Transient response of RC , RL circuits with step input

Explains Transient response of a RC , RL circuit with Step Input signal, RL , RC frequency response, TIME CONSTANT, LOW PASS FILTER,HIGH PASS FILTER, BAND-PASS FILTER frequency response, Passive Integrator, Passive Differentiator operation.

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Transient response of RC circuit with step input
Let us consider a circuit consisting of R and C in series with a Step input. A step signal is one
which has a voltage V0 at t < 0 and suddenly increase at t = 0 and remains constant Vs at t >>
0 as shown in the diagram.

By common sense we can understand that when t < 0, the capacitor will be charged to a constant
value V0 and the current in the circuit is zero.
At t = 0 , current will start flowing around circuit. This current will continue to flow as long
as there is a voltage difference across the Resistor.
Similarly at = infinity the input voltage rises to Vs and the voltage across the capacitor is Vs
which is a constant. So , the current through the circuit is zero.

Between the times t < 0 and t = infinity the current through the circuit increases gradually due
to the reactance of the Capacitor. This is also known as Transient response.
The total response of the circuit due to a step input signal can be considered as sum of two
responses. One is Natural response and the other is Forced response.
Total Response = Natural Response + Forced Response
Derivation:
At any instant of time t , let us assume that the voltage across the resistor is VR and voltage
across the Capacitor is Vc .

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Then VR= i.R and i = VR /R
The same current is passed through the Capacitor also. The current through the Capacitor is
i = C
????????????
????????????
where V is the voltage across the capacitor.

From the circuit VR = Vs – V
Therefore C
????????????
????????????
= VR /R =
Vs – V
&#3627408453;

C
????????????
????????????
=
Vs
&#3627408453;
-
V
&#3627408453;

C
????????????
????????????
+
V
&#3627408453;
=
Vs
&#3627408453;
----------------------------(1)
This is a non-homogeneous Equation whose solution consists of two parts. One is
Complementary Function part ( i.e natural response at t = 0) and the other is Particular
Integral Part(i.e Forced response when t = œ).
Natural Response: Natural response of the circuit means, the response of the circuit without
any external signal. This circuit is shown below. The voltage across the Capacitor at any
instant is given by VN(t) = K.&#3627408466;
−??????/&#3627408453;??????
--------------------------(2)
Here RC is called the Time Constant of the RC circuit. It has the units of time (sec)
Similarly the forced response means, the response of the circuitwhen external signal is
applied . In the present case the external input signal is Step Signal. At t = Infinity, the
voltage is equal to Vs .
Hence VF(t) = Vs -----------------------------(3)
So, the total response is VT = VN(t) + VF(t)
So, VT = K.&#3627408466;
−??????/&#3627408453;??????
+ Vs
To find the value of the constant K ,let us apply the initial conditions .At t = 0
VT = V0 = K + Vs where V0 is the voltage when t =0
Or K = V0 – Vs
Hence VT = (V0 – Vs ) &#3627408518;
−??????/????????????
+ Vs

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This is the total transient response of the RC circuit with Step Input Signal .This is
graphically shown as below. Here RC denotes the time constant.

Suppose the capacitor is initially discharged, Vo = 0.Sothe the above equation will beome
VT = (– Vs) &#3627408518;
−??????/????????????
+ Vs

VT = (Vs – Vs ) &#3627408518;
−??????/????????????
= Vs(1-&#3627408518;
−??????/????????????
)
So the total response of the RC circuit is
VT = Vs (1 - &#3627408518;
−??????/????????????
)


The response of the RC circuit with initially Capacitor discharged is shown in the above
diagram. It has both steady state and transient states.
Real life applications of RC circuits
i. Digital Camera Flash Light utilizes a RC circuit to create a short duration high current
pulse to energize the flash light.
ii. Neon Lights use RC delay timer circuit for periodic ON/OFF to avoid overheating.
iii. On computer mother boards for performing switching operations.
--------------XXX--------------

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Transient response of RL circuit with step input
Let us consider a circuit consisting of R and L in series with a Step input signal. A step signal
is one which has a voltage V0 at t < 0 and suddenly increase at t = 0 and remains constant Vs
at t >> 0 as shown in the diagram.

