Homework One
1. Calculate the channel capacity if the bandwidth is 1MHz and signal to noise ratio (S/N) is 127
with Shannons theorem.
[Answer]: data rate = 106 log2 (1 + 127) = 7000000 bps.
2. Usually we say a sine wave is of frequency fHz if it is can be written as sin(2πf t). Let
x(t) = sin(3πf t)cos(πf t). If x(t) is sent to a low-pass filter with cut-off frequency of 3f /2Hz,
meaning that all signals on frequency 3f /2Hz or higher will be filtered out, what is the output
of the filter?
[Answer]: x(t) = sin(3πf t)cos(πf t) =
only 21 sin(2πf t) will be left.
1
2 sin(4πf t)
+ 21 sin(2πf t). After passing the LPF,
3. Suppose there is a communication link where the receiver receives y = x + n, where x takes
−1 and 1 with equal probability and n is the noise. If | n |≤ 0.5, how would you design the
detector? Describe your detector by stating the output (either 0 or 1) for what values of y. If the
noise takes value from {−1.5, −0.5, 0.5, 1.5} with probability {0.4, 0.1, 0.1, 0.4}, respectively,
how would you design the detector? What is the probability that your detector makes a mistake?
[Answer]: If | n |≤ 0.5, if y < 0, output 0, else output 1. If the noise is the second case, y can
take values −2.5, −1.5, −0.5, 0.5, 1.5, 2.5. When y = −2.5, −1.5, output 0. When y = 2.5, 1.5,
output 1. No ambiguity here. When y = −0.5, there are two cases. First, x = −1 and n = 0.5,
which occurs with probability 0.5 × 0.1 = 0.05. Second, x = 1 and n = −1.5, which occurs
with probability 0.5 × 0.4 = 0.2. So, when receiving y = −0.5, we should output 1. Similarly,
when receiving y = 0.5, we should output 0. The detector will make mistakes with probability
0.1.
4. Consider the basic wireless communication system, where the sender sends I(t)cos(2πf t) and
the receiver multiplies the received waveform with its local carrier and then sends the signal to a
low pass filter (which will pass the baseband signals but not signals on higher frequencies.)
• If the receiver’s local carrier is not synchronized with the sender’s carrier and is cos(2πf t+
φ) where φ is the phase difference, please write down the signal after the low pass filter. If
φ = π/2, can the receiver receive any baseband signal?
• What if the receiver uses two receiving channels, one multiplying the input by cos(2πf t +
φ) and the other by sin(2πf t + φ)? If φ = π/2, can the receiver receive any baseband
signal?
[Answer]: 1. Note that
I(t)cos(2πf t)cos(2πf t + φ) = 1/2I(t)[cos(4πf t + φ) + cos(φ)]
1
r(t)
I(t)
LPF (B)
cos(2πft )
cos(2πft )
Figure 1: Problem 4.
After the low pass filter, the signal is I(t)cos(φ). If φ = π/2, the receiver receives nothing.
2. The receiver gets I(t)cos(φ) in the I channel and I(t)sin(φ) on the Q channel. Yes.
5. If to minimize the bit error ratio, which assignments of bits to the QPSK constellation makes
more sense, why?
01
00
01
00
11
10
10
11
Figure 2: Problem 5.
[Answer]: The one on the left, because the adjacent points differ by only one bit.
6. The (7,4) Hamming code we discussed in class has minimum distance 3. Using this fact, prove
that no two rows in the parity checking matrix H are the same. (Hint: Recall that the minimum
distance is the minimum weight of codewords, and any codeword multiplied with H is 0.)
[Answer]: If there are two rows in H that are equal to each other, there will be a codeword with
only two 1s.
7. the (7,4) cyclic we discussed in class has generator polynomial g(x) = x3 + x + 1. Find the
code polynomial for m(x) = x2 + x when C(x) = m(x)g(x). Find the code polynomial for the
same m(x) when using systematic code.
[Answer]: In the first case, C(x) = m(x)g(x) = x5 + x4 + x3 + x. In the second case,
r(x) = m(x)x3 mod g(x) = 1, so C(x) = x5 + x4 + 1.
2