EG1110 SIGNALS AND SYSTEMS
– A class of these inputs are sinusoidal inputs
Frequency response function
Intro
• Transfer functions give information about a system’s
– stability - very important
– transient behaviour
• Given a system with transfer function G(s), output can be computed using
Y (s) = G(s)U (s)
where U (s) is (almost) any input
• However, often interested in a system’s output due to a particular input.
1
2
Basic idea of frequency response function
The maths (optional)
• Frequency response function determines output of system if input is sinusoidal.
u(t)
y(t)
G
• It can be proved, via convolution that
Given a sinusoidal input an LTI system will produce a sinusoidal output, possible shifted in
amplitude and phase
• Let G be a stable system with transfer function G(s)
(poles in left half s-plane)
(in steady state)
• This means after initial conditions and transients have died away
• Let Y (s) and U (s) be Laplace Transforms of y(t) and u(t).
• Let u(t) = αsin(ωt)
the output is a sine wave appropriately scaled and shifted in phase
⇒ U (s) =
αω
s2 + ω 2
(1)
(From Laplace Transform tables)
• Let G(s) be
G(s) =
3
(s + z1)(s + z2) . . .
(s + p1)(s + p2) . . .
4
(2)
• We calculate the output as
• Hence taking inverse Laplace transforms...
e−p1t + A{z2e−p2t + . . .}
y(t) = β1e−jωt + β2ejωt + A
| 1
αω
Y (s) = G(s)U (s) = G(s) 2
s + ω2
limt→∞ =0
(as G is stable, p1, p2, . . . > 0)
• We know that, using partial fractions we can split up G(s) as
G(s) =
?
?
+
+ ...
s + p1 s + p2
(3)
Hence in steady state
yss(t) = β1e−jωt + β2ejωt
and we can separate
s2
αω
?
?
=
+
+ ω 2 s + jω s − jω
• Thus we can write an expression for Y (s) as
Y (s) =
To find β1, β2 note that
β1
A2
β2
A1
G(s)ωα
=
+
+
+
s2 + ω 2
s + jω s − jω s + p1 s + p2
β1(s − jω) + β2(s + jω) + (s + jω)(s − jω)[A1/(s + p1) + A2/(s + p2) + . . .]
=
(s + jω)(s − jω)
A2
β2
A1
β1
+
+
+
s + jω s − jω s + p1 s + p2
6
5
This implies that
Thus,
yss =
G(s)ωα = β1(s − jω) + β2(s + jω) + (s + jω)(s − jω)(A1/(s + p1) + A2/(s + p2) + . . .)
αG(−jω) −jωt αG(jω) jωt
+
e
e
−2j
2j
However, as G(jω) is a complex number we can write it as
Letting s = jω we get
G(jω) = |G(jω)|ejφ
G(jω)α = β2(2jω)
αG(jω)
⇒ β2 =
2j
where φ = 6 G(jω) = arctan(=(G(jω))/<(G(jω)))
Similarly
Letting s = −jω we get
G(−jω) = |G(jω)|e−jφ
G(−jω)α = β1(−2jω)
αG(−jω)
⇒ β1 =
−2j
7
Thus
−e−jωt−φ + e−jωt+φ
2j
= α|G(jω)| sin(ωt + φ)
yss = α|G(jω)|
8
(4)
Implications of frequency response function
• Given a system G(s) and an input α sin(ωt) the output (in steady state) is given by
y(t) = Y sin(ωt + φ)
6
G(jω) = 6
where
– ω - frequency of input sine (and output sine).
– α - amplitude of input sine
– Y = α|G(jω)| - amplitude of the output.
– φ = 6 G(jω) - phase difference between output and input sine
• G(jω) is known as the frequency response function or the sinusoidal transfer function.
• Can also calculate the frequency response function from input/output behaviour:
9
Y (jω) |G(jω)| =
U (jω) 10




Y (jω) 

U (jω)