Signal Operations and Transformations

Chapter 3 assembled a small catalogue of elementary signals and then, in its last section, showed that any signal at all can be written as a sum of shifted, scaled impulses. That sentence already contains the subject of this chapter. The catalogue on its own is inert: a step is a step, an exponential is an exponential. What makes the catalogue powerful is the freedom to place its members — to switch a step on at \(t=3\) rather than at the origin, to run an exponential backwards, to squeeze a pulse into half its duration.

These placements are the signal operations, and they are worth a chapter of their own for two reasons. The first is that they are used everywhere downstream. Convolution, the central operation of Chapter 7, is nothing but a reversal, a shift, a multiplication and an integration, applied in that order; get the reversal or the shift wrong and every convolution you ever compute will be wrong. Sampling in Part 6 is a multiplication by an impulse train. Modulation in Chapter 28 is a multiplication by a sinusoid. The Fourier and Laplace transforms of Parts 4 and 5 acquire most of their useful properties by asking what a shift or a scaling does in the transform domain.

The second reason is more practical: this is where students lose marks. The operations divide into two families, and the families are not equally difficult.

Operations on the dependent variable are the easy family, and they can be disposed of quickly. Amplitude scaling multiplies every value by a constant, \(y(t) = A\,x(t)\). The graph stretches vertically by \(|A|\), and if \(A\) is negative it also flips about the time axis. Nothing moves horizontally: every zero crossing, every corner, every discontinuity stays at exactly the instant it occupied before. In hardware this is an amplifier, or an attenuator, or an inverting op-amp stage.

Addition and subtraction are pointwise: \(y(t) = x_1(t) \pm x_2(t)\) means add the two ordinates at each instant. The support of the sum is the union of the two supports — the result is non-zero wherever either signal is. This is the operation behind every superposition argument in the book, and it is the operation a summing junction performs in a block diagram.

Multiplication is also pointwise, \(y(t) = x_1(t)\,x_2(t)\), but its support is the intersection: the product vanishes wherever either factor vanishes. That single observation explains two of the most useful constructions we already have. Multiplying by \(u(t-t_0)\) is a switch, because the step is zero on half the axis. Multiplying by a rectangular pulse is a window, because the pulse is zero outside a finite interval. Multiplying by \(\cos\omega_c t\) is amplitude modulation, and the fact that this is the same operation as windowing — only with a different second factor — is the reason Part 4 will be able to treat both with a single theorem.

The additive constant \(B\) deserves a remark. Adding a DC offset shifts the whole waveform vertically, and it is the one amplitude operation that can change a signal's classification from Chapter 2: an energy signal plus a non-zero constant is no longer an energy signal, because the constant contributes infinite energy over an infinite interval. Adding \(B\) also destroys oddness. Amplitude scaling by \(A\), by contrast, preserves every classification in Chapter 2 — periodicity, symmetry, energy or power type — and merely multiplies the energy by \(A^2\).

Now the difficult family. The time shift is defined by \(y(t) = x(t - t_0)\), and the perpetual question is which way it moves the graph. The answer is fixed by a one-line argument that is worth internalising, because the same argument settles reversal, scaling and the combined operation as well.

Suppose \(x\) has some identifiable feature — a peak, a corner, a jump — at the instant \(t = \alpha\). Where does \(y\) show that same feature? At whatever \(t\) makes \(y\)'s argument equal \(\alpha\), that is, at \(t - t_0 = \alpha\), or \(t = \alpha + t_0\). So every feature moves \(t_0\) units later. With \(t_0 \gt 0\), the graph moves to the right: the signal is delayed.

Shifting changes nothing about the waveform itself. Duration, peak value, area, energy and average power are all identical before and after, because the integral defining each of them is invariant under a change of origin. Only the position in time changes, which is why a shift can never turn an energy signal into a power signal or a periodic signal into an aperiodic one.

The discrete-time version is \(y[n] = x[n - n_0]\), and here \(n_0\) is forced to be an integer. There is no such thing as a shift by half a sample: the sequence has no values between its samples to move to. This restriction looks like a technicality and is not — recovering a fractional-sample delay from a sampled signal requires interpolation, which requires the sampling theorem, which is Part 6.

