System Property Explorer
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Description
Each system property is a statement that two signals must be equal. This MicroSim builds both signals for the system you select and overlays them on the output plot. When a property holds, the dashed orange curve lies exactly on the solid blue one. When it fails, the curves separate, and the panel below the plots reports the maximum difference and a one-line reason.
Use the System menu to choose one of eight example systems and the Test menu to run a linearity, time-invariance, causality or BIBO stability test. The Amplitude A, Frequency f and Time shift \(t_0\) sliders change the test input, so you can check whether a verdict depends on the input. The frequency slider is greyed out for the stability test, which uses a step. Five badges across the top give the true classification of the selected system, and the badge for the running test has a heavy outline.
The simulation illustrates superposition, time shifting, the rule that a causal output cannot depend on future input, and the definition of BIBO stability. It also makes a point of logic: one example can disprove a property but cannot prove it. It supports the section on system properties in Chapter 3: System Properties and Analysis.
Mathematical Concepts
| Test | What is compared | Input used |
|---|---|---|
| Linearity | \(\mathcal{T}[A x_1 + x_2]\) against \(A\,\mathcal{T}[x_1] + \mathcal{T}[x_2]\) | \(x_1\) is a sine burst, \(x_2\) is a 2 s pulse starting at \(t_0\) |
| Time-invariance | \(\mathcal{T}[x(t - t_0)]\) against \(y(t - t_0)\) | A sine burst of amplitude \(A\) |
| Causality | Outputs for two inputs that are identical up to \(t_0\) | A sine burst, and the same burst with 1 added after \(t_0\) |
| BIBO stability | Whether the output stays bounded | The bounded step \(A\,u(t - t_0)\) |
The sine burst is \(\sin(2\pi f t)\) for \(0 \le t \le 4\) s and zero elsewhere. The badges give the true classification of the selected system:
| System | Linear | Time-invariant | Causal | BIBO stable | Memoryless |
|---|---|---|---|---|---|
| Amplifier \(y(t) = 2x(t)\) | yes | yes | yes | yes | yes |
| Linear filter \(y(t) = \int_{t-1}^{t} x(\tau)\,d\tau\) | yes | yes | yes | yes | no |
| Time-varying multiplier \(y(t) = t\,x(t)\) | yes | no | yes | no | yes |
| Squarer \(y(t) = x^2(t)\) | no | yes | yes | yes | yes |
| Compressor \(y(t) = \tanh(x(t))\) | no | yes | yes | yes | yes |
| Predictor \(y(t) = x(t + 1)\) | yes | yes | no | yes | no |
| Integrator \(y(t) = \int_{-\infty}^{t} x(\tau)\,d\tau\) | yes | yes | yes | no | no |
| Offset \(y(t) = x(t) + 1\) | no | yes | yes | yes | yes |
One example can disprove a property, but cannot prove it
A property must hold for every input. A single pair of curves that separate is enough to show that a property fails. Curves that match only show that the property holds for those inputs. For some slider settings a nonlinear system passes the linearity comparison by coincidence (try the squarer with \(A = 1\) and \(t_0 = 4\) s, where the burst and the pulse do not overlap). The MicroSim then reports that the system is not linear and that these inputs do not show it.
How to Use
- Choose a System and a Test.
- Compare the solid blue and dashed orange curves on the output plot. Read the verdict and the one-line reason in the panel below the plots.
- Move Amplitude, Frequency and Time shift to change the test input and check whether the verdict depends on the input.
- For the causality test the vertical dashed line marks \(t_0\), where the two inputs begin to differ. A thick red trace marks any output difference that appears before \(t_0\).
- For the stability test the dotted red lines mark the input bound \(\pm A\).
Lesson Plan
Learning Objectives
By the end of this activity, students will be able to:
- Explain the superposition principle and test a system for linearity.
- Distinguish time-invariant systems from time-varying systems using a shifted input.
- Determine whether a system is causal by comparing outputs for inputs that agree up to a given time.
- Classify a system as BIBO stable or unstable from its response to a bounded input.
- Evaluate why a single matching example cannot prove that a system has a property.
Target Audience
Undergraduate students in an introductory signals and systems course. Plan 20-25 minutes for the activities below.
Prerequisites
- Signals as functions of time, time shifting
- The unit step function
- Basic integration
Activities
Activity 1: Predict first (5 min). Before running any test, fill in a table of the four tested properties for \(y(t) = 2x(t)\), \(y(t) = t\,x(t)\), \(y(t) = x^2(t)\) and \(y(t) = x(t+1)\). Then select each system and read the badges.
- The amplifier has all four properties.
- The multiplier is linear and causal but not time-invariant and not BIBO stable.
- The squarer is time-invariant, causal and stable but not linear.
- The predictor is linear, time-invariant and stable but not causal.
Activity 2: Put a number on each failure (10 min). Keep the defaults \(A = 2\) and \(t_0 = 2\) s unless stated.
- Select the time-varying multiplier and the time-invariance test. The maximum difference is 4.00. Explain this number from \(\mathcal{T}[x(t - t_0)] - y(t - t_0) = t_0\,x(t - t_0)\), whose peak is \(t_0 A\).
