Part 1 · Chapter 3

Elementary Signals and Singularity Functions

Chapter 1 promised a strategy: decompose a complicated signal into simple building blocks, respond to each, superpose. This chapter supplies the blocks. A handful of elementary signals — the step, the impulse, the ramp, the exponential — turn out to be enough to construct almost every waveform in engineering, and one of them, the unit impulse, is strange enough to deserve a section of its own.

Signals and Systems Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • The unit step \(u(t)\) and \(u[n]\) — the switch that turns a signal on, and the reason \(u(t)\) appears in almost every formula in this book.
  • The unit impulse \(\delta(t)\): why it is not a function at all, how to define it as a limit, and how it differs completely from its harmless discrete-time cousin \(\delta[n]\).
  • The sifting property — the one impulse identity you will use more than any other — together with the sampling and scaling properties.
  • The singularity family linked by differentiation and integration: doublet, impulse, step, ramp, parabolic.
  • The standard window and sign functions — rectangular, triangular, signum — and how to build them out of shifted steps.
  • Real and complex exponentials, damped sinusoids, and the peculiar fact that discrete-time frequency only exists over a range of \(2\pi\).
  • The impulse decomposition \(x[n] = \sum_k x[k]\delta[n-k]\) — literally the "decompose" step of Chapter 1, and the direct doorway to convolution.
Section 3-1

Why Elementary Signals

Chapter 1 set out a three-step strategy — decompose, respond, superpose — and observed that the whole art lies in choosing the building blocks. Chapter 2 sorted signals into classes. This chapter finally introduces the blocks themselves, and there are remarkably few of them.

The reason so few suffice is that engineering waveforms are overwhelmingly built from switching, decaying and oscillating. A voltage is applied at some instant: that is a step. A capacitor discharges: that is an exponential. An oscillator runs: that is a sinusoid. A signal is confined to a finite window: that is a rectangular pulse, which is just two steps. Combine, shift and scale these and you can write down almost any waveform an engineer encounters, in closed form, ready to be pushed through an integral.

One of the blocks is different in kind from the others. The unit impulse is not a signal you could ever generate or measure; it is an idealisation with infinite height, zero width and unit area. It is also the single most useful object in the subject, because a system's response to it — the impulse response — turns out to determine the system's response to everything else. Chapter 7 will prove that. This chapter builds the tool.

A note on the word "singularity". The step, impulse and their relatives are called singularity functions because each is discontinuous, or has a discontinuous derivative, at a single point. That awkwardness at one instant is the price of being able to describe switching exactly, and every one of their strange properties traces back to it.
Section 3-2

The Unit Step

The unit step is the mathematical description of a switch closing at \(t = 0\). It is zero before the origin and one after it, and it is the reason the phrase "and nothing happened before we started watching" can be written as a multiplication.

The unit step, continuous- and discrete-time
\[ u(t) = \begin{cases} 1, & t \gt 0 \\ 0, & t \lt 0 \end{cases} \qquad\qquad u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n \lt 0 \end{cases} \]
t u(t) 1 0
Continuous-time — a jump discontinuity at the origin
n u[n] 1 0
Discrete-time — no ambiguity, \(u[0] = 1\)

Notice the small asymmetry between the two definitions. In discrete time \(u[0]\) is simply defined to be 1 and there is nothing more to say. In continuous time the value at \(t=0\) is genuinely ambiguous — the function jumps there — and different books adopt \(u(0)=1\), \(u(0)=0\) or \(u(0)=\tfrac12\). Fortunately it almost never matters: changing a signal at a single isolated instant changes no integral, and integrals are what we compute. The symmetric choice \(u(0)=\tfrac12\) is the tidiest, because it makes \(u(t) + u(-t) = 1\) hold everywhere.

The step earns its keep as a switch. Multiplying any signal by \(u(t)\) annihilates everything before the origin and leaves the rest untouched, and shifting the step moves the switching instant: \(u(t - t_0)\) turns on at \(t = t_0\). Subtracting two shifted steps isolates a window, which is how every finite-duration pulse in this book is written.

