Reconstruction

Here are seventeen samples, 1 ms apart, of a smooth wave. A pulse goes at every sample, as tall as its sample, and the pulses are added one at a time. Decide first: where will the sum pass? Watch it at the dots, and between them.

A pulse at every sample, added

The page The sampling theorem (10.2) ended with a promise. If a signal has nothing at or above half the sample rate, an ideal low-pass filter can cut the middle copy out of the sampled signal’s spectrum, and that copy is the original. Here I show what that filter does in time.

In this lesson $T_s=1$ ms, so $f_s=1$ kHz and the Nyquist frequency is $f_N=500$ Hz. The ideal low-pass from Frequency response and Bode plots (8.4) keeps $\lvert f\rvert<f_s/2$. With time in milliseconds, its impulse response is $h(t)=\mathrm{sinc}(t/T_s)$, where $\mathrm{sinc}(x)=\sin(\pi x)/(\pi x)$. This sinc pulse is 1 at $t=0$ and 0 at every other whole number of milliseconds.

Now send the sampled signal’s spikes through that filter. By Continuous-time convolution (5.3), a spike leaves a copy of $h$ at its own position, scaled by the spike’s area. The area of the spike at $nT_s$ is the sample $x[n]$. So the filter’s output is a sinc pulse at every sample, each as tall as its sample, added together. I call that output the reconstruction $x_r(t)$, and I write it as

The wave in this lesson is $x_c(t)=0.8\cos(2\pi\cdot110\,t)+0.4\cos(2\pi\cdot270\,t-0.6)$, with $t$ in seconds. It has nothing above 270 Hz, well under $f_N=500$ Hz. I sample it at $f_s=1$ kHz. Think of a flexible ruler bent through pins on a board: the pins are the samples, and the ruler is the sum.

The picture at the top of the page adds these sinc pulses one at a time. Its readout “largest gap, −2 to 2 ms” is the largest distance between the sum and $x_c$ on the stretch from −2 to 2 ms. With no pulses, the gap is the largest value of $x_c$ itself, 1.182.

One pulse at sample 0 is as tall as that sample, and 0 at every other sample instant. The gap is then 0.804. One pulse is wrong between the dots, but it is right at every dot, because it is 1 at its own sample and 0 at the others. The other pulses join from the middle outward, each as tall as its own sample and 0 at every other dot. With all seventeen added, the sum passes through every dot and, in the middle, lies on the original: the gap is 0.016. Near the ends it drifts, because pulses from samples further out are missing.

Notice the order of events. Every pulse keeps the sum on every dot, and each new pulse changes only what happens between the dots. The gap on the middle stretch falls from 1.182 to 0.804, and then to 0.226, 0.083, 0.048, 0.018 and 0.016 after 1, 3, 5, 9, 13 and 17 pulses.

Why the sum passes through every dot

At the instant $t=mT_s$, each sinc in the sum is $\mathrm{sinc}(m-n)$. That is 1 when $n=m$ and 0 for every other whole number. So only one term survives, and

The sum passes through every sample, whatever the samples are. What happens between the dots is the tails of the pulses filling in. For a signal with nothing at or above $f_N$, 10.2 says that they fill in the original, and the middle of the instrument shows it.

Near the ends the sum drifts, because the pulses from samples beyond the ends are missing. The sinc has tails that never end, and a pulse far away still contributes. This is also the filter’s problem. By 8.4, the ideal low-pass is not causal: its response starts before the spike that causes it. A real converter delays the pulse and uses a few dozen neighbours, with the sinc tapered toward its ends, a method from Window-method FIR design (19.1). Smooth interpolators such as cubic and Lagrange interpolation work the same way, with a short pulse in place of the sinc.

The same recipe with other pulses

Nothing in the recipe needs the sinc. Replace it by any pulse $p(t)$, and the sum is $x_r(t)=\sum_n x[n]\,p(t-nT_s)$. Two cheap pulses are in everyday use. The hold is a rectangle one sample period long, starting at each sample. It makes the staircase that a digital-to-analog converter (DAC) makes, and the circuit is a zero-order hold. The straight lines pulse is a triangle two sample periods wide, which connects the dots.

