Is it linear, time invariant, causal, stable?

Pick a system and run each definition as an experiment: superpose two inputs, delay the input, change only the future, feed a bounded input for a long time. The two curves that must coincide are drawn on top of each other, so a failure is something you see, not a rule you recall.

Pick a system

Time invariant: $y(t)=x(t-2)$

click a box above to switch test

Delaying before or after a delay is the same delay.

Input, and the input delayed by t₀

x(t)x(t − t₀)

Delay then system, against system then delay

response to x(t − t₀)y(t − t₀)

Delay t₀1.5 s

Delaying the input just delays the output: the curves coincide.

How to use this in the exam

Linearity: zero in must give zero out, and any added constant, square, product of x with itself, or function like e^x breaks it. Multiplying by a fixed function of t (t, cos t, u(t)) does not.
Time invariance: look at every place t appears outside x(·). A gain like t or cos t, or a warped argument like x(2t) or x(−t), makes it time varying. Only x(t − constant) arguments are safe.
Causality: check the argument of x for both t > 0 and t < 0. x(t/2) looks harmless but at t = −2 it reads x(−1), which is in the future.
Stability: try u(t) and cos t. Any gain growing with t, or a pure integrator, fails.
Discrete time works the same way: x[2n], n·x[n], x[n]·x[n−1] and x[−n] get the same verdicts as their continuous twins.

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Is it linear, time invariant, causal, stable?

Time invariant: y(t)=x(t−2)y(t)=x(t-2)y(t)=x(t−2)

How to use this in the exam

Time invariant: y(t)=x(t−2)y(t)=x(t-2)y(t)=x(t−2)

How to use this in the exam

Time invariant: $y(t)=x(t-2)$

Time invariant: $y(t)=x(t-2)$