The step as a switch and a window
\[ x(t)\,u(t-t_0) \;=\; \begin{cases} x(t), & t \gt t_0\\ 0, & t \lt t_0\end{cases} \qquad\qquad u(t-a) - u(t-b) \;=\; \begin{cases} 1, & a \lt t \lt b\\ 0, & \text{otherwise}\end{cases} \]
Section 3-3

The Unit Impulse

Now the strange one. The unit impulse \(\delta(t)\) — also called the Dirac delta — is meant to represent an event of vanishing duration but finite effect: a hammer blow, a lightning strike, a switching spike. We want it to be zero everywhere except at the origin, and yet to have unit area. No ordinary function can do both, and \(\delta(t)\) is therefore not a function at all. It is a generalised function, or distribution, defined entirely by what it does inside an integral.

The honest way to introduce it is as a limit. Take a rectangular pulse of width \(\Delta\) and height \(1/\Delta\), so that its area is exactly 1 no matter how small \(\Delta\) is. Now shrink \(\Delta\). The pulse becomes taller and narrower while its area stays pinned at unity, and the impulse is what that sequence tends to.

t area = 1 area = 1 area = 1 width Δ → 0, height 1/Δ → ∞
The impulse as a limit — area held at unity while the width vanishes
t (1) 0 δ(t) (2) t₀ 2δ(t − t₀)
Arrow notation — the label is the area, not the height
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Defining properties of the unit impulse
\[ \delta(t) = 0 \ \ \text{for}\ t \ne 0, \qquad \int_{-\infty}^{\infty}\delta(t)\,dt = 1 \]

These two statements define \(\delta(t)\) completely. Its value at the origin is not a number and should never be asked for — the impulse is characterised by area, never by height.

This is why the arrow notation matters. An impulse is drawn as a vertical arrow, and the number written beside it is its weight — the area — not the height of the arrow on the page. An arrow labelled \((2)\) represents \(2\delta(t)\), an impulse of twice the area, and its being drawn taller is purely a drawing convention.

The discrete-time impulse is, by contrast, completely tame. There is no limiting process and no distribution theory: \(\delta[n]\) is an ordinary sequence which is 1 at \(n=0\) and 0 everywhere else. Every property we will labour over in continuous time is either obvious or trivial for \(\delta[n]\), which is one reason discrete-time analysis is often the easier place to build intuition.

The discrete-time impulse and its relation to the step
\[ \delta[n] = \begin{cases}1, & n = 0\\ 0, & n \ne 0\end{cases} \qquad \delta[n] = u[n] - u[n-1] \qquad u[n] = \sum_{m=-\infty}^{n}\delta[m] \]

Those last two relations are worth staring at, because they are the discrete counterparts of a calculus statement. The impulse is the first difference of the step, and the step is the running sum of the impulse. In continuous time, difference becomes derivative and sum becomes integral, giving the pair that ties the two most important elementary signals together.

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Impulse and step are a derivative-integral pair
\[ \delta(t) = \frac{d\,u(t)}{dt} \qquad\qquad u(t) = \int_{-\infty}^{t}\delta(\tau)\,d\tau \]

The step has zero slope everywhere except at the origin, where it jumps by 1 in zero time — an infinite slope of unit "strength". That is exactly an impulse. Differentiating a jump always produces an impulse whose weight equals the size of the jump, a fact used constantly when analysing switched circuits.

Is any of this legitimate? Treating \(\delta(t)\) as a function and differentiating a discontinuity are both, strictly, illegal in ordinary calculus. They were made rigorous by Laurent Schwartz's theory of distributions in the 1940s, well after engineers had been using them successfully for decades. For this course you may use the properties below freely — they are theorems, not conveniences — but remember that \(\delta(t)\) only ever means something when it sits inside an integral.
Section 3-4

Properties of the Impulse

Four properties do essentially all the work. The first is the one you will use more than any other identity in the entire course.

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The sifting property
\[ \int_{-\infty}^{\infty} x(t)\,\delta(t-t_0)\,dt \;=\; x(t_0) \]

An integral against a shifted impulse simply reads off the value of \(x\) at the impulse's location. The impulse "sifts out" one sample from the whole signal — which is why it is also the mathematical model of sampling.

The reason is easy to see from the limiting picture. The impulse is zero everywhere except in a vanishing neighbourhood of \(t_0\); over that tiny interval \(x(t)\) is essentially constant at \(x(t_0)\), so it comes out of the integral, and what remains is the impulse's area, which is 1.

The second property is the sampling or product property, and it is the pointwise version of the same idea. Multiplying a signal by a shifted impulse gives an impulse whose weight is the signal's value at that point.