Adding a pulse at every sample is convolution of the spikes with the pulse. So by Properties of the Fourier transform (8.2) the spectrum is 10.2’s copies, multiplied by the pulse’s spectrum. From series to transform (8.1) gives the rectangle’s spectrum as a sinc, and 8.2 gives the triangle’s, which is the rectangle’s squared. I measure each gain relative to 0 Hz, which means dividing the pulse’s spectrum by $T_s$:

• hold: $\lvert\mathrm{sinc}(f/f_s)\rvert$;
• straight lines: $\mathrm{sinc}^2(f/f_s)$;
• sinc: 1 for $\lvert f\rvert<f_s/2$ and 0 above, the brick wall of 8.4.

A good pulse keeps the middle copy whole and removes every other copy. Which of the three does both? Watch how much is kept at 270 Hz, and how much of the 730 Hz copy is left, for each pulse.

Hold each sample for one period and you get a staircase. Its spectrum droops, keeping 0.884 at 270 Hz, and lets 0.327 of the 730 Hz copy through. Join the dots with straight lines and less of the copy gets through, 0.107, but the droop is worse, 0.782 at 270 Hz. Sinc pulses keep everything below 500 Hz whole and remove every copy, 1.000 and 0.000, so only the sinc rebuilds the original exactly.

Switch pulse to each option yourself, and compare the two readouts. Notice that the hold droops a little and leaks a lot, and the lines leak less and droop more. Each cheap pulse trades one error for the other. Only the sinc has no droop and no leak, and only for a signal with nothing at or above $f_N$.

The hold’s droop and delay

The hold’s gain falls even inside the band you want to keep. At $f_N$, where $f/f_s=0.5$, the gain is $\lvert\mathrm{sinc}(0.5)\rvert=2/\pi=0.637$, which is $-3.92$ dB. For straight lines it is $0.405$, which is $-7.84$ dB, twice as far down in dB, as squaring doubles a dB value. This loss of the high end is called droop.

The hold also delays the signal. A staircase lags its curve by half a sample period, because each step starts at the sample and then holds it. At 44.1 kHz that is $T_s/2=11.34\ \mu\text{s}$. In a CD player the audio band reaches 20 kHz, so $f/f_s=20/44.1$. There the hold’s gain is $0.694$, which is $-3.17$ dB.

The fix is called compensation. Before the DAC, the player boosts the top of the band by $1/\lvert\mathrm{sinc}(f/f_s)\rvert$, which at 20 kHz is $1.440$, or $+3.17$ dB. After the DAC an analog low-pass smooths the steps. That filter is the subject of Anti-aliasing and practical converters (10.4).

Two staircases

A hold staircase is easy to confuse with a second staircase, which comes from a different step. Rounding a value to the nearest of a few levels also makes steps, as I showed in Kinds of signals (1.2). They differ in what is kept and what is changed.

A hold changes the time axis, because the curve between ticks is replaced by one flat step, and keeps each value exact. Rounding does the opposite. It keeps the time and changes the value, and Quantization and noise (11.1) is about what that costs.

Worked example

1. Gains. At $f/f_s=0.27$ the hold’s gain is $0.8843$ ($-1.07$ dB) and the lines’ gain is $0.7820$ ($-2.14$ dB). At $0.73$ they are $0.3271$ ($-9.71$ dB) and $0.1070$ ($-19.41$ dB). At $0.5$ they are $0.6366$ ($-3.92$ dB) and $0.4053$ ($-7.84$ dB).
2. A CD player. At 20 kHz and $f_s=44.1$ kHz the hold’s gain is $0.6944$ ($-3.17$ dB), the compensation boost is $1.440$ ($+3.17$ dB), and the hold’s delay is $T_s/2=11.34\ \mu\text{s}$.
3. The pulse sum. $x[0]=1.130$. The largest gap on $[-2,2]$ ms after 1, 3, 5, 9, 13 and 17 pulses is $0.804$, $0.226$, $0.083$, $0.048$, $0.018$ and $0.016$.

Where you’ll meet this

Every DAC, from a phone’s headphone output to a CD player, uses a hold followed by an analog filter, and the droop above is why the digital part boosts the top of the band first. Raising a sample rate by inserting samples is the same recipe done in the digital domain, in Upsampling and interpolation (22.2).

Reference card