Find the transfer function of the given network.

Option 3 : \(\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)

__Concept:__

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output] / L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

__Calculation:__

The Laplace transform equivalent network for the given circuit is,

By applying KVL in the Laplace equivalent circuit, we get

V_{i} (s) - R I(s) - \({{1} \over cS}\) I(s) = 0

Vi (s) = I(s) (R + \({{1} \over cS}\))

I(s) = Vi (s) / (R + \({{1} \over cS}\)) ......(1)

Now output voltage or voltage across the capacitor is

Vo (s) = \({{1} \over cS}\) I(s) ...... (2)

Put the value of I(s) from equation (1) in equation(2)

∴ \(V_o(s)= {\frac {1} {sC}} {{V_i (s)} \over {(R+\frac {1} {sC})}}\)

Now the transfer function of the given circuit is the ratio of the Laplace transform of output voltage to the Laplace transform of input voltage.

∴ \(TF =\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)

__Important Points__

Element |
It’s equivalent in Laplace domain |

Find the Transfer function of the network given below.

Option 3 : \(\frac{1}{RCs + 1}\)

__Concept:__

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output] / L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

__Calculation:__

The Laplace transform equivalent network for the given circuit is,

By applying KVL in the Laplace equivalent circuit, we get

Vi (s) - R I(s) - \({{1} \over cS}\) I(s) = 0

Vi (s) = I(s) (R + \({{1} \over cS}\))

I(s) = Vi (s) / (R + \({{1} \over cS}\)) ......(1)

Now output voltage or voltage across the capacitor is

Vo (s) = \({{1} \over cS}\) I(s) ...... (2)

Put the value of I(s) from equation (1) in equation(2)

∴ \(V_o(s)= {\frac {1} {sC}} {{V_i (s)} \over {(R+\frac {1} {sC})}}\)

Now the transfer function of the given circuit is the ratio of the Laplace transform of output voltage to the Laplace transform of input voltage.

∴ \(TF =\frac{V_0\left( s\right)}{V_i\left( s\right)}=\frac{\frac{1}{RC}}{s+\frac{1}{RC}}\)

\(TF =\frac{V_0\left( s\right)}{V_i\left( s\right)}=\frac{1}{sRC + 1}\)

__Important Points__

Element |
It’s equivalent in the Laplace domain |

Option 1 :

__Concept:__

A transfer function is defined as the ratio of the Laplace transform of output to the Laplace transform of input when initial conditions are zero.

\(T.F. = \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}}\)

Where, Vo = Output voltage

Vi = input voltage

**Calculation:**

Vo(t) = (1 - e-t / RC) ---(1)

V_{i}(t) = 1 V ---(2)

Applying Laplace transform to the (1) equation:

\(V_0(s)=\frac{1}{s}-\frac{1}{s \ + \ \frac{1}{RC}}\)

\(V_0(s)=\frac{1}{s(RCs \ + \ 1)}\) ---(3)

Now taking Laplace transform of equation (2):

\(V_i(s)=\frac{1}{s}\)

Now dividing V_{0}(s) by V_{i}(s) we get:

\(\frac{V_0(s)}{V_i(s)}=\frac{1}{1 \ + \ sRC}\) ---(4)

**Now consider the diagram given in option(1)**:

Finding V_{0} by using **voltage division rule:**

\(V_0(s)=\frac{V_i(s) \ \times \ \frac{1}{sC} }{R \ + \ \frac{1}{sC}}\)

\(\frac{V_0(s)}{V_i(s)}=\frac{1}{1 \ + \ sRC}\)

Hence it gives the same transfer function as calculated in equation (4)

Hence **option (1) is the correct answer.**

__Important Points__

__Voltage Division Rule__:

The voltage across a resistor in a series circuit is equal to the value of that resistor times the total voltage across the series elements divided by the total resistance of the series elements.

\( {V_{R1}} = \;\frac{{{R_1}}}{{{R_1} + {R_2}}}{V_s}\;\; \ and\)

\({V_{R2}} = \;\frac{{{R_2}}}{{{R_1} + {R_2}}}{V_s}\)

Option 2 : The transfer function is dependent on the input of the system.

__Transfer function:__

The transfer function is defined as the ratio of the Laplace transform of the output variable to the input variable with all initial conditions zero.

TF = \(\left. {\frac{{L\left[ {output} \right]}}{{L\left[ {input} \right]}}} \right|\) Initial conditions = 0

TF = C(s) / R(s)

The transfer function of a linear time-invariant system can also be defined as the Laplace transform of the impulse response, with all the initial conditions set to zero.

