Hey friends, welcome to the YouTube channel ALL
ABOUT ELECTRONICS. So in this video, we will understand the two terms related to the electrical
network. That is a driving point impedance and the transfer function. And we will see that if we have
been given any electrical network then how to find its driving point impedance and the transfer
function. So before understanding this driving point impedance, first of all, let us understand
what is the driving point. So for any electrical network, the driving point is the two terminals
of the circuit across which, we are connecting the energy source. And this energy source, like
a voltage or the current source that drives the entire network. So these two terminals of the
network where we are connecting the voltage or the current source are known as the driving point
of the network. Now in a one-port Network since there is only one port, so this standalone port
will act as a driving point. Because this energy source can only be connected to that particular
port. But in a multi-port network, depending on where the power supply is connected, the specific
port will act as a driving point. For example, if the power supply is connected to the port-1, then
that port will act as a driving point. Similarly, if it is connected to the second port, then in
that case, this port two will act as a driving point. So now, the impedance which is seen through
this driving point is known as the driving point impedance of the network. So basically, it is
the input impedance of the network which is seen through the voltage source. So for a one-port
network, if this V(s) is the voltage at the port and this I(s) is the incoming current to the port,
then the driving point impedance of the network is equal to V(s) divided by the I(s). So here both
V (s) and the I(s) are the voltage and the current in the S-domain. So here we are assuming that, our
network consists of a resistor, capacitor, and the inductor. And that's why these voltages and the
current are represented in the s-domain. But if it is purely resistive Network, then there is no
need to use this s-domain representation. So here, since both V(s) and I(s) are in the s-domain,
so this driving point impedance will also be in the s-domain. Now for a two port network, if the
voltage source is connected to the port 1, then we will get the driving point impedance corresponding
to that port. That means if V1(s) and the I1(s) are the s-domain voltage and the current at the
port-1 then the driving point impedance which is seen through the port 1 is equal to V1(s) divided
by the I1(s). So basically, it is the input impedance of the network which is seen through the
port-1. So similar to the impedance, if you want to find the driving point admittance, then it will
be the inverse of the Impedance. And as you know, the admittance is represented by the symbol
Y. That means the driving Point admittance at the Port-1, that is Y11 can be given as this I1(s)
divided by V1(s). So that is the expression of the driving point impedance and the admittance, when
the input is applied at the Port-1. But suppose, if the input is applied at the Port-2, then the
driving Point impedance at the Port-2 can be given as is V2(s) divided by the I2(s). And here
these V2 and I2 are the voltage and current at the Port-2. And similarly, if we see the driving
point admittance at the Port-2, then this Y22 will be the inverse of the impedance. That means
this Y22 is equal to I2 / V2. So in this way, if we know the voltage and current at the specific
port, then we can easily find the driving point impedance or the admittance of the network. So
basically, this driving point impedance of the network is the input impedance of the network
which is seen through the specific port. And even if we do not know the port parameters, like
the voltage and current at the specific port, then also it is possible to find the driving point
impedance. And of course, for that we should be aware about the internal circuit of the network.
So through a couple of examples, let us see how to find the driving point impedance of the network.
So in this example, we have been asked to find the driving point impedance of the given circuit.
So here we do not know the port parameters, like the input voltage and the input current.
But we know that, the driving point impedance is nothing, but the input impedance of the circuit.
And here if you see, then it is the Thevenin's equivalent impedance, which is seen across these
two terminals. So here to find that, first of all let us find the equivalent s-domain representation
for the given circuit. So for the capacitor, we know that the equivalent s-domain representation
is equal to 1/ Cs. And of course, here we are assuming that, all the initial conditions of
the circuit are 0. That means the s-domain representation for the capacitor is equal to 1/
Cs. And here the value of C is equal to 0.5 Farad. So we can say that, this 1/ Cs is equal to 2 /
s. Similarly for the inductor, the equivalent s-domain representation is equal to Ls. And here
the value of L is equal to 2 Henry. That means if we see the equivalent s-domain representation,
then that is equal to 2s. And for the resistor, the equivalent representation will remain the
same. That is equal to 2 ohm. That means now, if we see the equivalent s-domain circuit, then
it will look like this. So for the given circuit, now let us find the equivalent impedance, which
is seen across these two terminals. So here, this 2 ohm and the 2s are connected in the
series connection. And that combination is connected in parallel with the capacitor. So we
can say that the Z(s) is equal to 2 + 2s that is this combination in parallel with this capacitor.
