Represent inputs that switch or strike

Try

Let u(t−a) be zero before time a and one after time a. Its value exactly at the switching point does not change its ordinary integral. A delayed ramp is u(t−a)(t−a): it starts at zero when the switch turns on, rather than jumping to a.

For a=2, evaluate the delayed ramp at times one, two, three, and five. Then compare it with u(t−2)t. Both are zero before two, but they differ after switching because the second expression retains the original clock.

Predict

If f begins at time zero, delaying its whole graph by a should produce u(t−a)f(t−a). Predict why the transform gains a factor e^(−as). Use the substitution τ=t−a in the defining integral rather than memorizing the rule without its argument shift.

Experiment

For a≥0, transform the delayed function by integrating from a to infinity. Substituting t=τ+a factors out e^(−as), leaving the transform F(s). Therefore L[u(t−a)f(t−a)]=e^(−as)F(s), under the usual convergence assumptions.

Input	Rewritten in delayed time	Transform
u(t−2)(t−2)	u(t−2)τ	e^(−2s)/s²
u(t−2)t	u(t−2)(τ+2)	e^(−2s)(1/s²+2/s)

Solve y′+y=u(t−2), y(0)=0. The unit-step response beginning at zero is 1−e^(−t), so the delayed response is u(t−2)[1−e^(−(t−2))]. Verify zero response before two and the forced equation after two. The function is continuous at switching, while its derivative has different one-sided values.

Understand

Piecewise forcing can be assembled from step functions. A rectangular input that is one from a to b and zero elsewhere is u(t−a)−u(t−b), assuming b>a. Superposition then combines two delayed step responses. At switching times, a piecewise smooth solution is interpreted through continuity and one-sided derivatives, or an integral formulation; a classical derivative need not exist exactly at the switch.

An impulse δ(t−a) is different from a step. It is a distribution representing idealized unit input concentrated at one time, not an ordinary function with an infinitely tall graph. Its transform is e^(−as). A narrow pulse with fixed area can motivate the idealization, but a real finite pulse has duration and shape.

For my″+cy′+ky=Jδ(t−a), integrate across a shrinking interval around a. If displacement stays continuous and the other terms remain bounded in the required integral sense, the impulse causes m[y′(a+)−y′(a−)]=J. Thus velocity jumps by J/m. For y″+y=3δ(t−1), the velocity jump is three. A displacement jump would introduce additional distributional terms and is not the assumed response here.

Apply

Represent a pulse lasting from time two to four using steps, then describe its response as the difference of two delayed responses. Sketch input and output on separate labeled axes. Predict which quantity jumps and which remains continuous for a first-order state equation with a step input.

Compare a step, a ramp, and an impulse using duration, integrated input, and the effect on state continuity. Complete the practice with the shifted argument written explicitly. The factor e^(−as) records delay; it does not repair an incorrectly shifted polynomial.

Continue

Convolution next assembles a general input from accumulated impulse responses. It makes the distinction between initial-state motion and input-driven motion especially clear.

Continue with graphs, jumps and generalized derivatives to read ordinary slopes and signed impulse terms from a piecewise graph.