Handouts 6002 L18 Oei12 Gaps Annotated

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  • 1

    6.002x CIRCUITS AND ELECTRONICS

    Damped Second-Order

    Systems

  • Review

    C A B

    5V

    +!!

    5V

    CGS large loop

    2K 50

    2K S

    LvA 5

    0

    vB

    0

    vC

    0

    50

    t

    t

    t

    5

    5

  • 3

    In the last lecture, we started by analyzing the simpler LC circuit to build intuition

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    And for initial conditions

    We solved

    For input

    In the last lecture LC circuit

  • 5

    In the last lecture LC circuit

    Total solution

    +!! C L +!

    !)(tv

    )(tvI

    )(ti

  • 6 See A&L Section 13.6

    Today, we will close the loop on our observations in the demo by analyzing the RLC circuit

    +!! C

    L

    +!

    !

    i(t)

    v(t) vI (t) LC

    I v

    v(t) 2VI

    t

    VI

    0

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    Recall element rules

    Lets analyze the RLC network

    L:

    C:

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    Set up the differential equation vA

    +!! C L +!

    !

    R i(t)

    v(t) vI(t) Need to get rid of vA

    v

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    Set up the differential equation differently

    v +!!

    C

    L +!

    !

    R i

    vI

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    Solve IvLCvLCdtdv

    LR

    dtvd 112

    2

    =++

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    Lets solve IvLC

    vLCdt

    dvLR

    dtvd 112

    2

    =++

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    1 Particular solution IvLC

    vLCdt

    dvLR

    dtvd 112

    2

    =++

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    1 Particular solution IvLC

    vLCdt

    dvLR

    dtvd 112

    2

    =++

    IPPP V

    LCv

    LCdtdv

    LR

    dtvd 112

    2

    =++

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    Homogeneous solution 2 IvLC

    vLCdt

    dvLR

    dtvd 112

    2

    =++

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    Homogeneous solution 2 0vLC1

    dtdv

    LR

    dtvd

    HHH =++2

    2

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    0vLC1

    dtdv

    LR

    dtvd

    HHH =++2

    2

    Solution to Homogeneous solution 2

    Assume solution of the form 2A

    vH = AeST , A, S = ?

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    Homogeneous solution 2 0vLC1

    dtdv

    LR

    dtvd

    HHH =++2

    2

    Solution to

    02 =++LC1s

    LRs

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    Homogeneous solution 2 0vLC1

    dtdv

    LR

    dtvd

    HHH =++2

    2

    Solution to

    02 =++LC1s

    LRs

    Roots o1s22 +=

    os 222 =

    characteristic equation

    LC1

    o =

    LR2

    =

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    Total solution 3

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    v(t) =VI + A1ete22o( )t + A2ete

    22o( )t

    Lets stare at this a while longer 3 v(t) =VI + A1e+ 22o( )t + A2e

    22o( )t

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    v(t) =VI + A1ete22o( )t + A2ete

    22o( )t3

    Overdamped o

    >

    Underdamped o <

    Critically damped o

    =

    Overdamped o >

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    Underdamped o < 3 v(t) =VI + A1ete22o( )t + A2ete

    22o( )t

    Overdamped o

    >

    Underdamped o <

    Critically damped o

    =

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    Underdamped contd o < 3 teKteKVtv d

    td

    tI

    sincos)( 21 ++=

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    Underdamped contd o <

    Remember, scaled sum of sines (of the same frequency) are also sines! -- Appendix B.7

    v(t) =VI VIet cosdt VId

    et sindt3

    +=+

    1

    2122

    2121 tancossincos A

    AAAAA

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    Critically damped o =

    Section 13.2.3

    3

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    Remember this? We have now closed the loop

    vA 5

    0

    vB

    0

    vC

    0

    50

    t

    t

    t

    5

    5

  • 27 See Sec. 12.7 of A&L textbook

    ringing v(t)

    VI 0

    t

    Easy way: Characteristic equation tells the whole story!

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    Intuitive Analysis +!! C

    L +!

    !)(tvIv

    )(ti

    R

    )0(i)0(v

    given -ve +ve

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    Next, introduce R: RLC Circuits

    More in the next sequence! If you are impatient, see A&L Section 13.2

    +!! C

    L

    +!!vI (t)

    i(t)

    v(t)

    v(t)

    t

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    What about other variables?

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    Parallel RLC Characteristic equation says it all