Good morning friends, today we will be discussing
about a new topic locus diagram. I will just
give you a brief idea before we
start about the importance of this locus diagram
sometimes we are interested in studying the
behavior of the current in a circuit
when say one of the parameters or one of the
elements of the impedance is varied over a
wide range, this is applicable in many
circuits specially those who were you have
studied about induction machines will know
induction motor parameters the in the
equivalence circuit you have some variable
resistance. So you can play with this, you
can have maximum torque transfer
condition, maximum power transfer condition
and so on.
So let us see the basic principle involved
in this suppose we have a very simple circuit
consisting of a resistance and then the
inductance, the impedance is j X L and resistance
is R, this is the voltage that is fixed and
one of them either R or XL is varied,
say R is varied what would be the locus of
the current. Obviously, the total impedance
value is changing the changing so the
current will be varying. So what will be the
nature of variation of the current, so if
we see the impedance diagram suppose this
is
the real part, this is the imaginary part
then it is j X L which is fixed and we are
taking different values of R.
So R plus j X L will be say R 1 plus j X L,
R 2 plus j X L. So the impedance varies along
this line is it not its minimum when it
becomes purely inductive so that time its
value is X L okay resistance is 0 and its
maximum value it tends to infinity when R
is
infinity and it will have an angle will find
this angle will tend to 0 so at a very very
large value of resistance, it will be practically
resistive with an infinite magnitude okay.
So this is the locus of Z all right. Now what
will be the current, current is V by Z for
the
time being let us forget about the angle part
at this moment. So you can write Y into V
where Y is the admittance okay. So if the
locus of Z is known what will be the locus
of Y since V is fixed, so the locus of Y will
be a scaled version of current I is it not.
So
how is Y changing with different values of
R, okay? So for a given supply supply voltage
V, Y represents the locus of the locus of
Y represents the locus of I in some scale
Y is having the dimension of mho all right,
unit of mho, I is current ampere.
So obviously, it will be only a scaled version
with a different unit the locus of I with
a scale factor V, we can change the scale
that
is not a problem. So our first task is to
find the locus of a straight line like this
inverse of a straight line. Let us take any
straight
line and find its inverse, what is the inverse
of this straight line, this is the origin,
this is the straight line I want to know for
any
point on this straight line what will be the
corresponding inverse. So can you tell me
what will be the inverse of this, what will
be
the inverse of this?
Suppose we have O, R 1, Q 1, P 1, I have taken
3 arbitrary points one of them is a perpendicular
all right. I have taken 3 points
on this straight line, I would like to see
what will be the positions of their inverses
that is R 2, P 2 and say Q 2, what will be
the
location of R 2, P 2, Q 2 and how are these
points changing their positions okay. Now
OR 1, OR 1 into OR 2 is equal to 1, is it
not
because OR 2 is 1 by OR 2, OR 1 and that is
equal to OP 1 into OP 2 its equal to OQ 1
into OQ 2, is it not, is that all right. Now
P
1 is a perpendicular all right this is a perpendicular
that is this the shortest line.
Now from here, consider these points P 2,
R 2, Q 2 if you remember in geometry you have
studied if I draw 2 lines like this say
O, A 1, A 2, B 1, B 2, so O, A 1, A 2 is a
line cutting the circle at 2 point C 1 A 2
similarly, this one cutting at B 1, B 2 then
OA 1
into OA 2 is OB 1 into OB 2 is that all right.
So if I find OP 1 into OP 2 is equal to OR
2 into OR 1 is equal to OQ 2 into OQ 1,
what does it prove? These points A 1, A 2,
B 1, B 2, C 1, C 2 these are all lying on
a circle. So Q 1, Q 2 let us take these 4
points
Q 1, Q 2, P 2, P 1 these are lying on a circle
okay. Similarly these 4 points will also lie
on a circle is that all right?
Now once again I redraw it for this is a perpendicular
all right, this is 90 degree, OP 1 is perpendicular
all right. These 4 points are
lying on a circle, so opposite angles of a
quadrangle inside a circle will sum up to
180 degrees. So if this is 90 degrees, this
will
also be 90 degree all right similarly on this
side I have taken Q 1, Q 2 this will also
be 90 degree okay. So if this is 90 degree
this is
also 90 degree if this is 90 degree, this
is also 90 degree.