By common sense we can understand that when t < 0, the Inductor current is i0 .
At t=0 , current will start flowing around circuit. This current will continue to flow as long as
there is a voltage difference across the Resistance.
Similarly at t = infinity the input voltage rises to Vs and the voltage across the inductor is Vs
which is a constant. The current through the inductor is Is

Between the times t < 0 and t = infinity the current through the circuit increases gradually due
to the reactance of the Inductor. This is also known as Transient response.
The total response of the circuit due to a step input signal can be considered as sum of two
responses. One is Natural response and the other is Forced response.
Total Response = Natural Response + Forced Response
Or
Total Response = Transient Response + Steady state Response
We know that the transient response is always exponential.

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So at any instant t , the current through the Inductor is I(t) = i1(t) + i2(t)
Where i1(t) = K.&#3627408466;
−??????&#3627408453;/??????
where K is a constant whose value is to be found from
initial conditions.
Similarly the forced response is (at t = Infinity) i2(t) = Is
Hence the complete response is i(t) = i1 + i2
so, I(t) = K.&#3627408466;
−??????&#3627408453;/??????
+ Is ------------------ (1)
Applying the initial conditions, at t = 0
I(t) = I0 ; So , I0 = K + Is
Or K = I0 - Is
Substituting in equation 1
I(t) = (I0 -Is ).&#3627408518;
−????????????/??????
+ Is
So, the Response of a LR circuit for a Step input signal is given by
I(t) = (I0 – Is) &#3627408518;
−????????????/??????
+ Is -----------------------------(2)
This is the total transient response of the LR circuit with Step Input Signal .This is
graphically shown as below. Here L/R denotes the time constant.

Suppose the Inductor has no initial current I0 = 0. So the above equation (2) will become

I(t) = (0 – Is) &#3627408518;
−????????????/??????
+ Is

Or the total response of the RC circuit is

I(t) = Is (1 - &#3627408518;
−????????????/??????
)

This response can be shown graphically as below.

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The response of the LR circuit with initially Inductor current is zero is shown in the above
diagram. It has both steady state and transient states.
Real life applications of RL circuits
i. RL circuits are used in Relay coils. Relay coil is nothing but an RL circuit used to control
the switching of another circuit
ii. RL combination is used in Spark plugs of Vehicles to ignite fuel-air mixture in each
cylinder at proper times
iii. Also used in Fluorescent tubes and single phase motors etc.

Frequency Response of RC Circuit (Low Pass Filter)
Let us consider a circuit consisting of a Resistor R and a Capacitor C connected in series with
a source of AC as shown in the diagram below. The input voltage is Vi and the output
voltage is V0. The current in the circuit is I.
This circuit can act as a low–pass filter which allows the signals of only low frequency and
attenuates the high frequency signals.
To analyse the frequency response of the circuit, apply KVL to the circuit.

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Vi = IR + I.Xc where Xc is the capacitive reactance and I is the
total current in the circuit.
The capacitive reactance is given by Xc =
1
&#3627408471;????????????

So, Vi = I.R + I.
1
&#3627408471;????????????
---------------------------(1)
Or Vi = I.(R+1/jωC)
I =
??????&#3627408470;
( R+
1
&#3627408471;????????????
)
----------------------------------(2)
From the circuit, the output voltage or voltage across the Capacitor is
V0 = I.Xc =
??????&#3627408470;
( R+
1
&#3627408471;????????????
)
.Xc
So, V0 =
??????&#3627408470;
( R+
1
&#3627408471;????????????
)
.
1
&#3627408471;????????????

Or V0 =
??????&#3627408470;
( &#3627408471;????????????R+1)


The gain of the circuit is
V0
Vi
=
1
( &#3627408471;????????????R+1)

Here ω = 2πF , where F s the frequency of the applied input signal .

Let
1
2πRC
= Fc be the Cut-off frequency of the RC circuit.
So, is
V0
Vi
=
1
(1+ &#3627408471; 2πFRC)
=
1
(1+ &#3627408471;.
F
Fc
)

This gain is the complex gain. The magnitude of the gain is given by
A = |
V0
Vi
| =
1
√(1+( .
F
Fc
)2

This equation gives the frequency –Gain response of a CR circuit.
This also represents the operation of a Low-pass filter.
In the above equation, when the frequency of the applied signal F is small,
1
√(1+( .
F
Fc
)2
= 1
The gain is equal to 1. Or output voltage is equal to input voltage.
i.e V0 = Vi
which means , the circuit passes low frequency signals completely.

