V25

4×10 s

s2

500π( )2+( ) s

2350s 125000+ +( )

-----------------------------------------------------------------------------------------2.38 0.56∠

s 175 j307.2–+---------------------------------------

2.38 0.56–∠s 175 j307.2+ +---------------------------------------

2.078 2.91–∠s j1570.8–

--------------------------------2.078 2.91∠s j1570.8+----------------------------+ + += =

v2 t( ) 4.76e175t–

307.2t 0.56+( )cos 4.056 1570.8t 2.91–( )cos+{ }u t( )=

12


En régime permanent, on a: et


Méthode des mailles:

On calcule J1 et J2 par la méthode de Cramer.

51

175---------× 29ms=

i1 t( ) 0.326 1570.8t 1.4–( )cos= v2 t( ) 4.056 1570.8t 2.91–( )cos=

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035−0.4

−0.2

0

0.2

0.4

Cou

rant

i 1(t) ,

AExercice 8.6 − Excitation sinusoidale

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035−6

−4

−2

0

2

4

6

Tens

ion

v 2(t) ,

V

Temps , s

+

-Vs

I1

V2

Vs = 100/s

500+

-

13333.33/s

25

0.1s

J1 J2

0.1s 500+ 500–

500– 500 2513333.33

s----------------------+ +

J1

J2

Vs

0=

0.1s 500+ 500–

500– 52513333.33

s----------------------+

J1

J2

Vs

0=

13



→

→



J1

Vs 500–

0 52513333.33

s----------------------+

0.1s 500+( ) 500–

500– 52513333.33

s----------------------+

------------------------------------------------------------------------------------

52513333.33

s----------------------+

Vs

0.1s 500+( ) 52513333.33

s----------------------+

500( )2–

------------------------------------------------------------------------------------------------= =

J1 700.75s 19.05+

5.25s2

1383.33s 0.66666×10+ +

------------------------------------------------------------------------------- Vs××=

J2

0.1s 500+( ) Vs

500– 0

0.1s 500+( ) 500–

500– 52513333.33

s----------------------+

------------------------------------------------------------------------------------500Vs

0.1s 500+( ) 52513333.33

s----------------------+

500( )2–

------------------------------------------------------------------------------------------------= =

J250s

5.25s2

1383.33s 0.66666×10+ +

------------------------------------------------------------------------------- Vs×=

I1 J1 700.75s 19.05+

5.25s2

1383.33s 0.66666×10+ +

------------------------------------------------------------------------------- Vs×× 700.75s 19.05+

5.25s2

1383.33s 0.66666×10+ +

-------------------------------------------------------------------------------100

s---------××= = =

I17000 0.75s 19.05+( )

s 5.25s2

1383.33s 0.66666×10+ +( )

---------------------------------------------------------------------------------------=

V2 25J21250s

5.25s2

1383.33s 0.66666×10+ +

------------------------------------------------------------------------------- Vs× 1250s

5.25s2

1383.33s 0.66666×10+ +

-------------------------------------------------------------------------------100

s---------×= = =

V2125000

5.25s2

1383.33s 0.66666×10+ +( )

------------------------------------------------------------------------------------=

I10.2s

-------1.474 1.64–∠

s 131.75 j331.1–+----------------------------------------------

1.474 1.64∠s 131.75 j331.1+ +-----------------------------------------------+ +=

i1 t( ) 0.2 2.948e131.75t–

331.1t 1.64–( )cos+{ }u t( )=

V235.96 1.57∠

s 131.75 j331.1–+----------------------------------------------

35.96 1.57–∠s 131.75 j331.1+ +-----------------------------------------------+=

v2 t( ) 71.92e131.75t–

331.1t 1.57+( )cos{ }u t( )=

51

131.75----------------× 38ms=

14


On a:



0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−0.5

0

0.5

1

1.5

2

Cou

rant

i 1(t) ,

A


0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−50

−40

−30

−20

−10

0

10

20

Tens

ion

v 2(t) ,

V

Temps , s

Vs 100s

s2

400π( )2+

-------------------------------×=

I1 700.75s 19.05+

5.25s2

1383.33s 0.66666×10+ +

------------------------------------------------------------------------------- Vs×× 700.75s 19.05+

5.25s2

1383.33s 0.66666×10+ +

-------------------------------------------------------------------------------100s

s2

400π( )2+

-------------------------------××= =

I17000s 0.75s 19.05+( )

s2

400π( )2+[ ] 5.25s

21383.33s 0.6666

6×10+ +( )-------------------------------------------------------------------------------------------------------------------------=

