EE313 Linear Systems & Signals - Homework 7 Hints

• Problem 7.1(a).
  • i. "Difference equation" means the input-output relationship between y[n] and x[n]. Because the system is linear and time-invariant (LTI) ,

                           oo                 Nh-1
    y[n] = h[n] * x[n] =  Sum  h[k] x[n-k] =  Sum  h[k] x[n-k] 
                         k=-oo                k=0
    

    where "oo" means infinity and N_h is the length of x[n] in samples. Since the LTI system has a finite impulse response from n = 0 to n = N_h - 1, we can compute the summation for k = 0 to k = N_h - 1:

    y[n] = h[0] x[n] + h[1] x[n-1] + h[2] x[n-2] + ... + h[Nh - 1] x[n - (Nh - 1)]
    

    We can substitute the values for h[n] = 1 for n = 0, 1, ..., N_h - 1,

    y[n] = x[n] + x[n-1] + x[n-2] + ... + x[n - (Nh - 1)]
    

  • ii. Initial conditions can be discovered by computing the first several values of y[n] for n ≥ 0:

    y[0] = x[0] + x[-1] + x[-2] + ... + x[-(Nh - 1)]
    

    The initial conditions are x[-1], x[-2], ..., x[-(N_h - 1)]. They must be set to zero as necessary conditions for LTI system properties to hold.

  • iii. In this part, you'll develop a piecewise representation for the convolution result. The convolution of two finite-length signal will provide a piecewise representation of five intervals: no overlap, partial overlap, complete overlap, partial overlap, and no overlap.

    In discrete time, the convolution of two finite-length signals h[n] and x[n] produces a result that is as long as the sum of the lengths of h[n] and x[n] minus one.

    The convolution of two rectangular pulses of equal length will produce a triangle whose width is the twice the width of the rectangular pulse minus one. For more information, please see Handout E Convolving Two Rectangular Pulses.

    The convolution of two rectangular pulses of different lengths will produce a trapezoid whose duration is the sum of the durations of the two pulses minus one. The trapezoid corresponds to the five cases of convolving two finite-length signals when flipping-and-sliding one signal across the other: no overlap, partial overlap, complete overlap, partial overlap, and no overlap. The plateau of the trapezoid corresponds to the complete overlap case.

    Using the following relationships might be helpful in this problem:

    N_min = min(N_h, N_x)
    N_max = max(N_h, N_x)

  • iv. You can use the conv command in MATLAB.

  • v. Convolution of signals h[n] and x[n] in the discrete-time domain becauses the product of their z-transforms H(z) and X(z) in the z-domain. After computing Y(z) = H(z) X(z), use the inverse z-transform to find y[n].

• Problem 7.1(b).
  • i. The input-output relationship for the system is

    y[n] = 0.9 y[n - 1] + 0.1 x[n]

    for n ≥ 0. We can find the initial values by computing the first output values:

    y[0] = 0.9 y[-1] + 0.1 x[0]
    y[1] = 0.9 y[0] + 0.1 x[1]

    No initial conditions have arisen other than y[-1].

  • ii. To compute the impulse response h[n], we let the input be an impulse d[n]:

    h[n] = 0.9 h[n - 1] + 0.1 d[n]

    for n ≥ 0. We could compute output values manually and infer the impulse response as we did in LTI Example #2 on Lecture Slide 11-4, or use the z-transform.

  • iii. This problem involves convolving two causal exponential sequences. Handout F Convolution of Two Causal Exponential Sequences might be helpful.

  • iv. Here, h[n] = C aⁿ u[n] and x[n] = bⁿ u[n]. In the discrete-time domain, the output y[n] = h[n] * x[n].

    In the z-domain, Y(z) = H(z) X(z). The z-transform of x[n] = bⁿ u[n] is X(z) = 1 / (1 - b z^-1) for |z| > |b|. That is, the region of convergence is |z| > |b|.

