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Need Homework help!

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jacky2011

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Need help for home work..Some problems are pretty tough..Please find the attachment for problems..Any kind of help would be appreciated..Thank you all.
 

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Hello there,


You're asking a lot of questions in one post :)

I'll start you off with Task 1 and we'll see how it goes.

Task 1:

We need to find the optimal resistor value for R2 that provides a maximum
VoutLight-VoutDark with R1 changing value R_light to R_dark. Since we
need to maximize VoutLight-VoutDark, we need to find an equation for that
quantity first so we can maximize it when we change R2.
Lets first simplify the notation a little:
Va will be VoutLight,
Vb will be VoutDark,
Ra will be R1 Light,
Rb will be R1 Dark,
we'll keep R2 the same (R2).
Since a resistive voltage divider has output:
E*R2/(R1+R2)
we can set up two circuits one with Ra and one with Rb and then subtract the
two outputs, then use that equation in the maximization process. Since E
is common to both we can normalize it to 1 to simplify just a little.
First circuit:
Va=R2/(R2+Ra)
Second circuit:
Vb=R2/(R2+Rb)
Combined:
Vdiff=Vd=Va-Vb=R2/(R2+Ra)-R2/(R2+Rb)
Now we want to maximize Vd as R2 changes so we differentiate:
d(Vd)/dR2=d(R2/(R2+Ra)-R2/(R2+Rb))/dR2=((Ra-Rb)*(R2^2-Ra*Rb))/((R2+Ra)^2*(R2+Rb)^2)
Now we set that equal to zero:
0=((Ra-Rb)*(R2^2-Ra*Rb))/((R2+Ra)^2*(R2+Rb)^2)
and solve for R2.
Multiplying both sides by the denominator we get:
0=((Ra-Rb)*(R2^2-Ra*Rb))
Multiplying that out we get:
0=-Rb*R2^2+Ra*R2^2+Ra*Rb^2-Ra^2*Rb
which is a quadratic with no R2 term, so we solve for R2^2 and we get:
R2^2=Ra*Rb
the solutions are:
R2=-sqrt(Ra*Rb), and
R2=sqrt(Ra*Rb)
Since the quantity under the radical is positive the first solution yields a
negative resistance, but the second solution gives us a positive resistance so
we drop the first and use the second:
R2=sqrt(Ra*Rb)
and in terms of the original variables that gives us:
R2=sqrt(R_Light*R_Dark)
and that's the solution.

For example, if R_Light=1k and R_Dark=2k then R2=sqrt(1k*2k)=1.4142k approximately.
You can double check the result by incrementing that value of R2 by a small amount like 0.1 ohm
and checking it in the original equation, then decrementing that value of R2 by a small amount and
checking that in the original equation. Both of these new test values should yield smaller differences in
the difference voltage if we got it right.
 
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Hello again jacky,

Where are you from BTW?

I was thinking some others might jump in and offer some comments too. But anyway, on to Task 2.

Task 2:

Here we just have to find the time domain response to a network containing resistors and one capacitor. We could derive the equation using the impedances for the cap and R1 and for R2 and use the voltage divider formula again, but since they gave us the frequency domain response that saved us some work so we can start from there. The intent is to convert F(w) to F(s) and then use a table of Laplace Transforms to convert to the time domain. There are other ways to do the inverse Laplace Transform and it's a good idea to become familiar with some of these methods, but table lookup is quite simple for a small network so we're going to do it that way. There are a lot of Laplace Transform tables published on the web so finding one shouldnt be hard to do.

Starting with the equation given:
[note 'i' here is the imaginary operator usually denoted by 'j' in electrical work]
F(w)=R2*C*w/((R1+R2)*C*w-i)
if we multiply top and bottom by 'i' we get:
F(w)=R2*C*w*i/((R1+R2)*C*w*i+1)
and now if we replace every i*w with 's' we get:
F(s)=R2*C*s/((R1+R2)*C*s+1)
Since they didnt tell us what kind of input to use we use the step input,
so we multiply the equation by 1/s and we get:
(1/s)*F(s)=R2*C/((R1+R2)*C*s+1)
now we divide top and bottom of the right side by x=(R1+R2)*C and we get:
(R2*C/x)/(s+1/x)
and because x is a constant we can recognize this form as:
1/(s+a)
after it is multiplied by R2*C/x, and where
a=1/x
and looking up 1/(s+a) in a table of Laplace Transforms we find that:
1/(s+a) => e^(-a*t)
and since we have a constant multiplier we multiply both sides by R2*C/x and we get:
R2*C/x*1/(s+a) => R2*C/x*e^(-a*t)
or simply:
F(t)=R2*C/x*e^(-a*t)
and since a=1/x we have on the right:
R2*C/x*e^(-t/x)
and now subsitute the original value of x:
x=(R1+R2)*C
back into the equation we get:
F(t)=R2*C/((R1+R2)*C)*e^(-t/((R1+R2)*C))
and simplifying a little we get:
F(t)=R2/(R1+R2)*e^(-t/((R1+R2)*C))
and that's the solution for F(t).

