# starting condions..

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#### transgalactic

##### Banned
how to know if the starting conditions at t=0- and t=0+

without doing integration from t=0- to t=0+ method

??

#### MrAl

##### Well-Known Member
Hi,

That is a very general question, and usually you have to be more specific
to be able to find starting conditions, especially since you may want
derivatives or just the function at t=0.

One general method for finding f(0) is to use the initial value theorem, which
states that the initial value in the time domain is the F(s=inf) value in
the frequency domain. In this way, you can use Laplace transforms
and take the limit as s approaches infinity.

This really is a varied subject though, because sometimes we need f'(0)
or something like that. One of your other problems (the parallel RLC with
impluse circuit) required the derivative of iL at t=0, and that can come
from either Laplace or taking a good look at the physical layout and how
the various components (L,C,R) respond to the excitation.

Two very useful ideas are for the L and C, and state:

For an inductor, the current can not change by a finite amount in zero time,
and for a capacitor, the voltage can not change by a finite amount in zero time.
There is one exception though, and that is with an impulse. An impulse can cause
a finite change in a cap or inductor in zero time, but that is the only way this
can happen.

Because of the above, many times v(0+)=v(0-) and i(0+)=i(0-), unless there
is an impulse source present and then you have to look at the physical layout
and try to determine how to go about solving it, or use Laplace.

If there is an impulse, then you have to look at the way the circuit element
reacts to the impulse. A cap subject to an impulse will react by integrating it,
so you can say that the cap reacts to the 'area' of the impulse, for example.

In the parallel RLC + impulse circuit, first the cap integrates and causes a voltage,
then the inductor starts to react to the voltage, but even though the inductor current
is zero at t=0, the derivative is not zero. Thus, we have to solve for the derivative of iL
at t=0.
To get the initial cap voltage, we would get 1/C volts. This is because when we integrate
the impulse we get 1, and 1/C is outside the integrand. Then to solve for di/dt, we use
v=L*di/dt

so we get:
di/dt=v/L

and since v=vc and vc=1/C, we get:
iL'(0)=1/(L*C).

That should show how varied this kind of question can be with various circuits.

Last edited:

#### MrAl

##### Well-Known Member
Hello,

We can solve this by taking the physical layout into consideration.

The 0- inductor current is I0 and the application of a finite voltage can not change
the current in zero time, so iL0+=iL0-.
The 0- cap voltage is vC0, and the application of a finite current (from the inductor)
can not change the voltage in zero time, so vC0+=vC0-.

At t=0, the voltage starts to change due to the (reverse) inductor current iL0,
and a finite current though a capactor starts to change the voltage by the rate
of i/C, and it acts to discharge the cap, so the first derivative at t=0+ is
equal to -iL0/C.

At t=0, the inductor current starts to change due to the (forward) cap voltage vC0,
and a finite voltage across an inductor starts to change the current by the rate
of v/L, and it acts to increase the inductor current, so the first derivative at t=0+
is equal to vC0/L.

Thus,

vC(0+)=vC(0-)
iL(0+)=iL(0-)
dvC/dt(0)=-iL(0)/C
diL/dt(0)=vC(0)/L

#### transgalactic

##### Banned
can you explain this sentence further
"The 0- inductor current is I0 and the application of a finite voltage can not change
the current in zero time, so iL0+=iL0-."

what is "application of a finite voltage"
and why it cannot change "the current in zero time"
and why does it give us " iL0+=iL0-"

??

#### dknguyen

##### Well-Known Member
can you explain this sentence further
"The 0- inductor current is I0 and the application of a finite voltage can not change
the current in zero time, so iL0+=iL0-."

and why it cannot change "the current in zero time"
and why does it give us " iL0+=iL0-"

??
It just means that the function is continuous, smooth and therefore differentiable (deriviatives, should you need to use them, will be valid).

You already know that the current in an inductor must change smoothly which explains iL0+ = iL0-. For an inductor:
V = L*(dI/dt)->(di/dt) means the current in an inductor is always differentiable which means it always has a finite slope and is smooth.

Last edited:

#### transgalactic

##### Banned
how can i see it in the 0+ 0- integrating method

its a second order equation
how to find the I'(0-)
I'(0+)
?
because i was told that a circuit with a switch creates a jump in current and voltage in t=0
so its not continues

can you show how to find them using the integrating method from 0+ to 0-
?
?

