- The limiting argument that turns a Fourier series into a Fourier transform: hold the signal fixed, let \(T \to \infty\), and watch \(Ta_k\) become an envelope.
- The transform pair — analysis \(X(j\omega) = \int x(t)e^{-j\omega t}dt\) and synthesis \(x(t) = \frac{1}{2\pi}\int X(j\omega)e^{j\omega t}d\omega\) — and why \(X(j\omega)\) is a density, not an amplitude.
- When the transform exists, what it converges to at a jump, and which signals need impulses to be transformed at all.
- The handful of transforms everything else is built from, including the rectangular pulse and its sinc.
- Reciprocal spreading: short in time means wide in frequency, always, with a product that cannot be beaten.
- Spectra containing impulses — the constant, the sinusoid, the signum and the unit step — and why the transform of a periodic signal is a line spectrum made of impulses.
- Parseval's relation for energy signals and the energy spectral density \(|X(j\omega)|^2\).
Why the Series Is Not Enough
Part 3 built a complete theory for periodic signals. Chapter 10 decomposed them into harmonics, Chapter 11 gave the decomposition its algebra, and Chapter 12 showed that an LTI system acts on the harmonics by simple multiplication. It is a powerful and self-consistent story, and it applies to almost nothing that engineers actually measure.
A spoken word, a radar echo, a switching transient, a single data pulse, the impulse response of a filter — none of these repeats. Ask for their fundamental frequency and the honest answer is \(\omega_0 = 2\pi/T\) with \(T\) infinite, which is to say zero, which is to say the harmonics are packed infinitely close together and every coefficient \(a_k\) has been divided by an infinite \(T\) and is zero. The series machinery does not merely become inconvenient; it degenerates.
And yet the intuition it was built on plainly survives. A short pulse audibly contains high frequencies and a slow one does not; an oscilloscope's bandwidth decides whether an edge looks sharp; a radio channel passes some signals and mangles others. Nobody believes an aperiodic signal has no frequency content. What is needed is a decomposition that does not depend on periodicity.
Notice also how little of Chapter 12 actually used periodicity. The eigenfunction property — that \(e^{j\omega t}\) emerges from an LTI system multiplied by \(H(j\omega)\) — was proved straight from the convolution integral and never mentioned a period. Only the decomposition of the input into exponentials needed the signal to repeat. Repair that one step and the whole apparatus of frequency response, filtering and distortionless transmission carries over untouched.
From Series to Transform
Let \(x(t)\) be a signal of finite duration, zero outside \(|t| \le T_1\). It is not periodic, so build a periodic signal that agrees with it: let \(\tilde{x}(t)\) repeat \(x(t)\) every \(T\) seconds, with \(T \gt 2T_1\) so the copies do not overlap. Over the interval \(|t| \lt T/2\) the two signals are identical, and as \(T\) grows the neighbouring copies retreat towards infinity, so \(\tilde{x}(t) \to x(t)\) for every fixed \(t\).
Now compute the Fourier coefficients of \(\tilde{x}\). Over one period \(\tilde{x}\) equals \(x\), and outside the pulse \(x\) is zero, so the limits of the analysis integral may be pushed out to infinity without changing anything:
The integral on the right is a function of the single variable \(k\omega_0\) and involves \(x\), not \(\tilde{x}\), and not \(T\). Give it a name. Define
This is the pivot of the entire chapter, and it deserves a slow reading. The Fourier coefficients of the periodic extension, scaled by \(T\), are samples of one fixed function of frequency, taken at the harmonic frequencies \(k\omega_0\). The function \(X(j\omega)\) does not know or care what \(T\) is; it belongs to the pulse alone. Changing \(T\) does exactly two things: it changes how finely the envelope is sampled, since the spacing is \(\omega_0 = 2\pi/T\), and it scales every coefficient down by \(1/T\).
The figure shows this happening. Doubling the period halves the line spacing and halves every line's height, but the shape the lines trace out is identical, because it was never a property of the periodic signal in the first place.