Sampling (product) property
\[ x(t)\,\delta(t-t_0) \;=\; x(t_0)\,\delta(t-t_0) \qquad\qquad x[n]\,\delta[n-n_0] \;=\; x[n_0]\,\delta[n-n_0] \]

The third is time scaling, and it is the one most often got wrong. Compressing the time axis by a factor \(a\) makes the impulse narrower without making it taller, so its area shrinks by \(|a|\) — and to restore the definition we must divide by \(|a|\). The modulus matters: reversing time does not change the area.

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Scaling and symmetry
\[ \delta(at) = \frac{1}{|a|}\,\delta(t) \qquad\qquad \delta(-t) = \delta(t) \]

The impulse is an even signal. Setting \(a=-1\) in the scaling rule gives the symmetry statement immediately, since \(|-1| = 1\).

The fourth property is really a warning: whenever you meet \(\delta(at - b)\), factor before doing anything else. Write \(\delta(at-b) = \delta\!\big(a(t - b/a)\big) = \tfrac{1}{|a|}\delta(t - b/a)\), so the impulse sits at \(t = b/a\) and carries weight \(1/|a|\). Skipping this step is the single most common error in impulse problems, and Worked Example 1 shows it in action.

PropertyContinuous-timeDiscrete-time
Sifting\(\int x(t)\delta(t-t_0)dt = x(t_0)\)\(\sum_n x[n]\delta[n-n_0] = x[n_0]\)
Sampling\(x(t)\delta(t-t_0) = x(t_0)\delta(t-t_0)\)\(x[n]\delta[n-n_0] = x[n_0]\delta[n-n_0]\)
Scaling\(\delta(at) = \delta(t)/|a|\)Not applicable — \(n\) is an integer
Symmetry\(\delta(-t) = \delta(t)\)\(\delta[-n] = \delta[n]\)
Relation to step\(\delta(t) = du/dt\)\(\delta[n] = u[n]-u[n-1]\)
Section 3-5

The Singularity Family

The step and the impulse are two members of a chain. Integrate the impulse and you get the step; integrate the step and you get the unit ramp, a signal that is zero until the origin and then rises with unit slope; integrate again and you get the parabolic signal. Going the other way, differentiating the impulse gives the doublet \(\delta'(t)\), a pair of opposite impulses infinitesimally close together.

δ′(t) δ(t) u(t) r(t) p(t) DOUBLET IMPULSE STEP RAMP PARABOLIC integrate ⟶ ⟵ differentiate
One chain, linked by calculus — each step right integrates, each step left differentiates
The ramp and the parabolic signal
\[ r(t) = t\,u(t) = \int_{-\infty}^{t} u(\tau)\,d\tau, \qquad p(t) = \frac{t^2}{2}u(t) = \int_{-\infty}^{t} r(\tau)\,d\tau \]

The practical value of this chain is that piecewise-linear signals become sums of shifted ramps. Any waveform made of straight segments can be written as a combination of \(r(t - t_i)\) terms, one for each point where the slope changes, with the coefficient equal to the change in slope there. Differentiating that expression once gives steps, and once more gives impulses — which is exactly the sequence you need when computing the response of a circuit to a trapezoidal drive. Worked Example 3 does this end to end.

The doublet is rarer but appears whenever an impulse is differentiated, and it has its own sifting rule with a sign flip: \(\int x(t)\,\delta'(t-t_0)\,dt = -x'(t_0)\). Its discrete counterpart, \(\delta[n] - \delta[n-1]\), is entirely ordinary.

Section 3-6

Window and Sign Functions

Three more named signals appear often enough to be worth naming, and all three are built from steps. The rectangular pulse or gate function confines a signal to a finite window; the triangular pulse tapers instead of cutting sharply; and the signum function extracts the sign of a signal and nothing else.

t −τ/2 τ/2 1
Rectangular pulse (gate)
t −τ τ 1
Triangular pulse
t +1 −1
Signum function
SignalDefinitionIn terms of steps
Rectangular pulse\(1\) for \(|t| \lt \tau/2\), else \(0\)\(u(t+\tau/2) - u(t-\tau/2)\)
Triangular pulse\(1-|t|/\tau\) for \(|t| \le \tau\), else \(0\)Built from three ramps
Signum\(+1\) for \(t \gt 0\), \(-1\) for \(t \lt 0\)\(2u(t) - 1\), or \(u(t)-u(-t)\)
Looking ahead to the sinc. One more function will join this list in Part 4: \(\operatorname{sinc}(t) = \sin(\pi t)/(\pi t)\). It is not introduced here because it has no independent motivation yet — it earns its place as the Fourier transform of the rectangular pulse, and again as the interpolating function in the sampling theorem of Part 6. Keep the rectangular pulse in mind; the sinc is its shadow in the frequency domain.
Section 3-7

Exponentials and Sinusoids

If the step and the impulse describe switching, the exponential describes everything that happens afterwards. A capacitor discharging, a current decaying in an inductor, an unstable system running away, an oscillator ringing down — all of them are exponentials, and Chapter 1 already gave the deep reason they matter: complex exponentials pass through LTI systems unchanged in shape.

