Signal Reconstruction, Interpolation, DAC, Bandpass Sampling, and Oversampling

From Samples Back to a Continuous Signal

The sampling theorem promises exact reconstruction, but it does not say how. The answer is ideal interpolation: convolve the samples with a sinc kernel.

\[x_a(t)=\sum_{n=-\infty}^{\infty}x[n]\,\operatorname{sinc}\!\left(\frac{t-nT}{T}\right)\]

where

\[\operatorname{sinc}(u)=\frac{\sin(\pi u)}{\pi u}\]

Each sample contributes a sinc centred at $nT$ that is 1 at its own sample and 0 at every other sample. The sum is exact if $F_s>2F_{\max}$.

Why sinc?

In the frequency domain, reconstruction multiplies the sampled spectrum by an ideal lowpass of cutoff $F_s/2$. The impulse response of that brick-wall filter is a sinc. Reconstruction and ideal lowpass filtering are the same operation.

Samples ──► Impulse train ──► Ideal LPF (Fc = Fs/2) ──► Continuous signal

A real filter cannot be brick-wall, so real reconstruction is approximate — the error is the filter’s transition band and stopband.

The Sinc Has Infinite Support

Exact reconstruction needs every sample, past and future. That is non-causal and impossible online. Practical interpolators use a windowed sinc, a polynomial, or a spline — trading exactness for finite latency.

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Zero-Order Hold and the DAC

A digital-to-analog converter cannot output impulses. It outputs a staircase: each sample is held for one period $T$. That is a zero-order hold (ZOH).

The ZOH impulse response is a rectangle of width $T$:

\[h_{ZOH}(t)=\begin{cases}1, & 0\le t<T\\0, & \text{otherwise}\end{cases}\]

Its frequency response is

\[H_{ZOH}(f)=T\,e^{-j\pi fT}\operatorname{sinc}(fT)\] \[=T\,e^{-j\pi fT}\frac{\sin(\pi fT)}{\pi fT}\]

The sinc Droop

The magnitude $\vert\operatorname{sinc}(fT)\vert$ is not flat. At the band edge $f=F_s/2$:

\[\operatorname{sinc}(1/2)=\frac{\sin(\pi/2)}{\pi/2}=\frac{2}{\pi}\approx0.637\]

That is a −3.92 dB droop at Nyquist. The ZOH attenuates high frequencies inside the band, which is audible/visible as a dull top end.

The Images

Sampling creates copies of the spectrum at $kF_s$. The ZOH does not remove them; it only shapes them. A real DAC therefore needs an analog reconstruction filter (a “smoothing” or “anti-imaging” filter) after the hold stage to suppress the images.

Stage	Job
Digital interpolation	Raise the rate so images are far from the band
ZOH / DAC	Convert to a staircase
Analog lowpass	Remove the images and flatten the droop

Practical fix for the droop: a inverse-sinc digital filter ahead of the DAC pre-emphasizes the high end so the ZOH droop cancels out.

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Bandpass Sampling

You do not always need $F_s>2F_{\max}$. If the signal occupies only a band $F_L\le F\le F_H$, you can sample at a much lower rate and let the band alias down to baseband — as long as the shifted copies do not overlap.

The Condition

Let $B=F_H-F_L$ be the bandwidth. Sampling at $F_s$ is alias-free if

\[\frac{2F_H}{m}\le F_s\le\frac{2F_L}{m-1}\]

for some integer $m$ with

\[1\le m\le\left\lfloor\frac{F_H}{B}\right\rfloor\]

where $m$ is the number of the Nyquist zone the band is folded from.

Example. A radar band from 10 MHz to 12 MHz ($B=2$ MHz). Then

\[\left\lfloor\frac{12}{2}\right\rfloor=6\]

so $m$ can be 1 to 6. Choosing $m=6$:

\[\frac{24}{6}=4\le F_s\le\frac{20}{5}=4\]

So $F_s=4$ MHz works — one fifth of the naive $2F_H=24$ MHz.

Why It Works

Folding by $F_s$ maps the band to

\[F_{\text{base}}=F\bmod F_s\]

and the negative-frequency copy mirrors it. If the band folds into $[0,F_s/2]$ without overlapping its own image, no information is lost. The band is translated, not distorted.

The Catch

• The anti-alias filter must now be a bandpass filter, not a lowpass — often harder to build.
• Any noise or interference outside the band but inside a folding zone will alias into the band.
• You must know $F_L$ and $F_H$ precisely; a drifted band edge breaks the inequality.

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Oversampling

Oversampling means sampling well above the Nyquist rate: $F_s=L\cdot2F_{\max}$ with $L\gg1$.

