Electronics RCL time area

    When the about-face is closed, a voltage move is activated to the RCL circuit. Yield the time the about-face was bankrupt to be 0s such that the voltage afore the about-face was bankrupt was 0 volts and the voltage afterwards the about-face was bankrupt is a voltage V. This is a move action accustomed by $Vcdot\; u(t)$ area V is the consequence of the move and $u(t)=1$ for $tgeq0$ and zero

    otherwise.

    To analyse the ambit acknowledgment using brief analysis, a cogwheel blueprint which describes the arrangement is formulated. The voltage about the bend is accustomed

    $$

    Vu(t)=v_c(t)+fracL+Ri(t) mbox

    

    

    where $v\_c(t)$ is the voltage beyond the capacitor, $fracL$ is the voltage beyond the inductor and $Ri(t)$ the voltage beyond the resistor.

    Substituting $i(t)=frac$ into blueprint 1:

    

    $$

    Vu(t)=v_c(t)+fracLC+RfracC

    

    

    

    $$

    frac+fracfrac+fracv_c(t)=frac mbox

    

    

    The voltage $v\_c(t)$ has two components, a accustomed responce $v\_n(t)$ and a affected reponse $v\_f(t)$ such that:

    

    $$

    v_c(t)=v_f(t)+v_n(t)mbox

    

    

    substituting blueprint 3 into blueprint 2.

    

    $$

    igg+igg=0+frac

    

    

    when $t>0s$ then $u(t)=1$:

    

    $$

    igg=0 mbox

    

    

    

    $$

    igg=frac mbox

    

    

    The accustomed acknowledgment and affected band-aid are apparent separately.

    Solve for $v\_f(t):$

    Since $frac$ is a polynomial of amount 0, the band-aid $v\_f(t)$ haveto be a connected such that:

    

    $$

    v_f(t)=K

    

    

    

    $$

    frac=0

    

    

    

    $$

    frac=0

    

    

    Substituting into blueprint 5:

    

    $$

    fracK=frac

    

    

    

    $$

    K=V

    

    

    

    $$

    v_f=V mbox

    

    

    Solve for $v\_n(t)$:

    Let:

    

    $$

    frac=2alpha

    

    

    

    $$

    frac=omega_n^2

    

    

    

    $$

    v_n(t)=Ae^

    

    

    Substituting into blueprint 4 gives:

    

    $$

    frac+2alphafrac+omega_n^2Ae^=0

    

    

    

    $$

    s^2Ae^+2alpha Ae^+omega_2^2Ae^=0

    

    

    

    $$

    s^2+2alpha s+omega_n^2=0

    

    

    

    $$

    s=frac}=-alphapmsqrt

    

    

    Therefore $v\_n(t)$ has two solutions $Ae^$ and $Ae^$

    where $s\_1$ and $s\_2$ are accustomed

    $$

    s_1=-alpha+sqrt

    

    

    

    $$

    s_2=-alpha-sqrt

    

    

    The accepted band-aid is then accustomed

    $$

    v_n(t)=A_1e^+A_2e^

    

    

    Depending on the ethics of the Resistor, inductor or capacitor the band-aid has three posibilies.

    1. If $alpha\; >\; omega\_n$ the arrangement is said to be overdamped

    2. If $alpha\; =\; omega\_n$ the arrangement is said to be alarmingly damped

    3. If the arrangement is said to be underdamped

    Given the accepted solution

    

    

    

    

    

    

    

    

    

    

    

    

    

    

    

R	L	C	V
0.5H	1kΩ	100nF	1V

    

    

    $$

    alpha=frac=1000

    

    

    

    $$

    omega_n=fracapprox 4472

    

    

    

    $$

    s_1=-1000-4359j

    

    

    

    $$

    s_2=-1000+4359j

    

    

    

    $$

    v_n(t)=A_1e^+A_2e^

    

    

    Thus by Eulers blueprint ($e^=cos+jsin$):

    

    $$

    v_n(t)=e^ig

    

    

    

    $$

    v_n(t)=e^ig

    

    

    Let $B\_1=A\_1+A\_2$ and $B\_2=j(-A\_1+A\_2)$

    

    $$

    v_n(t)=e^ig

    

    

    Solve for $B\_1$ and $B\_2$:

    From blueprint
ef, $v\_f=1$ for a assemblage move of magnitude

    1V. Accordingly barter of $v\_f$ and $v\_n(t)$ into blueprint
ef gives:

    

    $$

    v_c(t)=1+e^ig

    

    

    for $t=0$ the voltage beyond the capacitor is zero, $v\_c(t)=0$

    

    $$

    0=1+B_1cos(0)+B_2sin(0)

    

    

    

    $$

    B_1=-1mbox

    

    

    for $t=0$, the accepted in the inductor haveto be zero, $i(0)=0$

    

    $$

    i(t)=fracC

    

    

    

    $$

    i(0)=100cdot10^ig

    

    

    

    $$

    0=100cdot10^ig

    

    

    substituting $B\_1$ from blueprint
ef gives

    

    $$

    B_2approx-0.229

    

    

    For $t>0$, $v\_c(t)$ is accustomed

    $$

    v_c(t)=1-e^ig

    

    

    $v\_$ is accustomed

    $$

    v_=V_-v_c(t)

    

    

    

    $$

    v_=Vu(t)-v_c(t)

    

    

    For $t>0$, $v\_$ is accustomed

    $$

    v_=e^ig