That is equal to 2 / s. So we can say that that is equal to (2 + 2 s) * 2 / s / (2 + 2 s) + 2
/ s. That means the Z(s) is equal to 4 *s + 1 / 2 * s + s² + 1. Or we can say that, that is equal
to 2 * s + 1 / s² + S + 1. So that is the driving point impedance of the circuit in the s-domain. So
in this way, we can easily find the driving point impedance. So similarly let us take the another
example. So here we have been given the driving point impedance of the network and we have been
asked to find the component values of the given Network. So in this question we have been given
this parallel RLC circuit and for this parallel RLC circuit we have been given the driving
point impedance. So here since the driving Point impedance is given in the s-domain. So
first of all, let us find the equivalent s-domain representation for the given circuit. So we know
that for the capacitor, the equivalent s-domain representation is equal to 1/ Cs. And similarly
for the inductor, that is equal to LS. And for the resistor, it will remain as it is. That means for
the given circuit, this is the equivalent s-domain circuit. Now if you see over here, then all the
elements in the network are connected in the parallel connection. And therefore it is easier to
find the admittance of the network compared to the impedance. So for the given circuit if you see the
driving point admittance, that is Y(s), then that can be given as Cs+ 1/ R + 1/ LS. So basically
it is the admittance of the each term. And since these components are connected in the parallel
connection, so the overall admittance of the network will be the summation of the individual
admittance. And here from the given expression of the driving point impedance, we can say that the
driving point admittance of the network or this Y(s) is equal to 1/ Z(s). So here that is equal to
s² +0.1 s + 2 / 0.2 s. So we can say that that is equal to this s /0.2+0.1/0.2 + 2 / 0.2 s. So this
will be the expression of the Y(s) from the given expression. So now if we compare this expression
with this expression, then we can say that over here the C is equal to 1/0.2. That is equal to
5 Farad. Similarly if you see, this 1/ R, then that is equal to 0.1 / 0.2. That is equal to 1/2.
So we can say that, the value of R is equal to 2 ohm. And likewise if we see this 1/ L, then that
is equal to 2 /0.2. That means the value of L is equal to 0.1 Henry. So in this way, for the given
circuit the value of C is equal to 5 Farad, while the value of R is equal to 2 ohm. And likewise,
the value of L is equal to 0.1 Henry. So in this way, from the expression of the driving point
impedance, we found the values of the component for the given circuit. So similar to the driving
point impedance, now let us understand what is the transfer function of the electrical network.
Now if you see the expression of the driving point impedance, or the admittance, then it relates
the voltage and current of the same port. That means in the driving but impedance both voltage
and current are of the input side. But if you see the transfer function, then it relates either
voltage or current of the one port to the voltage or current of the another port. So basically
this transfer function relates either voltage or current of the output port with the voltage or
current of the input port. That means to define the transfer function we require at least two
ports. And therefore it cannot be defined for the one port network. So for the two-port network,
let's say this V1(s) and the I1(s) are the voltage and current at the port one. Similarly, let's say
the voltage and current at the Port two are V2(s), and the I2(s). So as you can see over here all the
voltage and currents are defined in the s-domain. Now here for each input and output Port, we have
total two variables. That is voltage and the current. And for the two variables, we can have
total four different types of transfer functions between the input and the output Port. So one of
them is the voltage transfer function. So this voltage transfer function is the ratio of the
output voltage to the input voltage. So in this case, we have assumed that the input is applied
at the Port one, and the output is measured at the second Port. That means in this case this V1(s) is
the voltage on the input side while the V2(s) is the voltage on the output side. That means in this
case this voltage transfer function is equal to V2(s) / V1(s). But suppose if the input is applied
at the Port two and the output is measured at the Port one, then in that case this voltage transfer
function will become V1(s) / V2(s). And here it is represented as the S21. So here this first
subscript represents the input port and the second subscript represents the output Port. So as you
can see over here, the second port is our input port while the first port is our output port.
