- The band-limited interpolation formula \(x_r(t)=\sum_n x(nT)\,\mathrm{sinc}\!\big((t-nT)/T\big)\), derived by taking the impulse response of the ideal recovery filter of Chapter 20 seriously.
- Why that formula is exact at the samples but unrealisable everywhere else — non-causal, infinite in extent, and decaying only as \(1/t\).
- What aliasing physically is: overlapping replicas adding, high frequencies acquiring the identity of low ones, and information destroyed beyond recovery.
- The apparent-frequency rule \(f_a=|f-kf_s|\) and the folding diagram that turns it into a picture, including the phase inversion that accompanies a fold.
- Why an anti-aliasing filter must sit before the sampler and can never be retro-fitted after it, and how filter order trades against sampling rate.
- The zero-order hold: its \(T/2\) delay, its \(\mathrm{sinc}\) droop of \(-3.92\) dB at the Nyquist frequency, its useful nulls at multiples of \(f_s\), and the inverse-sinc equaliser that corrects it.
- How oversampling converts a nearly impossible analogue filter into an easy one, and where the first-order hold and digital interpolation fit in.
What Reconstruction Must Do
Chapter 20 ended with an existence proof. Given a band-limited signal and a fast enough sampler, the ideal lowpass filter of gain \(T\) applied to \(x_p(t)\) returns \(x(t)\) exactly. That argument was conducted wholly in the frequency domain, and while it settles the question of whether reconstruction is possible, it says nothing at all about what the reconstructor does to the numbers it is given.
The engineering question is concrete. A digital-to-analogue converter is handed a list \(x(0), x(T), x(2T),\dots\) and must produce a continuous voltage. At \(t = 1.37T\) — an instant at which nothing was ever measured — it has to output some particular value. Which value, and computed how? The frequency-domain proof guarantees that exactly one band-limited answer exists; this section extracts it.
There is a second question, equally practical and rather more dangerous, which Chapter 20 raised and left standing. Suppose the sampling condition was violated. The replicas overlapped, information was lost, and no filter can recover the original. But a converter presented with those samples will still produce something. What signal comes out, and what relation does it bear to the one that went in? Sections 21-4 and 21-5 answer that, and the answer is disquieting: the output is a perfectly clean, entirely wrong signal, with no trace in it of anything having gone amiss.
The Interpolation Formula
The reconstructor is an LTI system — an ideal lowpass filter — and Chapter 7 established what an LTI system does to any input: it convolves it with the impulse response. So the reconstruction rule is \(x_r(t) = x_p(t) * h(t)\), and everything hinges on finding \(h(t)\) for the filter Chapter 20 prescribed.
Take the cutoff at the natural place, halfway between the replicas, \(\omega_c = \omega_s/2 = \pi/T\), and invert the transform directly from the definition of Chapter 13.
The gain \(T\) and the cutoff \(\pi/T\) have cancelled each other perfectly, leaving a function of unit height at the origin: \(h(t) = \operatorname{sinc}(t/T)\), the sinc that Chapter 3 declined to introduce for want of a motive and that Chapter 14 produced as the transform of a rectangle. Now convolve. Since \(x_p(t)\) is a sum of impulses and convolution with a shifted impulse is a shift,
Centre a sinc on every sample instant, scale each by the sample value there, and add. If \(x(t)\) was band-limited to \(\omega_M \lt \omega_s/2\), this sum reproduces it exactly at every \(t\), not merely at the sample instants.
The formula answers the question posed at the start. The value at \(t = 1.37T\) is a weighted sum of every sample, with weights read off a sinc curve centred at that instant. It is not the nearest sample, not a straight line between neighbours, and not any local rule; it is a global average with a very specific set of coefficients.
Reading the Formula
Four features of the interpolation formula do all the explaining, and the first is the reason it works at all.
It passes through the samples. The sinc \(h(t)=\sin(\pi t/T)/(\pi t/T)\) equals 1 at \(t=0\) and vanishes at every other multiple of \(T\), because \(\sin(m\pi)=0\) for integer \(m\). So when \(t = mT\) exactly one term of the sum survives and \(x_r(mT) = x(mT)\). Every sinc in the sum is invisible at all the sample instants except its own — which is exactly what an interpolating function must do, and why the zero crossings in the figure line up with the ticks.
It is non-causal. The sinc extends to \(t \lt 0\) as far as it extends to \(t \gt 0\), so computing the output at any instant requires samples not yet taken. A real reconstructor must therefore delay: accept a latency of \(L\) sample periods, and it can use \(L\) future samples. Delay is the currency in which causality is bought, exactly as it was in the filter designs of Chapter 15.