The real exponential \(x(t) = Ce^{at}\) has three behaviours according to the sign of \(a\). For \(a \lt 0\) it decays; for \(a \gt 0\) it grows without bound; for \(a = 0\) it is a constant. When it decays, the quantity \(\tau = 1/|a|\) is the time constant: the signal falls to about 37% of its initial value in one time constant, and to under 2% in four, which is the usual engineering definition of "settled".

The general complex exponential
\[ x(t) = C e^{st}, \qquad s = \sigma + j\omega_0 \qquad\Longrightarrow\qquad x(t) = C e^{\sigma t}\big[\cos\omega_0 t + j\sin\omega_0 t\big] \]

That single expression contains the whole family. The real part \(\sigma\) sets the envelope — decaying, constant or growing — and the imaginary part \(\omega_0\) sets the oscillation. With \(\sigma = 0\) you get a pure sinusoid; with \(\omega_0 = 0\) a pure exponential; with both non-zero, a damped sinusoid \(Ce^{\sigma t}\cos(\omega_0 t + \phi)\), which is the natural response of essentially every second-order circuit you will ever analyse.

t σ < 0
Decaying — \(e^{\sigma t}u(t)\), \(\sigma \lt 0\)
t σ > 0
Growing — \(e^{\sigma t}u(t)\), \(\sigma \gt 0\)
t
Damped sinusoid inside its envelope

The discrete-time story runs in parallel with one genuine surprise. The natural form is \(x[n] = C\alpha^n\), which decays when \(|\alpha| \lt 1\), grows when \(|\alpha| \gt 1\), and — a behaviour with no continuous-time analogue — alternates in sign when \(\alpha\) is negative. Writing \(\alpha = e^{j\Omega_0}\) gives the discrete-time complex exponential \(e^{j\Omega_0 n}\), and here the surprise appears.

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Discrete-time frequency lives on a circle
\[ e^{j(\Omega_0 + 2\pi)n} = e^{j\Omega_0 n}e^{j2\pi n} = e^{j\Omega_0 n} \]

Because \(n\) is an integer, \(e^{j2\pi n} = 1\) always. Two discrete-time frequencies differing by \(2\pi\) produce identical sequences — so \(\Omega_0\) is only meaningful over any interval of length \(2\pi\), and there is no such thing as an arbitrarily high discrete-time frequency.

Where the highest frequency lives. In continuous time, larger \(\omega_0\) always means faster oscillation. In discrete time the fastest possible variation occurs at \(\Omega_0 = \pi\), where the sequence becomes \(e^{j\pi n} = (-1)^n\) — alternating every sample, which is as fast as samples permit. Increase \(\Omega_0\) beyond \(\pi\) and the sequence starts looking slower again. This wrap-around is aliasing seen from the frequency side, and it is the reason the sampling theorem of Part 6 exists.
Section 3-8

Building Any Signal from Impulses

We now have the payoff, and it is the reason this chapter sits where it does. Consider an arbitrary discrete-time signal \(x[n]\). By the sampling property, \(x[k]\delta[n-k]\) is a sequence that equals \(x[k]\) at \(n=k\) and is zero everywhere else — a single sample, isolated. Add up one such term for every \(k\) and you have reconstructed the entire signal, exactly.

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The impulse decomposition
\[ x[n] = \sum_{k=-\infty}^{\infty} x[k]\,\delta[n-k] \qquad\qquad x(t) = \int_{-\infty}^{\infty} x(\tau)\,\delta(t-\tau)\,d\tau \]

Every signal is a weighted sum — or integral — of shifted impulses, with the weights being the signal's own values. This is exact, not an approximation.

n x[n] x[k] · δ[n − k] k
Each sample is one scaled, shifted impulse — the signal is their sum

Read that alongside Chapter 1's strategy and the plan for the next few chapters writes itself. The decomposition step is done: the building blocks are shifted impulses, and the coefficients are the signal's own samples. The respond step asks what a system does to a single impulse — its impulse response \(h[n]\). The superpose step then adds those responses together, and because the system is linear and time-invariant, the sum takes one specific form. That form is convolution, and it is the whole of Chapter 7.

