H OMEWORK 2.1
I) Show that for the harmonic process H(t) = A cos (ωt + T )with A and T are independent
RVs, and T uniformly distributed in [−π, π] the following is true:
1 Rπ
g(T ) dT ;
(1) The expected value of a function g(T ) is E {g(T )} =
2π −π
(2) Mean: µH = 0;
1 2
(3) Variance: σH
= E |A|2 ;
2
1 2
(4) Auto-correlation: RHH (t1 , t2 ) = RHH (∆t) = E |A|2 cos ω∆t = σH
cos ω∆t;
2
Proof.
(1) By definition
Z
∞
1
E {g(T )} =
g(u) fT (u) du;
2π
−∞
Z
π
g(T ) dT,
−π
but since T uniformly distributed in [−π, π], also by definition, its PDF should be

1
−π ≤ u ≤ π
;
fT (u) =
2π
0
otherwise
Substituting, the expected value of g(T ) becomes
Z π
1
E {g(T )} =
g(T ) dT.
2π −π
(2) Since RVs A and T are independent (uncorrelated would also work) E {A g(T )} =
E {A} E {g (T )}, the mean is
Z π
1
µH = E {A cos (ωt + T )} = E {A} E {cos (ωt + T )} = E {A}
cos(ωt + T )dT,
2π −π
which, with the variable change θ = T + ωt, dθ = dT , T = ±π⇒ θ = ±π + ωt,
therefore
Z
E {A} π+ωt
E {A}
E {A}
µH =
cos θdθ =
[sin θ]π+ωt
[sin (π + ωt) − sin (−π + ωt)] ,
−π+ωt =
2π
2π
2π
−π+ωt
and using the trigonometric formula for the difference of sines
E {A}
2 sin π cos (ωt) = 0.
2π
(3) Using the same change of variables as before:
Z
2 2
E {A2 } π
2
2
σH (t) = E H (t) = E A cos (ωt + T ) =
cos2 (ωt + T )dT,
2π
−π
2 Z π+ωt
E {A }
2
σH
(t) =
cos2 θdθ.
2π
−π+ωt
µH =
1
An easy way to calculate the above integral is as follows: denote
Z π+a
1
cos2 θdθ,
I =
2π −π+a
Z π+a
1
J =
sin2 θdθ;
2π −π+a
then
Z π+a
Z π+a
1
1
2
2
cos θ + sin θ dθ =
dθ = 1,
I +J =
2π −π+a
2π −π+a
Z π+a
Z π+a
1
1
2
2
cos θ − sin θ dθ =
cos 2θdθ = 0;
I −J =
2π −π+a
2π −π+a
1
Therefore I = J = , hence the result.
2
(4) Auto-correlation:
RHH (t1 , t2 ) = E {H(t1 )H(t2 )} = E A2 cos (ωt1 + T ) cos (ωt2 + T ) =
Z π
2 1
= E A
cos(ωt1 + T ) cos(ωt2 + T )dT,
2π −π
where
1
2π
Z
π
Z
1
cos(ωt1 + T ) cos(ωt2 + T )dT =
2π
−π
π
−π
1
[cos(ω(t1 + t2 ) + 2T ) + cos ω(t1 − t2 )] dT ;
2
the integral of the firsty term is zero (see the calculation of the mean), and the
second term yields
Z π
Z π
1
1
1
1
1
cos ω(t1 − t2 )dT = cos ω(t1 − t2 )
dT = cos ω∆t,
2π −π 2
2
2π −π
2
with ∆t = t2 − t1 , thus
1 2
RHH (t1 , t2 ) = E A2 cos ω∆t = σH
cos ω∆t.
2
P
II) Let H(t) = N
n=1 Hn (t), where Hn (t) are harmonic processes, Hn (t) = An cos (ωn t + Tn )
with An and Tn mutually independent RVs, and Tn uniformly distributed in [−π, π] the
following is true. Show that:
(1) Processes Hn (t) are orthogonal, i.e.
hHn , Hm i = Rnm (t1 , t2 ) = Rnn (t1 , t2 )δnm
(2) The auto-correlation RHH (t1 , t2 ) =
of process Hn .
PN
n=1
2
(
Rnn (t1 , t2 ) for m = n
=
;
0
for m 6= n
σn2 cos ωn (t1 − t2 ), where σn2 is the variance
Proof.
(1) Orthogonality: from the definitions of the inner product and process H
hHn , Hm i = Rnm (t1 , t2 ) = E {H(t1 )H(t2 )} =
= E {An Am cos (ωn t1 + Tn ) cos (ωm t2 + Tm )} =
= E {An Am } E {cos (ωn t1 + Tn ) cos (ωm t2 + Tm )} .
Since for m 6= n Tn and Tm are independent, the mean factorizes into the means of
the factors
E {cos (ωn t1 + Tn ) cos (ωm t2 + Tm )} = E {cos (ωn t1 + Tn )} E {cos (ωm t2 + Tm )}
and the product cancels (see I.1), so in fact
E {cos (ωn t1 + Tn ) cos (ωm t2 + Tm )} = E {cos (ωn t1 + Tn )} E {cos (ωn t2 + Tn )} δnm ,
that is
hHn , Hm i = E {An Am } E {cos (ωn t1 + Tn )} E {cos (ωn t2 + Tn )} δnm ,
= E A2n E {cos (ωn t1 + Tn )} E {cos (ωn t2 + Tn )} δnm =
= Rnn (t1 , t2 )δnm .
(2) Auto-correlation: using the orthogonality of the processes
RHH (t1 , t2 ) =
N X
N
X
Rnn (t1 , t2 )δnm =
n=1 m=1
=
N
X
N
X
Rnn (t1 , t2 )
n=1
Rnn (t1 , t2 ) =
n=1
N
X
N
X
δnm
m=1
σn2 cos ωn ∆t.
n=1
III) Show that if a process X(t) is 2-order stationary and X 0 (t) is its derivative with respect
to time, for any given time t RVs X(t) and X 0 (t) are orthogonal and uncorrelated. Hint:
assume that you can exchange the order of the time differentiation and the expected value
operator.
Proof. Using the linearity of the mean
1 d 2 1 d 2
dX
dX
1d 2
X
=
E X =
σ =0
X,
=E X
=E
dt
dt
2 dt
2 dt
2 dt X
3