It never ends, and it decays slowly. The sinc envelope falls off only as \(1/t\), so a sample taken a hundred periods ago still contributes about one per cent of its value. Truncating the sum to a finite window is unavoidable and always introduces error, and worse, an abrupt truncation is a multiplication by a rectangular window, whose spectrum is itself a sinc — reintroducing exactly the ripple that Chapter 11 identified as the Gibbs phenomenon.
It is not bounded by the samples. Because the sinc has negative sidelobes, the reconstructed curve can overshoot every sample value it passes through. Worked Example 1 exhibits two unit samples whose band-limited interpolation reaches \(4/\pi \approx 1.27\) between them. Nothing is wrong: a band-limited signal simply cannot turn a corner, and the overshoot is the price of smoothness.
| Interpolator | Kernel \(h(t)\) | Frequency response | Verdict |
|---|---|---|---|
| Ideal (band-limited) | \(\operatorname{sinc}(t/T)\) | Brick wall of gain \(T\) | Exact, unrealisable |
| Zero-order hold | Rectangle, width \(T\) | \(T\,\mathrm{sinc}\) magnitude, delay \(T/2\) | Trivial, drooped, standard |
| First-order hold | Triangle, base \(2T\) | \(T\,\mathrm{sinc}^2\) magnitude | Smoother, more droop |
| Truncated sinc | Windowed \(\operatorname{sinc}(t/T)\) | Brick wall with ripple | What real converters approximate |
Aliasing
Everything above assumed the Nyquist condition held. Drop it, and the picture from Chapter 20 tells the whole story: the replicas overlap, and in the region of overlap \(X_p(j\omega)\) is the sum of a contribution from the baseband copy and a contribution from a neighbour.
Addition is the operative word. Two numbers have been added, and no operation whatever recovers the two from the one. This is not a subtle degradation like noise, which can sometimes be filtered or averaged away; it is arithmetic annihilation. Once a 7 kHz component and a 3 kHz component have been added together at 3 kHz by a 10 kHz sampler, the record contains a single number where two used to be.
Now pass those samples to an ideal reconstructor. The filter has cutoff \(\omega_s/2\) and dutifully outputs the band-limited signal whose spectrum is whatever it finds in \((-\omega_s/2,\ \omega_s/2)\) — corrupted sum and all. What emerges is a legitimate, perfectly smooth, band-limited signal. It contains no clicks, no discontinuities, nothing that looks like an error. It simply is not the signal that went in.
Aliasing is not distortion added to the signal; it is a relabelling of frequencies. The out-of-band component is not attenuated, not smeared, not made noisy — it arrives in the baseband at full amplitude, wearing a different frequency as a disguise.
The name is exact. In law an alias is a false name under which a person passes; here a 7 kHz sinusoid passes under the name of a 3 kHz one and is accepted without question, because at the sample instants the two are numerically identical. Worked Example 5 of Chapter 20 showed the whole family \(3, 7, 13, 17, 23,\dots\) kHz collapsing to a single sequence at \(f_s = 10\) kHz; sinc interpolation always returns the one member of that family lying below \(f_s/2\), since it is the only one the filter's passband admits.
Aliasing is also visible without any electronics. A cine camera is a sampler running at 24 frames per second, and a wheel whose spokes pass the camera slightly faster than 24 times a second appears to creep slowly backwards — the wagon-wheel effect of old Westerns, and the same phenomenon that makes a strobe-lit shaft seem to stand still. Worked Example 6 computes one. In every case the mechanism is the one above: a rate too high for the sampler is reported as a rate the sampler can represent.
Apparent Frequency and Folding
To predict where a given frequency lands, follow the replicas. The copy indexed \(k\) places the original content from \(f - kf_s\) at frequency \(f\); equivalently, an input component at \(f\) deposits copies at \(f - kf_s\) for every integer \(k\). Exactly one of those copies falls inside the reconstructor's passband \((-f_s/2,\ f_s/2)\), and that is the one that survives. Since the spectrum of a real signal is conjugate-symmetric, a copy landing at a negative frequency is heard as its positive mirror image.
Frequencies below \(f_s/2\) are unchanged. Above it they fold about \(f_s/2\), then about \(f_s\), then about \(3f_s/2\), and so on — the map is a triangle wave, not a sawtooth.
