6.02 Tutorial Problems: LTI Systems, Intersymbol Interference and Deconvolution
Problem 1.
The input sequence to a linear time-invariant
(LTI) system is given by
x[0] = 0,
x[1] = 1,
x[2] = 1 and
x[n] = 0 for all other values of n
and the output of the LTI system is given by
y[0] = 1,
y[1] = 2,
y[2] = 1 and
y[n]=0 for all other values of n.
Is this system causal? Why or why not?
The system is not causal because y becomes nonzero before x does,
i.e., y[0]=1 but x[0]=0.
What are the nonzero values of the output of this
LTI system when the input is
x[0] = 0,
x[1] = 1,
x[2] = 1,
x[3] = 1,
x[4] = 1 and
x[n] = 0 for all other values of n?
The easiest approach is by superposition:
So y = 1, 2, 2, 2, 1, 0, 0, ... when n ≥ 0, 0 otherwise.
Problem 2.
Determine the output y[n] for a system 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 3.
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.
Note that
δ[n] = u[n] - u[n-1]
Since the system is linear we can compute the response of the system to
the input δ[n] using the superposition of the appropriately scaled
and shifted v[n]:
h[n] = v[n] - v[n-1]
The result is shown in the figure below:
Problem 4.
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 6.
Consider the following eye diagram from a transmission where five samples were used per bit.
That is, a one bit was transmitted as five one-volt samples and a zero bit
was transmitted five zero-volt samples. The eye diagram shows the voltages at the receiver.
The channel is charcterized by the following unit-sample response.
Determine the eight unique voltage values for sample number 8 in the eye diagram.
At sample number 8 we can see that there are 4 samples above and 4
samples below the nominal threshold at the half-way voltage value. This
means the inter-symbol interference is carrying over from the two previous
transmitted bits (which can take on 4 possible values: 00, 01, 10, and 11).
The eye diagram is showing us what happens when transmiting a 0-bit or a 1-bit
in the current bit cell, given the 4 possible choices for the previous two
bits.
Looking at the eye diagram, we can see that the first sample in a bit
time occurs at receiver sample 1 and 6 (the diagram shows two bit times),
(note that h[0] = 0).
So sample 8 in the eye diagram corresponds to the third sample in the
transmision of a bit. Thinking in terms of the convolution equation
y[n] = Σh[k]x[n-k]
and recalling that we're using 5 samples/bit, we can determine that
bits start at n = 0, 5, 10, ... So we if want to evaluate what
happens in the third sample of a bit time in a channel that has two bits
of ISI, y[12] is the first such value, i.e., the third sample
of the third bit time.
To determine y[12], it's useful 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 y[12].
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 7.
This question refers to the LTI systems, I, II and III,
whose unit-sample responses are shown below:
In this question, the input to these systems are bit streams with
eight voltage samples per bit, with eight one-volt samples
representing a one bit and eight zero-volt samples representing a zero bit.
Which system (I, II or III) generated the following
eye diagram? To ensure at least partial credit for your answer, explain what
led you to rule out the systems you did not select.
The rise or fall time of a transition as seen in the eye diagram is only 3 samples,
so it can only be System I since it's the only system with a 3-sample unit-sample
response. Note that all the systems have some ISI, but System I's ISI is limited
to only 3 samples after which the received signal is stable for the remainder
of the bit cell.
This question refers to a fourth LTI system whose
unit-sample response, hIV[n], is given below:
where, just like in (A),
the input to this system is a bit stream with
eight voltage samples per bit, with eight one-volt samples
representing a one bit and eight zero-volt samples representing a zero bit.
Determine the voltage level denoted by D
in the eye diagram generated from the system with unit-sample response
hIV[n].
The lowest curve above threshold, i.e., the curve the arrow points at,
must be due to the transmission of an isolated 1-bit, preceded and followed
by 0-bits. This corresponds to a sample stream of 8 zeros, followed by
8 ones, followed by another 8 zeros. We can generate the entire received
waveform by convolving this sample stream with the given unit-sample
response. But since D is at the maximum value, we can compute its value
from the convolution sum when the 8 one samples overlap the 6 values of
0.1 in the response. So
For this problem, please consider three linear and time-invariant
channels, channel one, channel two, and channel
three. The unit sample response for each of these three channels
are plotted below. Please use these plots to answer all the parts of
this question.