One way to test this result would be to use a bunch of harmonics
in the F(w) equation and compare with the F(t) at a given point
in time, but even simpler would be to test with a circuit simulator
program that is known to give correct results.
 
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Thanks a lot Sir.
I am from Portsmouth, UK.
I don't know about others, but you are doing a lot for me. I am not good at Electronics, but with your help I am getting into the subject.

Thank you again. :)
 
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Hello again,


I couldnt help but wonder if you are just getting into electronics where did you get those questions from?
Also, where are the lectures mentioned in those questions?
 
Hi MrAl,

These questions are from our Electrical lectures homework..I don't have any .pdf or .doc file for lectures so I could not upload..Thanks for your help so far, any further assistance for rest of work would be really helpful..

Kind regards,
Jack
 
Hello again,

Ok, well some of the problems may not be solvable without seeing the lectures. I'll take a better look though.
In the mean time, what kind of circuit analysis have you already done in the past? Maybe you should start with some simpler problems first and work up to those problems?
 
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Hi MrAl,

I collected some lectures notes from one of my friend. These notes may be useful.

Regards,
Jack
 

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  • Lecture 5.pdf
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  • Lecture 6.pdf
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  • Lecture 7.pdf
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Hello again,

Ok i'll take a look.
BTW, what kind of circuits have you analyzed in the past already?
 
Hello again,

It would help if you tell me what kind of circuits you've done already.

In the mean time, here is a solution for Task 3, although i dont know what good this will do for you :)

The circuit is shown in the attachment. There's one inductor and one capacitor and two resistors. They didnt give us the input again, so we assume a step input.

We are already given the frequency domain equation:
F(w)=(w*C*R1*R2+i*(w^2*C*L-1)*R2)/(i*(w^2*C*L*R2-R2-R1)+w*C*R1*R2+w*L)
and we note that there are powers of w in this equation and if we multiply top and bottom by i we will get pairs of (i*w) and its powers again:
F(w)=(i*w*C*R1*R2+i^2*w^2*C*L*R2-i^2*R2)/(i*w*C*R1*R2+i^2*w^2*C*L*R2-i^2*R2-i^2*R1+i*w*L)
and now we simplify this and we get:
F(w)=(i*w*C*R1*R2+i^2*w^2*C*L*R2+R2)/(i*w*C*R1*R2+i^2*w^2*C*L*R2+R2+R1+i*w*L)
and now replace every i*w with s again we get:
F(s)=(s^2*C*L*R2+s*C*R1*R2+R2)/(s^2*C*L*R2+s*C*R1*R2+s*L+R2+R1)
and now transforming that to the time domain we get the solution shown below for F(t). This is a second order transfer function and this is usually what comes up when we have two energy storage components in the circuit ie one order per energy storage element. Sometimes we get lucky though and get a lower order equation when some elements combine in a certain way.
This equation is a bit harder to transform so you should look at ways to transform these equations on the web. The time domain equation here comes from using Inverse Laplace Transform methods. There are lots of sites showing these methods.

So the general solution comes out to be:

F(t)=E*B*(R2+e^(-a*t)*(cos(t*w)+A*(1/w)*sin(t*w)))
where
E is the input step amplitude,
b=2*C*L*R2,
a=(C*R1*R2+L)/b,
d=2*R2*b-C^2*R1^2*R2^2+b*R1-L^2,
w=sqrt(d)/b,
A=(C*R1^2*R2-2*L*R2-L*R1)/b, and
B=1/(R1+R2)

There are three distinct cases for this general solution so a couple notes about this solution are surely worth mentioning.

