Chapter2 Pedro

Exercises Chapter 2

”System Identification”

The solutions presented here are for the exercises:

• 2G.2

• 2G.4

• 2E.2

• 2E.4

• 2E.6

Pedro Xavier Miranda La Hera. Solutions. Chapter 2 Book: ”System Identification” (Umea, January, 2005) 1

2G.2 Let Φs(ω) be the (power ) spectrum of a scalar signal defined as in

(2.63). Show that

i Φs(ω) is real

ii Φs(ω) ≥ 0∀ω

iii Φs(−ω) = Φs(ω)

(2.63)

Φs(ω) =

∞∑

τ=−∞

Rs(τ )e−iτω


2G.2 Solution

The expression (2.63) represents the Fourier transform of the Autocorrelation

function. This Expression can be divided as:

Φs(ω) =

0∑

τ=−∞

Rs(τ )e−iτω +

∞∑

τ=0

Rs(τ )e−iτω − Rs(0)

rewriting this expression we get

Φs(ω) =

∞∑

τ=0

Rs(−τ )eiτω +

∞∑

τ=0


According to the properties of the Autocorrelation Function we know that it

is an even function which tell us:

Rs(τ ) = Rs(−τ )


We can rewrite the expression in the next form

Φs(ω) =

∞∑

τ=0

Rs(τ )eiτω +

∞∑

τ=0


=

∞∑

τ=0

Rs(τ )(eiτω + e−iτω

)− Rs(0)

by definition

cos(θ) =

(eiθ + e−iθ

)

2

Constructing this expression in our function we get:

Φs(ω) = 2∞∑

τ=0

Rs(τ )

(eiτω + e−iτω

)

2) − Rs(0)

Φs(ω) = 2∞∑

τ=0

Rs(τ )cos(τω) − Rs(0)


which can be also written as

Φs(ω) = Rs(0) + 2

∞∑

τ=1

Rs(τ )cos(τω)

By definition Rs(τ ) and cos(τω) are real functions, therefore using this

expression we can conclude that:

Φs(ω) is real

ii. By theory any matrix M is said to be positive-semidefinite if it has the

following equivalent property:

M = LL∗

s∗LL∗s ≥ 0

(L∗s)∗(L∗s) = ||L∗s||2 ≥ 0


The expression for the spectral density is defined as

Φs(ω) =

∞∑

τ=−∞

Rs(τ )e−iτω

where

Rs(τ ) = Es(t)s(t − τ ) =∞∑

t=−∞

s(t)s∗(t − τ )

Replacing in the expression for the spectral density

Φs(ω) =

∞∑

τ=−∞

Es(t)s(t − τ )e−iτω

=

∞∑

τ=−∞

∞∑

t=−∞

s(t)s∗(t − τ )e−iτω

=

∞∑

τ=−∞

∞∑

t=−∞

s(t)s∗(t − τ )e−iωteiω(t−τ)


if we make t − τ ∼= ξ then we can rewrite the last expression as

Φs(ω) =

∞∑

t=−∞

s(t)e−iωt

∞∑

ξ=−∞

sT (ξ)eiωξ

=

[∞∑

t=−∞

s(t)e−iωt

]

︸︷︷︸

L

∞∑

ξ=−∞

s(ξ)e−iωξ

∗

︸︷︷︸

L∗

if we apply the positive semi-definite theorem to this expression

z∗Φ(ω)z = z∗

[∞∑

t=−∞

s(t)e−iωt

] [∞∑

t=−∞

s(t)e−iωt

]∗

z


z∗Φ(ω)z =

([∞∑

t=−∞

s(t)e−iωt

]∗

z

)∗ ([∞∑

t=−∞

s(t)e−iωt

]∗

z

)

=

∣∣∣∣∣

∣∣∣∣∣

[∞∑

t=−∞

s(t)e−iωt

]∗

z

∣∣∣∣∣

∣∣∣∣∣

2

≥ 0

this last term tell us that the expression for ω will be always larger or equal

than zero due to the square exponent in the final expression

Φs(ω) ≥ 0 ∀ ω


iii. If we use the relation

Φs(ω) = 2

∞∑

τ=0

Rs(τ )cos(τω) − Rs(0)

instead of (2.63), we see that ω only affects the term cos(τω). By definition

of the cosine function we know that:

cos(ω) = cos(−ω)

This relations implies that

Φs(ω) = Φs(−ω)


2G.4. Let a continuous time system representation be given by

y(t) = Gc(p)u(t)

The input is constant over the sampling interval T . Show that the sampled

input-output data are related by

y(t) = GT (q)u(t)

where

GT (q) =1

2πi

∫ i∞

s=−i∞

Gc(s)esT − 1

s

1

q − esTds

note: correction for the exercise can be found at:http://www.control.isy.liu.se/ ljung/sysid/errata/


