16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Lecture 14
Last time: w(t ,τ ) ⇒ w(t − τ )
t
y (t ) =
∫ w(t − τ ) x(τ )dτ
−∞
⎧ τ ' = t −τ
Let: ⎨
⎩ − dτ = dτ ′
∞
y (t ) = ∫ w(τ ′) x (t − τ ′)dτ ′
0
For the differential system characterized by its equations of state, specialization
to invariance means that the system matrices A, B, C are constants.
x = Ax + Bu
y = Cx
For A, B, C constant:
y (t ) = Cx (t )
t
x (t ) = Φ (t − t0 ) x (t0 ) + ∫ Φ (t − τ ) Bu (τ )dτ
t0
The transition matrix can be expressed analytically in this case.
d
Φ (t ,τ ) = AΦ (t ,τ ), where Φ (τ ,τ ) = I
dt
This is a matrix form of first order, constant coefficient differential equation. The
solution is the matrix exponential.
Φ (t ,τ ) = e A( t −τ )
e A( t −τ ) = I + A(t − τ ) +
1 2
1
A (t − τ ) 2 + ... + Ak (t − τ ) k + ...
2
k!
Useful for computing Φ (t ) for small enough t − τ .
Page 1 of 6
16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
The solution is
y (t ) = Cx (t )
t
x (t ) = e A( t −t0 ) x (t0 ) + ∫ e A( t −τ ) Bu (t )dτ
t0
For t0 → ∞ :
t
x (t ) =
∫e
A( t −τ )
Bu (τ )dτ
−∞
∞
= ∫ e Aτ ′ Bu (t − τ ′)dτ ′
0
and for a single input, single output (SISO) system,
w(t ) = c T e At b
If x (t ) = e jωt for all past time
∞
y (t ) = ∫ w(τ )e jω ( t −τ ) dτ
0
⎡∞
⎤
= ⎢ ∫ w(τ )e − jωτ dτ ⎥ e jωt
⎣0
⎦
= F (ω ) x (t )
Since w(τ ) = 0 for τ < 0 for a realizable system, we see that the steady state
sinusoidal response function, F (ω ) , for a system is the Fourier transform of the
weighting function – where the coefficient unity must be used.
F (ω ) =
∞
∫ w(τ )e
− jωτ
dτ
−∞
and w(τ ) for a stable system is Fourier transformable.
Then
∞
1
w( t ) =
F (ω )e jωt d ω
∫
2π −∞
You can compute the response to any input at all, including transient responses,
having defined F (ω ) for all frequencies.
The static sensitivity of the system is the zero frequency gain, F (0) , which is just
the integral of the weighting function.
∞
F (0) = ∫ w(τ )dτ
0
Page 2 of 6
16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Stationary statistics
Invariant output statistics implies more than stationary inputs and invariant
systems; it also implies that the system has been in operation long enough under
the present conditions to have exhausted all transients.
Input-Output Relations for Correlation and Spectral Density
Functions
Derive autocorrelation of output in terms of autocorrelation of input
∞
y (t ) = ∫ w(τ 1 ) x (t − τ 1 )dτ 1
0
∞
y = ∫ w(τ 1 ) xdτ 1
0
∞
= x ∫ w(τ 1 )dτ 1
0
R yy (τ ) = y (t ) y (t + τ )
∞
∞
= ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) x (t − τ 1 ) x (t + τ − τ 2 )
0
0
∞
∞
= ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ + τ 1 − τ 2 )
0
0
∞
∞
0
0
y 2 = ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ 1 − τ 2 )
Rxy (τ ) = x (t ) y (t + τ )
∞
= x (t ) ∫ w(τ 1 ) x (t + τ − τ 1 )dτ 1
0
∞
= ∫ w(τ 1 ) Rxx (τ − τ 1 )dτ 1
0
Page 3 of 6
16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Transform to get power density spectrum of output.
