1
Time Response
The nature of the time response at the electrical system is the same as at the mechanical system. As
example in the figure 1 are
presented speed-time graph of the
i(t)
airplane is the same current time
v(t)
graph of electrical system with
R
one energy-storage element (in
this case kinetic). Such graphs are
L
characteristics of a first-order
system.
The energy storage
elements in the electrical
system are capacitance and
Fig.1. Time characteristics of the first order system
inductance and response of
circuits containing these elements may be determined using the relationships i = Cdv dt and
v = Ldi dt together with v = Ri for the resistor.
Constant force
t
Constant voltage
t
The series RC circuit
Such circuit is shown in the Fig. 2. Applying the Kirchoff’s
voltage law around the loop gives:
v R + vC = E (after the switch is closed)
V
R
t=0
Substituting
R
VC
i = Cdve dt
gives
v R = Ri = RC dvc dt
resulting the differential equation in v c :
dv e
+ ve = E
E
dt
E is the source e.m.f. and is known as the forcing function.
As a matter of convenience, the forcing function is written as
Fig.2. Series RC circuit
the right-hand side of the equation. The general forms of the
current and capacitor voltage responses are shown in the figure 3. These graphs are build
assuming that capacitor is initially uncharged.
After closing the switch, the voltage builds up from zero to its
i
final steady-state value E. At the end of the period v c is
constant and the capacitor behaves as an open circuit. The
complete response can be resolved into two components. The
t
forced response is that component of the total response which
v
is dependent as the forcing function and the circuit elements. It
is the response which would exist if the forcing function had
always been connected to the network. In this example the
capacitor-voltage forced function is therefore E. Forced
t
response is also called steady-state response, being that to
Fig.3. Time responses of the
which the capacitor voltage eventually settles. However, the
current and voltage in the RC
forcing function is not in fact connected until the switch is
circuit
closed and is unlikely that, at the point of switching, the initial
condition of the energy storage element in the network. To bridge this mismatch an additional
component of response arises. This is the natural response, existing for the period of adjustment
only. Its form is dependent on the network elements alone and its magnitude in dependent on the
mismatch between the initial conditions and the forced function at the time at switching. The
total response is the sum of the natural and forced components. A special case occurs when
C
CR
2
forcing function is zero. Then, the forced response does not exist and the total response is also
the natural response, which is dependent only on the network elements and initial conditions and
known as the zero input response. An example in the freely swinging pendulum which is given
and initial displacement and in then released.
The response is shown in the Fig. 4. The same graph represents the natural current response of an
RLC network.
Another term used in the zero-state response – this is the complete response (natural and
forced) of the network or system having all initial conditions at zero.
Determination of the natural response
R=0
R
R>0
L
t
C
Fig.4. Analogy between pendulum and RLC circuit
Because the form of the
natural
response
is
in
dependent of the forcing
function, it may be determined
by setting E to zero in the
differential
equation
developed for the circuit in the
Fig. 2. Circuit becomes that
shown in the Fig. 5. The
differential equation is now:
dvc
+ vc = 0 .
dt
The solution of this equation may be found by the standard procedure of forming an auxiliary or
characteristic equation.
A solution, v c = Vn e S nt where Vn is constant and S n is a circuit parameter is assumed which,
substituting in to the equation gives:
CR
S n CRV n e S nt + Vn e S nt = 0
R
(S n CR + 1)Vn e S t = (S n CR + 1)Vc
n
=0
C
Assuming that vc is not zero, S n CR + 1 = 0 and this is the auxiliary
equation. The root of this equation is S n = − 1
giving a solution
CR
Fig. 5. RC circuit
v c = Vn e
−t
CR
.
To determine the constant Vn initial conditions are used: it for
−t
example vc (0) = v0 , v n = v0 and v c becomes v0 e CR . The shape of this natural response will be
seen to be as exponential discharge (of a capacitor).
To determine the forced response the complete equation is considered:
dv
CR c + vc = E
dt
The standard method, known as the D-operator method could be used. In this method different
ion of a variable v c in this case, is represented by the operational notation Dvc . Then, Dvc
represents dvc dt ; D 2 v c represents d 2 v c dt 2 .
The equation then becomes
CRDvc + vc = E
or
(CRD + 1)vc
=E
3
whence
v
vc =
t
Natural
response
-E
E
CRD + 1
1
can be expanded by means of the binomial
CRD + 1
(− 1)(− 2)(CED )2 and the
−1
theorem (1 + CRD ) = 1 + (− 1)CRD +
2!
expression for v c becomes vc = (1 − CRD + K)E = E because the
first and all the higher derivatives are
equal to zero. The forced response
R
is therefore E.
The term
E
Forced
response
t
E
Complete
response
t=0
C
The complete response
t
The complete response is given by
the sum of the two component
responses: natural and forced.
For the natural response:
Fig. 6. Time responses of the
RC circuit
Fig.7. RC circuit
(S n CR + 1)vc = 0 ;
(DCR + 1)vc = E .
The complete response is, therefore vcnatural = Vn e
−t
CR
v cforced = E .
With the forced component is present, a new value for the constant v n may be now evaluated. It
vc (0) = 0 , substituting this into the equation gives 0 = E + Vn . Whence Vn = − E . Substituting
this value into the solution gives, finally
v c = E + Vn e
−t
CR
for t ≥ 0 ;
vc = E ⎛⎜1 − e CR ⎞⎟ for t ≥ 0 .
