- Why a delay costs phase and nothing else: \(x(t-t_0) \leftrightarrow e^{-j\omega t_0}X(j\omega)\), and why the phase it costs is linear in \(\omega\).
- The frequency-shift or modulation property, and how multiplying by \(\cos\omega_0 t\) splits a spectrum into two half-height copies — the whole of amplitude modulation in one line.
- The symmetry rules that let you halve the work for real signals, and the exact sense in which even signals have real spectra and odd signals imaginary ones.
- Time scaling \(x(at) \leftrightarrow \tfrac{1}{|a|}X(j\omega/a)\), which turns Chapter 13's reciprocal spreading from an observation into a theorem.
- Duality — the near-symmetry of the analysis and synthesis equations, which buys you a second transform pair for every one you derive.
- Differentiation and integration in both domains, including where the stray \(\pi X(0)\delta(\omega)\) in the integration rule comes from.
- The convolution property \(y = x * h \leftrightarrow Y = H(j\omega)X(j\omega)\) — the reason the transform exists at all — and its dual, the multiplication property that explains spectral leakage.
Why Properties Beat Integrals
Chapter 13 obtained every transform the honest way: write down the analysis integral, evaluate it, simplify. That is a reasonable method for \(e^{-at}u(t)\) and for a rectangular pulse, and it becomes unbearable almost immediately afterwards. Consider a pulse that has been delayed by three seconds, compressed by a factor of two, and multiplied by a cosine. Its analysis integral is a genuinely unpleasant object. Yet nothing new has happened: it is still the same pulse, acted on by three operations whose effect on the spectrum we can learn once and apply forever.
That is the change of outlook this chapter asks for. Each property is an entry in a dictionary between an operation on \(x(t)\) and an operation on \(X(j\omega)\). Once the dictionary is in your head, the analysis integral is almost never evaluated again. You look up a base pair in Chapter 13's table, apply two or three properties, and read off the answer — and in an examination you do it in a minute rather than a page.
There is a second reason to take the properties seriously, and it matters more than the arithmetic. Every one of them is a physical statement. Delay does not change which frequencies are present, only when they arrive, so it must leave the magnitude spectrum untouched and act on phase alone. Squeezing a signal in time must broaden it in frequency, because a faster event contains faster oscillations. Passing a signal through a filter must scale each frequency component independently, because Chapter 12 showed that complex exponentials are eigenfunctions. The properties are where the mathematics and the engineering meet.
The structure of what follows will feel familiar. Chapter 11 did exactly this for the Fourier series, and every property proved there has a counterpart here, obtained by the same substitution-and-swap manoeuvres with sums replaced by integrals. Where a proof is a direct copy of the Chapter 11 argument it is stated briefly; where the transform behaves differently — and there are a few places, notably duality and the impulse in the integration rule — the difference is spelled out.
Linearity and the Two Shifts
Linearity comes first because everything else leans on it. The analysis integral is an integral, and integration is linear, so a weighted sum of signals transforms to the same weighted sum of transforms. There is nothing more to prove.
Trivial as it looks, linearity is what makes a transform table useful: any signal you can decompose into table entries has a transform you can write down. It is also the property that fails first when a system is non-linear, which is why Chapter 12 could speak of "new frequencies appearing" only after abandoning linearity.
The time-shift property is the first genuinely informative one. Delay a signal by \(t_0\) and ask what happens to its spectrum. Substituting \(\tau = t - t_0\) in the analysis integral does the whole job:
Delaying a signal multiplies its spectrum by a unit-magnitude factor. Not one frequency component is created, destroyed or resized; each is merely retarded, and a retardation of \(t_0\) seconds costs a component at \(\omega\) exactly \(\omega t_0\) radians of phase.
The proportionality is the point. A delay adds a phase that is linear in frequency, with slope \(-t_0\), and the converse is equally useful: whenever you meet a system whose phase response is a straight line through the origin, you are looking at a pure delay. Chapter 15 builds the whole theory of distortionless transmission on that observation, and Section 15-4 turns the slope of the phase curve into a measurable quantity, the group delay.
