Prokis Solution Manual

Problem 8.17

a) If the power spectral density of the additive noise is Sn(f ), then the PSD of the noise at the
output of the prewhitening filter is

Sν(f ) = Sn(f )|Hp(f )|2
In order for Sν(f ) to be flat (white noise), Hp(f ) should be such that

1
Hp(f ) = Sn(f )

2) Let hp(t) be the impulse response of the prewhitening filter Hp(f ). That is, hp(t) = F −1[Hp(f )].
Then, the input to the matched filter is the signal s˜(t) = s(t) hp(t). The frequency response of
the filter matched to s˜(t) is

S˜m(f ) = S˜∗(f )e−j2πft0 == S∗(f )Hp∗(f )e−j2πft0

where t0 is some nominal time-delay at which we sample the filter output.

3) The frequency response of the overall system, prewhitening filter followed by the matched filter,

is S∗(f ) e−j2πft0
Sn(f )
G(f ) = S˜m(f )Hp(f ) = S∗(f )|Hp(f )|2e−j2πft0 =

4) The variance of the noise at the output of the generalized matched filter is

σ2 = ∞ ∞ |S(f )|2
df
Sn(f )|G(f )|2df =
−∞ Sn(f )
−∞

At the sampling instant t = t0 = T , the signal component at the output of the matched filter is

∞∞
Y (f )ej2πfT df =
y(T ) = s(τ )g(T − τ )dτ
=
−∞ −∞

∞ S∗(f ) ∞ |S(f )|2
−∞ S(f ) Sn(f ) df = −∞ Sn(f ) df

Hence, the output SNR is

y2(T ) ∞ |S(f )|2
SNR = = df
σ2 −∞ Sn(f )

Problem 8.18
The bandwidth of the bandpass channel is

W = 3300 − 300 = 3000 Hz

In order to transmit 9600 bps with a symbol rate R = 2400 symbols per second, the number of
information bits per symbol should be

k = 9600 = 4
2400

Hence, a 24 = 16 QAM signal constellation is needed. The carrier frequency fc is set to 1800 Hz,
which is the mid-frequency of the frequency band that the bandpass channel occupies. If a pulse

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with raised cosine spectrum and roll-off factor α is used for spectral shaping, then for the bandpass
signal with bandwidth W

R = 1200(1 + α) = 1500
and

α = 0.25
A sketch of the spectrum of the transmitted signal pulse is shown in the next figure.

1/2T

-3300 -1800 -300 300 600 1800 3300 f
3000

Problem 8.19

The channel bandwidth is W = 4000 Hz.

(a) Binary PSK with a pulse shape that has α= 1 . Hence
2

1
(1 + α) = 2000

2T

and 1 = 2667, the bit rate is 2667 bps.
T
1 1
(b) Four-phase PSK with a pulse shape that has α = 2 . From (a) the symbol rate is T = 2667 and

the bit rate is 5334 bps.

(c) M =8 QAM with a pulse shape that has α= 1 . From (a), the symbol rate is 1 = 2667 and
2 T
3
hence the bit rate T = 8001 bps.

(d) Binary FSK with noncoherent detection. Assuming that the frequency separation between the

two frequencies is ∆f = 1 , where 1 is the bit rate, the two frequencies are fc + 1 and fc − 1 .
T T 2T 2T
1 1
Since W = 4000 Hz, we may select 2T = 1000, or, equivalently, T = 2000. Hence, the bit rate is

2000 bps, and the two FSK signals are orthogonal.

(e) Four FSK with noncoherent detection. In this case we need four frequencies with separation

of 1 between adjacent frequencies. We select f1 = fc − 1.5 , f2 = fc − 1 , f3 = fc + 1 , and
T T 2T 2T
1.5 1 1
f4 = fc + T , where 2T = 500 Hz. Hence, the symbol rate is T = 1000 symbols per second and

since each symbol carries two bits of information, the bit rate is 2000 bps.

(f) M = 8 FSK with noncoherent detection. In this case we require eight frequencies with frequency

separation of 1 = 500 Hz for orthogonality. Since each symbol carries 3 bits of information, the
T

bit rate is 1500 bps.

Problem 8.20
1) The bandwidth of the bandpass channel is

W = 3000 − 600 = 2400 Hz

Since each symbol of the QPSK constellation conveys 2 bits of information, the symbol rate of

transmission is 2400
R = = 1200 symbols/sec

2

Thus, for spectral shaping we can use a signal pulse with a raised cosine spectrum and roll-off factor

α = 1, that is

Xrc(f ) = T [1 + cos(πT |f |)] = 1 cos2 π|f |
2 2400 2400

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