Finally, a shift is the operation that separates causal processing from the other kind. A delay can be built: a memory element, a length of cable, a shift register. An advance cannot, because it would require knowing the input before it arrives. Chapter 5 will make that observation into the formal definition of causality, and Chapter 8 will show why a system whose impulse response is shifted left can never be realised in real time.

Time reversal, or folding, is \(y(t) = x(-t)\). Applying the landmark question: a feature of \(x\) at \(t = \alpha\) appears in \(y\) where \(-t = \alpha\), that is at \(t = -\alpha\). The graph is reflected in the vertical axis. A tape played backwards; a right-sided signal becomes left-sided; a causal signal becomes anti-causal.

Reversal about the origin is the standard case, but reflection about an arbitrary instant \(t = c\) comes up often enough — particularly in convolution — to be worth a formula. Reflecting about \(c\) must send \(c\) to itself and \(c + d\) to \(c - d\), which is precisely what the argument \(2c - t\) does.

Reversal, like shifting, leaves duration, energy and average power alone; it is a relabelling of the axis, not a distortion of the waveform. What it does change is symmetry-related information. An even signal satisfies \(x(-t) = x(t)\) and is therefore its own reflection; an odd signal satisfies \(x(-t) = -x(t)\) and is its own reflection with a sign flip. Section 4-8 turns this pair of statements into a decomposition that applies to every signal.

In discrete time, \(y[n] = x[-n]\) reflects the sequence about \(n = 0\). The index that most often goes wrong is the origin itself: \(y[0] = x[0]\) always, so the sample at the origin is the pivot and does not move. If \(x[n]\) is non-zero for \(n = 0,1,2\), then \(x[-n]\) is non-zero for \(n = 0,-1,-2\), and the first sample of the original becomes the last sample of the reflection.

Time scaling is \(y(t) = x(at)\) with \(a\) real and non-zero, and it is the operation most often stated backwards. Run the landmark argument once more. A feature at \(t = \alpha\) in \(x\) appears in \(y\) where \(at = \alpha\), that is at \(t = \alpha/a\). If \(a \gt 1\), every landmark is pulled towards the origin by the factor \(a\): the signal is compressed, it happens faster, and it occupies a shorter interval. If \(0 \lt a \lt 1\), every landmark is pushed away from the origin: the signal is expanded and takes longer.

The intuition that resists this is the graph-drawing habit of reading \(a\) as a stretch factor. The cure is the tape-recorder picture: \(x(2t)\) is the tape played at double speed, which finishes in half the time. A signal of duration \(T\) becomes a signal of duration \(T/|a|\).

Negative \(a\) does both things at once: \(x(-2t)\) is compressed by two and reflected, because \(-2t = 2(-t)\). It is always safe to treat \(|a|\) as the scaling factor and the sign of \(a\) as a separate reversal.

Scaling is the first operation that changes the area under a signal, and the change follows from a substitution. Put \(\tau = at\), so \(d\tau = a\,dt\); for \(a \lt 0\) the limits swap and the two sign changes combine into a modulus.

Real problems rarely ask for a single operation. They ask for \(x(2t+3)\), or \(x(3-2t)\), or \(x(-t/2+1)\), and here the ordering question becomes real: do you shift first and then scale, or scale first and then shift? Both are possible, but the shift amount is not the same in the two routes, and using \(b\) where \(b/a\) belongs is the single most common error in this chapter.

The resolution is algebraic, not visual. Everything follows from one factorisation.

Spelled out, the two routes are these. Route A — shift first. Form \(v(t) = x(t+b)\), which is an advance by \(b\) when \(b \gt 0\). Then compress the result by \(a\): \(y(t) = v(at) = x(at+b)\). Route B — scale first. Form \(w(t) = x(at)\). Then shift by \(b/a\): \(y(t) = w(t + b/a) = x\big(a(t+b/a)\big) = x(at+b)\). Route A shifts by \(b\); Route B shifts by \(b/a\). Neither is more correct, and Route A is usually safer because the shift is read straight off the expression.

In practice neither route is necessary, because the landmark method sidesteps the ordering question entirely. If \(x\) is defined piecewise with break points at \(t = \alpha_1, \alpha_2, \dots\), then \(y(t) = x(at+b)\) has its break points wherever \(at + b = \alpha_i\), that is at \(t = (\alpha_i - b)/a\). Map every break point, carry the corresponding values across, and the sketch draws itself. If \(a \lt 0\) the break points come out in reversed order, which is the reversal announcing itself; that is a check, not a problem.