- Select the offset system and the linearity test. The maximum difference is 2.00. Set \(A = 1\) and then \(A = 3\): it becomes 1.00 and 3.00, so it equals \(A\) for every setting. Explain why from the two expressions in the reason line.
- Select the integrator and the stability test. The panel reports a magnitude of 16.0 at \(t = 10\) s. Compare it with \(A\,(10 - t_0) = 2 \cdot 8 = 16\).
- Select the predictor and the causality test. The thick red trace runs from \(t = 1\) s to \(t = 2\) s and the difference is up to 1.00. Why does the trace start 1 s before \(t_0\)?
Activity 3: Matching curves that prove nothing (5 min). Select the squarer and the linearity test.
- Set \(A = 1\) and \(t_0 = 4\) s. The curves coincide and the maximum difference is 0.00, but the verdict reads "Not linear, but these inputs do not show it".
- Raise \(A\) to 2. The curves separate and the maximum difference becomes 2.00 (it is 6.00 when \(t_0 = 2\) s).
- Switch to the compressor. With \(A = 1\) and \(t_0 = 4\) s it also passes; at \(A = 2\) the difference is 0.56.
Activity 4: Bounded or unbounded? (5 min). Run the stability test with \(A = 2\) and \(t_0 = 2\) s on four systems.
- Linear filter: the magnitude never exceeds 2.00, which is \(A\) times the 1 s window.
- Compressor: the magnitude never exceeds 0.96, which is \(\tanh(2)\).
- Integrator: the output is still growing at \(t = 10\) s and has reached 16.0.
- Time-varying multiplier: the output is still growing and has reached 20.0, which is \(A\,t\) at \(t = 10\) s.
Assessment
- State both parts of linearity and say which one the offset system \(y(t) = x(t) + 1\) violates. (Answer: additivity, \(\mathcal{T}[x_1 + x_2] = \mathcal{T}[x_1] + \mathcal{T}[x_2]\), and homogeneity, \(\mathcal{T}[a x] = a\,\mathcal{T}[x]\). The offset system violates both: \(\mathcal{T}[x_1 + x_2] = x_1 + x_2 + 1\) but \(\mathcal{T}[x_1] + \mathcal{T}[x_2] = x_1 + x_2 + 2\), and \(\mathcal{T}[a x] = a x + 1\) but \(a\,\mathcal{T}[x] = a x + a\). The test shows a difference of \(A\).)
- Why is \(y(t) = t\,x(t)\) linear but time-varying? (Answer: the gain \(t\) multiplies every input in the same way, so superposition holds. Delaying the input gives \(t\,x(t - t_0)\), but delaying the output gives \((t - t_0)\,x(t - t_0)\). The two differ by \(t_0\,x(t - t_0)\), which peaks at 4.00 for \(A = 2\) and \(t_0 = 2\) s.)
- Give a bounded input that produces an unbounded output for the integrator. (Answer: the step \(A\,u(t - t_0)\) integrates to the ramp \(A\,(t - t_0)\), which grows without limit. At \(t = 10\) s with \(A = 2\) and \(t_0 = 2\) s it has reached 16.)
- The squarer passes the linearity comparison for \(A = 1\) and \(t_0 = 4\) s. Is the squarer linear? (Answer: no. The burst and the pulse do not overlap, so the cross term \(2A\,x_1 x_2\) is zero and \(A = 1\) hides the \(A^2 - A\) term. A counterexample such as \(A = 2\) disproves linearity; matching curves cannot prove it.)
- For \(y(t) = x(t + 1)\), which property fails and what does the red trace show? (Answer: causality. The output at time \(t\) uses \(x(t + 1)\), so inputs that differ only after \(t_0\) already give different outputs from \(t_0 - 1\) to \(t_0\). The red trace marks that interval.)
Common Misconceptions
- "A system whose equation contains \(t\) multiplied by \(x(t)\) is nonlinear." Select the time-varying multiplier. The linearity curves coincide, but the time-invariance curves separate. The system is linear and time-varying.
- "Adding a constant keeps a system linear." Select the offset system and the linearity test. The curves separate by \(A\) at every setting, because a zero input does not give a zero output.
- "One passing example proves that a property holds." The squarer passes the linearity comparison at \(A = 1\) and \(t_0 = 4\) s. BIBO stability has the same limit: the MicroSim tests one bounded input, a step, whereas the definition requires a bounded output for every bounded input.
References
- Linear time-invariant system - Wikipedia - definition of linearity and time invariance and the properties of LTI systems
- BIBO stability - Wikipedia - bounded-input bounded-output stability and its test for linear systems
- Causal system - Wikipedia - definition of causality and examples of non-causal systems
- Linear system - Wikipedia - the superposition principle behind the linearity test
- The Scientist and Engineer's Guide to Digital Signal Processing - Steven W. Smith - free online textbook with chapters on linear systems and convolution
- Oppenheim, A. V., Willsky, A. S., & Nawab, S. H. (1997). Signals and Systems (2nd ed.). Prentice Hall. Chapter 1 covers the basic system properties: memory, invertibility, causality, stability, time invariance and linearity.