EE313 Linear Systems and Signals - Homework #6 Hints

• As mentioned in the course reader in Appendix C, you can use Mathematica to help you check your answers. For the difference equation

  y[n + 2] + 2 y[n + 1] + y[n] = 0
  

  with y[-1] = 1 and y[-2] = 1, the Mathematica code to compute the general form of the solution follows:

  Needs[ "DiscreteMath`Master`" ];
  RSolve;
  ysols = RSolve[ { y[n + 2] + 2 y[n + 1] + y[n] == 0, y[-1] == 1, y[-2] == 1 },
                  y[n], n ]
  

  Note that the second statement, RSolve, will force the recursive (difference) equation solver to be loaded. This would normally not be needed in Mathematica, but it is here to work around a bug in the RSolve package. You can use the solution to the difference equation to compute the values of y[n] for n = 0, 1, 2, 3, 4:

  Apply[Join, Table[ ysols, {n, 0, 4} ] ]
  

• For insight into stability for the discrete-time systems, it is a good idea to try to connect a discrete-time impulse response with a continuous-time impulse response.

  For the zero-input case in continuous time, characteristic modes have the form of e^{g t} for continuous-time systems (where g is a characteristic root) and aⁿ for discrete-time systems (where a is a characteristic root). In both cases, the characteristic modes are exponentials:

  aⁿ = e^{b n} = (e^b) ⁿ

  So, a = e^b, i.e. b = ln a.

  Let's consider the three possible cases for the zero-input solution for a positive real value of a:

  1. When 0 < a < 1, we have b < 0, so e^{b n} corresponds to an asymptotically stable response.
  2. When a = 1, we have b = 0, and e^{b n} is marginally stable.
  3. When a > 1, we have b > 0, and e^{b n} is unstable.
  More generally, the characteristic mode is asymptotically stable if |a| < 1.

• Problem 6.1: The system has two distinct characteristic roots, each of magnitude one. We can find the general form of the system response by taking a weighted combination of characteristic modes.

  Because the system in linear and time-invariant, the initial conditions must be zero. This means that the zero-input solution must be zero. The zero-state solution is simply the convolution of the input and impulse response.

  In this problem, the form of the impulse response matches the general form of the system response. We can then compute the impulse response by setting the input equal to an impulse and iteratively calculating h[0], h[1], etc. For the impulse response, we know that the initial conditions are zero, i.e. h[-1] = 0 and h[-2] = 0. From the calculated values of the impulse response, we can solve for the constants in the general form of the impulse response.

  Part (c) is quite similar to problem 1.5 on midterm #1.

• Problem 6.2: The discrete-time frequency domain is periodic with period 2 pi. Plots of discrete-time frequency responses are commonly made from -pi to pi. Appendix D in the course reader gives Matlab code on how to plot magnitude and phase responses.

• Problem 6.3: The problem says to use n to represent the nth day after January 1, 2000. The system "turns on" at n = 1 instead of at n = 0.

  The system has one distinct characteristic root. The general form of the system response is a weighted version of the characteristic mode. We can then solve for the zero-input solution using the initial condition given.

  The form of the impulse response matches the general form of the system response. We can then compute the first impulse response value by setting the input equal to an impulse. For the impulse response, we know that the initial condition is zero, i.e. h[0] = 0. From the calculated value of the impulse response at h[1], we can solve for the constant in the general form of the impulse response.

bevans@ece.utexas.edu