Determine the output y[n] for a systems with the input x[n]
and unit-sample response h[n] shown below. Assume h[n]=0 and
x[n]=0 for any times n not shown.
Problem 2.
A discrete-time linear system produces output v when the input is the
unit step u. What is the output h when the input is the unit-sample
δ? Assume v[n]=0 for any times n not shown below.
A signal is transmitted at 20 samples/bit and 8 bits/symbol.
How many voltage samples are sent for each symbol?
160 samples/symbol.
To convert the samples to bits, where should the receiver
sample? The first half of the samples? Or the second half of the
samples? Why?
Signal transitions over most channels take at least several sample times
to complete; for slow channels transitions may take many sample times.
So it is better to sample at the latter half of the samples, but not too
close to the end since slow transitions make us uncertain about exactly
where the sample sequence for a particular bit started, which
means we are also uncertain as to where the sequence ends.
According to your response in part b), pick a sample number
(out of the 20 samples/bit) at which to sample. A list of samples,
representing exactly one 8-bit character, has been transmitted. Write a python
procedure receive that converts the list of samples into an 8-bit
list representation of the 8 bits. Try to do this without using a for
loop. Hint: List comprehension.
def receive(samples, samples_per_bit, bits_per_symbol):
index = samples_per_bit*3/4
return samples[index::samples_per_bit]
Problem 4.
The following figure show plots of several received waveforms. The transmitter is
sending sequences of binary symbols (i.e., either 0 or 1) at some
fixed symbol rate, using 0V to represent 0 and 1V to represent 1.The
horizontal grid spacing is 1 microsecond (1e-6 sec).
Answer the following questions for each plot:
Find the slowest symbol rate that is consistent with the
transitions in the waveform.
Using your answer in question 1, what is the decoded bit string?
a) 1e6 symbols/sec, 101010101010101010
b) 333,333 symbols/sec, 101101
c) 1e6 symbols/sec, 101100111001110110
d) 5e5 symbols/sec, 100010001
Problem 5.
Ben Bitdiddle is doing a 6.02 lab on understanding the effect of
noise on data receptions, and is confused about the following
questions. Please help him by answering them.
In these questions, assume that:
The sender sends 0 Volts for a "0" bit and 1 Volt for a "1" bit
P_ij = Probability that a bit transmitted as "i" was received as a "j" bit (for all four combinations of i and j, 00, 01, 10, 11)
alpha = Probability that the sender sent bit 0
beta = Probability that the sender sent bit 1
and, obviously, alpha + beta = 1
The channel has non-zero random noise, but unless stated otherwise,
assume that the noise has 0 mean and that it is a Gaussian with finite
variance. The noise affects the received samples in an additive
manner, as in the labs you've done.
Which of these properties does the bit error rate of this channel depend on?
The voltage levels used by the transmitter to send "0" and "1"
The variance of the noise distribution
The voltage threshold used to determine if a sample is a "0" or a "1"
The number of samples per bit used by the sender and receiver
In general the bit error rate is a function of both noise and inter-symbol interference.
The noise is a function of
Φ[(vth - (signal_level + noise_mean))/noise_sigma] multiplied as appropriate
by alpha or beta. So the bit error rate clearly depends on the signal level,
the mean and variance of the noise and the digitization threshold.
The number of samples per bit doesn't enter directly into the bit
error calculation, but more samples per bit gives each transition more
time to reach its final value, reducing inter-symbol interference. This
means that the eye will be more open. In the presence of noise,
a wider eye means a lower bit error rate.
Suppose Ben picks a voltage threshold that minimizes the bit
error rate. For each choice below, determine whether it's true or false.
P_01 + P_10 is minimized for all alpha and beta
alpha * P_01 + beta * P_10 is minimized
P_01 = P_10 for all alpha and beta
if alpha > beta then P_10 > P_01
The voltage threshold that minimizes BER depends on the noise variance if alpha = beta
(b) is the definition of bit error rate, so that's clearly minimized. Thus
(a) is only minimized if alpha = beta.
The magnitude of the BER is, of course,
a function of the noise variance, but for a given noise variance, if alpha = beta,
the minimum BER is achieved by setting the digitization threshold at 0.5. So
(e) is false.