First, you'll note that the variable 'd' is found inside the square root, so we have to take the square root as part of the calculation for the angular frequency 'w'. 'd' is often called the "discriminant", and tells us some interesting facts about the solution. 'd' takes on three different classes of values, either negative, positive, or zero. You can note that the class of 'd' depends on the component values so depending on those values we'll get different values for 'd'. If we get a positive value for 'd', we note that sqrt(d) is real and that makes 'w' real also, so we can use the general solution directly. If we get a negative value for 'd' however we note that sqrt(d) is not real and so that makes 'w' imaginary, and if we get zero for 'd' sqrt(d) is zero and that makes 'w' zero also.
Now if 'w' comes out imaginary, that makes the sin and cos functions turn into sinh and cosh respectively, and the sinh term ends up being imaginary. However, since 'w' is also imaginary that makes the whole sinh term come out real. That's why i left the factor (1/w) out in front of the sin() term so that in the case where 'w' is imaginary we can reduce that whole term to a real before we proceed with the calculation. Now if 'w' comes out to zero we have another little problem. With that factor (1/w) again since the w is in the denominator that would make this factor (and thus the whole term) indeterminate, and so instead of calculating it directly we have to take the limit as 'w' approaches zero. That reduces that whole term to simply A*t. Also, with w=0, cos(w*t)=1 and that simplifies that term also.
So we have three separate cases:
1. d=positive and so w=real,
2. d=negative and so w=imaginary, and
3. d=0 and so w=0.
For case 1 we apply the equation directly, for case 2 we have to replace sin and cos with sinh and cosh and we have to reduce i*sinh() term to a real, and for case 3 we have to take the limit of the sin() term and cos() goes to 1.
Mathematically this happens automatically so we dont really have to do anything, but when we go to calculate it with a calculator or even a computer those programs usually can not handle the symbolic math required for the reductions so it's a good idea to know how this formula changes. Ultimately what i like to do is solve the general equation for the three separate cases and code them individually in a program and solve for the three different conditions of d and w.
This would look something like this:
if d>0 then
Do Case1
elsif d<0 then
Do Case 2
else
Do Case 3

Also, when d is real the response is called under damped, and when d is imaginary the response is over damped, and when d is zero the response is critically damped. I like to think of the under damped case as "wild and oscillatory", and the over damped case as "lazy and slow", and the critically damped case as "just barely over damped".
The "damping factor" can be found from examining the denominator of F(w), and that tells us something about the solution right away even before we transform it to F(t). Im sure there is much about this on the web too.
 

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Hi MrAl,

Thank You so much Sir. Sorry for late reply. In our last practical we did MI sensor circuit and magneto-impedance effect, sample tensile load network,current leakage sensor.

Thanking You,
Regards,
Jack
 
Hi again,

Task 11 was simple so i did that one next. I hope you get some use out of this :)

One way to solve for F(w) is to solve for the junction at the resistors. When we get to the second junction we can simply multiply by -1 and that gives us the output. F(s) therefore comes out to:
Vout(s)=-(Vin*R3)/(2*s^2*C1*C2*R1*R2*R3 + 2*s*C2*R2*R3 + 2*s*C2*R1*R3 + s*C1*R1*R3 + 2*s*C2*R1*R2 + R3+2*R1)
and to get that into F(w) just replace all 's' with "j*w" or "i*w".

This equation F(s) is in the general form:
Vout(s)=-(Vin*R3)/(s^2*A+s*B+C)

and the Inverse Laplace Transform for that with Vin a step is:
Vout=Vin*(R3/C)*(e^(-a*t)*(sin(t*w)*B/(2*w*A)+cos(t*w))-1)
where
A=2*C1*C2*R1*R2*R3
B=2*C2*R2*R3+2*C2*R1*R3+C1*R1*R3+2*C2*R1*R2
C=R3+2*R1
d=4*A*C-B^2
w=sqrt(d)/(2*A)
Note again that d is under the square root, so we again have three cases and we do the same as with the previous result for the three different cases.
 
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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 :)
 
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Hi again,

You're welcome.

Note that i had to edit a couple more typos though so the results changed slightly from before. I had w^2 in two places where it should have been just w. It's fixed now.
 
Hello again,

I went over that and simplified a few things, and here are the waveforms i promised. These waves show the process step by step and make it much more clear how this transform can be developed.



I thought it would be better to start with a regular stepped
wave (staircase) and show the transforms for them and how they
add in time.