__Properties of transfer function:__

- The transfer function is defined only for a linear time-invariant system. It is not defined for nonlinear systems.
- The transfer function is independent of the input and output.
- Because the transfer function of the system depends on the governing dynamic equation of the system only.
- If the transfer function is dependent on the input means the system will come under non-linear systems, but actually, the transfer function is defined for linear systems only.
- Transfer function analysis is not valid for the system that contains variables having initial values.

The transfer function \(\frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}}\) of the circuit shown below is

Option 4 : \(\frac{{s + 1}}{{s + 2}}\)

__Concept:__

The transfer function of a linear time-invariant function is defined to be the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable under the assumption that *all initial** conditions *

Order: The **highest power** of the complex variable ‘s’ in the denominator of the transfer function determines the order of the system.

In the Laplace domain, inductor impedance is written as “sL”

and capacitance impedance as:

\(\frac{1}{{sC}}\) .

For a resistor, there is no change.

__Calculation:__

From the given circuit capacitor is of 100 μ F value and resistor of 10 KΩ

Capacitive impedance is

\(\frac{1}{{sC}} = \frac{1}{{s100 \times {{10}^{ - 6}}}}\)

\( = \frac{1}{{s{{10}^{ - 4}}}}\)

\( = \frac{{{{10}^4}}}{s}\)

Assume current I is flowing in a circuit and the circuit in Laplace domain will be

Output is

\({V_2}\left( s \right) = I\left( s \right)\left( {10 \times {{10}^3} + \frac{{{{10}^4}}}{s}} \right)\)

Applying KVL to the loop we get

\( - {V_1}\left( s \right) + I\left( s \right)\left( {\frac{{{{10}^4}}}{s}} \right) + I\left( s \right)\left( {{{10}^4} + \frac{{{{10}^4}}}{s}} \right) = 0\)

\({V_1}\left( s \right) = I\left( s \right)\left( {2 \times \frac{{{{10}^4}}}{s} + {{10}^4}} \right)\)

Now the ratio of Laplace transform is:

\(\frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{I\left( s \right)\left( {\frac{{s + 1}}{s}} \right){{10}^4}}}{{I\left( s \right)\left( {\frac{{s + 2}}{s}} \right){{10}^4}}} \)

\( = \frac{{s + 1}}{{s + 2\;}}\)

__ Important points:__

NOTE: poles and zeroes are nothing but the negative of the inverse of time constant

Closed-loop system |
Open-loop system |

Type: Number of poles at the origin in the open-loop transfer function. NOTE: Feedback must be unity. i.e, |
Type: open-loop poles at the origin in the open-loop transfer function |

Order: Number of closed-loop poles in the closed-loop transfer function |
Order: Total number of open-loop poles in the open-loop transfer function |

Find the transfer function of the given network.

Option 3 : \(\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)

__Concept:__

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output] / L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

__Calculation:__

The Laplace transform equivalent network for the given circuit is,

By applying KVL in the Laplace equivalent circuit, we get

V_{i} (s) - R I(s) - \({{1} \over cS}\) I(s) = 0

Vi (s) = I(s) (R + \({{1} \over cS}\))

I(s) = Vi (s) / (R + \({{1} \over cS}\)) ......(1)

Now output voltage or voltage across the capacitor is

Vo (s) = \({{1} \over cS}\) I(s) ...... (2)

Put the value of I(s) from equation (1) in equation(2)

∴ \(V_o(s)= {\frac {1} {sC}} {{V_i (s)} \over {(R+\frac {1} {sC})}}\)

Now the transfer function of the given circuit is the ratio of the Laplace transform of output voltage to the Laplace transform of input voltage.

∴ \(TF =\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)

__Important Points__

Element |
It’s equivalent in Laplace domain |

Find the Transfer function of the network given below.