2G.4. Solution

If we express the system in terms of the Laplace transform we have

Y (s) = G(s)U(s)

where U(s) will be the Laplace transform of the constant step signal over the

period T

U(s) =

∫ T

0

u(t)e−st dt

which will lead to the solution

U(s) =1

se−st|(0,T )

=1 − e−sT

s

rewriting the expression for Y (s) we will have

Y (s) = Gc(s)1 − e−sT

s


To be able to use the discrete transform GT (q) it is needed first to take this

expression back into time domain

y(t) =1

2πi

∫ i∞

s=−i∞

Y (s)est ds

y(t) =1

2πi

∫ i∞

s=−i∞

Gc(s)1 − e−sT

sest ds

From this expression we can calculate GT (q) applying the definition of the

discrete transform

GT (q) =T∑

k=1

gT (k)q−k

=

T∑

k=1

1

2πi

∫ i∞

s=−i∞

Gc(s)1 − e−sT

seskT q−k ds


The expression only affects the last exponential term

GT (q) =1

2πi

∫ i∞

s=−i∞

Gc(s)1 − e−sT

s

T∑

k=1

eskT q−k ds

Analyzing the sumatory

T∑

k=1

eskT q−k =

T∑

k=1

(esT q−1

)k

If we apply the known expression

T∑

k=0

(aq−1

)k=

q

q − a

in the equation above

T∑

k=0

(esT q−1

)k= 1 +

T∑

k=1

(esT q−1

)k


As a result we will have that

T∑

k=1

(esT q−1

)k=

T∑

k=0

(esT q−1

)k − 1

=q

q − esT− 1

=esT

q − esT

Plugging the last expression in the general formula

GT (q) =1

2πi

∫ i∞

s=−i∞

Gc(s)1 − e−sT

s

esT

q − esTds


GT (q) =1

2πi

∫ i∞

s=−i∞

Gc(s)1 − 1

esT

s

esT

q − esTds

GT (q) =1

2πi

∫ i∞

s=−i∞

Gc(s)esT

−1esT

s

esT

q − esTds

which finally leads to the expression

GT (q) =1

2πi

∫ i∞

s=−i∞

Gc(s)esT − 1

s

1

q − esTds


2E.2. Suppose that {η(t)} and {ξ(t)} are two mutually independent

sequences of independent random variables with

Eη(t) = Eξ(t) = 0, Eη2(t) = λη, Eξ2(t) = λξ

Consider

ω(t) = η(t) + ξ(t) + γξ(t − 1)

Determine a MA(1) process

v(t) = e(t) + ce(t − 1)

where {e(t)} is white noise with

Ee(t) = 0, Ee2(t) = λe

such that {ω(t)} and {v(t)} have the same spectra; that is, find c and λe so

that Φv(ω) ≡ Φω(ω)


2E.2. Solution

If we want to achieve

Φv(ω) ≡ Φω(ω)

then the Fourier transform of both autocorrelation functions should be the

same∞∑

τ=−∞

Rv(τ )e−iτω =

∞∑

τ=−∞

Rω(τ )e−iτω

this reasoning leads to the fact that both autocorrelation functions should be

the same

Rv(τ ) ≡ Rω(τ )

or

Ev(t)v(t − τ ) ≡ Eω(t)ω(t − τ )


Separating the problem in two parts

Ev(t)v(t − τ ) = E {(e(t) + ce(t − 1)) (e(t − τ ) + ce(t − 1 − τ ))}

= E{

e(t)e(t − τ ) + ce(t)e(t − 1 − τ ) + ...

... + ce(t − 1)e(t − τ ) + c2e(t − 1)e(t − 1 − τ )}= Ee(t)e(t − τ ) + 2cEe(t)e(t − 1 − τ ) + ...

... + c2Ee(t − 1)e(t − 1 − τ )}

As we can see the expressions above represent only the Autocorrelation

functions of the white noise e(t) shifted or multiplied by a scalar

Ev(t)v(t − τ ) = Re(τ ) + 2cRe(τ ) + c2Re(τ )

= Re(τ )(1 + c)2

Under certain conditions we could conclude that

Re(τ ) = Ee2(t) = λe


Then the final expression for Rv(τ ) will be

Rv(τ ) = Ev(t)v(t − τ ) = λe(1 + c)2

to obtain a expression for Rω(τ ) we follow the same procedure

Eω(t)ω(t − τ ) = E{(η(t) + ξ(t) + γξ(t − 1))(η(t − τ ) + ...

... + ξ(t − τ ) + γξ(t − 1 − τ ))}

= E{

η(t)η(t − τ ) + η(t)ξ(t − τ ) + ...

... + γη(t)ξ(t − 1 − τ ) + ξ(t)η(t − τ ) + ...

... + ξ(t)ξ(t − τ ) + γξ(t)ξ(t − 1 − τ ) + ...