So O, P 2, R 2, Q 2, O, P 2, Q 2, R 2, in
this quadrangle opposite angles are 90 degree
so these 4 again will form a circle is that
all
right that means all the inverse points that
is inverse of the points lying on this will
lie on this circle touching the origin and
OP 2
is the diameter because this is 90 degree.
So the semi-circle this will be a semi-circle.
Similarly, this side also it will be a semi-
circle. So OP 2 will be the diameter okay
therefore given a straight line. Now I know
what will be the locus of its inverse what
will be its inverse draw a perpendicular take
its inverse and then draw a circle including
with this as the diameter.
So the origin is included all right sorry,
so this will be the circle is that all right.
So the inverse of a straight line is a circle
that will
include the origin. Conversely, the inverse
of a circle where the origin lies on that
circle will be a straight line okay. So if
I am
given a circle, what will be its inverse,
how to determine that straight line draw a
diameter through the origin then extend it
take
the inverse of the diameter whatever be that
you mean and then drop a perpendicular on
this line at this point. So this will be the
straight line which will be the inverse of
this circle all right. So inverse of a circle
is a straight line if the circle touches the
origin
okay.
Now let us take another situation what would
be the inverse of a circle lying outside the
origin that means the origin lies outside
the circle will it be again a straight line
so you join this okay you can take any other
line so inverse of this say OP 1 is may be
OP
2 O similarly, because the tangents are of
the same length OR 1 is OR 2 okay. So these
are the inverses, suppose this is Q 1 and
Q
1 dashed then Q 1 will have its counterpart
Q 2 and Q 2 dashed somewhere here. We will
find, we will prove it they will again
form another circle that is inverse of a circle
inverse of a circle when the origin is outside
the circle is another circle whereas the
inverse of a circle when the origin was on
the circle is a straight line is that all
right. Should I leave it as an exercise or
would you
like me to prove it okay I will give you the
hints. See OP 1 into OP 2 okay OQ 1 into OQ
1 dashed 
is equal to OP 1 squared or OR
1 square either way okay because the tangents
are of same length OP 1 squared this is a
constant.
Similarly, OQ dashed OQ 1 dashed into Q 2,
Q 2 dashed OQ 2 into OQ 2 dashed what will
it be equal to OQ 2 dashed into OQ 2
will be just inverse of this 1 by OQ 1 into
1 by OQ 1 dashed and 1 by a constant OP 1
squared which is again a constant okay. So
what does it prove, the product of these segments
is also a constant whenever you are having
a straight line drawn from a
particular point and there are 2 points the
product of these 2 segments is constant then
these points lying on such straight lines
will also lie on a circle okay. So, all these
points will lie on a circle, they are all
inverses of this okay. So the inverse of a
circle
when the origin is outside that circle is
also another circle, we will make use of this
relationship for the computation of currents,
the locus of currents in different situations.
Now so far we have not taken care of the angle
part all right if we have in electrical engineering
we deal with phasors or
impedances with angles. So when you write
V by Z or 1 by Z the angle associated with
Y will be just negative of the angle
associated with Z, 1 by Z theta will come
out as Y with an angle minus theta. So whenever
we are inverting a circle here if these
represent the phasors or impedances then this
should come in the other segment other quadrant.
So, first quadrant quantities will
be reflected in the 4th quadrant.
Similarly, if it is R minus J X kind of impedance
function 4th quadrant quantities will be taken
in the first quadrant all right. So
this is to be taken care of when we handle
a new problem on circuits. Now we will also
take the scaling factor suitably to get the
circles or the other loci of proper sizes.
So we will take a suitable scaling factor
and then take care of them while making the
final
computation okay.
Suppose, we have a resistance and then inductance,
the resistance is varied say between 5 and
50 ohms okay and inductance is
say given some particular value 4 ohms j 4
we are giving a supply of 100 volts, what
would be the location of the current that
is
what would be the locus of the current and
what will be the values of the currents at
these 2 values say 5 and 50 ohms or you can
take any other intermediate value, what will
be their specific magnitudes and angle.