Option 3 : \(\frac{1}{RCs + 1}\)

__Concept:__

TF = L[output] / L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

__Calculation:__

The Laplace transform equivalent network for the given circuit is,

By applying KVL in the Laplace equivalent circuit, we get

Vi (s) - R I(s) - \({{1} \over cS}\) I(s) = 0

Vi (s) = I(s) (R + \({{1} \over cS}\))

I(s) = Vi (s) / (R + \({{1} \over cS}\)) ......(1)

Now output voltage or voltage across the capacitor is

Vo (s) = \({{1} \over cS}\) I(s) ...... (2)

Put the value of I(s) from equation (1) in equation(2)

∴ \(V_o(s)= {\frac {1} {sC}} {{V_i (s)} \over {(R+\frac {1} {sC})}}\)

∴ \(TF =\frac{V_0\left( s\right)}{V_i\left( s\right)}=\frac{\frac{1}{RC}}{s+\frac{1}{RC}}\)

\(TF =\frac{V_0\left( s\right)}{V_i\left( s\right)}=\frac{1}{sRC + 1}\)

__Important Points__

Element |
It’s equivalent in the Laplace domain |

Option 1 :

__Concept:__

A transfer function is defined as the ratio of the Laplace transform of output to the Laplace transform of input when initial conditions are zero.

\(T.F. = \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}}\)

Where, Vo = Output voltage

Vi = input voltage

**Calculation:**

Vo(t) = (1 - e-t / RC) ---(1)

V_{i}(t) = 1 V ---(2)

Applying Laplace transform to the (1) equation:

\(V_0(s)=\frac{1}{s}-\frac{1}{s \ + \ \frac{1}{RC}}\)

\(V_0(s)=\frac{1}{s(RCs \ + \ 1)}\) ---(3)

Now taking Laplace transform of equation (2):

\(V_i(s)=\frac{1}{s}\)

Now dividing V_{0}(s) by V_{i}(s) we get:

\(\frac{V_0(s)}{V_i(s)}=\frac{1}{1 \ + \ sRC}\) ---(4)

**Now consider the diagram given in option(1)**:

Finding V_{0} by using **voltage division rule:**

\(V_0(s)=\frac{V_i(s) \ \times \ \frac{1}{sC} }{R \ + \ \frac{1}{sC}}\)

\(\frac{V_0(s)}{V_i(s)}=\frac{1}{1 \ + \ sRC}\)

Hence it gives the same transfer function as calculated in equation (4)

Hence **option (1) is the correct answer.**

__Important Points__

__Voltage Division Rule__:

The voltage across a resistor in a series circuit is equal to the value of that resistor times the total voltage across the series elements divided by the total resistance of the series elements.

\( {V_{R1}} = \;\frac{{{R_1}}}{{{R_1} + {R_2}}}{V_s}\;\; \ and\)

\({V_{R2}} = \;\frac{{{R_2}}}{{{R_1} + {R_2}}}{V_s}\)

Option 2 : The transfer function is dependent on the input of the system.

__Transfer function:__

The transfer function is defined as the ratio of the Laplace transform of the output variable to the input variable with all initial conditions zero.

TF = \(\left. {\frac{{L\left[ {output} \right]}}{{L\left[ {input} \right]}}} \right|\) Initial conditions = 0

TF = C(s) / R(s)

The transfer function of a linear time-invariant system can also be defined as the Laplace transform of the impulse response, with all the initial conditions set to zero.

__Properties of transfer function:__

- The transfer function is defined only for a linear time-invariant system. It is not defined for nonlinear systems.
- The transfer function is independent of the input and output.
- Because the transfer function of the system depends on the governing dynamic equation of the system only.
- If the transfer function is dependent on the input means the system will come under non-linear systems, but actually, the transfer function is defined for linear systems only.
- Transfer function analysis is not valid for the system that contains variables having initial values.

The transfer function V_{2}(s) / V_{1}(s) is

Option 1 : 1/ (s + 1)

Transform the network given in the question into Laplace Domain. we get,

**Voltage division rule:**

If two resistors R_{1} and R_{2} are connected in series across the supply voltage V then the voltage across the resistor R_{1} is given by,

\(V_1=\frac{{{\rm{V\;}}{{\rm{R}}_1}}}{{{{\rm{R}}_1} + {R_2}}}\)

And voltage across the resistor R_{2} is given by,

\(V_2=\frac{{{\rm{V\;}}{{\rm{R}}_2}}}{{{{\rm{R}}_1} + {R_2}}}\)

__Calculation:__

Now apply voltage division to the above circuit then we get,

\(V_{2}(s)=\frac{{{{\rm{V}}_1}\left( s \right) \times 1/s}}{{1 + 1/s}} = \frac{{{{\rm{V}}_1}\left( s \right)}}{{s + 1}}\)

**\(\frac{{{{\rm{V}}_2}\left( s \right)}}{{{{\rm{V}}_1}\left( s \right)}} = \frac{1}{{s + 1}}\)**

__Note:__