As we saw in PSet #3, when alpha ≠ beta, the noise is minimized when
the digitization threshold moves away from the more probable signal. Suppose
alpha > beta. The digitization threshold would increase so P_01 would get
smaller and P_10 larger. So (c) is not true and (d) is true.
Suppose alpha = beta. If the noise variance doubles, what
happens to the bit error rate?
When alpha = beta = 0.5, the minimum BER is achieved when the
digitization threshold is half-way between the signaling levels, i.e.,
0.5V. Using Φ(x), the cumulative distribution function for the unit normal
PDF, we can write the following formula for the BER:
BER = 0.5*(1 - Φ[.5/σ]) + 0.5*Φ[-.5/σ]
= Φ[-.5/σ]
Doubling the noise variance is the same as multiplying σ by
sqrt(2), so the resulting BER would be
BERnew = Φ[-.5/(sqrt(2)*σ)]
The change in the bit error rate is given by BERnew - BER.
Problem 6.
If a transmitter can generate signals between 0V and 5V, how many
digital signaling levels can we have if the noise limit is ±.2V and
the receiver requires a .1V forbidden zone for each thresholding
operation that it needs to implement?
Our digital signaling convention requires a ±.2V region around each
signaling level and the regions must be separated by a .1V forbidden zone.
That means we must have .2+.1+.2 = .5V between signaling levels. With
a 5V signaling range, we can get (5/.5)+1=11 digital signaling levels.
Problem 7.
In this problem, we will understand the impact of imperfect clocks and
clock drift on clock recovery in communication systems. A transmitter and
receiver are communicating using a perfect communication channel (that is,
ignore the effects of ISI and noise for now). The transmitter transmits
every bit using symbols of duration 10 samples each. The receiver also tries
to sample the channel at the same frequency as the transmitter to obtain 10
samples for each symbol, and then digitizes the 10th sample of the symbol to
recover the transmitted bit. However, the non-ideal clocks at the
transmitter and receiver "drift away" from each other, as a result of which
the receiver ends up sampling the channel 10 times in only 9 sample periods
of the transmitter. To cope with this clock drift, the receiver resets its
sample counter on transitions from 0 to 1 or vice versa. What is the maximum
number of contiguous 0s or 1s that the data can have (i.e., the maximum
number of bits without transitions in between) before one starts to see bit
errors due to the clock drift?
One should have no more than 9 consecutive bits without a
transition in order not to see bit errors. Bit #10 at receiver = Sample #100
at receiver = Sample #90 at sender = last sample of bit #9 at transmitter =>
bit error.
Problem 8.
Consider the problem of synchronization to identify the start of a
message between a transmitter and receiver. Suppose the designers of the
communication system decide to use the sequence of bits '01111110' to signal
the start of a message. What property should the designers enforce on the
data to ensure correct synchronization?
Care should be taken to ensure that the synchronization pattern
does not appear in the data. For example, one can get around this problem
encoding the data in such a way that the synchronization pattern does not
appear in the data stream (e.g., 8b/10b).
Problem 9.
The output of a particular communication channel is given by
y[n] = αx[n] + βx[n-1] where α > β
Is the channel linear? Is it time invariant?
To be linear the channel must meet two criteria:
if we scale the inputs x[n] by some factor k,
the outputs y[n] should scale by the same factor.
if we get y1[n] with inputs x1[n]
and y2[n] with inputs x2[n], then we
should get y1[n] + y2[n] if the input is
x1[n] + x2[n].
It's easy to verify both properities given the channel response above, so the
channel is linear.
To be time invariant the channel must have the property that if we shift
the input by some number of samples s, the output also shifts by
s samples. Again that property is easily verified given the channel
response above, so the channel is time invariant.
What is the channel's unit-sample response h?
The unit-sample input is x[0]=1 and x[n]=0 for n≠0.
Using the channel response given above, the channel's unit-sample reponse
can be computed as
h[0]=α,
h[1]=β,
h[n]=0 for all other values of n
If the input is the following sequence of samples starting at time
0:
x[n] = [1, 0, 0, 1, 1, 0, 1, 1], followed by all 1's.
then what is the channel's output assuming α=.7 and
β=.3?