Now run the synthesis equation through the same substitution. Writing \(a_k = X(jk\omega_0)/T\) and using \(1/T = \omega_0/2\pi\):
The last expression is a Riemann sum for an integral: the summand is a function evaluated at \(\omega = k\omega_0\), multiplied by the width \(\omega_0\) of the strip it represents. As \(T \to \infty\), \(\tilde{x}(t) \to x(t)\) on the left, the strip width \(\omega_0 \to 0\) on the right, and the sum becomes an integral over all \(\omega\). That limit is the synthesis equation of the Fourier transform, and the pair is complete.
The Transform Pair
The first is the analysis equation, which measures how much of \(e^{j\omega t}\) the signal contains; the second is the synthesis equation, which rebuilds the signal from a continuum of exponentials. We write \(x(t) \leftrightarrow X(j\omega)\), or \(X(j\omega) = \mathcal{F}\{x(t)\}\).
Compare the pair with the series pair of Chapter 10 and the correspondence is exact: the sum over \(k\) has become an integral over \(\omega\), the discrete coefficients \(a_k\) have become the continuous function \(X(j\omega)\), and the \(1/T\) of the analysis equation has migrated to the \(1/2\pi\) of the synthesis equation. That stray \(2\pi\) is pure bookkeeping — it appears because we chose to work with \(\omega\) in radians per second rather than \(f\) in hertz. Writing the pair in terms of \(f\) makes it disappear entirely and makes both directions symmetric, which is why communication engineers usually prefer it.
One point of interpretation must be got right, and it is the point students most often slide past. \(X(j\omega)\) is not the amplitude of the component at frequency \(\omega\). An aperiodic signal has no finite amount of any single frequency — the synthesis equation says the contribution from the band \([\omega, \omega + d\omega]\) is \(X(j\omega)e^{j\omega t}\,d\omega/2\pi\), which vanishes as \(d\omega\) does. \(X(j\omega)\) is a spectral density: an amount of signal per unit of frequency. That is why its units are those of \(x\) multiplied by time — volt-seconds for a voltage — and why a plot of \(|X(j\omega)|\) is a continuous curve rather than a set of lines.
As before, \(X(j\omega)\) is complex and needs two plots: the magnitude spectrum \(|X(j\omega)|\) and the phase spectrum \(\angle X(j\omega)\). Taking the conjugate of the analysis equation for real \(x(t)\) gives \(X^{*}(j\omega) = \int x(t)e^{j\omega t}dt = X(-j\omega)\), so every symmetry of Chapter 12 survives the transition intact.
Magnitude even, phase odd. If \(x(t)\) is real and even, the imaginary part of the integrand integrates to zero and \(X(j\omega)\) is real and even — no phase plot is needed at all.
Existence and Convergence
The analysis equation is an improper integral, so it need not converge, and the limiting argument of Section 13-2 was carried out with a light hand. The conditions that make everything legitimate are the direct descendants of the Dirichlet conditions of Chapter 11, with the integral over one period replaced by an integral over all time.
together with a finite number of maxima and minima, and a finite number of finite discontinuities, in any finite interval. Under these conditions \(X(j\omega)\) exists, is continuous and bounded, and the synthesis integral returns \(x(t)\) everywhere except at a jump, where it returns the midpoint of the two one-sided values.
The midpoint behaviour at a discontinuity is the same result Chapter 11 proved for the series, arrived at by the same mechanism, and it should not be surprising: the synthesis integral is a sum of continuous functions and cannot be asked to reproduce a jump exactly at the jump. The Gibbs overshoot survives the transition too, appearing whenever a spectrum is truncated at some finite \(\omega\) — which is what every real filter does.
Absolute integrability is sufficient but not necessary. A second class of signals is admitted by a different argument: any energy signal, one with \(\int|x(t)|^2dt \lt \infty\), has a transform in the mean-square sense, even if \(\int|x|dt\) diverges. The signal \(\sin(Wt)/(\pi t)\) is the standard example — it decays too slowly to be absolutely integrable but its square is integrable, and its transform, as Worked Example 4 shows, is a perfectly ordinary rectangle.