V2 25J21250s

5.25s2

1383.33s 0.66666×10+ +

------------------------------------------------------------------------------- Vs× 1250s

5.25s2

1383.33s 0.66666×10+ +

-------------------------------------------------------------------------------100s

s2

400π( )2+

-------------------------------×= = =

V2125000s

2

s2

400π( )2+[ ] 5.25s

21383.33s 0.6666

6×10+ +( )-------------------------------------------------------------------------------------------------------------------------=

I10.126 2.32∠

s 131.75 j331.1–+----------------------------------------------

0.126 2.32–∠s 131.75 j331.1+ +-----------------------------------------------

0.422 1.37–∠s j1256.6–

--------------------------------0.422 1.37∠s j1256.6+----------------------------+ + +=

i1 t( ) 0.252e131.75t–

331.1t 2.32+( )cos 0.844 1256.6t 1.37–( )cos+{ }= u t( )

V23.06 2.39∠

s 131.75 j331.1–+----------------------------------------------

3.06 2.39–∠s 131.75 j331.1+ +-----------------------------------------------

10.05 1.35–∠s j1256.6–

--------------------------------10.05 1.35∠s j1256.6+----------------------------+ + +=

v2 t( ) 7.12e131.75t–

331.1t 2.39+( )cos 20.1 1256.6t 1.35–( )cos+{ }u t( )=

15

En régime permanent, on a:



i1 t( ) 0.844 1256.6t 1.37–( )cos=

v2 t( ) 20.1 1256.6t 1.35–( )cos=

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−1

−0.5

0

0.5

1Exercice 8.7 − Excitation sinusoidale

Cou

rant

i 1(t) ,

A

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−20

−10

0

10

20

Tens

ion

v 2(t) ,

V

Temps , s

+

-

Vs

I1

V2

vs

0.05s

+

-

100

20066666.67/s

0 25t

ms

100 V

Vs100

s---------

100s

---------e0.025s–

–=

66666.67/s

V1 V2

16

On calcule V1 et V2 par la méthode de Cramer.

Le courant I1 et la tension V2 sont calculés à partir de V1 et V2.

On calcule en premier lieu les réponses à un échelon de 100V: Vs = 100/s.

On a:

1100---------

10.05s------------- 15

6–×10 s+ + 1–

0.05s-------------

1–0.05s------------- 1

200---------

10.05s------------- 15

6–×10 s+ +

V1

V2

Vs

100---------

0

=

V1

Vs

100---------

1–0.05s-------------

01

200---------

10.05s------------- 15

6–×10 s+ +

1100---------

10.05s------------- 15

6–×10 s+ + 1–

0.05s-------------

1–0.05s-------------

1200---------

10.05s------------- 15

6–×10 s+ +

------------------------------------------------------------------------------------------------------------------------------------------=

V1

Vs

100---------

1200---------

10.05s------------- 15

6–×10 s+ +

1100---------

10.05s------------- 15

6–×10 s+ + 1

200---------

10.05s------------- 15

6–×10 s+ + 1

0.05s-------------

2–

------------------------------------------------------------------------------------------------------------------------------------------------------------2000 3s

21000s 4

6×10+ +[ ]Vs

9s3

9000s2

2.67×10 s 1.2

10×10+ + +-----------------------------------------------------------------------------------------= =

V2

1100---------

10.05s------------- 15

6–×10 s+ + Vs

100---------

1–0.05s------------- 0

1100---------

10.05s------------- 15

6–×10 s+ + 1–

0.05s-------------

1–0.05s------------- 1

200---------

10.05s------------- 15

6–×10 s+ +

------------------------------------------------------------------------------------------------------------------------------------------=

V2

Vs

100---------

10.05s-------------

1100---------

10.05s------------- 15

6–×10 s+ + 1

200---------

10.05s------------- 15

6–×10 s+ + 1

0.05s-------------

2–

------------------------------------------------------------------------------------------------------------------------------------------------------------8

9×10 Vs

9s3

9000s2

2.67×10 s 1.2

10×10+ + +-----------------------------------------------------------------------------------------= =

I1

Vs V1–

100-------------------

0.01 9s3

3000s2

2.48×10 s 0.4

10×10+ + +( )

9s3

9000s2

2.67×10 s 1.2

10×10+ + +--------------------------------------------------------------------------------------------------------- Vs×= =