So let us draw first of all the impedance
Z, what would be the locus of Z, this is fixed
4 j, so a unit of 4 taken on this side, mind
you the scales chosen for reactance and resistance
should be identical otherwise a circle might
be distorted into an ellipse unless
the scales are identical both are ohms so
both of them should have the same scale. So
this is 4 and this is varied 5 ohms on this
side will be somewhere here. So this will
be 5 plus j 4. Similarly, 50 ohms down the
line here somewhere here I say, so 50 ohms
which has to be chosen to scale we are not
really following any scale here, it has to
be drawn on a graph sheet with scale.
So this is say 50, so this is 5 plus j 4,
this is 50 plus j 4. So this is the location
this is the portion of our interest okay had
it been
from 0 to 50 then this would have been the
segment, corresponding segment is it not.
So we are interested in the inverse of this
portion. So we will take the entire straight
line what will be the inverse of this straight
line this is 0, this is 4. So 1 by 4 is a
very
small quantity all right now, they are 2 different
entities this is 4 ohms, 1 by 4 will be more
I can choose any skill 1 by 4 of
admittance can be of much larger magnitude
it all depends on the choice of the unit okay
they are not same units.
So I can choose a suitable scale for fixing
of the magnitude say if we take 40 as the
scale factor I have taken as scale factor
say
40 then 40 into 1 by 4 will be 10 units I
can choose the units that way. So I can choose
10 units here and choose this as 1 by 4 all
right, what is the location of the, what will
be the circle lie corresponding to this Z
line a horizontal line that means this itself
is a
perpendicular all right. So the circle will
be with this as the diameter you draw the
circle like this okay. I will its getting
clumsier
here, I will draw it once again as I told
you this is may not be to the scale and a
free hand drawing and making.
So this was 5 plus j 4 okay OP 1, so its inverse
will be on this line actually it will be in
this quadrant if we remember that the angle
has to be taken care by transferring it on
this side, this will be the magnitude, is
it not. So this point let it be P 2, so OP
2 is 1 by
OP 1 in magnitude. Similarly 5 plus j 50was
here say, say say here 50 plus j 4, so its
inverse would have been here all right
wherever this circle is met with this Z line.
So this is the portion which represents that
segment 5 to 50 of the resistance plus j 4
okay. So this is represented by this segment
inverse of that is represented by this segment.
So you can now multiply for any particular
location. So if I want to calculate for 10
plus j 4 what is a current for 10 plus j 4
suppose this is 10 plus j 4, this is a point.
So draw a line wherever it intersects this
is the length of the vector which is already
magnified 40 times I have taking taken a scaling
factor of 40. So that has to be divided by
40 is that all right? So whatever length
you get on this semi-circle divided by 40
that gives you and multiplied by the voltage
100 volts that give you the current.
So the current I is say for example for P
1, OP 2 into 100 divided by 40 the scale factor
once you know that you can calculate the
current also power also, what will be the
corresponding power dissipated VI cos theta
okay. So this is the real axis, so cos theta
means if you drop a projection, if you drop
a projection cos theta into yes, this is our
semi-circle. Suppose for any position this
is
the current then what is I cos theta this
one, is it not. So measure of this distance
multiplied by voltage 100 is that all right.
So VI cos theta will be the horizontal segment
of that point on the semi-circle multiplied
by V 100 of course that divided by 40
scale factor, what I am measuring as I is
actually 40 times I. So you have to take that
into account. Let us take another example,
we have a resistance 
and a capacitance, this capacitance is varied,
this resistance is fixed. There is another
problem extension of
this, this is j 20, this is 10 and this is
C. The question is first of all show the locus
of this current, next see this locus is determined,
next when is the total current having a power
factor of 0.4 lagging, for 0.4 lagging power
factor 
total current for the total current,
what is a value of C, question is for what
value of C will this total current, total
current I have a power factor of point 4 lagging
have you understood the question.
So let us try to find out the locus of this
first all right R is fixed 10 ohms C is varied.