Finally there are the signals that satisfy neither condition and yet are indispensable: a constant, a sinusoid, the unit step, a periodic signal. Their analysis integrals diverge outright. Rather than exclude them we extend the theory to allow impulses in the frequency domain, exactly as Chapter 3 extended the class of signals to allow an impulse in time. Section 13-7 does this, and it is neither a trick nor a fudge — it is the same generalised-function idea, applied to the other axis.
The Transforms You Must Know
Four or five transforms, honestly derived once, will carry you through most of Part 4. Every one of them is a direct evaluation of the analysis integral.
The one-sided exponential. Take \(x(t) = e^{-at}u(t)\) with \(a \gt 0\). The step confines the integral to \(t \gt 0\):
The upper limit vanishes because \(|e^{-(a+j\omega)t}| = e^{-at} \to 0\), and this is precisely where \(a \gt 0\) is needed: a growing exponential has no transform. Splitting into magnitude and phase, \(|X| = 1/\sqrt{a^2+\omega^2}\) and \(\angle X = -\arctan(\omega/a)\) — the \(RC\) lowpass response of Chapter 12 all over again, which is no coincidence, since \(e^{-t/RC}u(t)/RC\) is that filter's impulse response.
The two-sided exponential. For \(x(t) = e^{-a|t|}\), split the integral at the origin and evaluate each half:
Real and even, as promised for a real even signal. The impulse is even easier: sifting gives \(\mathcal{F}\{\delta(t)\} = \int\delta(t)e^{-j\omega t}dt = e^{-j0} = 1\). An impulse contains every frequency in equal measure, with zero phase — the reason it is the ideal test signal, and the reason engineers call it "white". Shifting it gives \(\mathcal{F}\{\delta(t-t_0)\} = e^{-j\omega t_0}\): unit magnitude at every frequency and phase linear in \(\omega\), which is Chapter 12's statement that linear phase is a pure delay, seen from the other side.
The rectangular pulse is the most important of all. Let \(x(t) = 1\) for \(|t| \lt T_1\) and zero elsewhere:
Three checks confirm the result. At \(\omega = 0\) the limit is \(2T_1\), which is the area under the pulse — and indeed \(X(0) = \int x(t)dt\) always, straight from the analysis equation with \(\omega\) set to zero. The function is real and even, as it must be for a real even pulse. And it vanishes whenever \(\omega T_1\) is a non-zero multiple of \(\pi\), so the zero crossings sit at \(\omega = k\pi/T_1\).
Writing the answer with the sinc function of Chapter 3, \(\operatorname{sinc}(\theta) = \sin(\pi\theta)/(\pi\theta)\), gives \(X(j\omega) = 2T_1\operatorname{sinc}(\omega T_1/\pi)\). Books differ over whether the \(\pi\) belongs inside the definition; check which convention is in use before quoting a formula from one.
| Signal \(x(t)\) | Transform \(X(j\omega)\) | Remark |
|---|---|---|
| \(\delta(t)\) | \(1\) | All frequencies equally, zero phase |
| \(\delta(t-t_0)\) | \(e^{-j\omega t_0}\) | Flat magnitude, linear phase |
| \(e^{-at}u(t),\ a \gt 0\) | \(\dfrac{1}{a+j\omega}\) | Bandwidth grows with \(a\) |
| \(e^{-a|t|},\ a \gt 0\) | \(\dfrac{2a}{a^2+\omega^2}\) | Real and even |
| \(1\) for \(|t| \lt T_1\), else \(0\) | \(\dfrac{2\sin\omega T_1}{\omega}\) | Zeros at \(\omega = k\pi/T_1\) |
| \(\dfrac{\sin Wt}{\pi t}\) | \(1\) for \(|\omega| \lt W\), else \(0\) | The same pair, read backwards |
Reciprocal Spreading
The last two rows of that table are the same statement written twice, and the symmetry between them is not an accident — Chapter 14 will name it duality. Before that, there is a physical consequence of the rectangular pulse pair that deserves its own section, because it constrains every engineering system that has ever been built.
