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ECE 562 Fall 2008 Solutions to Midterm 1
- (a) i. A total number of k + 1 distinct voltages can be sent. This corresponds to a data rate of dlog 2 (1 + k)e bits per unit time. (1) ii. We can try putting different energies at different times, say Em at time m. The average power constraint says that
∑^ N
m=
Em ≤ N E. (2)
Now the number of distinct voltages at time m is,
Em σ
So the data rate is
1 N
log 2
N +
∑^ N
m=
Em σ
bits per unit time. (4)
Now, the logarithm is a so-called concave function (one that has the property of diminishing returns).So,
1 N
log 2
N +
∑^ N
m=
xm
≤ log 2
N
∑^ N
m=
xm
Substituting xm =
Em σ
into Equation (5) we see that the data rate with the average power constraint is not better than the one with peak power constraint (cf. Equation (1)). They are the same only when we choose
Em = E, m = 1... N. (7)
(b) i. The answer does not change since we are still asking for exactly fully reliable communication. ii. If we allow for a small probability of making an error (however small), then we have a chance at improving the data rate.
- (a) The capacity is W 2
log 2
2 P¯
N 0 W
bits/second. (8)
(b) The capacity is W 2
log 2
2 L P¯
N 0 W
bits/second. (9)
The point is that with L receive antennas, we are in a situation as if we got each symbol repeated L times. We can do a simple average to reduce the noise variance by a factor of L. This improves the SNR by a factor of L. (c) L receive antennas boost the SNR by a factor of L. Boosting SNR is critical at low SNR values. On the other hand, at high SNR values, this boost is of not much use – much better to go for an increase in the bandwidth.
- (a) Correct. This follows from the log expression (cf. Lecture 8).
(b) Incorrect. At low SNRs, the capacity gets doubled when the SNR is doubled. But since the capacity at low SNRs is already much lesser than 1, it is not at all likely the capacity will increase by 1 bit per channel use. (c) i. Correct. ii. Incorrect. Turns out that the capacity is a sharp threshold: at rates above capacity, communication of a large packet is nearly certainly unreliable. (d) Incorrect. Energy efficient communication occurs at very low powers (and corre- spondingly low data rates).
- (a) There are really many codes that work. A simple one is the parity check code. We set
xk= 1 , k = = 1... 4. (10) xk= 1 , k = 5, = 1... 4. (11) xk` = 0 , else. (12)
Please note that I am looking for an explicit answer; this is a simple setting after all. (b) For the code proposed above, a simple decoding algorithm is the following:
- Use the unerased c 1 ,... , c 4 to decide right away on the corresponding infor- mation bits (these depend on only one information bit each).
- If one of c 1 ,... , c 4 is erased, the corresponding information bit can be inferred by taking the binary sum (modulo 2) of all the unerased coded bits c 1 ,... , c 5.