So XC is varied from 0 to infinity if it is
not
specified then the variation can be over the
entire range. So this will tend to infinity,
so this is the straight line of which only
this is
the relevant portion because X is not varied
from plus infinity to minus infinity, it is
from 0 to infinity, 0 to minus infinity all
right.
So it is 0 to minus infinity, so what will
be the location of the what will be the locus
of the admittance function it is 0 to 100
and
10. So one 10th, one 10th of this 1 by 10,
I can choose any scaling factor all right.
Let us choose a scale factor of 200 all right,
this voltage is 200 or straight away we can
multiply by 200 because after all it is a
voltage multiplied by the admittance which
gives me the current.
So if I take straight away a scaling factor
equal to this voltage then that represents
a current. So 200 into 1 by 10 that gives
me 20,
so 20 amperes I will choose 20 units will
represents with this 20 as the diameter, I
will draw a circle I need not draw a circle,
I can
draw only a semi-circle because inverse of
this will come in the first quadrant, this
is in the 4th quadrant. So inverse will come
in
the first quadrant, so this is the had it
been varied over this range also X is varying
with an inductive to capacitive range then
this
entire circle would have come into picture
okay. So this is the locus of current I call
it I 1 because I will refer to this current
here
also I 1 is that all right what about this
current I 2, it is 100 ohms sorry 200 volts
divided by j 20 it is a fixed current.
So 10 ampere all right 200 by j 20 what would
be the location minus j 10. So that current
is minus j 10 is that all right? So this
much this is the location this is the locus
of I 2, this is the locus of I 1, I 1 varies
like this with respect to the origin, I 2
is fixed like
this. So what is I 1 plus I 2, I 1 plus I
2 shift this entire thing downward by 10.
So say minus 10 suppose, this is I 1, I 1
minus 10
will be somewhere here, so minus j 10.
So the same semi-circle you reproduce here
in this particular example it so happens this
is 10, this is also 10, so their 
that means
diameter was 20. So radius was 10 and this
side the shift is also 10, so it so happens
that it is touching this foot of the semi-circle
it need not be so in all the cases okay. Now
what is the question, question is for point
4 power factor lagging what is the value of
the capacitance for this current, now the
current locus is with respect to this origin,
this is the current locus, is it not, total
current,
the origin is fixed. So when the current I
1 is 0 the total current is just minus j 10
when there is no current through this this
is
minus j 10 and so on.
So this is the current for point 4 power factor
lagging for 0.4 lagging power factor what
is the angle cos inverse of point 4
whatever be that angle okay. So how much is
it approximately 76, 71 about 75 degrees okay
sin inverse of point 4 is 4 into 6 to a
approximately 23, 24 degrees all right. So
90 minus 24 it is about say 76 degrees very
crude approximation. So draw 76 degree
here that is tan inverse is a cos inverse
point 4. So this is the current is there any
other possibility for lagging power factor
is there
any other possibility, suppose we okay let
me complete this first, so this is the total
current.
So what is the corresponding current in I
1, when the total power factor is point 4
lagging, what is the current through I 1 I
know
that time what is the total current this is
the total current. So how much is the current
I 1 shift it vertically wherever it hits,
this is
the current I 1 is that all right measure
of this angle, suppose that angle is alpha,
how do we get alpha, first of all you have
got the
given power factor corresponding angle you
draw here and then find out the current from
there you draw a vertical line you reach
here. So that gives you the angle alpha of
this current all right then how do I get the
corresponding impedance.
So draw an angle alpha here angle alpha here
whatever be that all right and then stretch
it, this is the point X, is it not. If you
remember from the current that is from admittance
if you come back to impedance the same angle
first quadrant angle will
become 4th quadrant and 4th quadrant angle
will come to first quadrant. So Z to Y or
Y to Z if you keep on interchanging, so this
was the location of I that means corresponding
to Y, corresponding to admittance this was
angle. So corresponding to Z it will be
minus alpha, so at minus alpha you draw a
line that will give you the corresponding
reactance it is 10 plus j X.