The pulse occupies \(2T_1\) seconds and its transform's main lobe occupies \(2\pi/T_1\) radians per second, from the first zero on the left to the first on the right. Multiply the two:
Compressing a signal in time stretches its spectrum by exactly the same factor. The product cannot be reduced by any choice of pulse width, and a more careful argument shows it cannot be reduced by any choice of pulse shape either.
The exponential tells the same story with different arithmetic. For \(e^{-at}u(t)\) the time constant is \(\tau = 1/a\) and the half-power bandwidth is \(\omega = a\), since \(|X|\) falls to \(1/\sqrt2\) of its peak there. Their product is 1, whatever \(a\) may be. A fast-decaying signal is a wideband signal; there is no third option.
Impulses in Frequency
A constant signal \(x(t) = 1\) is not absolutely integrable and its analysis integral \(\int e^{-j\omega t}dt\) does not converge in any ordinary sense. But the constant is far too useful to abandon, so approach it as a limit, exactly as Chapter 3 approached the impulse itself.
Take the two-sided exponential \(e^{-a|t|}\), whose transform we know to be \(2a/(a^2+\omega^2)\), and let \(a \to 0\). In the time domain \(e^{-a|t|} \to 1\) for every \(t\). In the frequency domain something more interesting happens. For any fixed \(\omega \ne 0\) the transform tends to \(2a/\omega^2 \to 0\), so the limit is zero everywhere off the origin. Yet its total area never changes:
Zero everywhere except the origin, with constant area \(2\pi\): that is the definition of \(2\pi\delta(\omega)\). So \(1 \leftrightarrow 2\pi\delta(\omega)\), and the result can be checked instantly against the synthesis equation, since \(\frac{1}{2\pi}\int 2\pi\delta(\omega)e^{j\omega t}d\omega = e^{j0} = 1\) by sifting. It is also exactly what reciprocal spreading demands: a signal of infinite duration must have a spectrum of zero width.
From that one result the rest of the family follows. Shifting the impulse in frequency corresponds to multiplying by a complex exponential in time, which the synthesis equation verifies in one line, and Euler's formula then delivers the sinusoids.
and \(\sin\omega_0 t \leftrightarrow j\pi\big[\delta(\omega+\omega_0) - \delta(\omega-\omega_0)\big]\). A pure tone that has been running forever occupies a single frequency exactly, so its spectral density is infinite there and zero elsewhere — an impulse is the only object that can express that.
The signum function needs the same limiting treatment. Approximate \(\operatorname{sgn}(t)\) by \(e^{-a|t|}\operatorname{sgn}(t)\), which is absolutely integrable for \(a \gt 0\):
No impulse appears this time, because \(\operatorname{sgn}(t)\) has no average value to concentrate at the origin. The unit step does, and the cleanest way to see it is to write the step as an average value plus a signum, \(u(t) = \tfrac12 + \tfrac12\operatorname{sgn}(t)\), and transform each piece with the results just obtained.
The impulse term carries the step's DC content — its non-zero average — and the \(1/j\omega\) term carries the switching edge. Dropping the impulse is the single most common error in Part 4, and it always shows up as a missing constant somewhere downstream.
| Signal | Transform | Why it needs impulses |
|---|---|---|
| \(1\) | \(2\pi\delta(\omega)\) | Infinite duration, so zero bandwidth |
| \(e^{j\omega_0 t}\) | \(2\pi\delta(\omega-\omega_0)\) | All its content at one frequency |
| \(\cos\omega_0 t\) | \(\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]\) | Real, so the spectrum is even |
| \(\sin\omega_0 t\) | \(j\pi[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]\) | Real and odd, so the spectrum is imaginary |
| \(\operatorname{sgn}(t)\) | \(\dfrac{2}{j\omega}\) | No impulse — zero average value |
| \(u(t)\) | \(\pi\delta(\omega) + \dfrac{1}{j\omega}\) | Average value \(\tfrac12\) sits at \(\omega = 0\) |
Periodic Signals Again
With impulses available in frequency, periodic signals can be brought back inside the theory rather than left outside it. A periodic signal is a sum of everlasting complex exponentials, and we have just transformed one of those. Transform the sum term by term:
The line spectrum of Part 3 is the Fourier transform: each line is an impulse at the harmonic frequency, of area \(2\pi a_k\). One framework now covers periodic and aperiodic signals together.