The triangular shape is worth dwelling on, because it is the source of most sign errors. As the input frequency rises past \(f_s/2\) the apparent frequency comes back down; it reaches zero at \(f = f_s\), climbs again to \(f_s/2\) at \(f = 3f_s/2\), and repeats. A slow sweep through a fixed sampler is heard as a tone rising to the Nyquist frequency, turning round, descending to DC, turning round again. The reflection points at \(f_s/2, f_s, 3f_s/2,\dots\) are why \(f_s/2\) is called the folding frequency.
One detail is easy to miss and matters in practice. When a component folds an odd number of times, the copy that lands in the passband came from the negative-frequency side of a replica, and a real sinusoid reconstructed from it carries a reversed phase. A cosine is unaffected, being even; a sine emerges inverted. Worked Example 3 of Chapter 20 showed this algebraically, and Worked Example 2 below tabulates it.
Anti-Aliasing Filters
If aliasing is irreversible, the only defence is prevention, and prevention means guaranteeing that nothing above \(f_s/2\) ever reaches the sampler. That guarantee is the job of the anti-aliasing filter: an analogue lowpass filter placed before the converter, never after it.
The word "before" carries the whole argument. A digital filter applied to the samples cannot help, because by then the offending component is already sitting in the baseband under a false name, arithmetically identical to legitimate content. Removing it would remove the legitimate content too. Chapter 22 will build a great deal of signal processing out of digital filters, and this is the one job they categorically cannot do.
Real filters have finite transition bands, so the design is a three-way negotiation between the signal bandwidth \(f_M\), the sampling rate \(f_s\), and the filter order \(N\). The filter must be flat to \(f_M\) and sufficiently dead by \(f_s - f_M\), the frequency at which the first replica begins to intrude. For a Butterworth filter of order \(N\) with cutoff at \(f_M\) the requirement is easy to turn into a number.
The logarithm in the denominator is where the design pain lives. Sampling just above the Nyquist rate makes \((f_s-f_M)/f_M\) barely greater than 1, its logarithm nearly zero, and the required order enormous. Doubling or quadrupling the rate makes that ratio large, and the order collapses. Worked Example 3 finds a twelfth-order filter turning into a third-order one for the sake of a four-times faster converter — and in modern silicon a faster converter is far cheaper than a twelfth-order analogue filter with the matching and temperature stability such an order demands.
The Zero-Order Hold
No practical converter computes a sinc sum. What it does instead is the simplest thing imaginable: on receiving a sample it drives its output to that value and holds it there until the next sample arrives. The result is a staircase, and the device is the zero-order hold — the flat-top sampling of Section 20-6 with the aperture opened all the way, \(\tau = T\).
Because the staircase is built by placing one rectangle of width \(T\) at each sample instant, the ZOH is an LTI system whose impulse response is that rectangle, and its frequency response follows at once from Chapter 14.
Three things at once: a gain \(T\), a pure delay of half a sample period, and a \(\mathrm{sinc}\) magnitude that droops across the passband and has nulls at every multiple of \(\omega_s\).
Each factor deserves a moment. The delay of \(T/2\) is exact and frequency-independent, being a linear phase term — the staircase is on average half a sample period late, which is intuitively obvious from the picture and is often the dominant lag in a digital control loop. The nulls at \(\omega = k\omega_s\) are a gift: they sit exactly at the centres of the unwanted image replicas and suppress them substantially, which is why a bare ZOH followed by a mild analogue filter is usually good enough.
The droop is the cost. Writing the magnitude in terms of frequency ratio, \(|H_0| = T\,|\sin(\pi f/f_s)/(\pi f/f_s)|\), and evaluating at the Nyquist frequency \(f = f_s/2\) gives the number every converter datasheet quotes.
Nearly 4 dB of attenuation at the top of the band is far too much to ignore in audio or instrumentation, and the correction is a filter with the reciprocal response, \((\pi f/f_s)/\sin(\pi f/f_s)\), applied anywhere convenient — as an analogue equaliser after the converter, or more commonly as a short digital filter before it, since the droop is known exactly in advance. The technique is universally called inverse-sinc or \(1/\mathrm{sinc}\) compensation.
Interpolating with straight lines between samples instead of steps gives the first-order hold, whose impulse response is a triangle of base \(2T\) and whose response is therefore \(T\,\mathrm{sinc}^2\). Its output looks far smoother, and its images are suppressed harder because the sinc is squared, but its passband droop is worse for the same reason: \((2/\pi)^2 = 0.405\), or \(-7.85\) dB at the Nyquist frequency. It also needs the next sample before it can draw the current segment, so it costs one full period of extra delay. Smoother output, more droop, more latency — there is no free interpolation.