Reading the property backwards explains a puzzle from Chapter 13. Two pulses of the same width and height, one centred at the origin and one starting there, have identical magnitude spectra and differ only by a linear phase term — because they are the same signal, shifted. Magnitude alone can never tell you where a signal sits on the time axis; all of that information is stored in the phase.
The frequency-shift property is the mirror image, and its proof is even shorter. Multiplying by a complex exponential before transforming simply merges the two exponentials in the integrand:
Multiplication by \(e^{j\omega_0 t}\) translates the entire spectrum bodily to the right by \(\omega_0\). Multiplication by a real cosine, which is two exponentials of half amplitude, produces two half-height copies, one at \(+\omega_0\) and one at \(-\omega_0\).
The cosine version follows from the exponential version and linearity alone: write \(\cos\omega_0 t = \tfrac12 e^{j\omega_0 t} + \tfrac12 e^{-j\omega_0 t}\), transform each term, add. It is worth doing on paper once, because this single line is the mathematical content of amplitude modulation. A baseband signal occupying \(|\omega| \lt W\) — a voice, say, occupying a few kilohertz — is multiplied by a carrier and lands as two sidebands around \(\pm\omega_c\), where it can be radiated by an antenna of sensible size and where it does not collide with the neighbouring station. Chapter 28 develops the consequences; the mechanism is entirely contained in the property above.
Two details deserve attention. First, the modulated signal occupies a band of width \(2W\) centred on the carrier, twice the bandwidth of the message: the copies at \(+\omega_c\) and \(-\omega_c\) each carry both the positive and negative halves of \(X\). Second, the copies are half height, so modulation costs 6 dB of amplitude — a real and irritating fact in receiver design.
Conjugation and the Symmetries of Real Signals
Section 13-3 stated that a real signal has a conjugate-symmetric spectrum. The general statement behind it is the conjugation property, obtained by conjugating the analysis equation and noticing that \((e^{-j\omega t})^{*} = e^{+j\omega t}\), which is the integrand of \(X\) evaluated at \(-\omega\):
Now let \(x(t)\) be real. Then \(x^{*}(t) = x(t)\), so the two sides of the property describe the same signal and their transforms must agree: \(X(j\omega) = X^{*}(-j\omega)\), which is the conjugate symmetry already used. In magnitude-and-phase terms it says the magnitude spectrum is even and the phase spectrum is odd, which is why every spectrum plotted in this book for a real signal is symmetric about the vertical axis in magnitude and antisymmetric in phase — and why plotting the negative-frequency half is, strictly, redundant.
Pushing one step further pays off. Any signal splits into an even and an odd part, \(x(t) = \mathcal{E}v\{x\} + \mathcal{O}d\{x\}\), as Chapter 2 defined them. For real \(x\), those two parts transform into the two halves of \(X\):
A real even signal therefore has a purely real, even transform — no phase plot is needed — and a real odd signal has a purely imaginary, odd transform, whose phase is \(\pm\pi/2\) everywhere.
These rules are worth more than they appear, because they are free error checks. If you compute the transform of a real even signal and obtain something complex, you have made a mistake, and you know it before the marker does. Chapter 13's pair \(e^{-a|t|} \leftrightarrow 2a/(a^2+\omega^2)\) is real and even on both sides; \(\mathrm{sgn}(t) \leftrightarrow 2/j\omega\) is real and odd on the left, purely imaginary and odd on the right. Neither result could have come out any other way.
| Signal \(x(t)\) | Transform \(X(j\omega)\) | Consequence |
|---|---|---|
| Real | \(X(-j\omega) = X^{*}(j\omega)\) | \(|X|\) even, \(\angle X\) odd |
| Real and even | Real and even | No phase plot needed |
| Real and odd | Purely imaginary and odd | Phase is \(\pm\pi/2\) |
| Imaginary | \(X(-j\omega) = -X^{*}(j\omega)\) | Conjugate antisymmetry |
| \(x^{*}(t)\) | \(X^{*}(-j\omega)\) | The general rule above |
Time Scaling and Reversal
Chapter 13 observed, from the rectangular pulse, that narrowing a signal widens its spectrum, and gave the product of duration and main-lobe width as a constant. Scaling turns that observation into a theorem valid for every signal.
