http://i44.tinypic.com/2i7o3v6.jpg
how to solve it.
i made a differential equation for the first and second periuds

and i got their starting conditions

i dont know how to find the starting condition of the ZSR(differential equation) after a long time
like they ask


??

does any body has an idea

i can post the differential equation of the first an second period if it helps

 

Hi,


Im not sure if i understand your question but i'll take a shot.

It sounds like that when you start your analysis you're ok, but
when you need to switch equations (when the switch changes)
you dont know how to figure out the initial conditions for the next
period.
If this is the case, the answer is quite simple really. The initial
conditions of the next period come from the end conditions of the
previous period...and they are...the same!
In other words, the current flowing at the time of the last period
is taken as the initial current in the next period.
Of course the first period has initial condition of zero.

If you want to post your equations we can take a look at those too.

 

ok i will copy what i got

 

i dont know how to find the initial conditions
of time "n"

in time
I(L/R)=(-E/R )*e^(-1) + E/R
I(2L/R)=((-E/R )*e^(-1) +E/R)*e^(-1)

what now??

 

I'm used to seeing the Laplace Transforms ...
Can you write the two LT equations ... for each respective switch state?

 

Hi again,


I see you found out that the charging current starts out at E/R, and
defining that as Im, for the first half cycle we get:
i11=Im*(1-e^(-a*dt))
where
a=R/L, and
dt=one half cycle time period
The second half cycle is discharging, so we get:
i21=i11*(e^(-a*dt))

The third half cycle is again charging, only this time:
i12=(Im-i21)*(1-e^(-a*dt))+i21
and again the forth half cycle is again discharging, so we have:
i22=i12*(e^(-a*dt))

It becomes clear that for odd half cycles the equation is:
i1=(Im-i2)*(1-e^(-a*dt))+i2 (with initial i2=0 of course)
and for even half cycles:
i2=i1*e^(-a*dt)

Now you can simply keep calculating i1 and inserting into the equation for i2,
and then calculate i2 and insert that back into the equation for i1, until you
reach a point where i1 and i2 no longer change, or change very little.
When you reach that point, you know i1 and i2 are the upper and lower
limits of the current wave.

Doing this analytically, we end up with:

i1=Im*(1-e^(a*dt)+e^(-2*a*dt)-e^(-3*a*dt)+e^(-4*a*dt)-...+...)

which can be summed up as:

i1=Im*(1+sum[1 to N][e^(-k*a*dt)*(-1)^k])

for N the number of half cycles and N is odd, and i1 is the upper limit on the sawtooth.
You can derive a similar equation for i2, the lower limit on the sawtooth,
or use the fact that it is centered between Im and 0 amps (which would be Im/2 amps).

Worked example:
L=100uH
R=100 ohms
dt=10us (each half cycle is 10us)
square wave with period 20us.

The upper limit i1 is: 0.00731 amps approximately, and
the lower limit i2 is: 0.010-i1=0.00269 amps approximately.

 

why you say that "E/R=10ma"

we are not given E and R

 

Hi again,


Yes, sorry, that was a typo. I have corrected that now and it should read:

"I see you found out that the charging current starts out at E/R, and
defining that as Im, for the first half cycle we get:"

instead of:

"I see you found out that the charging current starts out at 10ma (E/R), and
defining that as Im, for the first half cycle we get:"

The problem came in where i accidentally typed the 10ma from the
worked example instead of just the more general "E/R".
There were a couple other typos too that apparently you didnt see because
i had corrected them already. I believe they are all corrected now.

Also note that this charging current is not the initial current, just the coefficient
in the equation which also happens to be the max current that could be obtainable.

That sure would be magic if E/R always equaled 10ma wouldnt it? ha ha :-)

 

Hi again,


I just thought i would mention that there is a closed form for that
series i gave you a while back (a few posts back). Would you
like to try to find it?