Practical Reconstruction
Assembling the pieces gives the reconstruction chain that every converter uses. The digital samples are optionally pre-compensated for droop; a zero-order hold turns them into a staircase; an analogue lowpass filter — called the anti-imaging or reconstruction filter — removes what remains of the replicas; and the result is the output waveform.
The anti-imaging filter faces the mirror image of the anti-aliasing problem. It must pass everything up to \(f_M\) and reject everything from \(f_s - f_M\) upwards, where the first image lives. The transition ratio available to it is again \((f_s-f_M)/f_M\), so once more the sampling rate buys filter order. At the compact-disc rate the ratio is \((44.1-20)/20 = 1.205\), which is brutal; the analogue filters of early CD players were ninth-order elliptic designs with audible phase distortion, and it is precisely this that oversampling was introduced to fix.
Raising \(f_s\) widens the gap between the signal band and the first image, so the analogue filter's skirt may be gentle. Four-times oversampling takes the CD ratio from 1.205 to \((176.4-20)/20 = 7.82\) and turns a ninth-order elliptic filter into a simple second- or third-order one.
The trick, of course, is that the extra samples do not have to be measured — they can be manufactured. Insert three zeros between every pair of samples and pass the result through a digital lowpass filter, and out comes a sequence at four times the rate representing the same band-limited signal. That digital filter is doing sinc interpolation, approximately and at finite length, in the discrete domain where a hundred-tap response is trivial to build and perfectly linear in phase. The impossible analogue filter has been traded for an easy digital one plus a fast converter, which is the modern arrangement almost universally. Chapter 22 develops the upsampling machinery properly, and Chapter 26 supplies the efficient algorithms.
| Stage | Purpose | Placed | Failure if omitted |
|---|---|---|---|
| Anti-aliasing filter | Enforce the band limit | Before the sampler | Irreversible aliasing; raised noise floor |
| Sampler / ADC | Discretise in time | — | — |
| Inverse-sinc equaliser | Undo ZOH droop | Either side of the DAC | Up to \(-3.92\) dB tilt across the band |
| Zero-order hold / DAC | Samples to a waveform | — | — |
| Anti-imaging filter | Remove residual replicas | After the hold | Audible images above the signal band |
Worked Examples
Problem. A band-limited signal has samples \(x(0) = 1\), \(x(T) = 1\) and \(x(nT) = 0\) for every other \(n\). Find the reconstructed value (a) at \(t = 0\), (b) at \(t = T/2\). Comment on the result.
Solution (a). At a sample instant only one sinc is non-zero, because \(\operatorname{sinc}(m) = 0\) for every non-zero integer \(m\). So \(x_r(0) = x(0)\cdot 1 = 1\). The interpolation reproduces the samples exactly, as it must.
Solution (b). At the midpoint both surviving samples contribute. The sinc argument is \((t-nT)/T\), which is \(+\tfrac12\) for \(n=0\) and \(-\tfrac12\) for \(n=1\):
The reconstructed signal rises to 1.273 between two samples both equal to 1 — an overshoot of 27%. Nothing has gone wrong. A signal band-limited to \(f_s/2\) cannot hold two equal values and then drop abruptly to zero on either side; the only band-limited curve through these points is the one that bulges. Linear interpolation would have returned 1.000 and would have been the wrong answer for a band-limited signal, which is worth remembering whenever an oscilloscope is set to "dots" rather than "sin x / x".
Problem. A converter runs at \(f_s = 8\) kHz with no anti-aliasing filter. Sinusoids at 1, 3, 5, 7, 9, 11 and 13 kHz are applied in turn. Tabulate the apparent frequency of each after reconstruction, and state which emerge phase-inverted if the inputs are sines.
Solution. Apply \(f_a = |f - kf_s|\) with \(k\) the nearest integer to \(f/f_s\). For 5 kHz, \(5/8 = 0.625\) rounds to 1, giving \(|5-8| = 3\) kHz; for 13 kHz, \(13/8 = 1.625\) rounds to 2, giving \(|13-16| = 3\) kHz.