Take \(a \gt 0\) first and substitute \(\tau = at\), so \(dt = d\tau/a\):
For \(a \lt 0\) the substitution also reverses the limits of integration, and undoing that reversal contributes a second minus sign; the two combine into a modulus. The result covers compression, expansion and reversal in one statement.
Compressing a signal in time (\(|a| \gt 1\)) stretches its spectrum by the same factor and shrinks its height by \(1/|a|\). Stretching in time compresses in frequency. Time reversal reverses the spectrum, nothing more.
Two readings of this result are worth keeping. The engineering reading is that speed costs bandwidth: play a recording at twice the speed and every frequency in it doubles, so the channel carrying it must be twice as wide. Transmit data twice as fast — halving the width of each pulse — and the required bandwidth doubles. There is no way around this; it is a property of the transform, not a limitation of any particular technology.
The mathematical reading is that the area under the curve is conserved in a specific sense. Setting \(\omega = 0\) in the scaling result gives \(\tfrac{1}{|a|}X(0)\), which is exactly the area of the compressed signal — narrower by \(|a|\), so smaller by \(|a|\). That is the same \(1/|a|\) that appeared in Chapter 3's impulse-scaling rule \(\delta(at) = \delta(t)/|a|\), and for the same reason.
Duality
Look at the analysis and synthesis equations side by side and the near-symmetry is impossible to miss. They differ only in the sign of the exponent and in a factor of \(2\pi\). That is not an accident of notation, and it can be exploited: every pair you own immediately gives you a second one, free.
Start from the synthesis equation, written with \(\nu\) as the dummy variable, and evaluate it at \(-\omega\) rather than \(t\):
The right-hand side is an analysis integral — the analysis integral of the signal \(X(jt)\), obtained by taking the transform you already have and reading it as a function of time. Its value is \(2\pi x(-\omega)\).
Take any pair, swap the roles of the two domains, insert a factor of \(2\pi\) and reverse the frequency axis. If \(x\) happens to be even the reversal does nothing and the rule reads simply \(X(jt) \leftrightarrow 2\pi x(\omega)\).
The classic application recovers, in two lines, a pair that Chapter 13 obtained with a separate calculation. The rectangular pulse of half-width \(T_1\) transforms to \(2\sin(\omega T_1)/\omega\). Duality says that the signal \(2\sin(T_1 t)/t\) — the same expression, read as a function of time — must transform to \(2\pi\) times the pulse. Dividing through by \(2\pi\) gives the sinc-to-rectangle pair exactly as tabulated.
Duality also organises the properties themselves. Time shifting and frequency shifting are dual statements; so are differentiation in time and differentiation in frequency; and so, as Section 14-8 will show, are convolution and multiplication. Whenever you learn a property, ask what its dual says — half the time you will find you have learned two.
One warning. Duality is a statement about the continuous-time transform, where both domains are continuous and infinite in extent. It does not carry over to the Fourier series, where one domain is a discrete index and the other a continuous variable on a finite interval. Chapter 16 will show that the discrete-time transform has its own broken symmetry for exactly this reason: there, the time variable is discrete and the frequency variable is periodic, so a signal and its transform can never be interchanged.
Differentiation and Integration
These two properties are why the Fourier transform can solve differential equations, and they are the reason Part 5 exists in the form it does. Differentiate the synthesis equation with respect to \(t\), moving the derivative inside the integral, and the exponential brings down a factor of \(j\omega\):
Differentiation becomes multiplication by \(j\omega\). Calculus in time has become algebra in frequency — which is the entire trick on which transform methods rest.
Read the factor \(j\omega\) as an instruction and it tells you what differentiators do to signals. Its magnitude \(|\omega|\) grows without bound, so a differentiator amplifies high frequencies in direct proportion to frequency and is therefore an excellent noise generator — the reason practical circuits differentiate only over a limited band. Its phase is a constant \(+\pi/2\), so every component is advanced by a quarter-cycle, which is exactly what the derivative of a sinusoid does: \(\tfrac{d}{dt}\cos\omega t = \omega\cos(\omega t + \pi/2)\).
