The transfer function
Let’s begin with a general nth-order, linear,
time-invariant differential equation,
dn
d
an n c(t) + ... + a1 c(t) + a0c(t)
dt
dt
m
d
d
= bm m r(t) + ... + a1 r(t) + b0r(t) (1)
dt
dt
where c(t) is the output, r(t) the input, and
ai’s, bi’s are coefficients in the differential equation.
Taking the Laplace transform of the both sides
and assuming that all initial conditions are zero,
(an sn + ... + a1s + a0)C(s)
= (bmsm + ... + b1s + b0)R(s)
(2)
1
or
C(s)
bmsm + ... + b1s + b0
=
G(s) =
R(s)
an sn + ... + a1s + a0
(3)
We call this ratio, G(s), the transfer function.
The transfer function can be represented as a
block diagram as above. Note that the denominator of the transfer function is just the characteristic polynomial of the differential equation. Also, we can find out the output C(S)
by using
C(s) = R(s)G(s)
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Example: (i) Find the transfer function represented by
d
c(t) + 2c(t) = r(t)
(4)
dt
(ii) Find the response, c(t), to an input, r(t) =
u(t), a unit step, assuming zero initial conditions.
Solution: (i) Taking the Laplace transform of
both sides, assuming zero initial conditions, we
have
sC(s) + 2C(s) = R(s)
(5)
The transfer function, G(s), is
1
C(s)
=
(6)
G(s) =
R(s)
s+2
(ii) Since r(t) = u(t) R(s) = 1/s. C(s) =
1
R(s)G(s) = s(s+2)
. Expanding by partial frac1/2
1/2
tions, we get C(s) = s − s+2
Finally, taking the inverse Laplace transform
1 e−2t
−
of each side yields c(t) = 1
2
2
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Electric network transfer functions
We now combine electrical components into
circuits, decide on the input and output, and
find the transfer function.
Example: Find the transfer function relating
the capacitor voltage, Vc(s), to the input voltage, V (s), in the Figure below:
L
R
+
v(t)
C
v c (t)
+
_
−
i(t)
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Solution: Summing the voltages around the
loop, assuming zero initial conditions, yields
the integrodifferential equations for this network as
Z
d
1 t
i(τ )dτ = v(t)
(7)
L i(t) + Ri(t) +
dt
C 0
Changing the variable from current to charge
d q(t) yields
using i(t) = dt
d2
d
1
L 2 q(t) + R q(t) + q(t) = v(t)
(8)
dt
dt
C
The voltage-charge relationship for a capacitor
is
q(t) = Cvc(t)
(9)
Substituting above to (8) yields
d2
d
LC 2 vc(t) + RC vc(t) + vc(t) = v(t) (10)
dt
dt
Taking the Laplace transform assuming zero
initial conditions
(LCs2 + RCs + 1)Vc(S) = V (s)
(11)
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so
Vc(s)
1/LC
G(s) =
=
1 )
V (s)
(s2 + R
s
+
L
LC
(12)
A technique for simplifying the solution
1. Redraw the original network showing all
time variables, such as v(t), i(t) and vc(t), as
Laplace transforms V (s), I(s) an Vc(s) respectively.
2. Replace the component values with their
impedance values.
1
I(s)
Cs
for the inductor V (s) = RI(s)
for the resistor V (s) = LsI(s)
for the capacitor V (s) =
3. Finally we bypass the writing of differential
equation, and applying kirchhof f ’s voltage law
to the transform circuits.
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Example: Repeat the same example.
Redrawn
Ls
L
R
R
+
+
v(t)
v c (t)
+
_
1
V(s)
C
Vc (s)
Cs
+
_
−
−
I(s)
i(t)
→
We obtain
1
(Ls + Rs +
)I(s) = V (s)
(13)
Cs
But the voltage across the capacitor, Vc(s), is
1
Cs
Hence we have the same result.
Vc(s) = I(s)
(LCs2 + RCs + 1)Vc (s) = V (s)
(14)
(15)
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Complex circuit via mesh analysis
1. Replace passive elements values with their
impedances.
2. Replace all sources and time variables with
their Laplace transform.
3. Assume a transform current and a current
direction in each mesh.
4. Write Kirchhoff’s voltage law around each
mesh.
5. Solve the simultaneous equations for the
output.
6. Form the transfer function.
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Example: Given the network of the Figure (a)
below, find the transfer function, I2(s)/V (s).
R
R
1
2
+
v(t)
+
_
C
L
i 1(t)
−
i 2(t)
Figure (a)
The first step is to convert the network into
Laplace transforms for impedances and circuits
variables. The result is shown in the Figure
(b).
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R
R
1
2
1
Cs
V(s)
+
_
+
Ls
I 1 (s)
−
I 2 (s)
Figure (b)
Around Mesh 1, where I1(s) flows,
R1I1(s) + LsI1(s) − LsI2(s) = V (s)
(16)
Around Mesh 2, where I2(s) flows,
1
LsI2(s)+R2I2(s)+ I2(s)−LsI1(s) = 0 (17)
Cs
Rearrange the two equations above,
"
R1 + Ls
−Ls
1
−Ls
Ls + R2 + Cs
#"
I1(s)
I2(s)
#
=
"
#
V (s)
0
(18)
10
Using Cramer’s rule
R1 + Ls V (s)
−Ls
0
I2(s) =
R1 + Ls
−Ls
1
−Ls
Ls + R2 + Cs
=
LsV (s)
1 ) − L2s2
(R1 + Ls)(Ls + R2 + Cs
(19)
(20)
Forming the transfer function,G(s)
I (s)
Ls
G(s) = 2
=
1 ) − L2s2
V (s)
(R1 + Ls)(Ls + R2 + Cs
LCs2
=
(R1 + R2)LCs2 + (R1R2C + L)s + R1
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Operational amplifiers
An operational amplifier is an electronic amplifier used as building block to implement transfer functions. It has the following characteristics; (i)Differential two inputs; v2(t) − v1(t);
(ii)High input impedance, Zi = ∞ (ideal); (iii)Low
output impedance, Zo = 0; ; and (iv)High constant gain amplification, A = ∞ (ideal)
The output vo(t) = A(v2(t) − v1(t)).
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Inverting operational amplifier
If v2(t) is grounded, the amplifier is called an
inverting operational amplifier. We have (i)
Ia(s) = 0 and I1(s) = −I2(s); (ii)V1(s) ≈ 0
I1(s) =
Vi(s)
Vo(s)
=−
= −I2(s)
Z1(s)
Z2(s)
Hence, the transfer function is
Vo(s)
Z (s)
=− 2
Vi(s)
Z1(s)
(21)
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(s)
Example: Find the transfer function, VVo(s)
, for
i
the circuit given below
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Solution: The transfer function of the circuit
is given by (21). Since the admittances of parallel components add, Z1(s) is the reciprocal of
the sum of the admittances, or
Z1(s) =
1
C1s + R1
1
=
1
1
5.6 × 10−6s + 360×10
3
360 × 103
=
2.016s + 1
For Z2(s), the impedances add, or
107
1
3
= 220 × 10 +
Z2(s) = R2 +
C2s
s
The transfer function is
Vo(s)
Z2(s)
= −
Vi (s)
Z1(s)
s2 + 45.95s + 22.55
= −1.232
s
The resulting circuit is called a PID controller
and can be used to improve the performance
of a control system.
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