EE 511 Solutions to Problem Set 7
1. (i) SXˆ (f ) = |H(f )|2 SX (f ) where H(f ) = −jsgn(f ). Therefore, SXˆ (f ) = SX (f ).
(ii) Suppose h(τ ) is the impulse response of the Hilbert transformer, −h(τ ) = h(−τ )
is the impulse response of the inverse Hilbert transformer. Therefore, we have
RXX
ˆ (τ ) = RX (τ ) ⋆ h(τ )
and
RX Xˆ (τ ) = RXˆ (τ ) ⋆ (−h(τ )) = RX (τ ) ⋆ (−h(τ )) = −RXX
ˆ (τ ),
where ⋆ denotes convolution. From the above result, we have
SXX
ˆ (f ) = −SX X
ˆ (f ).
ˆ t ] = R ˆ (0) and E[Xt X
ˆ t ] = R ˆ (0). Therefore, we have E[Xt X
ˆ t ] = 0.
(iii)E[Xt X
XX
XX
(iv)
ˆ t+τ )(Xt − j X
ˆ t )]
RZ (τ ) = E[(Xt+τ + j X
= RX (τ ) − jRX Xˆ (τ ) + jRXX
ˆ (τ ) + RX
ˆ (τ )
= 2RX (τ ) + 2jRXX
ˆ (τ )
Therefore
SZ (f ) =
=
=
2SX (f ) + 2jSXX
ˆ (f )
2SX (f ) + 2jSX (f )[−jsgn(f )]
2SX (f )[1 + sgn(f )]
(
4SX (f ) f > 0
=
0 else
2. See Figure 1. Re[SX I X Q (f )] = 0.
Im[S I Q (f)]
XX
S I(f)
X
2
1/2
1/2
−2 −1
0
1
2
−2 −1
f
Figure 1:
3. See Figure 2. Re[SX I X Q (f )] = 0.
1
0
1
−1/2
2
f
−3
−1
0
S I(f)
X
Im[S I Q (f)]
XX
1
1
1
3
−3
f
−1
1
3
f
−1
Figure 2:
4.
ˆt+τ sin 2πfc (t + τ ))(Nt cos 2πfc t + N
ˆt sin 2πfc t)]
RN I (τ ) = E[(Nt+τ cos 2πfc (t + τ ) + N
= RN (τ ) cos 2πfc (t + τ ) cos 2πfc t + RN Nˆ (τ ) cos 2πfc (t + τ ) sin 2πfc t
+RNˆ N (τ ) sin 2πfc (t + τ ) cos 2πfc t + RNˆ (τ ) sin 2πfc (t + τ ) sin 2πfc t
Since RNˆ (τ ) = RN (τ ) and RNˆ N (τ ) = −RN Nˆ (τ ), we get
RN I (τ ) = RN (τ ) cos 2πfc τ + RNˆ N (τ ) sin 2πfc τ = RN (τ ) cos 2πfc τ + RˆN (τ ) sin 2πfc τ
Similarly, we have
ˆt+τ cos 2πfc (t + τ ) − Nt+τ sin 2πfc (t + τ ))(N
ˆt cos 2πfc t − Nt sin 2πfc t)]
RN Q (τ ) = E[(N
= RNˆ (τ ) cos 2πfc (t + τ ) cos 2πfc t − RNˆ N (τ ) cos 2πfc (t + τ ) sin 2πfc t
−RN Nˆ (τ ) sin 2πfc (t + τ ) cos 2πfc t + RN (τ ) sin 2πfc (t + τ ) sin 2πfc t
= RN (τ ) cos 2πfc τ + RNˆ N (τ ) sin 2πfc τ
= RN (τ ) cos 2πfc τ + RˆN (τ ) sin 2πfc τ
ˆt+τ sin 2πfc (t + τ ))(N
ˆt cos 2πfc t − Nt sin 2πfc t)]
RN I N Q (τ ) = E[(Nt+τ cos 2πfc (t + τ ) + N
= RN Nˆ (τ ) cos 2πfc (t + τ ) cos 2πfc t − RN (τ ) cos 2πfc (t + τ ) sin 2πfc t
+RNˆ (τ ) sin 2πfc (t + τ ) cos 2πfc t − RNˆ N (τ ) sin 2πfc (t + τ ) sin 2πfc t
= RN (τ ) sin 2πfc τ − RNˆ N (τ ) cos 2πfc τ
= RN (τ ) sin 2πfc τ − RˆN (τ ) cos 2πfc τ
RN Q N I (τ ) = RN I N Q (−τ ) = −RN (−τ ) sin 2πfc τ − RNˆ N (−τ ) cos 2πfc τ
= −RN (τ ) sin 2πfc τ + RNˆ N (τ ) cos 2πfc τ = −RN I N Q (−τ )
2
5. Nt = Wt cos (2πfc t + Θ) − Wt sin (2πfc t + Θ).
RN (τ ) = E[(Wt+τ cos (2πfc (t + τ ) + Θ) − Wt+τ sin (2πfc (t + τ ) + Θ))
(Wt cos (2πfc t + Θ) − Wt sin (2πfc t + Θ))]
= RW (τ )E[cos (2πfc (t + τ ) + Θ) cos (2πfc t + Θ)]
−RW (τ )E[cos (2πfc (t + τ ) + Θ) sin (2πfc t + Θ)]
−RW (τ )E[sin (2πfc (t + τ ) + Θ) cos (2πfc t + Θ)]
+RW (τ )E[sin (2πfc (t + τ ) + Θ) sin (2πfc t + Θ)]
= RW (τ ) cos 2πfc τ − RW (τ )E[sin (2πfc (2t + τ ) + 2Θ)]
= RW (τ ) cos 2πfc τ
SN (f ) =
1
[SW (f − fc ) + SW (f + fc )] .
2
S (f)
N
a/2
−fc −W −fc
−fc +W
f − W fc
c
Figure 3:
6. (a) SX Q (f ) = SX I (f ) and SX Q X I (f ) = −SX I X Q (f ).
(b) See Figure 6c.
(c) See Figure 6c.
7. (a) See Figure 4.
(b) See Figure 4.
(c) SX Q (f ) = SX I (f ) and SX Q X I (f ) = −SX I X Q (f ).
3
fc + W
f
S (f)
X
1
−14
−11 −9
0
9 11
14
f (MHz)
SX˜ (f )
4
−1
−3
−1
0
0
1
4
f
S I(f)
X
Im[S I Q (f)]
XX
1
1
1
3
−3
f
−1
1
−1
Figure 4:
4
3
f

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