Last edited:

#### MrAl

##### Well-Known Member
Hi again,

A circuit with a switch behaves in different ways, not always the same way.
It depends what is connected to the switch, voltage or current source, and
what element, inductor or capacitor. We either get a jump in current or a
jump in voltage, but not both (only barring an impulse source again).

We should talk about non-impulse sources for now because that's where you
seem to need some information.

We can show how the inductor reacts to a jump in voltage by looking at
what happens when a voltage is suddenly switched across an inductor.
The inductor current does not jump up to any value because the inductance
depends not only on that voltage but also on time. In other words, the L
needs time to react to the v before any current can actually flow. This
doesnt mean that it doesnt start to flow, but starting to flow and actually
flowing are two different things.

If we simply look at the defining equation for the inductor:

v=L*di/dt

we only wish to find out what happens to di for very short periods of time dt,
with constant voltage (across the inductor) of v.

Solving for di:
di=v*dt/L

Now lets first look at what happens between two time periods very close
together. We will look at t0 and sometime later t1:

di=1/L*(v*(t1)-v*(t0))

and when we let t1 come very very close to t0, we end up with the same
equation:

di=v*dt/L

and now take the limit:

di=lim [dt->0] v*dt/L=0

In other words, when dt approaches zero so does the change in the current.
Stated another way, the current does not change in zero time (dt=0) for any finite
applied voltage v.
This means that iL(0+)=iL(0-) because the time between 0- and 0+ is considered
to be zero.

If you would like to see this done with another method, post that method next.

Here is a drawing of the inductor I response to a step change in voltage, and a
capacitor voltage response to a step change in current:

It becomes apparent that the first derivative is not zero, but that
the initial value of the function is because there is no time to allow
that change.

#### Attachments

• InitialConditions-LC-01.gif
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#### transgalactic

##### Banned
ok
1.i understand that for an inductor I(0-)=I(0+)
and for a capacitor V(0-)=v(0+)
but you said that it depends on the source type
does it depends if an inductor is connected in parallel or not for current continuity?(current source) ??

does it depends if a capacitor is connected in parallel or not for voltage continuity?(voltage source)

2.does a circuit without a switch is continues for starting condition?

3.when on a circuit with a switch i can see that there will not be a continuity on starting conditions?

4.what happens on impulse regarding the starting conditions?

5.how to find the I'(0-) and I'(0+)
on my original circuit
???

Last edited:

#### MrAl

##### Well-Known Member
Hello,


It depends on the source type, the types being either finite or impulse.
The elements act differently with impulses than with finite sources.


I guess if there are no other abrupt changes to the network, such as
no impulses switched in later.


When there is an impulse.


Starting conditions for an impulse current in a capacitor cause the cap voltage
to rise instantaneously. This is more of a theoretical aspect of the circuit
but it has practical application so it ends up being very important. There are
many times when we want to know the impulse response of a network.
As you know, a capacitor is:
i=C*dv/dt
and we get the voltage when we integrate.
With an impulse and we want v(0):
v=1/C*Intg i(t) dt
and since i(t)=delta(t) and using the sampling property of the impulse, we get:
v(0)=1/C*Intg delta(t) dt [integrated over 0- to 0+]
and since delta(t) integrated even over 0- to 0+ gives us unity (1) we get:
v(0)=1/C*1
which of course gives us:
v(0)=1/C

So, a capacitor driven by an impulse current source gives us the initial voltage
v(0)=1/C volts.

Note this is very different than when we use a finite source because with the impulse
v(0-)=0 and v(0+)=1/C, and with a finite source if v(0-)=0 then v(0+)=0 too.


The way to find I'(0-) depends on how the circuit is connected and any driving sources
prior to t=0. With your original circuit, you should look at what happens when the
switches are set to the state t=0-. To get this value you would keep the two outside
switches closed and the inside center switch open. You would then look at the final
value of the current (inductor) and voltage (capacitor). In that circuit this is
easy because after t=inf all the activity has died down, so all the derivatives
are zero. The thing to note however is that just because I'(0-)=0 that does not
mean that I'(0+)=0 because things start to change when the switches change state.
I suggest that you look at the picture i posted last time though because those
two circuits are much more simple, yet they help to illustrate the same thing...
that's why i posted them. Please take a little time and look at them first.