So once you get this height that is 1 by omega
C, so C is suppose this is P then OP is 1
by omega C, so you know that is add C is
that all right. Now my question, next question
was suppose I give you a power factor of point
8 or may be point 9. So you draw
when the total current is having a power factor
of point 9 lagging, power factor of point
9 lagging means this much this is one
solution this also could have been another
solution. So long as it is within this semi-circle
there can be 2 possible values one is
here, one is here all right whereas for the
other one it may not have a feasible value
because it is on the other side of the semi-
circle which is not there for a capacitance
reactance okay. So there could have been 2
possibility…… we will 
first discuss about
simple signals and then at a later stage we
shall be using these signals for different
analysis, network analysis or representation.
Now signals that we we shall be discussing
are of 2 types, one is a continuous, the other
one is discrete. There are some signals
which are by nature continuous most of the
natural signals are continuous okay. For example,
if you take the record of voltage if
you have a pen recorder you get continuously
recorded values with respect to time it is
recording the value continuously but there
are certain signals which are integrated over
time and then they are registered at regular
intervals for example your energy,
energy meter records okay or the population
that you count after every 10 years so these
are all integrated signals by nature they
are integrated and hence you take stock, you
take the count after certain time. So they
are all or say sale of tickets every day sale
if you count so after every 24 hours you see
how many tickets have been sold. So most of
the records maintained in our offices,
most of the records are of discrete nature
they are not continuous.
Now even continuous signals when you use in
computers you have descritised versions. So
discrete signals are also equally
important, signals the discrete signals are
say this is a continuous signal and if we
take at regular intervals the values recorded
this
will be the descritised version okay. So for
example you measure the power of a particular
substation say at Kharagpur, the
substation power you measure at every 1 hour
is at 12’ o clock, 1’ o clock, 2’ o
clock you record the values, in the log book
they
enter these values. So they are descritised
version not that between 12 to 1 there was
no power power was there we assume this
to be some average of this either the value
at 12’ o clock is maintained till the next
value is registered or sometimes for
convenience if somebody wants to make any
mathematical calculations, any computation
then you can take the average of the 2
okay somewhere in between.
So you join these by straight lines okay subsequent
the consecutive values you join by straight
lines or you have blocks like this,
this is an approximation of the curve that
is representing the actual signal okay. Now
how fast, how fast can we measure these
descritised values or should we measure these
values, how quickly should we take the samples.
Let us take a sample here and a
sample here, does it represent the signal
very correctly, it has gone through a trough.
Suppose there is another signal like this.
So I cannot distinguish between this signal
and this signal if the signals are taken with
intervals so widely separated all right the
samples so widely separated that means with
large intervals. I will be missing out this
bottom or the top point, this is that means
with the signal if the largest variation largest
variation that is taking place is represented
by some frequency, this represents half
that variation. Suppose this can form at the
most half of a cycle, half of a sinusoidal
function okay this a maximum possible
change and if I take half that value then
I will be missing out the trough or the peak.
So the sampling frequency should be such
that I do not miss out this, so half the length
means twice that frequency that is present.
So maximum frequency that is present that
means maximum number of changes that may occur
in the original signal, twice that
frequency if I sample the original signal
at that frequency then only I will be able
to just recover this point okay. There can
be
some more changes okay, there can be a peak
and trough both then that means it is varying
at a much higher frequency. So I must
reduce the sampling time further to trap that
change. Otherwise, the sample value will not
be representing the original signal is
that all right? So this is very famous theorem
Nyquist or Shannon’s theorem, sampling theorem
okay that means the sampling rate
should be atleast twice the maximum frequency
present in the signal okay, right now we are
not going to discuss further details
about this we will be taking up this thing
further when we discuss about discrete signals
later on.
Now what are the standard signals that we
come across, mathematically a unit step is
a very convenient signal, it can be tested
on
simple system, simple networks suppose you
have a battery of some fixed source voltage
10 volts, you are switching it on if the
battery is ideal one then there is no internal
drop 10 volts supply will appear here after
the switching instant and that remains
constant.
So before the switching, it was 0, if we count
the time from t equal to 0 at the switching
instant then this represents a step
function 10 volts when the magnitude is 1,
it is unit step okay. This is unit step then
we have unit ramp 
for unit step, we use this
notation ut that is equal to 1 volt when t
is greater than equal to 0, equal to 0 otherwise
or in the negative region of time okay this
is for the unit step, unit ramp is a ramp
function with unity slope. So ft equal to
t into ut, y into ut the function t is also
having
values in the negative region of time but
I want a function which will start from t
equal to 0 all right before that it is non-existent.
