Linearized Analysis of the Synchronous Machine for PSS Chapter 6 ...

ΔT
exc
=
− K
+
3 5 A
osc F
( )( )( ) ( Δ δ )
jω
+ σ + jω
jω
+ σ + jω
jω
+ σ + jω
osc
1
1
2K
K K ( 1
osc
2
jω
T
2
)
osc
3
3
On combining imaginary terms in the denominator, we get:
− K2K3K5K
A(
1+
jωoscTF
)
ΔTexc
=
σ1
+ j( ω1
+ ωosc
) σ
2
+ j( ω2
+ ωosc
) σ
3
+ j( ω3
+
We are interested in the phase of ΔT exc relative to Δδ.
( )( )( ) ( Δ δ ) ω )
Fact: When the generator is heavily loaded, it is possible for K 5 to
be negative. See Fig. 6.1and section 8.4.3 in your text. This makes
the numerator of the previous transfer function positive.
A simulation of such a case is shown in Fig. 8 below. The solid
curve represents generators with fast high-gain excitation systems,
but no PSS. The other two curves represent significantly fewer of
such generators.
osc
Fig. 8
14