4.6.31 · D5Ordinary Differential Equations
Question bank — Heaviside step function and Dirac delta function
Set-up: the pictures and symbols you need first
Before hunting misconceptions, look at the objects themselves. Everything on this page refers back to three sketches — figure s01 (the step ), figure s02 (the impulse ), and figure s03 (the shrinking boxes). Glance at whichever one a symbol comes from whenever it feels unfamiliar.
Figure s01 draws this step: flat at , then a clean jump to at (the small open dot marks the one ambiguous point). Keep figure s01 in mind for every step-function item below.
Figure s02 draws as an arrow of "area " sitting exactly at — its height is symbolic, its area is what is real. Keep figure s02 in mind for every impulse item.
Figure s03 draws three boxes for shrinking : each keeps area (shaded), and the arrow shows them marching toward the ideal spike of figure s02. Keep figure s03 in mind whenever a delta item talks about "area."
True or false — justify
The Heaviside function is before and after; the delta is its slope.
is a genuine function you can plug a number into and get an output.
True — it is defined at every (only the single point is left ambiguous, conventionally ), so it is an ordinary piecewise function, unlike .
is a genuine function you can plug a number into and get an output.
False — outside an integral it has no usable pointwise value (" at " is not a number); it is a distribution, i.e. a linear functional sending a test function to .
The area under equals regardless of where sits.
True — the total impulse is always by definition (see the area- boxes in figure s03); shifting moves where the spike is, never how much it carries.
and are always equal.
False — they agree only when ; if the spike sits outside so the second integral is while the first is .
The boxcar equals on the whole interval including endpoints.
False (harmlessly) — it is on the open interval and undefined at the two switch points, but since integrals ignore single points this never affects a Laplace transform.
means the delta contains "all frequencies equally."
True — a flat transform of over all is why probes a system uniformly, making its response the impulse response (see Impulse Response and Convolution).
Multiplying a function by can change the value of that function for .
False — for the step equals , so there; the step only deletes the part for , it never rescales the visible part.
The value chosen for changes the Laplace transform .
False — Laplace is an integral, and integrals ignore the contribution of a single point, so or all give the identical transform (valid on the ROC ).
Gating a signal with or delaying it can shrink the region of convergence.
False — the delay stamp (with ) has modulus on , so it never worsens convergence; the ROC of is the same as that of .
Spot the error
Someone writes .
Error — the second shifting theorem (restated above) needs the shifted function , but is not shifted. Rewrite first, giving .
Someone inverts to get .
Error — the gate is missing; the correct inverse is , which is for . Dropping makes the answer wrong before the impulse even arrives.
Someone claims delivers "twice the impulse."
Error — squaring a distribution is undefined; is a linear functional and lives only inside linear integrals, so has no meaning at all, let alone a doubled area.
Someone writes .
Error — sifting evaluates at the spike location, so the result is the number , not the function ; the integral has already erased the variable .
Someone says " is the derivative of , so is smoother than ."
Error — differentiating makes things rougher here: has a jump, and its distributional derivative is even wilder (an infinite spike), not smoother.
Someone computes by integrating and multiplying by... nothing, getting .
Error — the lower limit must become because on ; integrating from gives the correct , on the ROC , not .
Someone models "input starting at time " as alone.
Error — without the gate the shifted function is still nonzero for ; the correct "turn on at " object is .
Why questions
Why must be defined by an integral rather than a formula like ?
Because a pointwise height "" cannot be added or scaled coherently, so no ordinary function reproduces the required behaviour. Instead we define as a linear functional — a machine mapping each smooth test function to the number . Linearity plus the sifting rule is the object; the distributional framework (Distributions and Generalized Functions) gives it full rigour.
Why does shifting a function in time turn into multiplying by in the -domain?
Substituting in the Laplace integral peels off a constant factor from every term, so the whole transform just gets stamped by the delay factor ; because on the convergence is untouched (the First and Second Shifting Theorems).
Why is the response to special enough to earn its own name?
Because , so is the raw system transfer function, and its inverse — the impulse response — convolves with any input to give every other response (see Impulse Response and Convolution).
Why does an impulse produce an ongoing oscillation in , not just a single blip?
The kick lasts an instant but changes the state (it hands the system velocity); once the mass is moving, the spring keeps it oscillating forever afterward.
Why do we insist the box keep area as rather than fixed height?
Because "total impulse = 1" is the physically conserved quantity; height must blow up as width shrinks precisely so the area (the punch delivered) stays constant — exactly the area- boxes drawn in figure s03.
Why is the derivative link believable even though has no ordinary derivative at ?
The averaged box is literally the average slope of 's rise over width ; taking makes this the (distributional) derivative — a jump of has "infinite rate for an instant."
Edge cases
What is when is discontinuous exactly at ?
It is ambiguous — sifting needs continuous at ; if jumps there the "value " isn't well-defined, so the integral has no clean answer (a symmetric-limit reading would return the midpoint of the jump).
What happens to in the limit ?
It tends to , matching ; the spike at the very edge of the integration domain is still captured whole (transform valid for all , ROC the whole plane).
Is meaningful for inside a one-sided Laplace transform?
No usable effect — the spike sits at , outside , so ; the impulse happened "before the system started listening."
What does the boxcar become if ?
It collapses to everywhere (a zero-width window carries no area), consistent with subtracting identical steps.
For , what is at the switch instant ?
Effectively — the solution starts from at and only then rises toward ; nothing has accumulated yet at the instant the switch flips.
What is — does starting the integral at the spike count it?
This is genuinely ambiguous. The symmetric-limit convention replaces the sharp spike by the symmetric boxes centred on and takes ; then exactly half the area lies at , giving the value (the same reason ). We prefer strictly inside so no half-spike ever sits on the boundary.