Repeating a few definitions 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]

Refer to the three attachments to view the figures.

Starting with a rising staircase with step period 1
(Figure 1A, blue waveform):
1/((1-E1)*s)

Using time shift and inverting that wave we get the same
wave but it is delaying by a period of 1/2 and is going
down instead of up (Figure 1A, red waveform):
(-E2)/((1-E1)*s)

That gives us two waves shown in Figure 1A, one that is
going up and the other that is going down but delayed by
1/2 period. When we add the two waves we get a pulsed
wave that does not increase but stays level. That is
simply the algebraic sum and is shown in Figure 1B.
Thus, we can see that two waves that are increasing can
sometimes add up to a wave that looks normal and does
not increase but simply repeats.

Now on to the actual transform of the full wave rectified
sine wave.

We start with a sine wave who's transform is:
w/(w^2+s^2)

and with period 1 is:
W=(2*pi)/((2*pi)^2+s^2)

and that is not graphed as it is just a sine wave.

Using time periodicity we get a strange 'stepped'
sine wave (Figure 2A, the blue wave):
W/(1-E1)

Note that this wave does not increase smoothly, but it
makes abrupt steps (just as with the simple staircase wave before)
as each sine period is complete in itself and
for each period the sine amplitude increases by 1.

Next, we use time shift on that same wave above and this
results in the same wave but it is delayed by 1/2 period
(Figure 2A, the red wave):
W*(E2)/(1-E1)

Now when we take the above two waves and add them together
they partly cancel each other. For every positive half
cycle of the blue wave the red wave only partially cancels
the blue wave, and for every negative half cycle of the blue
wave the red wave completely cancels the blue wave. When
the red wave only partially cancels the blue wave what
remains is a half cycle of the original sine wave, and when
the red wave completely cancels the blue wave what remains
is zero. This new wave is shown in Figure 2B and it is
plain to see this is a half wave rectified sine wave.
The entire transform is simply the sum of the above two
waves, so we get:
W*(1+E2)/(1-E1)

That is the transform of a half wave rectified sine, where
the positive half cycles of the original sine appear in the
wave but the negative half cycles are canceled out.
It is shown in Figures 2B and 3A.

What we want is a full wave rectified sine, and noting the
only thing missing from the above wave (Figure 3A) is the
other half of the waveform and noting that if we were to delay
that wave by 1/2 period we would get the missing half, we do
just that. We delay the above wave by 1/2 period and that
gives us the other half (Figure 3B) and the transform is:
W*(1+E2)*E2/(1-E1)

Now that we have both halves, we simply add them together
to get the transform of the entire full wave rectified sine:
F(s)=W*(1+E2)/(1-E1)+W*(1+E2)*E2/(1-E1)

and that final wave is shown in Figure 3C.
The final transform can be simplified in various ways and
i leave that to the reader.
 

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Hello again,


Looking over Task 10, it seems not too hard to do either so i think i'll just outline the procedure and see how you make out with that. It's just a matter of a little electrical circuit analysis and a tiny amount of insight into the motor electromechanical operation itself.

We are asked to connect the stator windings and armature windings in two different ways:
1. In series.
2. In parallel.

Since these windings both consist of only one inductance and one resistance, we have two series RL circuits to work with.
Putting them first in series we end up with one network with one current I=Ia=Is and that's just the two RL circuits connected in series.
Putting them in parallel we end up with one network with two separate currents Ia and Is, and two RL circuits in parallel.
Both of these configurations only require connecting the two simple networks together and analyzing the current(s) through them with excitation of an external voltage course. These are still very simple networks.
The only trick is in the electromechanical aspect, where we have to include both currents in the mechanical equation for torque. The two currents will multiply so in one circuit we get I^2 and in the other we get Ia*Is, instead of just Ia or Is alone. This is what makes it non linear.
So for the torque for the individual circuits we would get:
T=Kt*Ia and T=Ks*Is
and in series we would get:
T=Kt*Ia*Ks*Is
but since I=Ia=Is and we can lump the constants we would get the simpler:
T=K*I^2
For the parallel circuits, we would have to keep the currents separate:
T=Kt*Ia*Ks*Is
but we can still lump the constants so we end up with:
T=K*Ia*Is

So the circuits themselves either look like two different RL in series or two different RL in parallel, and the electromechanical expressions are as those above.
 
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