Every operation so far transfers to discrete time by replacing \(t\) with \(n\) — and two of them break in the process. Shifting survives intact provided \(n_0\) is an integer. Reversal survives without qualification. Scaling does not survive at all, and the reason is that a sequence has nothing between its samples.

Consider \(y[n] = x[2n]\), called decimation or downsampling. Every index \(2n\) is a legitimate integer, so the expression is well defined — but it keeps only the even-numbered samples of \(x\) and discards the odd ones permanently. This is not the harmless compression of continuous time, where \(x(2t)\) contains every value that \(x\) ever took, merely at different instants. Decimation destroys information, and no operation can recover it.

Now consider \(y[n] = x[n/2]\). For odd \(n\) the index \(n/2\) is not an integer and \(x\) has no value there, so the expression is meaningless as written. What is meant is expansion or upsampling, and it is a definition rather than a consequence: put the samples of \(x\) at the even positions and insert zeros at the odd ones.

The pair is deliberately asymmetric. Expansion followed by decimation returns the original exactly, because the decimator simply removes the zeros that the expander inserted. Decimation followed by expansion does not, because the expander has no way to reconstruct the samples the decimator threw away; it can only put zeros where they were. In operator language, expansion is invertible and decimation is not — an idea Chapter 5 will formalise as invertibility, and Chapter 22 will exploit when it builds multirate systems out of these two blocks.

Chapter 2 defined an even signal as one satisfying \(x(-t) = x(t)\) and an odd signal as one satisfying \(x(-t) = -x(t)\). Most signals are neither. The useful fact, and the reason time reversal earns a place in this chapter, is that every signal is the sum of an even signal and an odd signal in exactly one way.

The construction is forced. Suppose \(x(t) = x_e(t) + x_o(t)\) with \(x_e\) even and \(x_o\) odd. Replace \(t\) by \(-t\) throughout: \(x(-t) = x_e(t) - x_o(t)\), because the even part is unchanged and the odd part flips sign. That is two linear equations in the two unknown functions, and adding and subtracting them solves the pair immediately.

A quick check confirms the labels. Replacing \(t\) by \(-t\) in the first expression leaves it unchanged, so it is genuinely even; doing the same in the second reverses its sign, so that one is genuinely odd. The discrete versions are identical with \(n\) in place of \(t\), and the sample at \(n=0\) is worth watching: because \(x_o[0] = (x[0]-x[0])/2 = 0\), an odd sequence always vanishes at the origin, and a continuous odd signal must vanish there too unless it is discontinuous.

The multiplication rules follow from the definitions in one line each, and they are used constantly when evaluating integrals: even × even = even, odd × odd = even, even × odd = odd. Combine the last one with the fact that any odd signal integrates to zero over a symmetric interval, and a genuinely useful consequence appears.

Now expand the energy of \(x\) using \(x = x_e + x_o\). The square produces three terms, and the cross term is exactly twice the integral just shown, so it vanishes. What remains is a statement that the two parts carry the total energy between them with no interaction.

Chapter 2 classified signals by energy and power. It is worth knowing which operations preserve those quantities and which do not, both because examiners ask and because the answers are physically meaningful.

Shifting and reversal change nothing. Both are relabellings of the time axis, and every integral over the whole axis is blind to a relabelling. Amplitude scaling by \(A\) multiplies area by \(A\), and energy and power by \(A^2\). Time scaling is the interesting case, and it treats energy and power differently.

Compressing a signal reduces its energy, because energy is a total and there is less time in which to accumulate it. Average power behaves differently: it is a total divided by the length of the averaging window, and time scaling shrinks the numerator and the denominator by the same factor. Compressing a periodic signal by \(a\) divides its period by \(a\) and divides the energy in one period by \(a\), so the ratio — the average power — is unchanged. A sinusoid of amplitude \(A\) has average power \(A^2/2\) regardless of its frequency, which is the same statement in different words.

Problems 1 to 3 drill the single operations; 4 to 6 combine them; 7 and 8 are about symmetry and energy. In every sketching problem, list the landmark map before you draw anything.