To get I'(0+) i had shown in the previous post. Please take another look at
that and let me know how it looks to you.

Last edited:

#### transgalactic

##### Banned
i read your last two posts for 2 times
i still have these problems:
1.i know that I_L(0-)=I_0
on what basis you say that I_L'(0-) is I_0 too??

i know that V_c(0-)=V_0
on what basis you say that V_c'(0-) is v_0 too??

2. at time t=0
the switches are flipped
and we get a different circuit now.
i know a law:
for an inductor: even if the voltage function threw it is not continues(but bounded)
the currect has to be continues.

for a capacitor: even if the current function threw it is not continues(but bounded)
the voltage has to be continues.
now i dont know what change happens in t=0
as i see it
we have a change both in current and in voltage
for all the components on the new circuit

3.how to spot if there is some change in a certain component??

4.
on what basis i can say that v_c(0+)=v_c(0-)
so every time i see a capacitor ill say
aoutomatickly v_c(0+)=v_c(0-) because of this law

on what basis i can say that I_L(0+)=I_L(0-)
so every time i see a inductor ill say
aoutomatickly I_L(0+)=I_L(0-) because of this law

is that correct??

5.
"The thing to note however is that just because I'(0-)=0 that does not
mean that I'(0+)=0 because things start to change when the switches change state."

i know that there is a change of state but in this question i lack
the mathematical knowledge
of how to get to I_L'(0+) and V_c'(0+)
??
i guess that i use I_c=C(v_c)' (and i input the current of the cap on 0+)
and V_L=L(I_L)' (and i input the voltage of the inductor on 0+)

currect?

#### MrAl

##### Well-Known Member

I_L'(0) is not really equal to I_0, it is equal to Vc_0/L.
Vc'(0) is not really v_0, it is -I_L(0)/C.


Yes the circuit is different when the switches are flipped.
Yes, the current in the inductor is continuous.
Yes, the voltage in the capacitor is continuous also.

At t=0+ we have no change in the cap voltage or inductor current
because both are continuous and can not change in zero time.
We dont have an *immediate* change in V_C or I_L, although they
do change later. It's like having a wrist watch that can stop time...
nothing can happen when time is stopped.
We do have a change in both first derivatives however, from
zero to something else as noted above in .


You have to look at the physical considerations of the circuit,
in the way that the capacitor works and the way that the inductor
works as to continuous quantities current or voltage.
You have to do the same for the first derivatives too.
You use the defining equations for the inductor and capacitor to
do this, or just realize that the quantities are continuous.


The basis is the physical way in which the capacitor works.
The cap voltage can not change in zero time because it actually
takes at least some small amount of time to effect a change
in the capacitor voltage. In other words, the capacitor voltage
is a *constant* at t=0. The inductor current is also a *constant*
at t=0. The only time this doesnt apply is when there is an impulse,
and then you have to look at how the element reacts to an impulse.


Yes.
In your circuit the cap current comes from the inductor
initial current, and the inductor voltage comes from the capacitor
initial voltage. In those two ciruits i posted the same thing is
happening, the only difference is that i used two separate circuits
instead of one (of course the results later in time may be different
for the two circuits than for your one circuit).

[...]
I really think you should work out the solutions for those two circuits i
posted, because they involve the same principles only they are a little
easier to understand. Then later you can apply those principles to your
circuit with the three switches.

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#### transgalactic

##### Banned
so by the concepts that i learned here
i tried to solve it:
Ic(0-)=0
Vc(0-)=V_0

I_L(0-)=I_0
V_L(0-)=0

V_c(0-)=V_c(0+)=V_0

I_L(0-)=I_L(0+)=I_0 (because of the continuety)

so
by
(V_c)'=I_c/C

(I_L)'=V_L/L
(V_c(0-))'=I_c(0-)/C
(I_L(0-))'=V_L(0-)/L
(I_L(0-))'=0/L=0
(V_c(0-))'=0/C=0

V_c(0-)=V_c(0+)=V_0

I_L(0-)=I_L(0+)=I_0

(I_L(0+))'=0/L=0
(V_c(0+))'=0/C=0

where is the mistake?

Last edited:

#### transgalactic

##### Banned
i understand my mistake

i assumed that Ic(0-)=Ic(0+)=0
but the capacitor doesnt keep continuety on the current
thanks

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