V28

9×10

9s3

9000s2

2.67×10 s 1.2

10×10+ + +----------------------------------------------------------------------------------------- Vs×=

I10.01 9s

33000s

22.4

8×10 s 0.410×10+ + +( )

9s3

9000s2

2.67×10 s 1.2

10×10+ + +---------------------------------------------------------------------------------------------------------

100s

---------× 9s3

3000s2

2.48×10 s 0.4

10×10+ + +

s 9s3

9000s2

2.67×10 s 1.2

10×10+ + +( )------------------------------------------------------------------------------------------------= =

17

→ (réponse à un échelon de 100V)

On a:

→ (réponse à un échelon de 100V)

Les réponses au signal carré sont déduites à partir des réponses obtenus pour unéchelon de 100V:

I10.333

s-------------

0.709s 505.3+----------------------

0.102 1.78∠s 247.4 j1605.5–+----------------------------------------------

0.102 1.78–∠s 247.4 j1605.5+ +-----------------------------------------------+ + +=

i1 t( ) 0.333 0.709e505.3t–

0.204e247.4t–

1605.5t 1.78+( )cos+ +{ }u t( )=

V28

9×10

9s3

9000s2

2.67×10 s 1.2

10×10+ + +-----------------------------------------------------------------------------------------

100s

---------× 811×10

s 9s3

9000s2

2.67×10 s 1.2

10×10+ + +( )------------------------------------------------------------------------------------------------= =

V266.67

s-------------

66.53s 505.3+----------------------–

10.48 1.577∠s 247.4 j1605.5–+----------------------------------------------

10.48 1.577–∠s 247.4 j1605.5+ +-----------------------------------------------+ +=

v2 t( ) 66.67 66.53e505.3t–

– 20.96e247.4t–

1605.5t 1.577–( )cos+{ }u t( )=

100u t( ) 100u t 0.025–( )–{ }

i1 t( ) 0.333 0.709e505.3t–

0.204e247.4t–

1605.5t 1.78+( )cos+ +{ }u t( )

0.333 0.709e505.3 t 0.025–( )–

0.204e247.4 t 0.025–( )–

1605.5 t 0.025–( ) 1.78+[ ]cos+ +{ }u t 0.025–( )+

=

v2 t( ) 66.67 66.53e505.3t–

– 20.96e247.4t–

1605.5t 1.577–( )cos+{ }u t( )

66.67 66.53e505.3 t 0.025–( )–

– 20.96e247.4 t 0.025–( )–

1605.5 t 0.025–( ) 1.577–[ ]cos+{ }u t 0.025–( )+

=

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−1

−0.5

0

0.5

1

Cou

rant

i 1(t) ,

A

Exercice 8.8 − Excitation carree

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−20

0

20

40

60

80

Tens

ion

v 2(t) ,

V

Temps , s

18

b) Cas où vs est l’excitation montrée dans la figure suivante.

On a dans ce cas:

- On détermine en premier lieu les réponses i1 et v2 pour le cas où vs est une rampe 4000r(t) .

→ [réponse à une rampe 4000r(t)]

→ [réponse à une rampe 4000r(t)]

Les réponses au signal sont déduites à partir des réponses obtenus pour unerampe 4000r(t):

vs

0 25t

ms

100 V

vs t( ) 4000r t( ) 4000r t 0.025–( )–=

Vs4000

s2

------------4000

s2

------------e0.025s–

–=

Vs4000

s2

------------=

I10.01 9s

33000s

22.4

8×10 s 0.410×10+ + +( )

9s3

9000s2

2.67×10 s 1.2

10×10+ + +---------------------------------------------------------------------------------------------------------

4000

s2

------------× 40 9s3

3000s2

2.48×10 s 0.4

10×10+ + +( )

s2

9s3

9000s2

2.67×10 s 1.2

10×10+ + +( )----------------------------------------------------------------------------------------------------= =

I113.33

s2

-------------0.05

s----------

0.056s 505.3+----------------------–

0.0025 0.055∠s 247.4 j1605.5–+----------------------------------------------

0.0025 0.055–∠s 247.4 j1605.5+ +-----------------------------------------------+ + +=

i1 t( ) 13.33t 0.05 0.056e505.3t–

– 0.005e247.4t–

1605.5t 0.055+( )cos+ +{ }u t( )=

V28

9×10

9s3

9000s2

2.67×10 s 1.2

10×10+ + +-----------------------------------------------------------------------------------------