Applied to Chapter 9's differential equations, the property is decisive. A constant-coefficient equation \(\sum a_k y^{(k)} = \sum b_k x^{(k)}\) becomes, term by term, a polynomial identity in \(j\omega\), and the ratio of the two polynomials is the frequency response. Chapter 15 opens with exactly that calculation.
Integration is the reverse operation, and it must be handled with a little more care. If \(y(t) = \int_{-\infty}^{t}x(\tau)d\tau\), then \(y'(t) = x(t)\), so \(j\omega Y(j\omega) = X(j\omega)\) and \(Y = X/(j\omega)\) — but only away from \(\omega = 0\). Differentiation annihilates constants, so dividing by \(j\omega\) cannot possibly recover the signal's DC content; that information has to be restored by hand.
Restoring it is straightforward once you ask what \(y(t)\) does at the two ends of the time axis. It starts at zero as \(t \to -\infty\) and settles at \(y(\infty) = \int_{-\infty}^{\infty}x\,dt = X(0)\). A signal that rises from \(0\) to \(X(0)\) has an average value of \(X(0)/2\), and a constant \(c\) transforms to \(2\pi c\,\delta(\omega)\) by Section 13-7. The missing term is therefore \(2\pi \cdot \tfrac12 X(0)\,\delta(\omega)\).
The impulse at the origin carries the DC component that the division by \(j\omega\) cannot represent. It vanishes when \(X(0) = 0\) — that is, when the signal has zero net area and its running integral returns to zero.
Test the rule on the pair you know best. The impulse has \(X(j\omega) = 1\) and \(X(0) = 1\), and its running integral is the unit step, so the property predicts \(U(j\omega) = 1/(j\omega) + \pi\delta(\omega)\) — precisely the transform derived in Section 13-7 by a much longer route through the signum function. Section 14-7 gives a third derivation in a single line.
The dual property, obtained by differentiating the analysis equation with respect to \(\omega\) instead, brings down a factor of \(-jt\) and gives a fast way to transform any signal multiplied by \(t\).
Applied to \(e^{-at}u(t) \leftrightarrow 1/(a+j\omega)\), it produces \(t\,e^{-at}u(t) \leftrightarrow 1/(a+j\omega)^{2}\) in one step, since \(j\,\tfrac{d}{d\omega}(a+j\omega)^{-1} = j\cdot(-j)(a+j\omega)^{-2}\). Repeated \(n\) times it gives the general pair \(\tfrac{t^{n}}{n!}e^{-at}u(t) \leftrightarrow (a+j\omega)^{-(n+1)}\), which is the family you will meet again as repeated poles in Chapter 19.
The Convolution Property
Everything so far has been useful. This one is the reason the transform is taught at all.
Chapter 7 established that an LTI system's output is the convolution of its input with its impulse response, \(y(t) = \int h(\tau)x(t-\tau)d\tau\), and Chapter 8 catalogued how awkward that integral is to evaluate — a fresh integration for every value of \(t\), with limits that change as the two signals slide past each other. Now transform it. Take the transform of \(y\), interchange the order of integration, and recognise the inner integral as the time-shift property:
The single most consequential identity in the subject. An operation that costs an integral for every instant in time costs one multiplication at every frequency.
Nothing about this should be surprising by now, because Chapter 12 reached the same conclusion from the other direction. Complex exponentials are eigenfunctions of LTI systems: feed in \(e^{j\omega t}\) and out comes \(H(j\omega)e^{j\omega t}\), the same exponential scaled by a complex number. The synthesis equation writes any signal as a superposition of exponentials; each is scaled by its own \(H(j\omega)\) on the way through; the output is the superposition of the scaled versions. That is \(Y = HX\) restated in words, and the convolution property is the proof that the two accounts agree.