This is a genuine unification, not a relabelling. It means Chapter 12's rule \(b_k = H(jk\omega_0)a_k\) and the transform-domain rule that Chapter 15 will prove are the same statement, and it means a signal made of a pulse plus a hum can be handled in one calculation instead of two.
The example that matters most is the impulse train \(p(t) = \sum_n \delta(t-nT)\), the signal that models sampling. Its Fourier coefficients follow from sifting over one period, where the only impulse present is the one at the origin:
An impulse train in time is an impulse train in frequency, and the spacings are reciprocal: crowding the impulses closer together in time drives them further apart in frequency. Chapter 20 will show that sampling a signal is multiplication by \(p(t)\), that multiplication in time is convolution in frequency, and that convolving a spectrum with this comb therefore makes copies of it spaced \(2\pi/T\) apart. The whole sampling theorem is contained in this one transform pair; we are simply not yet equipped to unpack it.
Parseval and Energy Spectral Density
Chapter 11's Parseval relation accounted for the power of a periodic signal, harmonic by harmonic. Aperiodic signals of the kind treated here have finite energy and zero average power, so the corresponding statement must be about energy, and it comes out of substituting the synthesis equation into the energy integral.
Exchanging the order of integration, the inner integral over \(t\) is \(\int x(t)e^{-j\omega t}dt = X(j\omega)\), and what remains is the result.
Energy may be computed in either domain. The quantity \(|X(j\omega)|^2\) is the energy spectral density: integrating it over a band of frequencies gives the energy the signal carries in that band.
Its practical use is the same as the power spectrum's in Chapter 12. Because energy is additive across frequency with no cross terms, "how much of this pulse lies below 10 kHz" is a well-posed question with a numerical answer, and the essential bandwidth of an energy signal is the band containing an agreed fraction — usually 90%, 95% or 99% — of the total. Worked Example 5 computes one from end to end.
Notice that the phase spectrum has disappeared from this accounting entirely: \(|X|^2\) discards it. Two signals with identical magnitude spectra and wildly different phase spectra have exactly the same energy and the same energy distribution, while looking nothing alike in the time domain. Energy is blind to phase; shape is not.
Worked Examples
Problem. Find the transform of \(x(t) = e^{-3t}u(t)\), evaluate its magnitude and phase at \(\omega = 0\) and \(\omega = 3\), and compute its energy in both domains.
Solution. With \(a = 3\) the standard result gives \(X(j\omega) = 1/(3+j\omega)\), so
At \(\omega = 0\): \(|X| = 1/3 = 0.333\) with zero phase. At \(\omega = 3\): \(|X| = 1/\sqrt{18} = 0.2357\), which is \(1/\sqrt2\) of the peak, and \(\angle X = -45^\circ\). The half-power bandwidth is therefore 3 rad/s, equal to \(1/\tau\) — reciprocal spreading in one line.
In the time domain the energy is \(E = \int_0^{\infty}e^{-6t}dt = 1/6\). In the frequency domain, using \(\int_{-\infty}^{\infty}d\omega/(a^2+\omega^2) = \pi/a\):
The two agree, as they must. When an integral in one domain looks unpleasant, this identity is often the fastest way to evaluate it — that is how the frequency-domain integral above was done in reverse.
Problem. Find and compare the transforms of a unit-height pulse with \(T_1 = 1\) and one with \(T_1 = 2\). Evaluate the first at \(\omega = \pi/2\).