That means in general if you are taking the ratio of the output voltage to the input voltage, then
it is defined as the voltage transfer function. Now instead of the voltage, if you are taking
the ratio of the output and the input current, then it is defined as the current transfer
function. So if we assume that, if the port one is the input port and the port two is the output
port, then the current transfer ratio α12 is equal to I2 / I1. So like I said earlier, this first
subscript represents the input port, while the second subscript represents the output port. That
means here this α 1 2 is the ratio of the output current I2 / the input current I1. But on the
other hand if the port two is our input port and the port one is our output port, then in that case
this current transfer ratio or this α 2 1 is equal to I1 / I2. So basically depending on where we
are applying the input in the network and on which Port we are measuring the output the expression
of the current transfer ratio will change. But in general this current transfer ratio for the
given network is the ratio of the output current to the input current. So similarly if we take the
ratio of the output voltage to the input current, then it is known as the transfer impedance.
Because if you see the unit of this ratio, then it is same as the unit of the impedance. So
if the input is applied at the Port one, and the output is measured at the Port two, then this
transfer impedance Z12 can be given as V2 / I1. Where the V2 is the voltage on the output side,
while this I1 is the current on the input side. But on the other hand if we apply the input at the
Port two, and if we measure the output on the port one, then this transfer impedance can be given as
this V1(s) / I2(s). Because now this V1(s) is the output voltage, while this I2(s) is the input
current. And now it can be represented as the Z21. So in short this transferred impedance is the
ratio of the output voltage to the input current. So similarly, if we take the ratio of the output
current to the input voltage, then it is known as the transfer admittance. So for the two part
network, if the input is applied at the Port one, and the output is measured at the Port two, then
this transfer impedance Y12 can be given as this I2 / V1. That means here this I2 is the output
current, while this V1 is the input voltage. And the other way around, if the input is applied at
the Port two, and the output is measured on the port one, then this transferred impedance Y21 can
be given as this I1 / V2. That means in general for any given electrical network, the transfer
admittance is the ratio of the output current to the input voltage. So in this way, we have total
four different types of transfer functions between the input and the output Port. So now let us take
few examples based on the transfer function, and let us find out how to find the transfer function
for the given Network. So in this question, we have been asked to find the transfer function that
is Vout / Vi for the given circuit. So as you can see this V(s) is the output voltage across this 1
ohm resistor. So to find that, first of all let us find the equivalent s-domain representation for
the given circuit. So here we are assuming that, the initial conditions in the circuit are 0.
And with that assumption for the capacitor the equivalent s-domain representation is equal to 1/
Cs. So here since C is equal to 1 Farad. So we can say that, for these two capacitors, the equivalent
s-domain representation is equal to 1/ s. Similarly for the inductor the equivalent s-domain
representation is equal to Ls. And here since L is equal to 1 Henry, so we can say that, that is
equal to s. That means for these two capacitors, the equivalent s-domain representation is equal
to 1/ s, while for this inductor that is equal to s. And we know that for the resistors the
equivalent representation will remain as it is, so overall if you see the equivalent s-domain
representation for the given circuit, then this is how it will look like. So now if you see, then
we have total two nodes in the circuit. And the voltage at this node is same as is the V(s). So
similarly let's say the voltage at this first node is equal to Vx. So now by applying the KCL at
these two nodes we can find the relation between the Vout and the Vi, and eventually from that we
can easily find the transfer function. So first of all let us apply the KCL at this second node.