The consequences run through the rest of the book. A cascade of systems has an overall frequency response equal to the product \(H_1H_2\), so cascading is multiplication and the order cannot matter — the commutativity of convolution that Chapter 8 proved with a change of variable is now obvious. A parallel connection adds, giving \(H_1 + H_2\). An inverse system, if it exists, has \(H_{\text{inv}} = 1/H\), which shows immediately why a system whose frequency response vanishes at some frequency cannot be inverted: the information at that frequency has been destroyed, and no amount of processing recovers it. Equalisers work only where \(|H|\) is small, never where it is zero.
The property also gives the promised one-line derivation of the step transform. Since \(u(t) = \int_{-\infty}^{t}\delta(\tau)d\tau\) is the convolution \(\delta * u\), and since the running integral of any \(x\) is \(x * u\), the integration property of Section 14-6 is nothing but the convolution property applied with \(h = u\): \(Y = X(j\omega)\big[1/(j\omega) + \pi\delta(\omega)\big] = X(j\omega)/(j\omega) + \pi X(0)\delta(\omega)\), the last step by the sampling property of the impulse from Chapter 3.
Multiplication and Windowing
Duality now earns its keep. If convolving in time multiplies in frequency, the dual statement must be that multiplying in time convolves in frequency, and the factor of \(2\pi\) lands where duality says it must.
Multiplying two signals smears each spectrum with the other. The result is generally wider than either factor: convolution of a band of width \(W_1\) with a band of width \(W_2\) occupies \(W_1 + W_2\).
The modulation property of Section 14-2 is the simplest instance. Take \(p(t) = \cos\omega_0 t\), whose transform is the impulse pair \(\pi\delta(\omega-\omega_0) + \pi\delta(\omega+\omega_0)\) from Section 13-7. Convolving anything with an impulse at \(\omega_0\) shifts it to \(\omega_0\), so the result is two copies scaled by \(\pi/2\pi = 1/2\) — the same half-height sidebands obtained earlier by a different route. Consistency between two derivations of the same result is the best evidence that the framework is sound.
The second instance is more sobering, and it governs every measurement you will ever make. No real observation lasts forever: you record a signal for \(T_w\) seconds, which is to say you multiply the true signal by a rectangular window. By the multiplication property, the spectrum you obtain is not the true spectrum but the true spectrum convolved with the window's transform — a sinc whose main lobe has width \(4\pi/T_w\) and whose side lobes decay only as \(1/\omega\).
A third instance is the most important of all, and it is worth flagging now even though its consequences belong to Part 6. Sampling a signal is multiplication by an impulse train. The transform of an impulse train is another impulse train, as Section 13-8 showed, and convolving a spectrum with an impulse train produces periodic replicas of that spectrum, spaced by the sampling frequency. Every question about aliasing, every statement of the sampling theorem, and the entire justification for anti-alias filtering follows from that one application of the multiplication property. Chapter 20 does it properly.
Parseval and the Property Table
Section 13-9 stated Parseval's relation and used it to define the energy spectral density. The multiplication property lets us prove the general form in two lines. Consider \(\int x(t)y^{*}(t)dt\) and replace \(y^{*}(t)\) by the conjugate of its synthesis integral:
Setting \(y = x\) gives the familiar statement: total energy is the same quantity whether you accumulate it over time or over frequency. The transform redistributes energy across the axes; it never creates or destroys it.
Two further identities are not properties in the usual sense but are the fastest checks available, and both are obtained by setting a variable to zero in one of the two equations of the pair. Put \(\omega = 0\) in the analysis equation and \(t = 0\) in the synthesis equation:
The DC value of the spectrum is the area under the signal, and the initial value of the signal is \(1/2\pi\) times the area under the spectrum. Use the first every time you finish a transform: compute the signal's area by eye and check it against \(X(0)\). Worked Example 2 is built around exactly that check, and it catches the commonest error in the differentiation method.