⎝
⎠
Fig. 6 shows two component responses and the resultant complete response together with the
step-input voltage for reference. If it is required, the current response may be determined by
i = Cdvc dt . Thus
−t
⎛ 1 ⎞⎛ −t CR ⎞ E −t CR
.
i = EC ⎜ −
⎟⎜ − e
⎟= e
⎠ R
⎝ CR ⎠⎝
Example: Determine the voltage varieties across the capacitor in the circuit in the Fig. 7
following closure of the switch. C is initially charged to a +5 V.
Solution: there is no forcing function and the forced response is zero. The complete response is
given by the natural response:
v c = Vn e
−t
CR
, t ≥0.
Initially, vc = +5 the t = 0 and Vn = +5 . Substituting numerical values
4
v c = 5e −5t .
Example: Determine the complete capacitor voltage response for the circuit shown in the Fig. 8.
The capacitor is initially charged to +5V
and a forcing function is a linearly
v(t)
increasing ramp voltage having a store
of 5 × 10 6 V s for t > 0 .
The natural response is therefore given
−t
by Vcn = Vn e CR where Vn is constant
yet to be determined.
The forced response is given by
Natural response
Complete response
v cf = (1 − CRD + L)C
Forced response
t
Fig. 9. Time responses of the RC circuit
Where C is the ramp forcing function
given by C = kt (k-slope)
Then
v cf = (1 − CRD + L)kt = kt − KCR
.
In this case first derivative is not zero, although the higher order derivatives are.
v cn
Combining
and
and
substituting
values
of
C,
v cf
R
and
L
vc = Vn e −5⋅10 ⋅t + 5 ⋅ 10 6 ⋅ t − 5 ⋅ 10 6 ⋅ 2 ⋅ 10 −6 for t ≥ 0 .
Substituting the initial condition vc = +5V (t = 0 ) gives + 5 = Vn − 10
from which Vn = 16V .
6
CR = 10 4 ⋅ 2 ⋅ 10 −10 ⋅ 2 ⋅ 10 −6
Complete response is
5⋅ t
vc = 15e −5⋅10 + 5 ⋅ 10 6 ⋅ t − 10V (t ≥ 0)
This response is sketched in the figure 9.
R
t=0
e
v(t)
C
E
e(t)
τ
t
Fig.8. RC circuit
transient
period
steady state
period
t
Fig.10. Parameters of the time
response
Transient and steady-state periods: time
constant
It is common practice to call the natural response the
transient response and forced response-steady state
response. There is no problem with the latter because only the forced response exist in the
steady-state period. However, diving the transient period, both forced and natural responses
exist. A problem may arise in determining just when the transient period finishes and the steadystate period starts: now small does the natural component need be in order to be neglected?
There is no absolute answer, the division depending as the context of system in which the circuit
used. However, a measure of this time period is provided by the circuit time constant, τ . For
the single-energy storage system under discussion, time constant is defined as the time which
would be needed for the variable (for example capacitor voltage) to reach its final value it its
initial rate of change were maintained. Fig. 10 illustrates this idea.
5
Time constant of CR circuit. Then, at t = 0 , the initial rate of increase of ve = E CR . The
diagram shows, that the time required to reach the final value is E (E CR ) = CR . The unit is
second.
As a rule of thumb, a period equal to 5T is sometimes used. Thus, substituting t = τ into
vc = 0,993E when t = 5τ .
Forced response and exponential forcing functions: the phasor method.
The relationship
1
vc =
E
CRD + 1
was developed for the forced response in this case E is forcing function. It is now proposed that a
s t
general forcing function Fe f (where F and S f may be complex in the basis of a generalized
phasor method. Returning back to the general forcing function Fe
upon it is effectively to replace D by S f .
Sft
the effect of D operator
d
S t
S t
Fe f = S f Fe f
dt
1
1
S t
S t
vc =
Fe f =
Fe f
CRD + 1
S f CR + 1
DFe
Sft
=
The following example illustrates the procedure by which v c may be determined for a sinusoidal
forcing function and how the corresponding time function v c is then found.
Example: determine the capacitor voltage response if the ramp voltage is replaced by
cosinusoidal forcing function 50 cos ωt , where ω = 10 6 rad s .
The important step in the application of the phasor method is the representation of the forcing
function tg the exponential form. In this case
50 cosωt = 25e jωt + 25e − jωt
Setting S f = ± jω in the phasor operator 1 (S f CR + 1) yields
vc =
25
25
+
jωCR + 1 − jωCR + 1
Substituting values and working in the polar form
vc =
25
25
+
= 10,3∠ − 66 o + 10,3∠ + 66 o .
o
o
2,42∠66
2,42∠ − 66
To express v c as a function of time, vc must be multiplied by e
v c = 10,3∠ − 66 o e jωt + 10,3∠ + 66 o e − jωt
and noting that ∠66 o is equivalent to e j 66
o
(
)
vc = 10,3e j (ωt −66 ) + 10,3e − j (ωt −66 ) = 20,6 cos ωt − 66 o V.
o
o
Sft
, where S f = ± jω . Then