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When frequency F is large in the term
1
√(1+( .
F
Fc
)2
the denominator is very large or gain is
very small. So only very small part of the input is passed through the circuit or the output is
almost zero.
This response can be shown in the graph below and it is called low-pass filter response.

When the frequency of the input signal is equal to cut-off frequency Fc
i.e at F = Fc ,
Gain A =
1
√2

So, the cut-off frequency of the CR circuit can also be defined as the frequency at which the
gain becomes to
1
√2
times the maximum value or the output voltage is 0.707 times to the
input voltage.
This cut-off frequency is also known as 3dB frequency
The cut-off Frequency of the RC circuit is Fc =
&#3627409359;
&#3627409360;??????????????????

The frequencies over which the input signal is totally passed is called pass band and where
the gain is minimum is called stop band.
RC LOW-PASS FILTER AS PASSIVE INTEGRATOR
RC low –pass filter acts as a passive Integrator when the time constant of the circuit is very
large.
An integrator circuit is one whose output voltage is proportional to the Integral of input
Voltage. i.e V0 α ∫??????
&#3627408470; &#3627408465;?????? .
Let us consider a circuit consisting of Resistor® and Capacitor ( C ) connected in series with
an alternating source of voltage Vi as shown in the below diagram.

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The important condition is that the time constant (RC) of the circuit is very large.
When the time constant is very large, the Capacitor takes more time to charge and hence the
entire input voltage appears across Resistor only. Because voltage across the Capacitor will
be very small.
So, from the circuit Vi = I.R ……………….. (1).
and I =
??????
&#3627408470;
&#3627408453;
……………….. (2)
Now, from the circuit the output voltage across the Capacitor V0 =
&#3627408452;
??????

And we know that Q = ∫?????? &#3627408465;??????
Hence V0 =
1
??????
∫?????? &#3627408465;?????? ………………….(3).
Substituting (2) in the above equation

V0 =
1
??????
∫
??????
&#3627408470;
&#3627408453;
&#3627408465;??????
Or V0 =
&#3627409359;
????????????
∫ ??????
?????? &#3627408517;??????

Hence V0 α ∫??????
?????? &#3627408517;?????? as R & C are constants.
i.e the output voltage is proportional to the integral of input voltage.
Hence, a RC low –pass filter circuit with high time constant will act as an Integrator.
Frequency Response of CR Circuit (High Pass Filter)
Let us consider a circuit consisting of a Capacitor (C) and a Resistor R connected in series
with a source of AC as shown in the diagram below. The input voltage is Vi and the output
voltage is V0. The current in the circuit is I. This circuit can act as a High –pass filter which
allows the signals of only high frequency and attenuates the low frequency signals.

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To analyse the frequency response of the circuit, apply KVL to the circuit.
Vi = IR + I.Xc where Xc is the capacitive reactance and I is the
total current in the circuit.
The capacitive reactance is given by Xc =
1
&#3627408471;????????????

So, Vi = I.R + I.
1
&#3627408471;????????????
------------------------------(1)
Or Vi = I. (R+ 1/jωC)
I =
??????&#3627408470;
( R+
1
&#3627408471;????????????
)
------------------------------------(2)
From the circuit, the output voltage or voltage across the Resistor R is
V0 = I.R =
??????&#3627408470;
( R+
1
&#3627408471;????????????
)
.R
So, V0 =
??????&#3627408470;
( R+
1
&#3627408471;????????????
)
?????? .
Or V0 =
??????&#3627408470;
( 1+
1
&#3627408471;????????????R
)

The gain of the circuit is
V0
Vi
=
1
( 1+
1
&#3627408471;????????????R
)

Here ω = 2πF, where F s the frequency of the applied input signal.

Let
1
2πRC
= Fc be the Cut-off frequency of the RC circuit.
So, is
V0
Vi
=
1
(1+ 1/ &#3627408471; 2πFRC)
=
1
(1+ &#3627408471;.
FC
F
)

This gain is the complex gain. The magnitude of the gain is given by
A = |
V0
Vi
| =
1
√ (1+ .
FC
F
)
2


This equation gives the frequency – Gain response of a RC circuit.
This also represents the operation of a High-pass filter.

