The transfer function

Let’s begin with a general nth-order, linear,

time-invariant differential equation,

andn

dtnc(t) + ... + a1

d

dtc(t) + a0c(t)

= bmdm

dtmr(t) + ... + a1

d

dtr(t) + b0r(t) (1)

where c(t) is the output, r(t) the input, and

ai’s, bi’s are coefficients in the differential equa-

tion.

Taking the Laplace transform of the both sides

and assuming that all initial conditions are zero,

(ansn + ... + a1s + a0)C(s)

= (bmsm + ... + b1s + b0)R(s) (2)

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or

G(s) =C(s)

R(s)=

bmsm + ... + b1s + b0ansn + ... + 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 denom-

inator of the transfer function is just the char-

acteristic polynomial of the differential equa-

tion. 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 repre-

sented by

d

dtc(t) + 2c(t) = r(t) (4)

(ii) Find the response, c(t), to an input, r(t) =

u(t), a unit step, assuming zero initial condi-

tions.

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

G(s) =C(s)

R(s)=

1

s + 2(6)

(ii) Since r(t) = u(t) R(s) = 1/s. C(s) =

R(s)G(s) = 1s(s+2)

. Expanding by partial frac-

tions, we get C(s) =1/2s −

1/2s+2

Finally, taking the inverse Laplace transform

of each side yields c(t) = 12 −

12e−2t

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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 volt-

age, V (s), in the Figure below:

−c

_+

i(t)

v(t)

L

R

C

+

v (t)

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Solution: Summing the voltages around the

loop, assuming zero initial conditions, yields

the integrodifferential equations for this net-

work as

Ld

dti(t) + Ri(t) +

1

C

∫ t

0i(τ)dτ = v(t) (7)

Changing the variable from current to charge

using i(t) = ddtq(t) yields

Ld2

dt2q(t) + R

d

dtq(t) +

1

Cq(t) = v(t) (8)

The voltage-charge relationship for a capacitor

is

q(t) = Cvc(t) (9)

Substituting above to (8) yields

LCd2

dt2vc(t) + RC

d

dtvc(t) + vc(t) = v(t) (10)

Taking the Laplace transform assuming zero

initial conditions

(LCs2 + RCs + 1)Vc(S) = V (s) (11)

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so

G(s) =Vc(s)

V (s)=

1/LC

(s2 + RLs + 1

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) respec-

tively.

2. Replace the component values with their

impedance values.

for the capacitor V (s) =1

CsI(s)

for the inductor V (s) = RI(s)

for the resistor V (s) = LsI(s)

3. Finally we bypass the writing of differential

equation, and applying kirchhoff ’s voltage law

to the transform circuits.

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Example: Repeat the same example.

Redrawn

−c

_+

i(t)

v(t)

L

R

C

+

v (t)

→

_

1

Cs cV (s)

I(s)

R

Ls

V(s)

−

+

+

We obtain

(Ls + Rs +1

Cs)I(s) = V (s) (13)

But the voltage across the capacitor, Vc(s), is

Vc(s) = I(s)1

Cs(14)

Hence we have the same result.

(LCs2 + RCs + 1)Vc(s) = V (s) (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).

+

R1R

_+

2

1i (t)

i (t)

v(t) CL

−

2

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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2

1

Cs

R 1

+_

R 2

Ls

I (s)I (s)

V(s)

+

−

1

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,

LsI2(s)+R2I2(s)+1

CsI2(s)−LsI1(s) = 0 (17)

Rearrange the two equations above,[

R1 + Ls −Ls

−Ls Ls + R2 + 1Cs

] [

I1(s)I2(s)

]

=

[

V (s)0

]

(18)

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Using Cramer’s rule

I2(s) =

R1 + Ls V (s)−Ls 0

R1 + Ls −Ls

−Ls Ls + R2 + 1Cs

(19)

=LsV (s)

(R1 + Ls)(Ls + R2 + 1Cs) − L2s2

(20)

Forming the transfer function,G(s)

G(s) =I2(s)

V (s)=

Ls

(R1 + Ls)(Ls + R2 + 1Cs) − L2s2

=LCs2

(R1 + R2)LCs2 + (R1R2C + L)s + R1

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Operational amplifiers

An operational amplifier is an electronic ampli-

fier used as building block to implement trans-

fer functions. It has the following character-

istics; (i)Differential two inputs; v2(t) − v1(t);

(ii)High input impedance, Zi = ∞ (ideal); (iii)Low

output impedance, Zo = 0; ; and (iv)High con-

stant 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)

Z1(s)= −

Vo(s)

Z2(s)= −I2(s)

Hence, the transfer function is

Vo(s)

Vi(s)= −

Z2(s)

Z1(s)(21)

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Example: Find the transfer function, Vo(s)Vi(s)

, for

the circuit given below

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Solution: The transfer function of the circuit

is given by (21). Since the admittances of par-

allel components add, Z1(s) is the reciprocal of

the sum of the admittances, or

Z1(s) =1

C1s + 1R1

=1

5.6 × 10−6s + 1360×103

=360 × 103

2.016s + 1

For Z2(s), the impedances add, or

Z2(s) = R2 +1

C2s= 220 × 103 +

107

s

The transfer function is

Vo(s)

Vi(s)= −

Z2(s)

Z1(s)

= −1.232s2 + 45.95s + 22.55

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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