Transfer Function of Physical Systems
Transfer Function of Physical Systems
Transfer Function of Physical Systems
d 2 d d
T( t ) J D (4) ( t ) (5)
dt 2 dt dt
vin(t) is the applied voltage, Kb is the back-emf constant of the motor, Kt is the torque constant of the
motor, J is the polar MI of the armature and D is the damping coefficient.
T (s )
From equations (6), (7), and (8), Vin(s) = (Ra+ sLa) + Kb(s)
Kt
(Js 2 Ds )(s)
Substituting equations (9) and (10), Vin(s) = (Ra+ sLa) + Kbs(s)
Kt
(s) Kt
Simplifying, T(s)
Vin (s) JL s 3 (JR DL )s 2 (DR K K )s
a a a a t b
Physical Systems/Page 1 of 3
Method 2: Arrange the equations so that the right-hand side of each equation is either 0 or the input
variable. Now, arrange them in matrix form, Ax = B, where vector x consists of all the variables
except the input variable. Solve for the output variable.
(Ra+ sLa) Ia(s)+ Vb(s) = Vin(s) From (6)
Eb(s)-Kb(s) =0 From (7)
T(s)-KtIa(s) = 0 From (8)
T(s)-(Js2+Ds)(s) = 0 From (9)
(s)-s(s) = 0 From (10)
In matrix form:
Ra sLa 1 0 0 0 I a ( s) Vin ( s )
0 1 Kb 0 0 V ( s ) 0
b
Kt 0 0 1 0 (s) 0
0 0 0 1 ( Js 2 Ds) T ( s) 0
0 0 1 0 s ( s) 0
T (s) = Kt Ia(s)
T(s)
(s)
Js 2 Ds
(s) = s(s)
Vb(s) = Kb(s)
Draw the signal flow graph starting with the input node Vin(s); introduce the node Eb(s), form Ia(s)
and proceed to complete. Since (s) is the output node, a gain of 1 may be introduced as shown to
clearly show the output node.
(s)
1 1 1
Vin(s) Kt
Ra sLa Js2 Ds s Kb
Vb(s
)
1 Ia(s) T(s) (s)
-1
Find the derivatives of these variables from the time-domain equations (1) through (5).
di a 1 d
From equations (1), (2), and (5), ( v in ( t ) R a i a ( t ) K b )
dt La dt
dx1 R K 1
In terms of state variables, a x1 b x 3 v in ( t ) (11)
dt La La La
d 2 Kt D d
From equations (4) and (3), i a (t )
dt 2 J J dt
dx 3 K t D
In terms of state variables, x1 x 3 (12)
dt J J
dx 2
Also, x3 (13)
dt
Physical Systems/Page 3 of 3