h47478 s 00000/00001/00103 d D 1.3 24/11/17 00:10:49 bevans 3 2 c Updated e s 00001/00000/00103 d D 1.2 23/11/18 09:37:46 bevans 2 1 c Up e s 00103/00000/00000 d D 1.1 23/11/11 10:07:44 bevans 1 0 c date and time created 23/11/11 10:07:44 by bevans e u U f i f e 0 t T I 1 EE313 Linear Systems & Signals - Homework 7 Hints

EE313 Linear Systems & Signals - Homework 7 Hints

Homework #8 assignment

Visual Dictionary of Fourier transform pairs from M. Roberts Fundamentals of Signals and Systems (available on Canvas only) E 3 E 2
Problem 8.1.
- 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 8.2(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 8.2(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.

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