EE313 Linear Systems & Signals - Homework 6 Hints

• Problem 6.1(a). A linear time-invariant (LTI) system with input signal x[n] and output signal y[n] is described by

  y[n] = x[n] + x[n - 1] for n ≥ 0

  As a necessary condition for LTI properties to hold, the LTI system must be at rest. This means that the initial condition x[-1] = 0.

  The LTI system is a version of the two-point averaging filter in which the coefficients of 1/2 and 1/2 have been scaled by 2.

  1. The impulse response of a system is the system response (output) when the input is an impulse signal. Denoting the impulse response as h[n] and letting the input signal be the impulse signal d[n],

     h[n] = d[n] + d[n - 1] for n ≥ 0

  2. The step response is the response an input signal that is the unit step signal u[n] where the unit step signal is 1 when n ≥ 0 and 0 otherwise. Let x[n] = u[n],

     y_step[n] = u[n] + u[n - 1] for n ≥ 0

  3. The frequency response of the LTI system is the Fourier transform of the impulse response, where oo means infinity:

        j w       oo          -j w n    1         -j w n                -j w        -j w  
     H(e   ) =   Sum   h[n] e       = Sum  h[n] e       = h[0] + h[1] e     = 1 + e     
               n = -oo                n=0                                               
     

     We can factor the frequency response into magnitude-phase form per lecture slide 9-9 by factoring out the phase shift corresponding to the delay equal to the index of the midpoint of the impulse response, which is at an index of 1/2:

        j w     -j (w/2)      j (w/2)    -j (w/2)      -j (w/2)               
     H(e   ) = e          (  e        + e         ) = e          ( 2 cos(w/2) )
     
        j w                -j (w/2)
     H(e   ) = 2 cos(w/2) e
     

     • Magnitude response: 2 cos(w/2) which is non-negative for -pi ≤ w < pi.
     • Phase response: -w/2 which is a line of slope -1/2 that intersects the origin.

  4. The MATLAB command

     freqz( [1 1] );
     

     will plot the magnitude and phase response for an LTI system with impulse response d[n] + d[n - 1]. The magnitude response, by default, will be plotted in decibels vs. discrete-time frequency, where A_dB = 20 log₁₀ | A |.

• Problem 6.1(c). Please see the hints above for problem 6.1(a).

  As in problem 6.1(a), we can factor the frequency response into magnitude-phase form per lecture slide 9-9 by factoring out the phase shift corresponding to the delay equal to the index of the midpoint of the impulse response, which is at an index of 1 and hence the phase shift is exp(-j w).

• Problem 6.2(a). As we had seen on lecture slide 8-8 and in Signal Processing First Section 5-6, the output of an LTI system y[n] is the convolution of the input signal x[n] and the LTI system's impulse response h[n]

  When we cascade two LTI systems with impulse responses h₁[n] and h₂[n], as we had seen on lecture slide 8-11 and in Signal Processing First Section 5-7,

  y[n] = h₁[n] * h₂[n] * x[n]

  We can write the input-output relationship for the cascade of two LTI systems as a single LTI system

  y[n] = h[n] * x[n]

  where

  h[n] = h₁[n] * h₂[n]

bevans@ece.utexas.edu