The complete dictionary follows. Learn the derivations rather than the rows — every one of them is a two-line substitution, and a property you can rederive in ten seconds is worth more under examination pressure than a row you half-remember.
| Property | Signal | Transform | What it says |
|---|---|---|---|
| Linearity | \(ax_1(t)+bx_2(t)\) | \(aX_1(j\omega)+bX_2(j\omega)\) | Decompose freely |
| Time shift | \(x(t-t_0)\) | \(e^{-j\omega t_0}X(j\omega)\) | Delay is linear phase |
| Frequency shift | \(e^{j\omega_0 t}x(t)\) | \(X(j(\omega-\omega_0))\) | Modulation translates |
| Conjugation | \(x^{*}(t)\) | \(X^{*}(-j\omega)\) | Real \(\Rightarrow\) \(|X|\) even |
| Time reversal | \(x(-t)\) | \(X(-j\omega)\) | Both axes flip together |
| Scaling | \(x(at)\) | \(\dfrac{1}{|a|}X(j\omega/a)\) | Speed costs bandwidth |
| Duality | \(X(jt)\) | \(2\pi x(-\omega)\) | Every pair gives two |
| Convolution | \(x(t)*h(t)\) | \(X(j\omega)H(j\omega)\) | Filtering is multiplying |
| Multiplication | \(x(t)p(t)\) | \(\dfrac{1}{2\pi}X(j\omega)*P(j\omega)\) | Windowing smears |
| Differentiation | \(dx/dt\) | \(j\omega X(j\omega)\) | Calculus becomes algebra |
| Integration | \(\int_{-\infty}^{t}x(\tau)d\tau\) | \(\dfrac{X(j\omega)}{j\omega}+\pi X(0)\delta(\omega)\) | Restore the DC term |
| Frequency derivative | \(t\,x(t)\) | \(j\,dX(j\omega)/d\omega\) | Dual of differentiation |
| Parseval | \(\int|x|^{2}dt\) | \(\dfrac{1}{2\pi}\int|X|^{2}d\omega\) | Energy is conserved |
Worked Examples
Problem. Let \(x(t) = 2\) for \(1 \lt t \lt 5\) and zero elsewhere. (a) Find \(X(j\omega)\) using the shift property. (b) Find the transform of \(x(2t)\) using the scaling property, and check the answer against the area rule.
Solution (a). The pulse is 4 wide, height 2, centred at \(t = 3\). Start from the centred pulse \(p(t) = 2\) for \(|t| \lt 2\), whose transform is \(2 \cdot 2\sin(2\omega)/\omega\) from the Chapter 13 table, and note that \(x(t) = p(t-3)\):
Check: \(X(0) = \lim_{\omega\to 0} 4\sin(2\omega)/\omega = 8\), and the area of the pulse is \(2 \times 4 = 8\). The magnitude has nulls where \(\sin 2\omega = 0\) with \(\omega \ne 0\), that is at \(\omega = k\pi/2\); the delay contributes phase \(-3\omega\) and nothing else.
Solution (b). Apply \(x(at) \leftrightarrow X(j\omega/a)/|a|\) with \(a = 2\):
Verify directly: \(x(2t)\) equals 2 when \(1 \lt 2t \lt 5\), that is for \(0.5 \lt t \lt 2.5\) — a pulse of width 2, height 2, centred at \(t = 1.5\), of area 4. The result gives \(X(0) = 4\) and a delay term \(e^{-j1.5\omega}\), both correct. Notice what compression did: the nulls moved from \(\omega = \pi/2\) out to \(\omega = \pi\), so the spectrum is twice as wide and half as tall, exactly as the property promises.
Problem. Find the transform of the triangular pulse \(x(t) = 1 - |t|/T\) for \(|t| \le T\), zero elsewhere, without evaluating a single integral.
Solution. The signal is piecewise linear, so differentiating once gives a piecewise-constant signal and twice gives impulses — the technique of Chapter 3, run in reverse. The slope is \(+1/T\) on \((-T,0)\) and \(-1/T\) on \((0,T)\), so the first derivative is a pair of rectangles and the second derivative has one impulse at each slope change, weighted by the change in slope:
Transform both sides. The left becomes \((j\omega)^{2}X(j\omega) = -\omega^{2}X(j\omega)\); the right is a sum of shifted impulses, each contributing a complex exponential:
so that \(X(j\omega) = \dfrac{4}{T\omega^{2}}\sin^{2}\dfrac{\omega T}{2}\), which is tidier written as
Now the check that the method demands. Dividing by \((j\omega)^{2}\) discards any DC content, so the answer is only correct up to a term concentrated at \(\omega = 0\); we must confirm the DC value independently. The area of the triangle is \(\tfrac12(2T)(1) = T\), and the expression above tends to \(T\) as \(\omega \to 0\). They agree, so no correction is needed. The result is also real and even, as it must be for a real even signal, and non-negative — the squared sinc never dips below zero, so this spectrum has no phase reversals at all, unlike the rectangular pulse's.
