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Unformatted text preview: )vo
+ 3is = 144
The transfer function is
36s + 3
is 144s2 + 48s + 1 Transfer Functions
Eqn (1) in operator notation is . . .
(ansn + an−1sn−1 + . . . + a1s1 + a0)y = (bmsm + bm−1sm−1 + . . . + b1s1 + b0)u
= n=0 i
= G (s ) This is called the system’s transfer
Since y = G(s)u, the output is the transfer
function times the input. System’s Perspective u ✲ G (s ) y ✲ Think of u as the cause, y as the eﬀect.
u is the forcing function, y is the response
u is the system’s input, y is its output Transfer Function Terminology
A transfer function is said to be . . .
rational if it’s the ratio of two
proper if m ≤ n strictly proper if m < n
improper if m ≥ n strictly improper if m > n
biproper if m = n
A complex number z is a zero of the
transfer function if G(z ) = 0.
Zeros are the roots of the numerator
A complex number z is a pole of the
transfer function if G(z ) = ∞.
Poles are the roots of the denominator
polynomial. Forced Response
Suppose that u(t) = Aept, where A and p
are given complex numbers.
Signals of this form are called complex
To solve eqn (1), assume a solution of the
form y (t) = Bept. Then,
= pAept = pu,
dt2 dk u
= pk u
and similarly for y ,
= pk u
dtk etc. Substituting into
dk y dk u
ak k =
k =0 k =0 gives n
ak pk Bept = k =0 m
bk pk Aept k =0 n
= Bept ak pk = Aept bk p k k =0 k =0
⇒ B = A n=0
k =0 = AG(p) y = Bept = G(p)Aept = G(p)u Theorem 1
If u(t) = Aept, then one solution of eqn (1)
is y (t) = G(p)u(t).
Proof As above. This solution is called the forced response.
If the input is a complex exponential, the
output has a complex exponential
exponential of the same frequency. Since
y = Bept = G(p)Aept = G(p)u
⇒ = = G (p )
the transfer function represents the gain of
The gain varies with th...
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This note was uploaded on 01/22/2014 for the course ENG 282IN taught by Professor Haines during the Spring '12 term at Pima CC.
- Spring '12