... + γξ(t − 1)η(t − τ ) + γξ(t − 1)ξ(t − τ ) + ...

... + γ2ξ(t − 1)ξ(t − 1 − τ ))}


Separating the function we will get

Eω(t)ω(t − τ ) = Eη(t)η(t − τ ) + Eη(t)ξ(t − τ ) + ...

... + γEη(t)ξ(t − 1 − τ ) + Eξ(t)η(t − τ ) + ...

... + Eξ(t)ξ(t − τ ) + γEξ(t)ξ(t − 1 − τ ) + ...

... + γEξ(t − 1)η(t − τ ) + γEξ(t − 1)ξ(t − τ ) + ...

... + γ2Eξ(t − 1)ξ(t − 1 − τ )

applying the data given we can reduce the system to

Eω(t)ω(t − τ ) = Eη(t)η(t − τ ) + Eξ(t)ξ(t − τ ) + ...

... + γEξ(t)ξ(t − 1 − τ ) + γEξ(t − 1)ξ(t − τ ) + ...

... + γ2Eξ(t − 1)ξ(t − 1 − τ )


Applying the same reasoning as before, the terms in the last expression

represent only the Autocorrelation functions of Rη(τ ) and Rξ(τ ), therefore

we can rewrite as

Eω(t)ω(t − τ ) = Rη(τ ) + Rξ(τ ) + 2γRξ(τ ) + γ2Rξ(τ )

= Rη(τ ) + Rξ(τ )(1 + γ)2

Under certain conditions we could conclude that

Rξ(τ ) = γξ

Rη(τ ) = γη

Then the final expression for Rω(τ ) will be

Rω(τ ) = Eω(t)ω(t − τ ) = γη + γξ(1 + γ)2


By the equivalence

Rv(τ ) ≡ Rω(τ )

we obtain

λe(1 + c)2 = γη + γξ(1 + γ)2

λe + 2cλe + cλ2e = λη + λξ + 2γλξ + γ2λξ

we can form the equalities

λe = λη + λξ

cλe = γλξ

And finally we can conclude that

λe = λη + λξ

c =λξ

λη + λξ

γ


2E.4. Consider the ”state space description”

x(t + 1) = fx(t) + ω(t)

y(t) = hx(t) + v(t)

where x, f , h, ω and v are scalars. {ω(t)} and {v(t)} are mutually

independent white Gaussian noises with variances R1 and R2, respectively.

Show that y(t) can be represented as an ARMA process:

y(t)+a1y(t−1)+· · ·+any(t−n) = e(t)+c1e(t−1)+· · ·+cne(t−n)

Determine n, ai, ci and the variance of e(t) in terms of f , h, R1 and R2.

What is the relationship between e(t), ω(t) and v(t)?


Taking the discrete transform of the state space model:

qX(q) = fX(q) + ω(q)

Y (q) = hX(q) + v(q)

X(q) =ω(q)

q − f

Y (q) =hω(q)

q − f+ v(q)

Y (q)(q − f) = hω(q) + (q − f)v(q)

yk+1 − fyk = hωk + vk+1 − fvk

what we want to do now is to make valid the next equivalence

hωk + vk+1 − fvk∼= αek+1 + βek


as we see in the ARMA process, we are missing the values of α and β, but

one fact that we have is that the ARMA process is monic, which means that

the term for α = 1, thus

yk+1 − fyk∼= ek+1 + βek

if we name the expression as follow:

ξk = hωk + vk+1 − fvk

ζ = ek+1 + βek

To fulfill the equivalence of the process we will find the next equivalences:

Eξk = Eζk = 0

Eξkξk = Eζkζk

Eξkξk−1 = Eζkζk−1


replacing the expression

E{hωk + vk+1 − fvk}{hωk + vk+1 − fvk} =

= E{ek+1 + βek}{ek+1 + βek}E{hωk + vk+1 − fvk}{hωk−1 + vk − fvk−1} =

= E{ek+1 + βek}{ek + βek−1}

making the internal operations for Eξkξk = Eζkζk we end up in

h2Eω2k + Ev2

k+1 + f2Ev2k = Ee2

k+1 + β2Ee2k

h2R1 + R2 + f2R2 = Re + β2Re

the expression for Eξkξk−1 = Eζkζk−1

−fEv2k = βEe2

k

−fR2 = βRe


We need to solve the next system of equations for β and Re

h2R1 + R2 + f2R2 = Re(1 + β2)

−fR2 = βRe

getting Re from the second equation

Re =−fR2

β

Replacing into the first equation

1 + β2

β= −h2R1 + R2 + f2R2

fR2

as we see the numerator of the right hand side will be always positive and

equal to some constant value, for simplicity we can rewrite the expression like:

1 + β2

β= −2C


where

2C =h2R1 + R2 + f2R2

fR2

the solution for β will be then

β = C ±√

C2 − 1

So finally we can write the ARMA process in the form

yk+1 − fyk = ek+1 + (C ±√

C2 − 1)ek

now we need to check the stability of the ARMA process, to do that we

analyze first the square root in the expression for β:

C2 − 1 > 0


which tell us that C has to be always −1 < C < 1 in order to avoid complex

values. As we see in the expression for 2C this requirement is going to be

fulfilled whit a good choice of the value f . Another requisite will be

|C ±√

C2 − 1| < 1

and again this decision can be taken with a good choice of the parameter f .