Problem. Find the transform of \(x(t) = e^{-2t}\cos(5t)\,u(t)\), and say what the spectrum looks like.
Solution. Do not expand the cosine into the integral. Recognise the signal as the causal exponential \(g(t) = e^{-2t}u(t)\), whose transform is \(G(j\omega) = 1/(2+j\omega)\), multiplied by \(\cos 5t\). The modulation property applies at once:
Combining over a common denominator gives \(X(j\omega) = (2+j\omega)\big/\big[(2+j\omega)^{2}+25\big]\), which is the form a table would list, but the two-term version is the one that explains the picture: the lowpass resonance of \(g\), which peaks at \(\omega = 0\) with value \(1/2\), has been copied to \(\omega = \pm 5\) at half height. The magnitude spectrum therefore has two bumps near \(\pm 5\) rad/s, each with a half-power width set by the decay rate \(a = 2\).
This is the spectral signature of a damped oscillation, and it is worth committing to memory: a fast decay makes a broad bump, a slow decay a sharp one. A lightly damped resonator rings for a long time and produces a needle-thin peak, which is precisely the correspondence Chapter 15 exploits when it builds bandpass filters out of second-order sections.
Problem. A system with impulse response \(h(t) = e^{-3t}u(t)\) is driven by \(x(t) = e^{-2t}u(t)\). Find \(y(t)\).
Solution. Chapter 7 would set up \(\int_0^t e^{-2\tau}e^{-3(t-\tau)}d\tau\) and grind through it. Instead, transform, multiply, and invert:
The product is not in the table, but a partial-fraction expansion turns it into two terms that are. Writing \(v = j\omega\), the residues are \(1/(3-2) = 1\) at \(v = -2\) and \(1/(2-3) = -1\) at \(v = -3\):
Two checks. At \(t = 0\) the answer gives \(y(0) = 0\), which is right: the convolution of two causal signals starts from zero. And \(Y(0) = 1/6\) should equal the area of \(y\), which is \(\int_0^{\infty}(e^{-2t}-e^{-3t})dt = \tfrac12 - \tfrac13 = \tfrac16\). The pattern — transform, multiply, expand in partial fractions, invert term by term — is the standard method of Chapter 19, met here for the first time.
Problem. Given \(e^{-|t|} \leftrightarrow 2/(1+\omega^{2})\), find the transform of \(x(t) = 1/(1+t^{2})\).
Solution. The target is the known transform, read as a function of time — precisely the situation duality is built for. With \(g(t) = e^{-|t|}\) and \(G(j\omega) = 2/(1+\omega^{2})\), duality states \(G(jt) \leftrightarrow 2\pi g(-\omega)\), and \(g\) is even so \(g(-\omega) = e^{-|\omega|}\):
The direct evaluation of \(\int (1+t^{2})^{-1}e^{-j\omega t}dt\) requires contour integration; duality reduces it to dividing by two. Check the result with the area rule: \(x(0) = 1\) should equal \(\tfrac{1}{2\pi}\int \pi e^{-|\omega|}d\omega = \tfrac{1}{2\pi}\cdot\pi\cdot 2 = 1\). It does.
Problem. A pure cosine \(\cos\omega_0 t\) is observed only over \(|t| \lt T\). Find the spectrum of what you actually recorded, and deduce how close two tones may be before the record cannot distinguish them.