4000

s2

------------× 3.215×10

s2

9s3

9000s2

2.67×10 s 1.2

10×10+ + +( )---------------------------------------------------------------------------------------------------= =

V22666.67

s2

-------------------5.77

s----------–

5.27s 505.3+----------------------

0.258 0.15–∠s 247.4 j1605.5–+----------------------------------------------

0.258 0.15∠s 247.4 j1605.5+ +-----------------------------------------------+ + +=

v2 t( ) 2666.67t 5.77– 5.27e505.3t–

0.516e247.4t–

1605.5t 0.15–( )cos+ +{ }u t( )=

4000r t( ) 4000r t 0.025–( )–{ }

i 1 t( ) 13.33t 0.05 0.056e505.3t–

– 0.005e247.4t–

1605.5t 0.055+( )cos+ +{ }u t( )

13.33 t 0.025–( ) 0.05 0.056e505.3 t 0.025–( )–

–

0.005e247.4 t 0.025–( )–

1605.5 t 0.025–( ) 0.055+[ ]cos

+

+

{

}u t 0.025–( )

+

=

v2 t( ) 2666.67t 5.77– 5.27e505.3t–

0.516e247.4t–

1605.5t 0.15–( )cos+ +{ }u t( )

2666.67 t 0.025–( ) 5.77– 5.27e505.3 t 0.025–( )–

0.516e247.4 t 0.025–( )–

1605.5 t 0.025–( ) 0.15–[ ]cos

+

+

{

}u t 0.025–( )

+

=

19

c) Cas où vs est une tension sinusoïdale vs=100cos(500ππt)u(t)

Dans ce cas, on a:

→

→

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050

0.1

0.2

0.3

0.4

Cou

rant

i 1(t) ,

A

Exercice 8.8 − Excitation rampe

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050

10

20

30

40

50

60

70

Tens

ion

v 2(t) ,

V

Temps , s

Vs100s

s2

500π( )2+

-------------------------------=

I10.01 9s

33000s

22.4

8×10 s 0.410×10+ + +( )

9s3

9000s2

2.67×10 s 1.2

10×10+ + +---------------------------------------------------------------------------------------------------------

100s

s2

500π( )2+

-------------------------------×=

I1s 9s

33000s

22.4

8×10 s 0.410×10+ + +( )

s2

500π( )2+[ ] 9s

39000s

22.6

7×10 s 1.210×10+ + +( )

----------------------------------------------------------------------------------------------------------------------------------=

I10.187 0.16–∠s j1570.8–

--------------------------------0.187 0.16∠s j1570.8+----------------------------

0.066s 505.3+----------------------

0.337 0.576∠s 247.4 j1605.5–+----------------------------------------------

0.337 0.576–∠s 247.4 j1605.5+ +-----------------------------------------------+ + + +=

i1 t( ) 0.374 1570.8t 0.16–( )cos 0.066e505.3t–

0.674e247.4t–

1605.5t 0.576+( )cos+ +{ }u t( )=

V28

9×10

9s3

9000s2

2.67×10 s 1.2

10×10+ + +-----------------------------------------------------------------------------------------

100s

s2

500π( )2+

-------------------------------×=

V28

11×10 s

s2

500π( )2+[ ] 9s

39000s

22.6

7×10 s 1.210×10+ + +( )

----------------------------------------------------------------------------------------------------------------------------------=

V233.85 2.613–∠

s j1570.8–-----------------------------------

33.85 2.613∠s j1570.8+

-------------------------------6.24

s 505.3+----------------------–

34.75 0.374∠s 247.4 j1605.5–+----------------------------------------------

34.75 0.374–∠s 247.4 j1605.5+ +-----------------------------------------------+ + +=

v2 t( ) 67.7 1570.8t 2.613–( )cos 6.24e505.3t–

– 69.5e247.4t–

1605.5t 0.374+( )cos+{ }u t( )=

20

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−1

−0.5

0

0.5

1

Cou

rant

i 1(t) ,

A


0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−100

−50

0

50

100

Tens

ion

v 2(t) ,

V

Temps , s

Réponses des exercices du chapitre 8 (partie 1)w3.gel.ulaval.ca/~lehuy/gel16132/sol_ch8_a.pdf · 2...

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Transcript of Réponses des exercices du chapitre 8 (partie 1)w3.gel.ulaval.ca/~lehuy/gel16132/sol_ch8_a.pdf · 2...