Which channel (1, 2, or 3) has the following step response, and
what is the value of maximum value of the step response?
In this step response, the largest change in value happens on the
first sample, with successively smaller steps in subsequent samples.
This corresponds to the unit sample response for Channel 2.
The maximum value happens after the transition is complete, so
we just need to sum all the h[n] values:
Σh[n] = 0.5 + 0.4 + 0.3 + 0.2 + 0.1 = 1.5
Which channel (1, 2, or 3) produced the pair of transmitted and
received samples in the graph below, and what is the value of voltage
sample number 24 (assuming the transmitted samples have the value of
either one volt or zero volts)?
Looking at the transitions in the response, the largest change in value
happens in the middle of the transition, corresponding to the unit
sample response for Channel 3.
Using the convolution sum and noticing that h[m] = 0, m ≥ 5:
Which channel (1, 2, or 3) produced the eye diagram below
(based on 4 samples per bit), and how wide open is the eye at its
widest (lowest voltage associated with a transmitted ’1’ bit - highest
voltage associated with a transmitted ’0’ bit)?
Looking at the transitions in the response, the largest change in value
happens at the end of the transition (starts slow, finishes fast),
corresponding to the unit sample response for Channel 1.
The eye is at its "widest" (measuring vertically) at samples 4 and 8
in the eye diagram. We'll need to calculate several values to get
a numeric value for the "width".
We can sum the h[n] values in the unit sample response for
channel 1 to get the height of a completed transition:
Σh[n] = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 = 1.5
The max "width" happens just one sample into a transition -- the green
curve shows a 1-to-0 transition, the blue curve a 0-to-1 transition.
The size of the first step in a transition is determined by h[0],
which is 0.1. So one sample into a transition the response value of
the 1-to-0 transition is (1.5)-(0.1)=1.4 and the response value of the
0-to-1 transition is (0)+(0.1)=0.1.
So the max "width" is difference between these two values: (1.4)-(0.1)=1.3.
Problem 10.
In this problem you will be answering questions about a causal linear
time-invariant channel characterized by its response to a five-sample
pulse, denoted p5[n].
Suppose the input to the channel is as plotted below. Plot the
output of the channel on the axes provided beneath the input.
Noticing that
x[n] = pulse[n] + pulse[n-10]
We can use superposition to determine the response:
y[n] = p5[n] + p5[n-10]
which gives us the following picture:
The unit sample response, h[n], can be related to the
step response, s[n] by the formula h[n] = s[n] - s[n-1]. Please
derive a similar formula for h[n] in terms of the five-sample pulse
response p5[n] (an infinite series is an acceptable form for the
answer).
We can construct a unit step waveform by adding together copies of the 5-sample
pulse spaced at offsets of 5 samples:
u[n] = Σk pulse[n - 5k] for k = 0, 1, ..., ∞
So that means the unit step response is a scaled, time-shifted sum of p5[n]:
Suppose the perfect deconvolving difference equation for a linear time-invariant channel is
That is, if x[n] is the input to the channel
and y[n] is the channel output, in the noise-free case, w[n]
will be exactly x[n].
Give the unit sample response, h[n], for the channel.
Multiplying both sides of the equation above by 2 and rearranging terms we get
y[n] = 2w[n] + 2w[n-1] + w[n-2]
If the deconvolver is perfect, this must be the convolution sum, so we can simply
read off the h[n] values as the scaling factors for the w[n]:
h[0] = 2 (the scaling factor for w[n])
h[1] = 2 (the scaling factor for w[n-1])
h[0] = 1 (the scaling factor for w[n-2])
Suppose a one-sample 1V noise spike is added to the output of the channel
at time 0 and then the deconvolppupution proceeds as before. Here's the modified
equation for w[n] that includes the noise spike:
where δ[n] is the unit sample, i.e., it has the value
1 when n=0 and is 0 otherwise. Suppose the input to the
channel, x[n], is as plotted below. Please determine the first three
values of the deconvolver output, w[0], w[1], and w[2]. Because the calculation
of w[0] is corrupted by noise, we no longer expect the deconvolution
to exactly reproduce x[n].