Convolving x[n] with h[n] we get
y[n] = [.7, .3, 0, .7, 1, .3, .7, 1], followed by all 1's.
Again let α=.7 and β=.3.
Derive a deconvolver for this channel and compute the input sequence
that produced the following output:
y[n] = [.7, 1, 1, .3, .7, 1, .3, 0], followed by all 0's.
Each of the following eye diagrams is associated with transmitting bits
using one of the four wires, where five samples were used per bit.
That is, a one bit is five one-volt samples and a zero bit
is five zero-volt samples. Please determine which wire was used
in each case.
The eye diagram is from h2. Note that the signal transitions take three samples
to complete and that the transitions occur in 3 steps with a larger slope
in the middle step, an indicator of a response with 3 taps with a larger middle
tap.
The eye diagram is from h4. Note that the signal transitions take more than five samples
to complete and hence result in considerable inter-symbol interference. Response h4
is the only response that's non-zero for more than 5 taps.
The eye diagram is from h1. Note that the signal transitions take four samples
to complete and that the transitions have constant slope, an indicator of
a response with 4 equal taps.
The eye diagram is from h3. Note that the signal transitions take five samples
to complete and that the transitions occur in 5 steps with larger slopes
in the middle of the transition, an indicator of a response with 5 taps with larger middle
taps.
Problem 11.
Consider the second of the four eye diagrams in Problem 10. Determine the eight unique voltage values
for sample number 8.
If we convolve h4 with a unit step we get the unit-step response:
Now if we consider all possible values of the current bit and the previous
two bits (listed earliest-to-latest in the table below) we can use superposition
of hstep to compute the possible values at sample time 8, which is the same
as asking for y[12]. Note that you have to think about how the sample
numbers in the eye diagram align with the sample numbers of the bits -- the
eye diagram is not necessary aligned with bit boundaries (e.g., it isn't
in this case).
bits
decomposed into unit steps
computation for y[12]
1 1 1
u[n]
y[12] = hstep[12] = 1
1 1 0
u[n] - u[n-10]
y[12] = hstep[12] - hstep[2] = 1 - .12 = .88
1 0 1
u[n] - u[n-5] + u[n-10]
y[12] = 1 - .84 + .12 = .28
1 0 0
u[n] - u[n-5]
y[12] = 1 - .84 = .16
0 1 1
u[n-5]
y[12] = .84
0 1 0
u[n-5] - u[n-10]
y[12] = .84 - .12 = .72
0 0 1
u[n-10]
y[12] = .12
0 0 0
y[12] = 0
So the eight unique values for y[12] are 0, .12, .16, .28, .72, .84, .88 and 1.
Problem 12.
Messages are transmitted along a noisy channel using the following protocol:
a "0" bit is transmitted as -0.5 Volt and a "1" bit as 0.5 Volt.
The PDF of the total noise added by the channel, H, is shown below.
Compute H(0), the maximum value of H.
The area under the PDF is 1, so (0.5)*H(0)*(1+0.5) = 1 from which we get H(0) = 4/3.
It is known that a "0" bits 3 times as likely to be transmitted
as a "1" bit. The PDF of the message signal, M, is shown below. Fill
in the values P and Q.
We know that P=3Q and that P+Q=1, so P=0.75 and Q=.25.
If the digitization threshold voltage is 0V, what is the bit error rate?
The plot below shown the PDF of the received voltage in magenta. For a threshold
voltage of 0, there is only one error possible: a transmitted "0" received as a
"1". This error is equal to the area of the triangle formed by the dotted black
line and the blue line = 0.5*0.5*0.5 = 0.125.
What digitization threshold voltage would minimize the bit error rate?
We'll minimize the bit error rate if the threshold voltage is chosen
at the voltage where the red and blue lines intersect.
By looking at the plot from the previous answer, let the ideal threshold
be x and the value of the
PDF at the intersection point be y. Then y/x=2/3 and y/(0.5-x)=1,
so x = 0.3V.