Solution. Substituting into \(X(j\omega) = 2\sin(\omega T_1)/\omega\):
The peaks are \(X_1(0) = 2\) and \(X_2(0) = 4\), each equal to the area of its pulse. The first zeros are at \(\omega = \pi\) and \(\omega = \pi/2\) respectively: the pulse that is twice as long has a spectrum that is half as wide and twice as tall. Doubling the duration has therefore quadrupled the peak of the energy density \(|X|^2\) while halving the bandwidth, and the two changes together leave the energy merely doubled — as it must be, since the pulse itself is twice as long.
At \(\omega = \pi/2\), \(X_1 = 2\sin(\pi/2)/(\pi/2) = 4/\pi = 1.273\). Since \(x(t)\) is real and even, \(X_1\) is real, so the phase is \(0\) wherever \(X_1 \gt 0\) and \(\pi\) wherever \(X_1 \lt 0\) — a sinc has no phase in between, only sign changes at its zeros.
Problem. Find the transform of (a) \(x(t) = 1 + 2\cos 3t\); (b) \(x(t) = 3\delta(t-2)\); (c) \(x(t) = 4u(t)\).
Solution (a). Transform each term with the results of Section 13-7 and add:
The factor is \(2\pi\) rather than \(\pi\) on the cosine terms because the cosine carries a coefficient of 2. Three impulses, at \(0\) and \(\pm3\) rad/s — a line spectrum, exactly as Chapter 12 would have drawn it, now expressed as a transform.
Solution (b). Sifting gives \(X(j\omega) = 3\int\delta(t-2)e^{-j\omega t}dt = 3e^{-j2\omega}\). Magnitude 3 at every frequency, phase \(-2\omega\): flat magnitude and linear phase, which is Chapter 12's distortionless channel with \(K = 3\) and \(t_d = 2\).
Solution (c). Linearity on the step result: \(X(j\omega) = 4\pi\delta(\omega) + 4/(j\omega)\). The impulse must be kept. A step has a non-zero average value of \(2\) over all time, and that average has nowhere to live except at \(\omega = 0\).
Problem. Find the signal whose transform is \(X(j\omega) = 1\) for \(|\omega| \lt W\) and \(0\) beyond. Interpret the answer as the impulse response of an ideal lowpass filter.
Solution. Put the rectangle into the synthesis equation, where it merely truncates the limits:
At \(t = 0\) the limit is \(W/\pi\), and the zeros sit at \(t = k\pi/W\). This is the same rectangle-and-sinc pair as Section 13-5 with the roles of the two axes exchanged — the first sighting of duality.
Read as a system, \(X(j\omega)\) is the frequency response of an ideal lowpass filter and \(x(t)\) is its impulse response. It is non-zero for \(t \lt 0\): the filter responds before it is struck. That settles the debt from Chapter 12 — an ideal filter is non-causal and therefore cannot be built — and it also shows the price of trying, since the impulse response decays only as \(1/t\) and would have to be truncated.
Problem. For \(x(t) = e^{-2|t|}\), find the total energy and the fraction of it that lies in the band \(|\omega| \lt 2\) rad/s.
Solution. The transform is \(X(j\omega) = 4/(4+\omega^2)\). Total energy is quickest in the time domain:
For the band, integrate the energy spectral density using \(\displaystyle\int\frac{d\omega}{(a^2+\omega^2)^2} = \frac{\omega}{2a^2(a^2+\omega^2)} + \frac{1}{2a^3}\arctan\frac{\omega}{a}\) with \(a = 2\):
so the fraction is \(0.4092/0.5 = 0.818\). About 82% of the signal's energy lies below 2 rad/s, which is the frequency at which \(|X|\) has fallen to half its peak. Widening the band to \(|\omega| \lt 4\) captures 96%; the tail contributes little, which is why truncating it is usually harmless.
Problem. A signal is sampled every \(T = 1\) ms, modelled by \(p(t) = \sum_n \delta(t - nT)\). Find \(P(j\omega)\), state the spacing and weight of its impulses, and say what happens to the spectrum if the sampling rate is doubled.