Clearly it can be seen in the process transfer function:

Y (q)

E(q)=

q + β

q − f

if the parameter f in the denominator is chosen in a way that |f | > 1 then

we will have an unstable system. This means that nothing can be said about

the sign that we must choose for the expression β in order to make the system

stable. The value for the variance of e(t) will be

Re =−fR2

C ±√

C2 − 1


2E.6. Consider the system

d

dty(t) + ay(t) = u(t)

Suppose that the input u(t) is piecewise constant over the sampling interval

u(t) = uk, kT ≤ t < (k + 1)T

(a) Derive a sample data system description for uk, y(kT ).

(b) Assume that there is a time delay of T seconds so that u(t) in the

expression is replaced by u(t − T ). Derive a sample data system

description for this case.

(c) Assume that the time delay is 1.5T so that u(t) is replaced by

u(t − 1.5T ). Then give the sampled data description.


2E.6. Solution

a). The solution to the differential equation which includes the effects of the

input and initial conditions is given by:

y(t) = e−a(t−t0)y(t0) +

∫ t

t0

e−a(t−τ)u(τ )dτ

where y(t0) is the initial condition on the state variables. Based on this

solution the sampled state response is given by

y[(k + 1)T ] = e−aT y[kT ] +

∫ T

τ=0

e−a(T −τ)u(τ + kT )dτ

we can write

y[(k + 1)T ] = Ay[kT ] + Bu[kT ]

where

A = e−aT , B =

∫ T

τ=0

e−a(T −τ)dτ =1 − e−aT

a


if we use Z transform

z(Y (z) − y[0]) = AY (z) + BU(z)

Y (z)(z − A) = BU(z) + zy[0]

Y (z) =B

z − AU(z) +

z

z − Ay[0]

or expressed in geometric series

y(t) =B

A

∞∑

k=0

Ak+1u(t − k) +∞∑

k=0

Akq−ky[0]


(b). Introducing the time delay into the system

d

dty(t) + ay(t) = u(t − T )

The solution to the differential equation after applying the delay in the

equation including the effects of the input and the initial conditions is given

by:

y(t) = e−a(t−t0)y(t0) +

∫ t

t0

e−a(t−τ)u(τ − T )dτ



y[(k + 1)T ] = e−aT y[kT ] +

∫ T

τ=0

e−a(T −τ)u(τ + kT − T )dτ

we can write

y[(k + 1)T ] = Ay[kT ] + Bu[kT − T ]


where

A = e−aT , B =

∫ T

τ=0

e−a(T −τ)dτ =1 − e−aT

a

if we use the Z transform

z(Y (z) − y[0]) = AY (z) + Bz−1U(z) + Bu[−1]

Y (z)(z − A) = Bz−1U(z) + Bu[−1] + zy[0]

Y (z) =Bz−1

z − AU(z) +

B

z − Au[−1] +

z

z − Ay[0]

Y (z) =B

z(z − A)U(z) +

B

z − Au[−1] +

z

z − Ay[0]


(c). Introducing the time delay into the system

d

dty(t) + ay(t) = u(t − 1.5T )

The solution to the differential equation after applying the delay in the

equation including the effects of the input and the initial conditions is given

by:

y(t) = e−a(t−t0)y(t0) +

∫ t

t0

e−a(t−τ)u(τ − 1.5T )dτ



y[(k + 1)T ] = e−aT y[kT ] +

∫ T

τ=0

e−a(T −τ)u(τ + kT − T )dτ

where

T = 1.5T


The sampled state response will be given by:

y[(k + 1)T ] = Ay[kT ] + Bu[kT − T ]

where the parameters A and B will be:

A = e−1.5aT

B =

∫ 1.5T

τ=0

e−a(1.5T −τ)dτ =1 − e−1.5aT

a

if we use the Z transform

z(Y (z) − y[0]) = AY (z) + Bz−1U(z) − Bu[−1]

Y (z)(z − A) = U(z)(Bz−1) + zy[0] − Bu[−1]

Y (z) =Bz−1

z − AU(z) +

z

z − Ay[0] − B

z − Au[−1]

Y (z) =B

z(z − A)U(z) +

z

z − Ay[0] − B

z − Au[−1]


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