| Input \(f\) (kHz) | \(k\) | \(f-kf_s\) (kHz) | Apparent \(f_a\) (kHz) | Sine inverted? |
|---|---|---|---|---|
| 1 | 0 | \(+1\) | 1 | no |
| 3 | 0 | \(+3\) | 3 | no |
| 5 | 1 | \(-3\) | 3 | yes |
| 7 | 1 | \(-1\) | 1 | yes |
| 9 | 1 | \(+1\) | 1 | no |
| 11 | 1 | \(+3\) | 3 | no |
| 13 | 2 | \(-3\) | 3 | yes |
Seven distinct inputs produce two distinct outputs. The sign of \(f - kf_s\) decides the phase: a negative value means the surviving copy came from the negative-frequency half of a replica, and since \(\sin(-\theta) = -\sin\theta\) the reconstructed sine is inverted while a cosine, being even, is not. Sweeping the input from 0 to 16 kHz would produce an output that rises to 4 kHz, falls back to 0, rises again to 4 kHz and falls again — the folding diagram traced out audibly.
Problem. A signal of interest occupies 0 to 4 kHz but arrives accompanied by interference extending well beyond. A Butterworth anti-aliasing filter with cutoff at 4 kHz must provide at least 40 dB of attenuation at the frequency where the first replica begins. Find the required order for (a) \(f_s = 10\) kHz and (b) \(f_s = 40\) kHz.
Solution (a). The first replica begins at \(f_s - f_M = 10 - 4 = 6\) kHz, so 40 dB is needed at 6 kHz, that is \(|H|^2 \le 10^{-4}\):
so \(N = 12\). A twelfth-order analogue filter means six cascaded second-order sections whose component values must hold their ratios over temperature — expensive, and with a group delay that varies sharply near cutoff.
Solution (b). At 40 kHz the first replica begins at \(40 - 4 = 36\) kHz, nine times the cutoff:
so \(N = 3\). Sampling four times faster has replaced a twelfth-order filter with a third-order one. Since the excess samples can be thrown away digitally after a cheap digital filter has done the sharp band-limiting, the fast converter costs almost nothing and the analogue design becomes routine. This single trade is why essentially every modern converter oversamples.
Problem. A DAC runs at \(f_s = 10\) kHz with a zero-order hold. (a) What is the droop, in decibels, at 4 kHz? (b) At the Nyquist frequency? (c) What gain must the equaliser supply at 4 kHz, and what delay does the hold introduce?
Solution (a). The normalised magnitude is \(\sin(\pi f/f_s)/(\pi f/f_s)\). At \(f = 4\) kHz the argument is \(\pi(0.4) = 1.2566\) rad:
Solution (b). At \(f = f_s/2 = 5\) kHz the argument is \(\pi/2\) and the ratio is \(2/\pi = 0.6366\), giving the standard \(-3.92\) dB. The band therefore tilts by nearly 4 dB from DC to Nyquist, which is a gross error by any audio or instrumentation standard.
Solution (c). The equaliser must supply the reciprocal, \(1/0.7568 = 1.321\), or \(+2.42\) dB at 4 kHz, rising to \(+3.92\) dB at 5 kHz. The hold also contributes the linear phase term \(e^{-j\omega T/2}\), a delay of \(T/2 = 50\ \mu\)s independent of frequency — harmless in an audio path, but a genuine and often dominant lag if this converter sits inside a feedback loop, since Chapter 19 showed that added delay eats directly into phase margin.
Problem. Samples taken at \(f_s = 10\) kHz are found to be \(x[n] = \cos(0.6\pi n)\). (a) What does an ideal reconstructor produce? (b) List three continuous-time signals that could have produced these samples. (c) Can any measurement on the samples decide between them?
Solution (a). The reconstructor's passband is \(|f| \lt 5\) kHz, so it outputs the unique band-limited signal with that spectrum. Inverting \(\Omega = 2\pi f/f_s\) gives \(f = 0.6\pi \times 10\,000/(2\pi) = 3\) kHz, so the output is \(\cos(2\pi\,3000\,t)\).
Solution (b). Any frequency whose normalised value is congruent to \(0.6\pi\) modulo \(2\pi\), or to its negative, will do: \(f = 3\) kHz, \(f = 10-3 = 7\) kHz, \(f = 10+3 = 13\) kHz, and generally \(|10m \pm 3|\) kHz for integer \(m\). Sampling any of them at 10 kHz yields exactly this sequence.
Solution (c). No. The samples are numerically identical in every case, and the samples are all the reconstructor has. This is the whole content of aliasing stated as a negative result: the choice of 3 kHz is not a deduction from evidence but a convention, imposed by the reconstruction filter's cutoff, and it is correct only because the anti-aliasing filter was supposed to have guaranteed in advance that nothing above 5 kHz could be present. Remove that guarantee and the output is a guess.