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In the above equation, when the frequency of the applied signal F is very small, then
√ (1+ .
FC
F
)
2
= 0
The gain is equal to 0. Or output voltage is equal to 0.i.e the signal is not allowed through the
circuit. Or the signal is attenuated.
i.e V0 = 0
which means , the circuit do not passes low frequency signals completely.
When the when the frequency of the applied signal F is very high, then
√ (1+ .
FC
F
)
2
= 1
The gain is equal to 1. Or output voltage is equal to Vi. i.e the input signal is completely
allowed through the circuit. Or the signal is passed.
i.e V0 = Vi
which means , the circuit passes high frequency signals completely.
This is the working of a High-Pass filter. The response of the CR circuit is shown below.


When the frequency of the input signal is equal to cut-off frequency Fc. i.e at F = Fc ,
Gain A =
1
√2

So, the cut-off frequency of the CR circuit can also be defined as the frequency at which the
gain becomes to
1
√2
times the maximum value or the output voltage is 0.707 times to the
input voltage.
This cut-off frequency is also known as 3dB frequency
The cut-off Frequency of the CR circuit is Fc =
&#3627409359;
&#3627409360;??????????????????

The frequencies over which the input signal is totally passed is called pass band and where
the gain is minimum is called stop band.

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CR HIGH - PASS FILTER AS PASSIVE DIFFERENTIATOR
CR High –pass filter acts as a passive differentiator when the time constant of the circuit is
very small.
A differential circuit is one whose output voltage is proportional to the differential of input
Voltage. i.e V0 α
????????????
&#3627408470;
????????????

Let us consider a circuit consisting of Capacitor ( C ) and Resistor (R ) connected in series
with an alternating source of voltage Vi as shown in the below diagram.



The important condition is that the time constant (RC) of the circuit is very low.
When the time constant is very low, the Capacitor takes very less time to charge and hence
the entire input voltage appears across Capacitor only when compared to Resistor. As the
voltage across the Resistor is very small
from the circuit Vi = Q/C ………………..(1).
and Q = ????????????
&#3627408470; ………………..(2)
Now, from the circuit the output voltage across the Resistor R is V0 = IR = R
??????&#3627408452;
????????????

From equation (2), we know that Q = ????????????
&#3627408470;
Hence V0 = ??????.
??????????????????
&#3627408470;
????????????

or V0 = ????????????
????????????
&#3627408470;
????????????
… ------------------- (3)
therefore V0 α
&#3627408517;????????????
&#3627408517;??????
as R & C are constants.
i.e the output voltage is proportional to the differential of input voltage.
Hence, a CR High –pass filter circuit with low time constant can act as a
Differentiator.

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RL LOW - PASS FILTER AS PASSIVE DIFFERENTIATOR
RL Low–pass filter acts as a passive differentiator when the time constant of the circuit is
very large.
A differentia circuit is one whose output voltage is proportional to the differential of input
Voltage. i.e V0 α
????????????
&#3627408470;
????????????

Let us consider a circuit consisting of Resistor (R ) & Inductor (L) connected in series with
an alternating source of voltage Vi as shown in the below diagram.

Since the time constant is very large, the inductor takes more time to build or develop a
voltage in it. So, the maximum input voltage will be across the resistor only.
So, the input voltage Vi = I R
Or I =
??????
&#3627408470;
&#3627408453;
………………………….(1)
The output voltage or the voltage across the Inductor is V0 = L
????????????
????????????
---------------(2)
Substituting the value of I from equation (1)
V0 = L
??????
??????
&#3627408470;
??????
????????????
=
??????
&#3627408453;
????????????&#3627408470;
????????????

Since
??????
&#3627408453;
, which is the time constant of RL circuit is a constant, it can be written as
V0 α
????????????
&#3627408470;
????????????

i.e the output voltage is proportional to the differential of Input voltage .So this circuit acts as
a differentiator circuit.

L R HIGH - PASS FILTER AS PASSIVE INTEGRATOR
LR high –pass filter acts as a passive Integrator when the time constant of the circuit is very
small.

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An integrator circuit is one whose output voltage is proportional to the Integral of input
Voltage. i.e V0 α ∫??????
&#3627408470; &#3627408465;?????? .
Let us consider a circuit consisting of Inductor (L ) and Resistor ( R ) connected in series
with an alternating source of voltage Vi as shown in the below diagram.


