EE313 Linear Systems & Signals - Homework 7 Hints
Homework #8 assignment
- 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
- Tmin = min(Th, Tx)
- Tmax = max(Th, Tx)
- Ty = Th + Tx
- As we flip and slide one rectangular pulse against the other,
partial overlap occurs from 0 to Tmin seconds,
complete overlap from Tmin to Tmax seconds, and
partial overlap from Tmax to Ty seconds.
- We can check the points at the boundaries between intervals for a sanity check.
For Th = 4 and Tx = 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.
Last updated 11/17/24.
Send comments to
bevans@ece.utexas.edu