This is the hinge of Part 1. Everything before this section was vocabulary; everything after it is machinery. If one idea from these three chapters survives into your exam, make it this: a signal is a sum of impulses, so knowing how a system answers one impulse tells you how it answers everything.
Section 3-9

Worked Examples

1 Sifting integrals, including a scaled impulse

Problem. Evaluate (a) \(\displaystyle\int_{-\infty}^{\infty}(t^2+3)\,\delta(t-2)\,dt\); (b) \(\displaystyle\int_{0}^{5}\cos(\pi t)\,\delta(t-1)\,dt\); (c) \(\displaystyle\int_{-\infty}^{\infty} e^{-t}\,\delta(2t-4)\,dt\).

Solution (a). Straight sifting — read off the integrand at \(t = 2\): \((2)^2 + 3 = 7\).

Solution (b). The impulse sits at \(t=1\), which lies inside the interval \([0,5]\), so sifting applies as usual: \(\cos(\pi) = -1\). Had the impulse fallen outside the limits, the integral would have been zero — always check that the impulse is inside the range before sifting.

Solution (c). This is the one that catches people. Factor the argument before sifting:

Working — part (c)
\[ \delta(2t-4) = \delta\big(2(t-2)\big) = \tfrac{1}{2}\,\delta(t-2) \]

so the integral becomes \(\tfrac12\displaystyle\int e^{-t}\delta(t-2)\,dt = \tfrac12 e^{-2} \approx 0.0677\). Sifting without factoring would have given \(e^{-2}\) — twice the correct answer.

2 A pulse from steps, and its derivative

Problem. Write the pulse that equals 3 for \(1 \lt t \lt 4\) and zero elsewhere in terms of unit steps, then differentiate it.

Solution. Switch on at \(t=1\) and off at \(t=4\), scaling by the height:

Working
\[ x(t) = 3\big[u(t-1) - u(t-4)\big] \]

Differentiating term by term, and using \(du(t-t_0)/dt = \delta(t-t_0)\):

Working
\[ \frac{dx}{dt} = 3\delta(t-1) - 3\delta(t-4) \]

Two impulses, of weight \(+3\) at the rising edge and \(-3\) at the falling edge. The general rule is visible here: differentiating a waveform produces one impulse at every jump, with weight equal to the size of the jump and sign matching its direction.

3 A trapezoid built from ramps

Problem. A signal rises linearly from 0 to 1 over \(0 \le t \le 1\), stays at 1 until \(t=3\), then falls linearly to 0 at \(t=4\). Express it using ramps and verify by differentiating.

Solution. Work with the slope changes, since each one needs exactly one ramp. The slope goes \(0 \to +1\) at \(t=0\); \(+1 \to 0\) at \(t=1\); \(0 \to -1\) at \(t=3\); and \(-1 \to 0\) at \(t=4\). The ramp coefficients are those changes:

Working
\[ x(t) = r(t) - r(t-1) - r(t-3) + r(t-4) \]

To verify, differentiate once. Since \(dr(t-t_0)/dt = u(t-t_0)\),

Working — check
\[ \frac{dx}{dt} = u(t) - u(t-1) - u(t-3) + u(t-4) \]

which is \(+1\) on \((0,1)\), zero on \((1,3)\), \(-1\) on \((3,4)\) and zero thereafter — exactly the slopes described. The method generalises: any piecewise-linear signal is a sum of shifted ramps weighted by the slope change at each break point.

4 A finite sequence written with impulses

Problem. A discrete-time signal takes the values \(2, -1, 3\) at \(n = 0, 1, 2\) and is zero elsewhere. Write it using unit impulses, and evaluate \(\sum_n x[n]\,\delta[n-1]\).

Solution. Apply the impulse decomposition, keeping only the non-zero terms:

Working
\[ x[n] = 2\delta[n] - \delta[n-1] + 3\delta[n-2] \]

The sum is discrete-time sifting, which reads off the value at \(n=1\): \(\sum_n x[n]\delta[n-1] = x[1] = -1\). Notice how much simpler this is than the continuous-time version — no limits, no distributions, just picking a term out of a finite sum. That simplicity is why convolution is usually taught in discrete time first.