So if I multiply this by a unit step by a
unit step this type then on the left hand
side if gets 0.
So the usual notation for this will be t into
ut, you will find in many books they also
write rt, rt means it is a ramp function like
this 0 and then it is like this with a unit
is slope if it is having any other slope then
it will be k into rt or kt into ut okay where
k is
the slope then we define unit impulse, what
you mean by unit impulse if you, if you have
a step function like this applied for
some time tau if the magnitude is A okay this
is a pulse of width tau, what is the area
of the pulse A into tau all right.
Now if we make tau 10 into 0 that is the duration
of the pulse is very very small 10 into 0
but the product is finite A tau is finite
then A must tend to infinity. So you are having
a situation A is very large not measurable
tau is very small not measurable but
their product is measurable that is finite
and when that finite quantity is equal to
1, we call it a unit impulse okay. You hit
a cricket
ball, you hit a cricket ball with a large
amount of force all right P is tending to
infinity, the duration of contact is very
very small
tau is very small but P into t, P into tau
that is equal to change in momentum that is
finite. So momentum change in momentum is
this product A into tau something like A into
tau where you can measure the magnitude of
the impulse all right. So the magnitude
of the impulse can be measured in terms of
the product that is area under this curve
neither by the magnitude of the force A nor
by the duration of contact tau.
So this is something like if you switch on
a battery and then immediately switch off,
you give a kick to a circuit in a
galvanometer; you just apply a voltage and
then withdraw. Now the amount of voltage say
it may be very large or suppose it is 10
volts and you apply it for say 1 milli second
then 10 into point 1 that will be representing
a very small strip that will be almost
equivalent to an impulse of that magnitude
okay we will stop here for today and we will
discuss about this in the next class, next
class is now, thank you okay.
Okay, so gentlemen we will continue with the
discussions on signals. Last time we were
discussing about impulse functions, the
unit impulse will be denoting by delta t,
this is a function whose value is 1 at t equal
to 0 and equal to 0 otherwise where this
function is representing the force, so this
is from 0 minus to 0 plus or you can write
infinity, it really does not matter that means
the function is coming here its magnitude
is very very large and duration is very very
small and then it is going and this thick
line
represents something equivalent to this area
all right.
So the integral is finite when this is 1 it
is a unit impulse okay this is time t and
this is the function we normally show it by
an
arrow and a vertical line, impulse appears
at this moment t equal to 0. So delta t minus
3 represents what, this is a shifted version,
shifted version of the same impulse that is
at t equal to 3, this function appears okay.
Similarly what will be ut minus 4 it is the
shifted version of unit step 
okay ut minus 4 represents a function which
starts after t
equal to 4 or at t equal to 4, it is unity
after that before that it is 0. So any function
if we shift ft is a function, so like this
then
what is ft minus tau if I want to represent
the same function which is starting after
an interval tau if it is identical okay then
what
will be the mathematical representation of
this function ft minus tau does it represent
this cosine bt plus j sin bt okay.
So on this side if I separate out the real
and imaginary parts will be s plus a by s
plus a whole square plus b square minus jb
by s
plus a whole square plus b square equate the
real parts and the imaginary parts then e
to the power minus at cosine bt will give
me s plus a by s plus a whole square plus
b square e to the power minus at sin bt will
be b by s plus a whole square plus b square
c
to the power minus at into cosine bt is a
damp sinusoid starting with a maximum value,
this is e to the power minus at cosine bt.
Similarly, e to the power minus at sin bt,
it is like this you take a pendulum give it
an oscillation all right give it a displacement
and
then let it oscillate, it will oscillate and
die down is it not and you have a recorder
all right. You just record the image of this
point
the bob then it will be damp sinusoid if you
start your camera when it is in the maximum
position then it will be a cosine function
all right you move that you move the paper
it will be describing this if you start from
the central position then it will be this
one
okay so both are representing basically a
damp sinusoid. So today we will stop here
for today will continue with this in the next
class.