    For Y(z), the valid values of z must be those that are valid for H(z) and X(z). Hence, the region of convergence for Y(z) is the intersection of the regions of convergence for H(z) and X(z):

                     C              
    Y(z) = -----------------------   for |z| > max{ |a|, |b| }
                   -1          -1 
           (1 - a z  ) (1 - b z  ) 
    

    We can use partial fractions decomposition to express the transfer function as a sum of two first-order terms and then apply the inverse z-transform to each term. See Example 8-10 on page 219.

• Problem 7.2.
  • i. For Pole-Zero Plot #1, I would estimate the pole location ("X") to be at p0 = -0.9 and the zero location ("O") to be at z0 = 1. That would give a transfer function in the z-domain of

             z - z0
    H(z) = C ------   for |z| > |p0|
             z - p0
    

    where C is a non-zero constant.

  • ii. The question should have read "why can we convert from the transfer function in the z-domain to a frequency response directly?". Here's more information about converting from z to frequency domain

  • iii. Please provide the plots of the magnitude response to validate the transfer function in the z-domain. Here's example Matlab code bandstopfilter.m to design a bandstop filter and plot its magnitude response.

  • iv. The magnitude response in (A) passes low frequencies around 0 rad/sample and attenuates high frequencies above say pi/4 rad/samples. Frequencies at -pi and pi rad/samples are zeroed out. This is a lowpass filter. Here's an example of a bandstop filter.

• Problem 7.3.
  • Assume that we can observe the system for all time -oo < t < oo.
  • Linearity. When checking each system for linearity, we can use the quick test of input signal of 0 for all time, which is a by-product of the homogeneity property when the input signal is scaled by a = 0. If the output is not zero for all time, then the system is not linear. If the output is zero for all time, then we'll have to then apply the mathematical definitions for homogeneity and additivity.
    • Systems (a), (b), and (c) are not linear. All fail the all-zero input signal test.
    • System (d) is linear.
  • Time-Invariance. For time-invariant system, shift of the input signal by any real-valued tau causes same shift in output signal, i.e. x(t- tau) means y(t - tau) for all tau.
    • System (a) is time-invariant.
    • Systems (b), (c), and (d) are time-varying.
  • Stability. A stable system will always produce a bounded amplitude output signal when given a bounded amplitude input signal.
    • All four systems are stable.
  • Causality. A causal system depends only on current and previous input values and/or previous output values to compute an output value.
    • Systems (a) and (d) are not causal.
    • Systems (b) and (c) are causal.
    • For System (d), we can compute different output values.
      • When t = 0, y(0) = ( x(0) + x(0) ) / 2 = x(0). Causal.
      • When t = 2, y(2) = ( x(2) + x(-2) ) / 2. Causal.
      • When t = -2, y(-2) = ( x(-2) + x(2) ) / 2. Not Causal. Depends on future input value.

• Problem 7.4(a)
  • Convolving two rectangular pulses of different widths gives a trapezoid whose width is the sum of the widths of the two rectangular pulses
  • Let
    • T_min = min(T_h, T_x)
    • T_max = max(T_h, T_x)
    • T_y = T_h + T_x
  • As we flip and slide one rectangular pulse against the other, partial overlap occurs from 0 to T_min seconds, complete overlap from T_min to T_max seconds, and partial overlap from T_max to T_y seconds.
  • We can check the points at the boundaries between intervals for a sanity check. For T_h = 4 and T_x = 9, the transition point between partial overlap and complete overlap occurs at t = 4 seconds. At t = 4 seconds, the amplitude for partial overlap and complete overlap should be the same.

• Problem 7.4(b)
  • The integrator was operating before t = 0 seconds, but we aren't able to observe that. We are observing the system starting at t = 0 seconds.
  • Since we are starting the integration at t = 0 seconds, there is ambiguity as to whether Dirac delta signal would be included at time 0. A better notation for the lower limit on the integration might have been 0− to indicate that the integration starts at time 0 before the impulse occurs.  Clunky notation, but one that is used.

bevans@ece.utexas.edu