University of Illinois
Spring 2016
ECE 361: Problem Set *: Problems and Solutions
ISI practice
Released:
Due:
Reading:
Tuesday, January 19
Thursday, January 28 (in class)
361 Course Notes Lectures 1-3, and S1.
1. [Matched Filter Performance]
by Yu Meng
We have a 2-tap ISI channel
y[m] = h0 x[m] + h1 x[m − 1] + w[m], m ≥ 1
where w[m] ∼ N (0, σ 2 ). Define SNR = E/σ 2 .
√
(a) Suppose we only transmit one symbol (i.e. x[1] = ± E equiprobably and x[m] = 0 for any
m 6= 1). Use matched filter receiver strategy to detect the symbol transmitted. Specify the ML
decision rule and compute the corresponding error probability.
Solution: The matched filter scheme yields
ŷ[1] = h0 y[1] + h1 y[2]
= (h20 + h21 )x[1] + h0 w[1] + h1 w[2].
ML rule:
√
ŷ[1]
E
R
√
− E
0.
The probability of error
√ 1
√ 1
P r ŷ[1] > 0x[1] = − E + P r ŷ[1] < 0x[1] = E
2
2
q
=Q
(h20 + h21 )SNR .
P r[ε] =
√
(b) Suppose we send two symbols (i.e. x[1], x[2] = ± E equiprobably and x[m] = 0 for any m 6= 1, 2).
Does the ML decision rule differ that in (a)? Compute the error probability. Does the error
increase or decrease? Explain your answer intuitively.
Solution: The ML decision rule does not change. By matched filter
ŷ[1] = h0 y[1] + h1 y[2]
= (h20 + h21 )x[1] + h0 h1 x[2] + h0 w[1] + h1 w[2].
ML rule:
√
ŷ[1]
E
R
√
− E
0.
The probability of error
√ 1
√ √
√
1
P r[ε] = P r ŷ[1] > 0x[1] = − E, x[2] = − E + P r ŷ[1] < 0x[1] = E, x[2] = − E
4
4
√ 1
√ √
√
1
+ P r ŷ[1] > 0x[1] = − E, x[2] = E + P r ŷ[1] < 0x[1] = E, x[2] = E
4
4
!
!
2
2
2
√
1
h0 + h1 + h0 h1
1
h0 + h21 − h0 h1 √
p
p
= Q
SNR + Q
SNR .
2
2
h20 + h21
h20 + h21
The error increases because x[2] introduces extra interference.
2. [ISI Channel with Nonuniform Prior]
by Alican Akman
Consider the same 2-tap ISI channel as above.
√
(a) Suppose
only one symbol is sent, but with non-uniform probability. x[1] = E w.p. 3/4, x[1] =
√
− E w.p. 1/4. Use matched filter receiver strategy to detect the symbol transmitted. Specify
the ML decision rule and compute the corresponding error probability.
Solution: The matched filter scheme yields
ŷ[1] = h0 y[1] + h1 y[2]
= (h20 + h21 )x[1] + h0 w[1] + h1 w[2].
The conditional PDF is
√
√
E) = N (h20 + h21 ) E, (h20 + h21 )σ 2
√
√
p(ŷ[2]|x[1] = − E) = N −(h20 + h21 ) E, (h20 + h21 )σ 2 .
p(ŷ[2]|x[1] =
According to ML rule,
√
√
p(ŷ[2]|x[1] =
E)
E
R
√
− E
which simplifies to
√
p(ŷ[2]|x[1] = − E),
√
ŷ[1]
E
R
√
− E
0.
The probability of error
√
√
1
3
P r[ε] = P r ŷ[1] > 0x[1] = − E + P r ŷ[1] < 0x[1] = E
4
4
q
=Q
(h20 + h21 )SNR .
(b) Suppose we still use matched filter receiver strategy to detect the symbol transmitted, but with
MAP decision rule. Specify the MAP decision rule and compute the corresponding error probability. Is the error probability smaller or larger? Why?
