Hello again,
Ok sure, you're welcome.
Next, Task 9 wasnt too hard so i did that one next.
They are asking for the mathematical model of the full wave rectifier with sine input first, and also the model with a capacitive filter with some series resistance and some load resistance.
In the following, a 'stepped' sine wave is a stepped wave similar to a staircase except for each step we have a sine wave of one period that increases by 1 for each step. Using a stepped sine wave allows us to use the time shift and time periodicity theorems to develop a full wave rectified sine. This stepped wave is not to be confused with a sine wave that is approximated using small steps for tiny fractions of the wave, but instead each step is a complete perfect sine where its amplitude increases by 1 for each additional step. This is a rather unusual looking wave that has abrupt transitions between each period, but it enables us to use some common theorems that are the same for all waves. The wave itself looks first like a sine of amplitude 1, then the next period it is a sine of amplitude 2, then for the next period an amplitude of 3, etc., etc., and if left by itself it would increase forever, but we add to it another wave which cancels part of it out so we end up with a half wave rectified sine and that's when it gets easy. I'll post some waveforms as soon as possible as this is really quite neat to see graphically.
We start with a sine wave who's transform is:
w/(w^2+s^2)
and we'll use a frequency of 1 Hz for now so we get:
2*pi/((2*pi)^2+s^2)
Next we'll define a few things to make the notation simpler:
W=(2*pi)/((2*pi)^2+s^2) [note period is 1]
E1=e^(-s/1) [note period is 1]
E2=e^(-s/2) [note period is 1/2]
Using the time periodicity theorem 1/(1-e^(-s*Period)) we get a 'stepped'
sine wave starting at zero and amplitude increasing by 1 every period:
W/(1-E1)
Using a time shift of 1/2 period we get a stepped sine wave that is the same as above
but is delayed by 1/2 period:
W*(E2)/(1-E1)
Adding that to the original stepped sine that was not delayed we get:
W*(1+E2)/(1-E1)
and because the inverted stepped sine wave cancels every other cycle of
the original sine, that gives us a half wave rectified wave above. Thus,
a half wave rectified sine looks like this:
W*(1+E2)/(1-E1)
Now if we delay that by 1/2 period we get:
W*(1+E2)*e^(-s/2)/(1-E1)
or in simpler notation:
W*(1+E2)*E2/(1-E1)
and that gives us another half wave rectified sine, except that is now delayed
by 1/2 period from the first half wave rectified sine. This would look just
like the original half wave sine but would be delayed by 1/2 period.
Now that we have expressions for a half wave rectified sine and another half wave
delayed by 1/2 period, all we have to do is add them together and we get a
full wave rectified sine:
F(s)=W*(1+E2)/(1-E1)+W*(1+E2)*E2/(1-E1)
This works because every time the original half wave was active the other
half wave was zero, and vice versa.
The above simplifies to:
F(s)=((E2+1)^2*W)/(1-E1)
and all we have to do now is replace the W, E1 and E2 with the original
definitions and we have the required transform for the full wave rectified sine,
and then we can use an identity to simplify further.
The resulting exponential terms can be simplified into a trig form with:
coth(a*s/2)
using the identity:
coth(b)=(e^(2*b)+1)/(e^(2*b)-1)
So we end up being able to simplfy this transform into:
F(s)=((2*pi)/((2*pi)^2+s^2))*coth(a*s/2)
and again noting that the transform of a sine is:
w/(w^2+s^2)
we replace that in F(s) and we get:
F(s)=(w/(w^2+s^2))*coth(a*s/2)
and that is the transform of a full wave rectified sine wave.
Note you dont have to convert to the coth() form if you dont want too, but
if you do you'll have to solve for 'a', and that isnt too hard to do.
Now with capacitive filtering and some series resistance and a load as
shown in Fig 3, the transform of the filter alone is:
H(s)=R2/(s*C*R1*R2+R1+R2)
and so the filtered and loaded full wave rectified sine comes out to be
pulsating DC and the transform is:
Filtered(s)=F(s)*H(s)
where the '*' above stands for 'convolution', so we get:
Filtered(s)=(w/(w^2+s^2))*coth(a*s/2)*(R2/(s*C*R1*R2+R1+R2))
So there we have the output of the filter after a full wave rectification of a sine wave of amplitude equal to 1.
Note that this is quite theoretical in nature, as in a real circuit we would need to account at least for the
behavior of the diodes too, but they dont ask for that so we are done.
Looking back i see that they really wanted the differential equations for the circuit, but up to now they wanted the frequency domain equations so the above is still of interest.
We'll have to look at the differential equations next, but that should just be the filter network excited by the absolute value of the sine wave. If you have already done small resistor and capacitor networks this should not be too difficult at all.
Also, if i get some extra energy i'll post a few waveforms to illustrate the process of developing the half wave and full wave rectified sine transforms as that makes the derivation much easier to understand for sure.
BTW, Happy Easter