6.02 Tutorial Problems: LTI Systems, Intersymbol Interference and Deconvolution

Problem .

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.

1. 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.

2. 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 .

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.

y[n] = Σx[k]h[n-k] = x[0]h[n] + x[1]h[n-1] + x[2]h[n-2]
     = δ[n+1] + 4δ[n] + 8δ[n-1] + 8δ[n-2] + 3δ[n-3]

Problem . 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.

Problem .

The output of a particular communication channel is given by

y[n] = αx[n] + βx[n-1] where α > β

1. 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.

2. 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

3. 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.

4. 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.

   w[n] = (1/h[0])(y[n] - h[1]w[n-1]) = y[n]/.7 - (.3/.7)w[n-1]

   so

   w[n] = [1, 1, 1, 0, 1, 1, 0, 0], followed by all 0's

Problem .

Suppose four different wires {I,II,III,IIII} have four different unit sample responses:

h1 = .25, .25, .25, .25, 0, ...

h2 = 0, .25, .5, .25, 0, ...

h3 = .11, .22, .33, .22, .11, 0, ...

h4 = .04, .08, .12, .16, .20, .12, .12, .12, .04, 0, ...

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.

Problem .

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.

n           0    1     2     3     4     5     6     7     8     9
hstep[n] = 0.0, 0.04, 0.12, 0.24, 0.40, 0.60, 0.72, 0.84, 0.96, 1.00, ...

Problem .

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.

1. 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, h_IV[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.

2. Determine the voltage level denoted by D in the eye diagram generated from the system with unit-sample response h_IV[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

   D = (1)(0.05) + (1)(0.1) + (1)(0.1) + (1)(0.1) + (1)(0.1) + (1)(0.1) + (1)(0.1) + (1)(0.05)
     = 0.7

   where we've left out the terms in the convolution sum where x[n] is 0.

Problem .

Suppose a linear time-variant channel has a unit sample response given by

h[n] = 1/2  n = 0, 1, 2
h[n] = 0    otherwise

x[n] = 3/2  n = 2, 3, 4
x[n] = 0    otherwise

y[n] = x[n]h[0] + x[n-1]h[1] + x[n-2]h[2]

y[4] = x[4]h[0] + x[3]h[1] + x[2]h[2]
     = (3/2)(1/2) + (3/2)(1/2) + (3/2)(1/2)
     = 9/4

Problem .

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.

1. 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
   

2. 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:

   y[24] = x[24]h[0] + x[23]h[1] + x[22]h[2] + x[21]h[3] + x[20]h[4]
         = (0)(0.1) + (1)(0.2) + (1)(0.4) + (1)(0.2) + (0)(0.2)
         = 0.2 + 0.4 + 0.2
         = 0.8
   

3. 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 .

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].

1. 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:

   

2. 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]:

   s[n] = Σk p5[n - 5k]  for k = 0, 1, ..., ∞
   

   Now we can derive an expression for h[n]:

   h[n] = s[n] - s[n-1] = Σk (p5[n - 5k] - p5[n - 5k - 1])
   

Problem .

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].

1. 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])
   

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:

   w[0] = (0.5)(y[0] + δ[0]) = (0.5)(2 + 1) = 1.5
   w[1] = (0.5)(y[1] + δ[1]) - (0.5)(2w[0]) = (0.5)(0 + 0) - (0.5)(2*1.5) = -1.5
   w[2] = (0.5)(y[2] + δ[2]) - (0.5)(2w[1] + w[0]) = (0.5)(1 + 0) - (0.5)(2*-1.5 + 1.5) = 1.25
   

Problem .

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.

1. 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:

   

2. 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[∞]:

   sI[∞] = 2
   sII[∞] = 1
   sIII[∞] = 2
   sIV[∞] = 2.5
   sV[∞] = 4.5
   

   So the channels are I and III, which both have the final value of 2.

3. 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.

   For channel V, we have

   y[15] = h[0]x[15] + h[1]x[14] + ... + h[15]x[0]
         = h[6] + h[7] + ... + h[15]
         = 0.5 + 0.5 + 0.5 + 0.5 + 0.4 + 0.3 + 0.2 + 0.1
         = 3
   

4. 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

    hIV[3] + hIV[4] + hIV[5] + hIV[6] + hIV[7] = 0.3 + 0.4 + 0.5 + 0.4 + 0.3 = 1.9.
   

   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 .

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.

1. 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]?
   First, let's determine x[n]:

   n:     0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
   x[n]:  0  0  0  0  0  0  0  0  1  1  1  1  1  1  1  1  0  0  0  0  1  1  1  1  0
   

   Next we can determine h[n] by computing s[n] - s[n-1], reading the values of s[n] from the diagram above:

   n:        0   1   2   3   4   5
   s[n]:    0.2 0.4 0.7 1.0 1.0 1.0 ...
   s[n-1]:  0.0 0.2 0.4 0.7 1.0 1.0 ...
   h[n]:    0.2 0.2 0.3 0.3  0   0  ...
   

   Now we can determine y[24] using the convolution sum, keeping only the non-zero terms of h[n]:

   y[24] = x[24]h[0] + x[23]h[1] + x[22]h[2] + x[21]h[3]
         = (0)(0.2) + (1)(0.2) + (1)(0.3) + (1)(0.3)
         = 0.8
   

2. Suppose the samples received from the channel are used as input to Deconvolver A, described by the following equation:

   

   Assume that w_A[n] = 0 for n < 0. For what value of α will w_A[n] = x[n] where x[n] is the input to the channel?

   Using the convolution sum and the h[n] calculated above:

   y[n] = x[n]*0.2 + x[n-1]*0.2 + x[n-2]*0.3 + x[n-3]*0.3
   

   Reorganizing the deconvolver expression for y[n]:

   y[n] = wA[n]*α + wA[n-1]*0.2 + wA[n-2]*0.3 + wA[n-3]*0.3
   

   If we want w_A[n] = x[n] then α must be 0.2.