Section 1 introduced x(t) for a continuous-time signal and x[n] for a discrete-time one. It did not dwell on why the bracket matters. This section makes that convention exact. It also introduces the handful of basic sequences that chapter 1 needs: the unit impulse and the unit step. Chapter 1 builds every other discrete-time signal out of those two.
Think of x[n] as a numbered stack of index cards. There is one card for each whole number, and each card has a number written on it. Card 0, card 1, card 2, and so on up. Between card 3 and card 4 there is nothing at all.
That is exactly what the square bracket is telling you. It says "look up card number n." The letter n is the card number, so the question only makes sense for whole numbers. Asking "give me shelf number 4" works the same way. There is no shelf number 4.5 for anyone to hand over.
A parenthesis asks a completely different kind of question. In x(t) the letter t stands for time, measured as an exact real number. The question becomes "what is the signal's value at this exact instant t?" That question has an answer at every single point along a line, with no gaps anywhere.
Now for the two named stacks. The unit impulse is written \delta[n], using the Greek letter delta. It is the simplest card stack there is. Every card is blank, meaning zero. The one exception is card 0, which has a single 1 written on it.
The unit step, written u[n], is a switch. Every card before card 0 is blank. Card 0 has a 1 on it, and so does every card after it, forever.
Both stacks are building blocks. Chapter 1 shows that any discrete-time signal at all can be built out of copies of \delta[n] alone. You scale each copy by some number and slide it over to a new card. Whole numbers work the same way: any whole number can be built by adding up copies of 1.
A discrete-time signal is written x[n], with the index n enclosed in square brackets, never parentheses. This is not a stylistic preference. The bracket signals at a glance that n ranges only over the integers, n = \ldots, -1, 0, 1, 2, \ldots, so x[n] is fundamentally a sequence, an ordered list of numbers. The parentheses in x(t) signal something else, a real-valued argument t that can take any real value. Writing x(n) or x[t] is a genuine notational error in this course. It is the same category of mistake as writing a vector with the wrong number of components. It hides which of the two objects is actually meant, a sequence or a function of a real variable.
Two sequences appear constantly enough to earn their own names and symbols. The first is the unit impulse, also called the unit sample, \delta[n] = \begin{cases}1 & n=0\\0 & n\ne 0\end{cases} which is exactly one nonzero sample, at n=0. The second is the unit step, u[n] = \begin{cases}1 & n\ge 0\\0 & n<0\end{cases} which switches on at n=0 and stays on forever after. Both are shifted by writing \delta[n-n_0] or u[n-n_0]. That moves the nonzero sample, or the switch-on point, to n=n_0.
The unit impulse: exactly one nonzero sample
Drag n0. The single nonzero sample moves to n = n0, and every other sample stays at exactly zero -- confirm that no matter where n0 sits, the sequence never has a second nonzero sample.
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Worked example
Reading a shifted impulse
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Write the formula for a unit impulse located at n=3, and state its value at n=3 and at n=5. Shifting \delta[n] to land at n_0 means writing \delta[n-n_0] and substituting n_0=3.
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\delta[n-3] equals 1 exactly when n-3=0, i.e. n=3, and 0 everywhere else.
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3\boxed{\delta[n-3];\quad \delta[3-3]=\delta[0]=1,\quad \delta[5-3]=\delta[2]=0}
At n=5 the argument of \delta[\cdot] is 2, not 0, so the impulse has already switched off by then.
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Worked example
Reading a shifted step
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1
For u[n-3], state its value at n=1, n=3, and n=6. u[n-3] equals 1 whenever n-3\ge 0, i.e. n\ge 3, and 0 before that.
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n=1: 1-3=-2<0, so u[-2]=0. n=3: 3-3=0\ge 0, so u[0]=1. n=6: 6-3=3\ge 0, so u[3]=1.
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3\boxed{u[1-3]=0,\quad u[3-3]=1,\quad u[6-3]=1}
Once a shifted step switches on, at n=n_0, it stays on for every later index. The impulse behaves differently. It is only ever nonzero at one point.
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Where this shows up
Radar frequency analysis
A radar system sends out a pulse. The natural model for that pulse is a scaled, shifted rectangular sequence built from step functions. The expression A\big(u[n-n_0] - u[n-n_0-w]\big) switches on at n_0 and back off w samples later. Here A sets the height of the pulse, and w is its width in samples. That is exactly the pulse shape chapter 4 comes back to, when it asks what such a signal looks like in the frequency domain.
Check your understanding (counts toward your progress)
Once a shifted unit step u[n-n_0] switches on, at n=n_0, what happens for every later index?
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That describes the unit impulse \delta[n-n_0], not the unit step — the step never switches back off.
The unit step has no oscillating behavior anywhere in its definition.
The step's on/off pattern is the same regardless of where n_0 places the switch-on point.
Why
By definition u[n-n_0]=1 for every n\ge n_0 — once the step switches on, it stays on forever after, unlike the one-sample-only unit impulse.
Compare this to the unit impulse's behavior after its one nonzero sample -- the step behaves differently by design.
What is \delta[n] at n=0?
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Why
The unit impulse \delta[n] is defined to equal 1 at n=0 and 0 everywhere else.
Correct answer
Also accepted:
Why
Which notation correctly denotes a discrete-time signal, in this course's convention?
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Parentheses are reserved for a continuous-time argument in this course's convention — using them with the integer index n mixes the two conventions.
Square brackets are correct, but t denotes a continuous-time argument, not the integer index this notation needs.
Curly braces aren't used for this at all in this course.
Why
This course writes a discrete-time signal with square brackets around its integer index, x[n], deliberately distinct from x(t).
What is the smallest value of n at which u[n-4] first equals 1?
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Why
u[n-4] switches on exactly when its argument n-4 first reaches 0, i.e. at n=4 — the smallest n satisfying n-4\ge 0.
Correct answer
Also accepted:
Why
Set the step's own argument, n-4, equal to 0 and solve for n.
Practice: Brackets, shifts, and elementary sequences (2 problems · doesn’t affect your progress)
Evaluate \delta[n-6] when n=6.
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Why
The shifted impulse \delta[n-6] equals 1 exactly when its argument is 0, i.e. n-6=0, i.e. n=6.
Correct answer
Also accepted:
Why
What is u[n-2] - u[n-5] at n=3? (This difference is the rectangular-pulse construction from the lesson's applied example.)
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Why
Substituting n=3 gives u[3-2]=u[1]=1 (since 1\ge 0) and u[3-5]=u[-2]=0 (since -2<0), so the difference is 1-0=1 — the pulse is on at n=3.
Correct answer
Also accepted:
Why
Evaluate each shifted step separately at n=3 before subtracting.