Solution. From Section 13-8, \(P(j\omega) = (2\pi/T)\sum_k\delta(\omega - 2\pi k/T)\). With \(T = 10^{-3}\) s:
The frequency-domain comb has teeth every 1 kHz, each of area 6283.2. Doubling the sampling rate to 2 kHz halves \(T\), which doubles both the spacing and the weight: the teeth move to every 2 kHz and grow.
That widening spacing is the whole reason a fast sampler is a good sampler. Chapter 20 will show that sampling replicates a signal's spectrum at every tooth of this comb, so the further apart the teeth, the more room each copy has before it overlaps its neighbour — and overlap is aliasing.
Chapter Summary
For a pulse repeated with period \(T\), \(Ta_k = X(jk\omega_0)\). As \(T \to \infty\) the lines close up under a fixed envelope and the synthesis sum becomes an integral.
\(X(j\omega) = \int x(t)e^{-j\omega t}dt\) and \(x(t) = \frac{1}{2\pi}\int X(j\omega)e^{j\omega t}d\omega\). \(X(j\omega)\) is a spectral density, not an amplitude.
Absolute integrability plus the Dirichlet conditions suffices; energy signals qualify too; everything else needs impulses in frequency.
A pulse of half-width \(T_1\) transforms to \(2\sin(\omega T_1)/\omega\), with peak \(2T_1\) and zeros at \(k\pi/T_1\). Reading it backwards gives the ideal filter's impulse response.
Duration times bandwidth is a constant. Short pulses are wideband, everlasting signals are impulses in frequency, and no shape evades the trade.
\(1 \leftrightarrow 2\pi\delta(\omega)\), \(u(t) \leftrightarrow \pi\delta(\omega)+1/j\omega\), and periodic signals become impulse trains. Energy is \(\frac{1}{2\pi}\int|X|^2d\omega\).
Problems
Problems 1 to 4 are direct evaluations of the analysis integral; 5 and 6 work in the other direction; 7 and 8 ask you to reproduce derivations from the chapter. Sketch every magnitude spectrum you compute — the sketch catches sign errors that algebra hides.
- Find \(X(j\omega)\) for \(x(t) = e^{-5t}u(t)\). Sketch the magnitude and phase spectra, and find the frequency at which the magnitude has fallen to \(1/\sqrt2\) of its peak.
- Find the transform of the causal pulse \(x(t) = A\) for \(0 \lt t \lt \tau\) and zero elsewhere. Show that \(|X(j\omega)|\) is identical to that of the centred pulse of the same width and height, and that the two phases differ by a term linear in \(\omega\). Explain why, in one sentence.
- Find and sketch the transform of \(x(t) = \delta(t+1) + \delta(t-1)\), and show that it is \(2\cos\omega\). Why is the result real?
- Compute the transform of \(x(t) = e^{-a|t|}\) for \(a = 1\) and \(a = 4\), sketch both magnitude spectra on the same axes, and state how the half-power bandwidth depends on \(a\).
- Find the inverse transform of \(X(j\omega) = 2\pi\delta(\omega) + \pi\delta(\omega-4) + \pi\delta(\omega+4)\).
- Find the inverse transform of \(X(j\omega) = 1\) for \(2 \lt |\omega| \lt 4\) and zero elsewhere, by writing it as the difference of two ideal lowpass spectra.
- Starting from \(e^{-a|t|}\operatorname{sgn}(t)\), derive \(\mathcal{F}\{\operatorname{sgn}(t)\} = 2/j\omega\), then deduce the transform of \(u(t)\). Explain in words why the step needs an impulse at \(\omega = 0\) and the signum does not.
- Verify Parseval's relation for \(x(t) = e^{-t}u(t)\) by computing \(\int|x|^2dt\) and \(\frac{1}{2\pi}\int|X|^2d\omega\) separately. Then find the fraction of the energy lying below \(\omega = 1\) rad/s.