So applying the KCL we can write this Vout/ 1/ s, that is this current + Vout/ 1 ohm, that is this
current + Vout - Vx / s. That is this current should be equal to 0. That means the summation of
all the outgoing current should be equal to 0. So from this, if you further simplify it, then we
can write it as this s * V out + Vout + Vout/ s, that is equal to this Vx / s. Or if we further
simplify it, then we can write it as this Vout/ s * s² + s+ 1. That is equal to this Vx / s. So
here on both sides this s will get cancel out. That means now we will have this Vx is equal to
this s² + s + 1 * Vout. So let's say this is the equation number one. So similarly now let us apply
the KCL at this second node. So applying the KCL at this node, we can write this Vx/ 1/ s. That is
this current plus Vx - Vi / 1 ohm. That is this current plus Vx - Vout / s. That is this current
should be equal to 0. That means the summation of all the outgoing current should be equal to 0. So
if we further simplify it, then we can write it as this s * Vx + Vx - Vi + Vx / s - Vout/ s.
That is equal to 0. So we can say that, this Vx * s + 1 + 1 / s - Vout /s. That is equal to Vi.
Or if you further simplify it, then we can write it as this Vx * s² + s + 1 / s - Vout/ s. That is
equal to Vi. And earlier we have seen that, this Vx is equal to this s² + s + 1 * Vout. So in this
expression if we put this values, then we can say that this Vout / s * s² + s + 1² - V out/ s. That
is equal to Vi. So further we can write it as this Vout / s * [s² +( s + 1)² - 1] = Vi. So now let us
expand this term and let us try to simplify it. So if you expand this term, then we can write it as
this s ^ 4 + s² + 1 + 2 * s Cube + 2 s + 2 * s². And here we will also have this -1. That means
overall, this entire term times Vout, that is equal to S *Vin. So here this + 1 and the -1 will
get cancel out. That means, now we will have this s ^ 4 + 2 s Cube + 3 s² + 2 s * Vout. That is
equal to S * Vin. So now if you bring this s in the denominator and further we can write it as
this s cube + 2 s² + 3 s + 2 * Vout, that is equal to Vin. Or we can say that, this Vout / Vin, that
is = 1/ s + 2 s² + + 3 s + 2. That means this will be the transfer function for the given network.
So in this way we can easily find the transfer function for the any network. So similarly, let
us take another example. So in this question, we have been given the transfer function of the
given network, and here we have been asked to find the value of this load resistance RL, so that
we can get this specific transfer function. That means here this is the given network, and this
is the transfer function of the given network. So first of all let us find the equivalent s domain
representation for the given network. So as you know for the resistors, the equivalent s domain
representation will remain as it is. While for the capacitor, the equivalent s domain representation
is equal to 1/ Cs. That means overall if we see the equivalent s domain circuit, then this is
how it will look like. So now this capacitor and the load resistor are connected in the parallel
connection. Let's say that equivalent impedance is equal to Z1(s). That means this Z1(s) is the
parallel combination of these two elements. That is equal to this 1/ Cs * RL / 1/ Cs + RL. So that
is equal to RL / 1 + RL * Cs. And now, using the voltage divider rule, we can say that this Vout
= Z1(s) / Z1(s) + r * Vi. Or we can say that, this Vout / Vi = Z1(s) / Z1(s) + R. So now in this
expression let us put the value of this Z1(s). That means now this Vout / Vi = RL / (1 + RL *
Cs) / ( RL / 1 + RL * Cs + R. Or we can say that, that = RL / RL + R * (1 + RL * Cs). Or further we
can write it as, RL/(RL + R + R * RL * Cs). So now let us divide both numerator and the denominator
by the RL. So if we do so, then the RL in the numerator and the denominator will get cancel out.
And we will have this Vout/ Vi, that is equal to this 1/ 1 + R/ RL + R * Cs. So now if we compare
this expression with given expression, then we can say that over here this 1+ R/ RL should be equal
to 2. And it can happen whenever this R is equal to RL. So from this we can say that, to get this
transfer function, the value of the load resistor should be same as the R. And with this value we
can get this specific transfer function. So in this way, we can find the transfer function of the
given network, or if we know the transfer function of the network, then from that we can find the
component values. So I hope in this video, you understood what is the driving point impedance and
what is the transfer function of the electrical network. And if you have been given an electrical
network, then how to find these parameters. So if you have any question or suggestion, then do let
me know here in the comment section below. If you like this video hit the like button and
subscribe the channel for more such videos.