∞
y = x ∫ w(τ )dτ
0
= F (0) x
S yy (ω ) =
∞
∫R
yy
(τ )e − jωτ dτ
−∞
∞
=
∫
−∞
∞
∞
0
0
dτ ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ + τ 1 − τ 2 )e − jωτ1
∞
=
∞
∞
− jω (τ +τ −τ )
− jωτ
jωτ
∫−∞ dτ Rxx (τ + τ 1 − τ 2 )e 1 2 ∫0 dτ 1w(τ 1 )e 1 ∫0 dτ 2 w(τ 2 )e 2
first integral
⎧τ ′=τ +τ 1 − τ 2
In first integral only, let ⎨
⎩ dτ ′ = dτ
S yy (ω ) =
∞
∫
−∞
∞
dτ ′Rxx (τ ′)e − jωτ ′ ∫ dτ 1w(τ 1 )e jωτ1
−∞
∞
∫ dτ
2
w(τ 2 )e − jωτ 2
−∞
= S xx (ω ) F ( −ω ) F (ω )
2
= F (ω ) S xx (ω )
The power spectral density thus does not depend upon phase properties.
The cross-spectral density function can be derived similarly, to obtain:
S xy (ω ) = F (ω ) S xx (ω )
Mean squared output in time and frequency domain
∞
∞
0
0
y = R yy (0) = ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ 1 − τ 2 )
2
=
1
2π
∞
∫S
yy
(ω )d ω
−∞
∞
1
=
∫ F (ω ) F (−ω )S xx (ω )dω
2π −∞
Generally speaking, with linear invariant systems it is easier to work in the
transform domain than the time domain – so we shall commonly use the last
expression to calculate the mean squared output of a system. However, control
engineers are more accustomed to working with Laplace transforms than with
Fourier transforms. By making the change of variables s = jω we can cast this
expression in that form.
Page 4 of 6
16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
y2 =
=
1
2π
j∞
⎛ s ⎞ ⎛ s ⎞ ⎛ s ⎞ ds
F ⎜ ⎟F ⎜ − ⎟ S xx ⎜ ⎟
⎝ j⎠ ⎝ j⎠ ⎝ j⎠ j
− j∞
1
2 jπ
∫
∞
∫ F ′( s) F ′(− s)S ′ ( s)ds
xx
−∞
We know that S xx (ω ) is even. If it is a rational function of ω , and we will work
exclusively with rational spectra, it is then a rational function of ω 2 . So only
⎛s⎞
even powers of ω appear in S xx (ω ) and thus S xx ⎜ ⎟ which we may call S xx ( s ) is
⎝ j⎠
derived from S xx (ω ) by replacing ω 2 by − s 2 .
F ′( s ) is the ordinary transfer function of the system – the Laplace transform of its
weighting function. Because w(t ) = 0, t < 0 .
We shall drop the primes from now on.
j∞
1
y2 =
F ( s ) F ( − s ) S xx ( s )ds
2π j −∫j∞
ω 2 = − s 2 ⎫⎪
⎬ in S xx ( s )
ω 4 = s 4 ⎪⎭
Integrating the output spectrum
General method
Cauchy Residue Theorem
>∫ F ( s)ds = 2π j ∑ ( residues of F ( s) at the poles enclosed in the contour C )
C
If F ( s ) has a pole of order m at z = a ,
Res( a ) =
⎧ d m −1
⎫
1
m
⎨ m −1 ⎡⎣( s − a ) F ( s ) ⎤⎦ s =a ⎬
( m − 1)! ⎩ ds
⎭
F ( s ) has a pole of order m at s = a if m is the smallest integer for which
Page 5 of 6
16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
lim ⎡( s − a ) F ( s ) ⎤
⎦
s →a ⎣
is finite.
m
If F ( s ) is rational and has a 1st order pole at a ,
N ( s)
F ( s) =
D( s)
N ( s)
=
( s − a )( s − b)...
then
Res( a ) m =1 = lim [( s − a ) F ( s )]
s→a
=
N (a )
( a − b)( a − c )...
Page 6 of 6