Solution. Observing over a finite interval means multiplying by the window \(p(t) = 1\) for \(|t| \lt T\), whose transform is \(P(j\omega) = 2\sin(\omega T)/\omega\). Apply the modulation property to \(x(t) = p(t)\cos\omega_0 t\):
The two impulses that an everlasting cosine would have produced have become two sinc lobes of height \(T\), centred at \(\pm\omega_0\). Their first nulls lie at \(\omega - \omega_0 = \pm\pi/T\), so each lobe has a main-lobe width of \(2\pi/T\), and as \(T \to \infty\) the lobes grow taller and narrower with constant area — Chapter 3's limiting definition of the impulse, arriving from the frequency side.
Two tones can be told apart only if their lobes do not merge, which requires their separation to exceed roughly the main-lobe half-width \(\pi/T\). Writing the total record length as \(T_w = 2T\), the condition is \(\Delta\omega \gtrsim 2\pi/T_w\), or in hertz \(\Delta f \gtrsim 1/T_w\). Concretely: a 20 ms recording cannot resolve tones closer than about 50 Hz, whatever processing is applied afterwards. The remedy is not a cleverer algorithm but a longer observation — a limitation, not of the method, but of the information present in the data.
Chapter Summary
\(x(t-t_0) \leftrightarrow e^{-j\omega t_0}X(j\omega)\). Magnitude untouched; phase falls linearly with slope \(-t_0\).
\(x(t)\cos\omega_0 t\) puts half-height copies of \(X\) at \(\pm\omega_0\). Bandwidth doubles; amplitude halves.
\(x(at) \leftrightarrow X(j\omega/a)/|a|\). Compress in time, expand in frequency — speed always costs bandwidth.
\(X(jt) \leftrightarrow 2\pi x(-\omega)\). Every derived pair yields a second one, and properties come in dual pairs too.
\(dx/dt \leftrightarrow j\omega X\); integration divides by \(j\omega\) and restores DC with \(\pi X(0)\delta(\omega)\).
\(Y = HX\) — the identity that makes filtering a design problem, with windowing as its dual and its price.
Problems
Problems 1 to 3 exercise the shifting and scaling rules; 4 and 5 use differentiation and duality to avoid integrals altogether; 6 to 8 apply convolution and multiplication. In every case, quote the property you are using and check the result with \(X(0) = \int x\,dt\) before moving on.
- Given \(x(t) \leftrightarrow X(j\omega)\), find the transforms of \(x(t-2)\), \(x(2-t)\), \(x(3t-6)\) and \(x(t)e^{j4t}\). For the third, factor the argument before applying any property.
- A pulse of height 3 occupies \(0 \lt t \lt 2\). Find \(X(j\omega)\), sketch \(|X(j\omega)|\), and state the frequency of the first null. Then write down the transform of the same pulse delayed by 5 s and explain why the sketch does not change.
- Show that if \(x(t)\) is real, the signal \(x(t)\cos\omega_0 t\) has a magnitude spectrum symmetric about \(\omega_0\) for \(\omega \gt 0\), provided \(\omega_0\) exceeds the bandwidth of \(x\). What goes wrong when it does not?
- Find the transform of the trapezoidal pulse that rises linearly from 0 to 1 over \(-2 \lt t \lt -1\), holds at 1 for \(|t| \le 1\), and falls linearly to 0 over \(1 \lt t \lt 2\), by differentiating twice. Verify \(X(0)\) against the area.
- Use duality on the pair \(\delta(t) \leftrightarrow 1\) to find the transform of the constant signal \(x(t) = 1\), and confirm that it agrees with Section 13-7.
- Use the convolution property to evaluate \(e^{-t}u(t) * e^{-t}u(t)\), and confirm the answer by the frequency-differentiation property applied to \(1/(1+j\omega)\).
- A signal band-limited to \(|\omega| \lt 100\) rad/s is multiplied by \(\cos(1000t)\) and the product is passed through an ideal bandpass filter that keeps \(900 \lt |\omega| \lt 1100\). Sketch the spectrum at each stage, and state the bandwidth of the output.
- Prove the multiplication property directly from the analysis equation, then use it to show that the transform of \(\cos^{2}(\omega_0 t)\,p(t)\), where \(p\) is a rectangular window, contains energy near \(\omega = 0\) and near \(\omega = \pm 2\omega_0\) but nowhere near \(\pm\omega_0\).