Problem. A shaft with a single painted mark rotates at 1740 rpm and is lit by a stroboscope flashing 1800 times per minute. (a) How does the mark appear to move? (b) At what flash rates does it appear stationary? (c) What is the fastest rotation that this strobe can measure unambiguously?
Solution (a). Convert to hertz: the shaft turns at \(1740/60 = 29\) rev/s and the strobe samples at \(f_s = 1800/60 = 30\) Hz. The apparent frequency is \(f - kf_s\) with \(k=1\):
The negative sign is the direction, not merely a magnitude to be discarded: the mark appears to rotate backwards at 1 revolution per second, or 60 rpm. Each flash catches the mark \(1/30\) of a turn short of where it was, and the eye reads the short-fall as reverse motion. This is the wagon-wheel effect, and it is why a strobe is such a sensitive instrument — a 3% error in speed shows up as an obvious slow crawl.
Solution (b). The mark stands still when the shaft completes a whole number of turns between flashes, that is when \(29/f_s\) is an integer: \(f_s = 29\) Hz (1740 flashes/min), \(14.5\) Hz, \(29/3\) Hz, and so on. Only the highest of these, matching the true speed, shows a single mark; the slower ones do too, which is exactly the ambiguity a careful operator must guard against by starting fast and slowing down.
Solution (c). The sampling theorem applies unchanged: unambiguous measurement requires the rotation rate to be below \(f_s/2 = 15\) Hz, that is 900 rpm. Above that the reading folds, and the 1740 rpm shaft is reported as \(|1740-1800| = 60\) rpm in the wrong direction — a spectacular error produced by an instrument working perfectly.
Chapter Summary
\(x_r(t)=\sum_n x(nT)\operatorname{sinc}\big((t-nT)/T\big)\) — one sinc per sample, exact for a properly band-limited signal.
Non-causal, infinite in extent, decaying as \(1/t\). Real reconstructors truncate, delay, and approximate.
Two unit samples interpolate to \(4/\pi\) between them. A band-limited signal cannot turn a corner.
Overlapping replicas add. The output is a clean band-limited signal that is simply the wrong one.
\(f_a=|f-kf_s|\), a triangle wave of period \(f_s\) reflecting about \(f_s/2\). Odd folds invert a sine's phase.
The anti-aliasing filter must precede the sampler. No digital processing can undo a fold, and the filter also limits folded noise.
Gain \(T\), delay \(T/2\), sinc droop of \(-3.92\) dB at \(f_s/2\), and useful nulls on the image centres.
Widening \((f_s-f_M)/f_M\) collapses the analogue filter order — twelfth to third for a four-times rate increase.
Problems
Problems 1 to 3 exercise the interpolation formula; 4 to 6 are aliasing calculations; 7 and 8 concern practical reconstruction hardware. Draw the replica picture or the folding diagram before reaching for a formula.
- A band-limited signal has samples \(x(-T)=0\), \(x(0)=2\), \(x(T)=0\) and all others zero. Compute \(x_r(T/2)\) and \(x_r(T/4)\), and sketch the reconstruction over \(-2T \le t \le 2T\).
- Prove that band-limited interpolation reproduces every sample exactly, by evaluating the interpolation sum at \(t=mT\) and using \(\sin(m\pi)=0\).
- The interpolation kernel is truncated to the nearest eight samples on each side. Estimate the largest weight discarded, and explain why simply cutting the sum off is worse than tapering it towards zero.
- A 12 kHz sinusoid is sampled at 9 kHz. Find the apparent frequency, state whether a sine input is inverted, and give two other input frequencies that would be indistinguishable from it.
- An input is swept slowly from 0 to 30 kHz through a converter running at 12 kHz with no anti-aliasing filter. Describe the output frequency throughout the sweep, and mark on a folding diagram the points at which it reverses direction.
- Explain, using the replica picture, why an anti-aliasing filter placed after the sampler cannot remove an aliased component. Then explain why a digital filter can nevertheless remove out-of-band noise in an oversampled converter.
- A zero-order hold runs at 48 kHz. Tabulate its droop in decibels at 1, 5, 10, 20 and 24 kHz, and design the required inverse-sinc gain at each. What extra delay does the hold add to a control loop?
- A signal band-limited to 20 kHz is reconstructed at 44.1 kHz and again at 352.8 kHz. Compute the transition ratio available to the anti-imaging filter in each case, and use the Butterworth order formula of Section 21-6 to compare the orders needed for 60 dB of image rejection.