We're given that
x[n] = 1, -1, 1, -1, 0 for n = 0, 1, 2, 3, 4
Using the unit sample response computed in part (A) and the convolution
sum, we can compute y[n]:
y[n] = 2, 0, 1, -1, -1 for n = 0, 1, 2, 3, 4
Now we complete the calculation of w[n] using the equation given
above:
For all parts of this problem, please consider five linear and
time-invariant channels, cleverly titled channel I, channelII, channel
III, channel IV and channel V. The unit sample response for each of
these five channels is plotted below, with the values outside the
interval 0 to 14 being zero. Please use these plots to answer all the
parts of this problem.
Please note:
All the voltage values in the five plots are integer multiples of 0.1 volt.
A particular channel can be the answer to more than one part.
Plot the unit step response s[n] for Channel I for 0 ≤ n ≤ 14.
Recall that the unit step response is the cumulative sum of the
unit sample response:
s[n] = Σk h[k] for k = 0, 1, ..., n
This gives the following plot:
Which two channels have step responses, s[n], that approach the same
value as n → ∞ and what is that value?
Recall that the unit step response is the cumulative sum of the
unit sample response:
s[n] = Σk h[k] for k = 0, 1, ..., n
As n → ∞, we'll add up all the non-zero h[n] for
each channel to get the value for s[∞]:
So the channels are I and III, which both have the final value of 2.
Suppose the input to each of the channels is x[n] = 1
for 0 ≤ n ≤ 9 and zero otherwise. Which channel has the
output y[n] plotted below, and what is value of the n = 15 output
sample (not plotted)?
Answer: channel V, with y[15]=3.
Explanation: since all five channels under consideration have unit
sample responses h[n] satisfying the condition h[n] =
0 for n < 0, the channels are causal. Since x[n]
equals u[n] (the unit step sequence) for n <
10, y[n] equals the unit step response of the channel
for n < 10. Since the increments of the unit step
response are the values of the unit sample response, and since the
increments of x[n], as shown, are first increasing (until n =
5), and then not decreasing much (until n = 10), the unit sample
response of the channel is increasing until n = 5 and then not
decreasing much until n = 10, which leaves channel V as the only
option.
Suppose the transmitter sends bits using five samples per bit,
meaning a sequence of 5 one-volt samples is used to transmit a "1"
bit, and a sequence of 5 zero-volt samples is used to transmit a "0"
bit. Which channel produced the eye diagram below, and how wide open
is the eye (as shown in the figure)?
Answer: channel IV, with the eye width of 0.8.
Explanation: since all five channels under consideration have
non-negative unit sample responses, the steepest path from the bottom
to the top of the eye diagram yields the samples of the unit step
response (possibly shifted in time). Thus, the increments of the unit
step response have the pattern of increasing for five samples, then
decreasing for five samples, which is only matched by channel IV.
Hence the top line of the eye diagram corresponds to 2.5 volts. The
top of the eye corresponds to the maximal value of the channel
response to the input encoding the bit sequence [1, 0], which is
encoded as [1, 1, 1, 1, 1, 0, 0, 0, 0]. The response of channel IV to
this input is the moving average of 5 consecutive samples of the unit
sample response, achieving its maximal value at the sum
Since all eye diagrams are symmetric with respect to the
horizontal line in the middle, the bottom of the eye corresponds to
the value 0.6 = 2.5 - 1.9. Hence the eye width is 1.9 - 0.6 = 1.3.
Problem 13.
The unit step response of a particular channel is shown below:
Note: all the voltage values in the step response plot are integer
multiples of 0.1 volt.
Suppose the bit sequence 0011010 is transmitted starting with
the leftmost bit and working right. That is, first a "0" bit is
transmitted, followed by a "0", followed by a "1", etc. If four
samples per bit are used, what is the value of y[24]?