Oversampling in the ADC

Quantization noise power is spread uniformly over the whole band $[0,F_s/2]$. Its density is

\[S_q(f)=\frac{\Delta^2}{12}\cdot\frac{1}{F_s/2}\]

If you oversample and then digitally lowpass to the signal band $B$, you remove most of the noise:

\[P_{q,\text{in-band}}=S_q\cdot2B=\frac{\Delta^2}{12}\cdot\frac{2B}{F_s}\]

With $F_s=2BL$:

\[P_{q,\text{in-band}}=\frac{\Delta^2}{12}\cdot\frac1L\]

So oversampling by $L$ reduces in-band quantization noise power by a factor $L$ — a $10\log_{10}L$ dB improvement.

Combined with noise shaping (a sigma-delta modulator pushes noise to high frequencies), the improvement becomes much larger — first-order noise shaping gives roughly $9$ dB per doubling, and higher orders give more.

Oversampling in the DAC

The reverse: interpolate to a high rate so the images land far above the band and a gentle analog filter can remove them. This is why CD players upsample $44.1$ kHz to $352.8$ kHz or higher before the DAC.

The SNR Formula

For an ideal $B$-bit ADC with full-scale sine input, the classic result is

\[\text{SQNR}=6.02B+1.76\ \text{dB}\]

Oversampling by $L$ adds

\[10\log_{10}L\ \text{dB}\]

so the effective number of bits gained is

\[\Delta B=\frac{10\log_{10}L}{6.02}\approx\frac{L\text{ in dB}}{6.02}\]

Oversampling by 16 ($12$ dB) buys about 2 effective bits. That is why early sigma-delta converters used very high oversampling ratios to reach 16–20 bits from a 1-bit quantizer.

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Decimation and Interpolation in the Reconstruction Chain

The full path from analog to analog through digital processing:

Analog in
  → anti-alias LPF
  → ADC at Fs
  → digital processing (filter, decimate, interpolate)
  → interpolate to Fdac
  → DAC (ZOH)
  → anti-image LPF
Analog out

Every rate change needs a filter:

Operation	Filter required	Purpose
Decimation by $M$	Lowpass at $F_s/(2M)$ before downsampling	Prevent aliasing
Interpolation by $L$	Lowpass at $F_s/(2L)$ after upsampling	Remove images
DAC	Analog lowpass	Remove spectral images
ADC	Analog lowpass	Prevent aliasing

The golden rule: filtering always happens at the higher rate. In decimation, filter first (still at the high rate) then throw samples away. In interpolation, insert zeros first then filter.

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Worked Example: Reconstruction of a Sampled Cosine

$x_a(t)=3\cos(100\pi t)$, so $F=50$ Hz and $F_s=200$ Hz (from the earlier post). Samples:

\[x[n]=3\cos\left(2\pi\frac{50}{200}n\right)=3\cos\left(\frac{\pi}{2}n\right)\] \[=\{3,\ 0,\ -3,\ 0,\ 3,\ 0,\ -3,\dots\}\]

Reconstruct:

\[x_a(t)=\sum_n 3\cos\left(\frac{\pi}{2}n\right)\operatorname{sinc}\left(\frac{t-nT}{T}\right)\]

At $t=nT$ only the $n$-th sinc is 1, so the samples are reproduced exactly. Between samples, the sincs add up to the original cosine. Since $F_s=200>2(50)=100$, the reconstruction is exact.

The Same Signal Undersampled

Sample at $F_s=75$ Hz:

\[x[n]=3\cos\left(2\pi\frac{50}{75}n\right)=3\cos\left(\frac{4\pi}{3}n\right)\]

$\frac{4\pi}{3}$ is above $\pi$, so alias it: $\frac{4\pi}{3}-2\pi=-\frac{2\pi}{3}$. The reconstructed analog frequency is

\[F_{\text{alias}}=\frac{2\pi/3}{2\pi}\cdot75=25\ \text{Hz}\]

The output is $3\cos(50\pi t)$ — half the original frequency. That is aliasing, visible as a slow beat in the reconstruction.

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Summary

1. Ideal reconstruction is sinc interpolation, equivalent to a brick-wall lowpass at $F_s/2$.
2. A ZOH DAC produces a sinc-shaped droop of −3.92 dB at Nyquist and leaves images that need analog filtering.
3. Bandpass sampling can drop the rate far below $2F_H$ if the folded band does not overlap its image.
4. Oversampling reduces in-band quantization noise by $10\log_{10}L$ dB, about $1.66$ effective bits per doubling.
5. Filter at the higher rate: before decimation, after interpolation.