Hint: Write down the conditional PDF of ŷ[1] and follow the standard MAP procedure. The
MAP decision rule will produce a threshold on ŷ[1] which is not 0.
Solution: The matched filter scheme yields
ŷ[1] = h0 y[1] + h1 y[2]
= (h20 + h21 )x[1] + h0 w[1] + h1 w[2].
The conditional PDF is
√
√
E) = N (h20 + h21 ) E, (h20 + h21 )σ 2
√
√
p(ŷ[2]|x[1] = − E) = N −(h20 + h21 ) E, (h20 + h21 )σ 2 .
p(ŷ[2]|x[1] =
According to MAP rule,
√
E 1
√
√
3
R
p(ŷ[2]|x[1] = E) √
p(ŷ[2]|x[1] = − E),
4
− E 4
2
which simplifies to
√
ŷ[1]
E
R
√
− E
−σ 2 ln 3
√
.
E
The probability of error
√
√
1
3
−σ 2 ln 3 −σ 2 ln 3 √
x[1]
=
−
x[1]
=
P r[ε] = P r ŷ[1] > √
E
+
E
P
r
ŷ[1]
<
4
4
E E !
!
q
q
ln 3
3
ln 3
1
2
2
2
2
+ Q
.
= Q
(h0 + h1 )SNR − p 2
(h0 + h1 )SNR + p 2
4
4
(h0 + h21 )SNR
(h0 + h21 )SNR
The error probability is smaller because MAP minimizes the error probability.
3. [Choosing Equalizing Schemes]
by Ryan Stark
TRUE OR FALSE: As far as receiver-side approaches go, matched filter performs the best in low SNR,
the ZFE performs best at high SNR and MMSE performs best when SNR is in between.
Solution: FALSE, MMSE will produce the best result (lowest SINR) across different SNR.
4. [Whitening a 4-Tap ISI Channel]
by Xuanzhen Cao
Consider the following 4-tap ISI channel
1
1
1
y[m] = x[m] + x[m − 1] + x[m − 2] + x[m − 3] + w[m], m ≥ 1.
3
4
5
√
We send 2 bits through sequential signaling, i.e. x[1], x[2] = ± E equiprobably, and x[m] = 0, ∀m 6=
1, 2. w[m] are i.i.d N (0, σ 2 ) and indenpedent of the transmitted signal.
Suppose we want to detect x[1] from y[1] through y[4] with an MMSE equalizer. Find a linear transform
of y[m] that whiten the noise and interference.
Solution: Let
y[1] = x[1] + z[m]
1
y[2] = x[1] + z[2]
3
1
y[3] = x[1] + z[3]
4
1
y[4] = x[1] + z[4],
5
where
z[1] = w[1]
z[2] = x[2] + w[2]
1
z[3] = x[2] + w[3]
3
1
z[4] = x[2] + w[4].
4
z[2], z[3] and z[4] are all correlated. Let
ŷ[3] = y[3] − a32 y[2],
such that
ẑ[3] = z[3] − a32 z[2]
3
is uncorrelated with z[2].
a32 =
cov(z[3], z[2])
SNR/3
E/3
=
=
.
2
var(z[2])
E+σ
SNR + 1
Let
ŷ[4] = y[4] − a43 ŷ[3] − a42 y[2],
such that
ẑ[4] = z[4] − a43 ẑ[3] − a42 z[2]
is uncorrelated with ẑ[3] and z[2].
SNR/4
E/4
cov(z[4], z[2])
=
=
var(z[2])
E + σ2
SNR + 1
cov(z[4], ẑ[3])
SNR/4 · (1/3 − a42 )
E/4 · (1/3 − a42 )
=
=
=
.
var(ẑ[3])
(1/3 − a42 )E + σ 2
(1/3 − a42 )SNR + 1
a42 =
a43
4