Noninverting operational amplifier
We see that
Vo(s) = A(Vi(s) − V1(s))
But by voltage division
V1(s) =
Z1(s)
Vo(s)
Z1(s) + Z2(s)
simplifying
A
Vo (s)
=
Vi (s)
1 + AZ1(s)/(Z1 (s) + Z2(s))
Recall A is large, so the transfer function is
Vo(s)
Z1(s) + Z2(s)
=
Vi (s)
Z1(s)
(1)
1
(s)
Example: Find the transfer function, VVo(s)
, for
i
the circuit given below
2
Solution: The transfer function of the circuit
is given by (1).
Z1(s) =
1
1 + R1C1s
+ R1 =
C1s
C1s
(2)
For Z2(s), the admittances (reciprocal of the
impedances) add, followed by taking the reciprocal of the sum
1
Z2(s) = 1
R + C2s
2
R2
=
R2C2s + 1
(3)
The transfer function is
=
Vo(s)
Z1(s) + Z2 (s)
=
Vi(s)
Z1 (s)
C2C1 R2 R1s2 + (C2 R2 + C1R2 + C1 R1)s + 1
C2C1 R2 R1s2 + (C2R2 + C1 R1)s + 1
3
(s)
Example: Find the transfer function, VVo(s)
, for
i
the circuit given below
4
Solution:
1
3
Z1(s) = 200 × 10 +
=
1
+ 10−6s
500×103
s+7
× 105
1 + 0.5s
1
3
Z2(s) = 500 × 10 +
1
+ 10−6s
100×103
0.5s + 6
× 105
1 + 0.1s
The transfer function is
=
=
Vo(s)
Z1(s) + Z2(s)
=
Vi(s)
Z1(s)
3.5s2 + 52s + 130
s2 + 17s + 70
5
Mechanical system transfer functions
Transfer function – one equation of motion
Problem: Find the transfer function, X(s)/F (s),
for the system below.
Solution: Begin the solution by drawing the
free body diagram shown in Figure (a). Place
on the mass all forces felt by the mass.
6
Only the applied force points to the right. All
other forces impede the motion and act to oppose it. Hence the spring, viscous damper, and
force due to acceleration point to the left.
Using Newton’s law
d2x(t)
dx(t)
M
+ fv
+ Kx(t) = f (t)
2
dt
dt
Taking Laplace transform, assuming zero initial conditions,
M s2X(s) + fv sX(s) + KX(s) = F (s)
7
Solving for the transfer function yields
G(s) =
X(s)
1
=
F (s)
M s2 + fv s + K
Can we circumventing the writing of differential equations by defining impedances for mechanical components? Yes.
For the spring
F (s) = KX(s)
For the viscous damper
F (s) = fv sX(s)
For the mass
F (s) = M s2X(s)
Replace each force in Figure (a) we obtain Figure (b)
8
Transfer function – two degrees of freedom
Problem: Find the transfer function, X2(s)/F (s),
for the system below.
Solution: The system has two degrees of freedom, since each mass can be moved in the
horizontal direction while the other is held still.
Thus two simultaneous equations of motion
will be required to describe the system, each
of which comes from the free body diagrams
of each mass.
9
M1 :
(a) Forces on M1 due only to motion of M1;
(a) Forces on M1 due only to motion of M2;
(a) All forces on M1.
10
M2 :
(a) Forces on M2 due only to motion of M2;
(a) Forces on M2 due only to motion of M2;
(a) All forces on M2.
11
The Laplace transform of the equations of motion can be rewritten as
[M1 s2 + (fv1 + fv3 )s + (K1 + K2 )]X1 (s) − [fv3 s + K2]X2 (s) = F (s)
−[fv3 s + K2 ]X1 (s) + [M2 s2 + (fv2 + fv3 )s + (K2 + K3 )]X2 (s) = 0
From this, the transfer function, X2(s)/F (s),
is
X2(s)
fv3 s + K2
G(s) =
=
F (s)
△
where △ equals to
[M1 s2 + (fv1 + fv3 )s + (K1 + K2 )]
−[fv3 s + K2 ]
[M2
s2
−[fv3 s + K2 ]
+ (fv2 + fv3 )s + (K2 + K3 )]
For example, if M1 = 1kg, M2 = 2kg,
fv1 = fv2 = fv3 = 1N − s/m,
K1 = 1N/m, K2 = 2N/m, K3 = 3N/m,
△=
s2 + 2s + 3
−(s + 2)
−(s + 2)
= 2s4 + 6s3 + 20s2 + 20s + 19
(2s2 + 2s + 5)
Thus
s+2
G(s) = 4
2s + 6s3 + 20s2 + 20s + 19
12