5 Two discrete frequencies that are the same signal

Problem. Show that \(x_1[n] = e^{j0.2\pi n}\) and \(x_2[n] = e^{j2.2\pi n}\) are the same sequence, and identify which of \(\Omega = 0.5\pi\) and \(\Omega = 1.5\pi\) oscillates faster.

Solution. The two frequencies differ by exactly \(2\pi\), so factor that out:

Working
\[ x_2[n] = e^{j(0.2\pi + 2\pi)n} = e^{j0.2\pi n}\,e^{j2\pi n} = e^{j0.2\pi n}\cdot 1 = x_1[n] \]

since \(e^{j2\pi n} = 1\) for every integer \(n\). They are not merely similar — they are the identical sequence, sample for sample.

For the second part, remember that discrete-time oscillation is fastest at \(\Omega = \pi\) and slows again beyond it. So \(\Omega = 0.5\pi\) is at distance \(0.5\pi\) from \(\pi\), and \(\Omega = 1.5\pi\) is also at distance \(0.5\pi\) — they oscillate at the same rate. Indeed \(1.5\pi\) is equivalent to \(1.5\pi - 2\pi = -0.5\pi\), the same frequency with opposite rotation. In continuous time, tripling \(\omega_0\) would have tripled the rate; in discrete time it does nothing of the kind.

Review

Chapter Summary

The step

\(u(t)\) switches a signal on; \(u(t-a)-u(t-b)\) windows it. In discrete time \(u[0]=1\) with no ambiguity.

The impulse

Zero everywhere but the origin, unit area, not a function. Defined as a limit and used only inside integrals.

Sifting

\(\int x(t)\delta(t-t_0)dt = x(t_0)\). Factor first when the argument is scaled: \(\delta(at)=\delta(t)/|a|\).

The chain

Doublet → impulse → step → ramp → parabolic, linked by integration one way and differentiation the other.

Exponentials

\(Ce^{st}\) with \(s=\sigma+j\omega_0\) covers decay, growth, oscillation and damped sinusoids in one expression.

Decomposition

\(x[n]=\sum_k x[k]\delta[n-k]\) — every signal is a sum of shifted impulses, which is what makes convolution possible.

Practice

Problems

Problems 1 to 4 drill the definitions; 5 to 8 build and manipulate signals; 9 and 10 ask you to prove the properties you have been using. Sketch before you compute wherever a waveform is involved.

  1. Sketch \(u(t)\), \(u(-t)\), \(u(t-2)\) and \(u(2-t)\) on separate axes, and say in words what each one switches on or off.
  2. Evaluate \(\displaystyle\int_{-\infty}^{\infty}(t^3+2t)\,\delta(t+1)\,dt\).
  3. Evaluate \(\displaystyle\int_{-\infty}^{\infty}\sin(t)\,\delta(3t-\pi)\,dt\). Factor the impulse argument before sifting.
  4. Write the pulse that equals 5 for \(2 \lt t \lt 6\) and zero elsewhere in terms of unit steps, then find and sketch its derivative.
  5. Sketch \(x(t) = r(t) - 2r(t-2) + r(t-4)\) and describe the resulting waveform. What is its peak value, and where does it occur?
  6. Express the signal \(x[n]\) taking values \(1, 2, 0, -1\) at \(n = -1, 0, 1, 2\) as a sum of shifted unit impulses. Then write the same signal using unit steps only.
  7. Sketch \(x(t) = e^{-t/2}\cos(4\pi t)\,u(t)\). Identify the time constant, the oscillation frequency in hertz, and the approximate time at which the envelope has fallen below 2% of its initial value.
  8. Classify \(x(t) = 2\big[u(t) - u(t-3)\big]\) on the eight axes of Chapter 2, and compute its total energy.
  9. Prove that \(\delta(-t) = \delta(t)\) by applying the sifting property to \(\int x(t)\delta(-t)\,dt\) with the substitution \(\lambda = -t\).
  10. Show that \(u[n] = \sum_{k=0}^{\infty}\delta[n-k]\) and \(\delta[n] = u[n]-u[n-1]\). Explain why the continuous-time versions of these two statements require calculus while the discrete-time versions require only algebra.
Tip: when an impulse appears with a scaled or shifted argument, the first move is always to write it as \(\tfrac{1}{|a|}\delta(t-b/a)\). When a waveform is piecewise linear, the first move is always to list the slope changes. Those two habits will carry you through most of the impulse questions you will ever be set.