The important condition is that the time constant (RC) of the circuit is very small.
When the time constant is very small, the Inductor takes less time to build or develop the
voltage and hence the entire input voltage appears across Inductor only. So, voltage across
the Resistor is very small.
So, from the circuit Vi = L
????????????
????????????
………… ……..(1).
and
????????????
????????????
=
??????
&#3627408470;
??????

dI =
??????
&#3627408470;
??????
.dt
or I = ∫
??????
&#3627408470;
??????
&#3627408465;?????? -----------------------------(2)
Now, from the circuit the output voltage across the Resistor V0 = I.R
And substituting the value of I from the equation (2)
V0 = ??????.∫
??????
&#3627408470;
??????
&#3627408465;??????
Or V0 =
&#3627408453;
??????
∫???????????? &#3627408465;??????
As R/L is a constant (Time Constant) of LR circuit we can write the above equation as
V0 α ∫??????
?????? &#3627408517;?????? .
i.e the output voltage is proportional to the integral of input voltage.
Hence, a LR High –pass filter circuit with small time constant can act as an
Integrator.
BAND –PASS FILTER (Passive)

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A band-pass filter is one which allows only a band or group of frequencies and rejects
(attenuates) all other frequency signals outside this band.
A passive band pass filter can be constructed by cascading (connecting) a High-pass filter
with a Low-pass filter s shown in the diagram.
Circuit:

In the above circuit, the first section (I) acts like a High pass filter with a cut-off frequency of
Fc1 and the second section (II) acts like a Low-pass filter with the cut-off frequency Fc2.
When the action of both the filters is combined the operation is Band-pass filter. i.e a Band
Pass Filters passes signals of certain “band” or “ group” of frequencies without distorting the
input signal or adding any extra noise. This band of frequencies is commonly known as the
Bandwidth of the Filter.
Band-width is also defined as the difference between the upper cut-off frequency and Lower
cut-off frequency. BW = Fc2 - Fc1
This type of Band-pass filters are widely used in noise cancellation circuits in communication
systems.
A band pass filter is considered as a second-order (two-pole) type filter because it has “two”
reactive components in its construction. Also, the phase angle will be twice that of the first-
order filters, ie, 180
o
.
The frequency response of a Band-pass filter is shown in the graph below.

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The upper cut-off frequency is Fc2 =
1
2??????&#3627408453;2??????2

And the lower cut-off Frequency is Fc1 =
1
2??????&#3627408453;1??????1

The centre frequency is F = √Fc1Fc2
F = √
1
2??????&#3627408453;1??????1.2??????&#3627408453;2??????2
. =
1
2??????
1
√&#3627408453;1&#3627408453;2??????1??????2

Suppose R1=R2=R and C1 = C2 = C
The centre Frequency of the Band-pass filter is F =
1
2??????
1
√&#3627408453;&#3627408453;????????????
=
1
2??????&#3627408453;??????


So, the centre frequency of the Band-Pass filter is F =
&#3627409359;
&#3627409360;??????????????????

Time constant of a circuit.
Time Constant of a circuit always gives the time response of the circuit when an input
voltage or signal is applied. The time constant of any electronic circuit mainly depend on the
reactive components (capacitive or inductive) connected in the circuit.
The time constant is normally denoted by Tau (τ) with units secs.
Time Constant of RC circuit.
The time constant of a RC circuit is τ = RC
The response of RC circuit is V(t) = V0(1-&#3627408518;
−??????/????????????
) = V0(1-&#3627408518;
−??????/τ
)
When t = τ
V(t) = V0(1- &#3627408518;
−&#3627409359;
) = V0(1- 0.37) = 0.63V0 (nearly) or Q(t) = Q0(1-&#3627408518;
−??????/τ
) = 0.63Q0

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So, the Time constant of a RC circuit can be defined as the time in which the change or voltage
on a Capacitor increases to 63% of its maximum value or decays to 37% of the maximum
value.
Time Constant of RL Circuit
The time constant of a RL circuit is τ = L/R
The response of RL circuit is V(t) = V0(1-&#3627408518;
−????????????/??????
) = V0(1-&#3627408518;
−??????/τ
) or
I(t) = I0(1-&#3627408518;
−??????/τ
) here τ = L/R is the Inductive time constant.
When t = τ
V(t) = V0(1- &#3627408518;
−&#3627409359;
) = V0(1- 0.37) = 0.63V0 (nearly) or I(t) = I0(1-&#3627408518;
−??????/τ
) = 0.63I0
So, the Time constant of a RL circuit can be defined as the time in which the current or voltage
on a Inductor increases to 63% of its maximum value or decays to 37% of the maximum value.