# Circuit Theory/2Initially Excited

!full circuit with charging section, discharging section and two
switches
!charging circuit (sw1 closed, sw2
open)")
!discharging circuit (sw1 open, sw
closed)")

## Charging

The source voltage could be DC. In this case the initial capacitor
voltage is going to be V~s~ and the derivative of the voltage is going
to be zero.

The source voltage could be AC in which case the exact time or more
precise phase angle of the switch over is going to determine V~s~ and
the derivative of the voltage values.

## Discharging

The initial inductor current is going to be zero and derivative is going
to be zero.

After a long time nothing will be going on. The steady state particular
solution is going to be zero.

The current is going to be the same throughout. Goal should be to find
an expression for the current. However, any voltage could also be found
and then the current. Doesn\'t matter.

In summary, there will be no initial current in the discharging circuit,
there will be a voltage across the capacitor. The inductor is going to
act like an open at t=0~+~ so this entire voltage will appear across the
inductor and instantly (between t=0~-~ and t=0~+~) cause a change in the
derivative of the current through the inductor.

The equation for the current for first order circuits would be:

$$V_c(t) = A e^{-\frac{t}{\tau}}$$ However, this is a 2nd order circuit
(with both capacitors and inductors), thus:

$$V_c(t) = A e^{-st}$$ The complex frequency s has replace tau. The time
constant concept disappears. An intuitive interpretation of s exists,
but is more complicated. Remember:

$$s = \sigma + j\omega$$

## The differential equation

Go back to the complex frequency domain to find the time constant. Start
with the loop equation:

$$V_c + V_r + V_L =0$$ Substitute terminal equation for V~r~ and V~L~
but not for V~c~ because want to avoid an integral:

$$V_c + i(t)R + L{d i(t) \over dt} = 0$$ From the terminal relation of a
capacitor, know that:

$$i(t) = C {d V_c \over dt}$$ So substituting in, have a differential
equation for V~c~:

$$V_c + RC{d V_c \over dt} + LC{d^2 V_c \over dt^2} = 0$$ Can see the
second order now.

## Finding the time constant or zeros

A zero is like a time constant, but there are more than one zero\'s
possible. They might each influence the circuit differently.

Need to find V~c~. The particular solution (steady state) is going to be
zero. (Where the name zero comes from).

The homogeneous solution is going to be of this form:

$$V_c(t) = Ae^{st}$$ Plugging into the differential equation:

$$Ae^{st} + RCAse^{st} + LCAs^2e^{st} = 0$$ Dividing out Ae^st^ have:

$$1 + RCs + LCs^2 = 0$$

$$s^2 + \frac{Rs}{L} + \frac{1}{LC} = 0$$ The zero\'s of this equation
are:

$$s_{1,2} = -\frac{R}{2L} \pm \sqrt{(\frac{R}{2L})^2 - \frac{1}{LC}} = -(\sigma \pm j\omega)$$
From your experience with quadratics you know that ~1,2~ come in three
types of pairs:

  Pair Type         Revised equation for V~c~                            interpretation
  ----------------- ---------------------------------------------------- -------------------------------------------------------------------------------------
  real, not equal   $A_1e^{s_1t} + A_2e^{s_2t}$                          overdamped
  equal             $A_1e^{st} + A_2te^{st}$                             critically damped
  complex           $e^{-\sigma t}(A_1\cos\omega t + A_2\sin\omega t$)   underdamped

The revised equations (guesses) come from math proofs, and some courses
derive these from Laplace transformations. Just remember this table.

There are two storage elements with two initial conditions each, so
there could be four equations and four constants. A~1~ and A~2~ are two
of them. !for undamped (blue), under-damped (green), critically damped
(red), and over-damped (cyan)
cases, under-damped (green), critically damped (red), and over-damped (cyan) cases")
To understand damping, think of a car driving
over a curb. Undamped means the car bounces up and down for ever.
Under-damped is what happens when the shocks start leaking (car bounces
up and down for a while \... like riding in a boat). Critically damped
is why people hate sports cars (hurts their bottom). An over damped
response in a car gives feedback to a person driving a car. Circuits
like that above have been used to model shock obsorbers. These circuits
are called analog computers.

## Almost Resonance (Underdamped)

!Impact of R and L on
resonance

The most interesting of the above three is Underdamped. Something like
resonance occurs.

Satellites expand and contract depending upon which side is facing the
sun or the 2.7 kelvin of space. Repeated expansion and contraction
creates an oscillation that needs to be damped out by a motor that
flings a weight around that dampens the vibration. Otherwise everything
that can wiggle will wiggle more and more until parts begin to break
off. (A car on earth is no different).

Most control circuit input is an unbalanced situation. The output models
the desired behavior. The correcting (feedback control) circuits (not
talked being studied here) kick in to make sure the actual behavior
matches the desired behavior. Most often the objective is to design a
critically damped circuit, rather than merely predict what is going to
happen.

Resonance (or close to it) occurs when s~1~ and s~2~ are purely
imaginary.

$$s_{1,2} = -\frac{R}{2L} \pm \sqrt{(\frac{R}{2L})^2 - \frac{1}{LC}} = -(\sigma \pm j\omega)$$

So:

$$\sigma = \frac{R}{2L}$$ What is interesting is the radical term. It
determines the nature of the 2nd order response.

Pure resonance occurs when the real part of the quadratic solution is
zero which would only happen when R \<\< L.

!Crystals are modeled with a capacitor in parallel with a series RLC
network
The contents of the radical don\'t have to be zero, only the real part
has to be zero in order to be overdamped:

$$j\omega = \sqrt{(\frac{R}{2L})^2 - \frac{1}{LC}}$$

$$j\omega = j\sqrt{\frac{1}{LC} - (\frac{R}{2L})^2}$$ The j\'s cancel,
and we have already assumed R \<\< L, so:

$$\omega = \sqrt{\frac{1}{LC} - (\frac{R}{2L})^2}$$ Which called the
damped, ringing, or undamped frequency.

When R \<\< L, then this turns into the resonate frequency:

$$\omega = \sqrt{\frac{1}{LC}}$$

Mechanical engineers try to predict ω for a given structure and
materials. It could be a building or motor mount. The math proves that ω
exists. Finding the exact ω usually requires testing. Instead of
capacitors and inductors, mechanical engineers model everything with
springs and dampers. Finding ω can be hard. This why there is an entire
industry of shake tables.

The spring and damper values in a structure determine how narrow or
muted the resonance is. When designing a building, mechanical engineers
want the envelope to be as wide and flat (muted) as possible. When
designing a notch filter the electrical engineer\'s design goal is the
opposite: a tall and narrow envelope. Electrical engineers call this
quality.

Most oscillators today are built from
crystals.

