# EE471 HW 9 HW 9A: 5.12, 5.13, 5.14 HW 9B: 5.11, 5.15 ```EE471
HW 9
HW 9A: 5.12, 5.13, 5.14
HW 9B: 5.11, 5.15
HW 9C: 5.16, 5.17, 5.18
5.12)
A)
E b / N 0=9.1 W/W=9.6 dB
B) S n ( f )=2.894&times;10−21 W/Hz
C)
−15
P S =5.267&times;10
W=−112.8 dBm
D) SNR= P S /P N =4.55 W/W=6.58 dB
5.13)
A)
dk
bk
(0)
1
0
1
0
1
1
0
0
0
1
1
0
0
0
1
0
B)
m(t)
t
C)
dk
5.14)
1
0
1
0
1
1
0
0
bk
(0)
0
1
1
0
0
0
1
0
ak
-1
-1
+1
+1
-1
-1
-1
+1
-1
akak-1
+1
-1
+1
-1
+1
+1
-1
-1
dk
1
0
1
1
1
1
0
0
E b / N 0 (BPSK)=6.915 W/W=8.40 dB
E b / N 0 (DPSK)=8.517 W/W=9.30 dB
difference=0.9 dB
5.11)
A) S x ( f )=
A2
2
A2
2
|G ( f − f c )| +
|G( f + f c )|
4T s
4Ts
B) B = 150 kHz, Nyquist bandwidth = 100 kHz
5.15)
For BPSK, Pe = 10-7 implies Eb / N0 = 13.52
For QPSK with the same BW, the pulse width Ts will be the same as for BPSK. Using the same
average power implies that Es / N0 for QPSK will also equal 13.52 where Es is the QPSK pulse
energy. Since each pulse contains two bits, Eb / N0 = (13.25)/2 for QPSK and Pe = 1.17 x 10-4.
Note that the binary rate for QPSK will be twice that for BPSK in this case.
If we kept the binary rate the same for QPSK and BPSK we could attain the same Pe at the same
average power level in 1/2 the bandwidth required for BPSK.
5.16)
Δ f =10 kHz ,
f 1=10.010 MHz ,
f 0=9.990 MHz
5.17)
Coherent: Eb / N0 = 22.56 = 13.53 dB
Noncoherent: Eb / N0 = 26.24 = 14.19 dB
Penaly: 0.66 dB
5.18)
A) B = 700 kHz
B) h = 1.5
C) η = 0.286 bits/s/Hz
D) B = 1 MHz
E) B = 1.1 MHz
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