FOUNDATION · CHAPTER 7 OF 8

Sampling from first principles

Sampling at a steady rate copies a spectrum at every multiple of the sampling frequency. This chapter derives it, shows aliasing as two tones with equal samples, derives the ideal sinc kernel and links analog to digital frequency. These facts set how fast a converter must sample.

Updated 12 Oct 2026

What you will be able to do

You will show that sampling copies the spectrum of a signal at every multiple of the sampling frequency, and say exactly when the copies overlap. You will see aliasing as two different sinusoids that give identical samples. You will derive why the sinc is the ideal interpolation kernel. You will convert between analog frequency $ω$ and digital frequency $Ω$ through $T_{s}$ .

Before you start

Chapter 6, From series to transform: one family, five names
The Fourier transform, its inverse with the factor $\frac{1}{2 π}$ , and linearity
The exponential Fourier series; the unit impulse and sifting; $e^{j ω_{0} t} \leftrightarrow 2 π δ (ω - ω_{0})$ and $δ (t - t_{0}) \leftrightarrow e^{- j ω t_{0}}$
$x (t) y (t) \leftrightarrow \frac{1}{2 π} X (ω) * Y (ω)$
Euler's formula; $cos θ = cos (θ + 2 π k)$ for integer $k$ ; $cos (- θ) = cos θ$

Sampling copies the spectrum

Sampling means keeping the values of $x (t)$ only at $t = n T_{s}$ . The samples form the list $x [n] = x (n T_{s})$ . The cleanest way to describe that is to multiply $x (t)$ by a train of unit impulses spaced $T_{s}$ apart. Multiplication in time is convolution in frequency, so the question becomes: what is the spectrum of the impulse train? Here $T_{s}$ is the sampling period, $f_{s} = 1/ T_{s}$ and $ω_{s} = 2 π / T_{s}$ . (The z-transform notes write $T$ for $T_{s}$ .)

THE SAMPLED SIGNAL

x_{s} (t) = x (t) n = - \infty \sum \infty δ (t - n T_{s}) = n = - \infty \sum \infty x (n T_{s}) δ (t - n T_{s})

Holds for any $x (t)$ continuous at each $n T_{s}$ ; $n$ is an integer and $T_{s}$ is in s. The impulse at $n T_{s}$ has weight $x (n T_{s}) = x [n]$ , because $x (t) δ (t - t_{0}) = x (t_{0}) δ (t - t_{0})$ .

SPECTRUM OF THE IMPULSE TRAIN

p (t) a_{k} P (ω) = n = - \infty \sum \infty δ (t - n T_{s}) = k = - \infty \sum \infty a_{k} e^{j k ω_{s} t} = \frac{1}{T _{s}} \int_{- T_{s} /2}^{T_{s} /2} δ (t) e^{- j k ω_{s} t} d t = \frac{1}{T _{s}} = \frac{2 π}{T _{s}} k = - \infty \sum \infty δ (ω - k ω_{s})

$p (t)$ has period $T_{s}$ , so its Fourier series has $ω_{0} = ω_{s}$ , and $a_{k}$ is an integral over any one period. On $- T_{s} /2 < t < T_{s} /2$ the train is just $δ (t)$ , and sifting gives $a_{k} = 1/ T_{s}$ . By linearity, transform the series term by term with $e^{j k ω_{s} t} \leftrightarrow 2 π δ (ω - k ω_{s})$ . Units: $ω$ , $ω_{s}$ in rad/s, $T_{s}$ in s.

SPECTRUM OF THE SAMPLED SIGNAL

X_{s} (ω) = \frac{1}{2 π} X (ω) * P (ω) = \frac{1}{T _{s}} k = - \infty \sum \infty X (ω) * δ (ω - k ω_{s}) = \frac{1}{T _{s}} k = - \infty \sum \infty X (ω - k ω_{s})

Holds for any $x (t)$ with transform $X (ω)$ . The $\frac{1}{2 π}$ cancels the $2 π$ in $P (ω)$ ; convolving with $δ (ω - k ω_{s})$ shifts $X$ by $k ω_{s}$ . If $X (ω) = 0$ for $∣ ω ∣ > ω_{m}$ (rad/s; $f_{m} = ω_{m} /2 π$ Hz), the copies stay apart when $ω_{s} - ω_{m} > ω_{m}$ : $ω_{s} > 2 ω_{m}$ , or $f_{s} > 2 f_{m}$ . Without a band limit, no finite $ω_{s}$ guarantees this.

Read the result as a picture in words. Each copy is the original spectrum, scaled by $1/ T_{s}$ . One copy sits around $ω = 0$ and an identical one around every multiple of $ω_{s}$ . When the copies do not touch, the baseband copy can be cut out by a filter, and $x (t)$ is recovered. When they overlap, they add, and the baseband is corrupted.

WATCH OUT FOR

Do not take $f_{s} = 2 f_{m}$ as enough. Test: sample $sin (2 π 500 t)$ at $1000$ Hz. Every sample is $sin (π n) = 0$ , so the tone vanishes, while $cos (2 π 500 t)$ survives. The outcome depends on phase, so only $f_{s} > 2 f_{m}$ works for every signal.

Aliasing seen in the time domain

The overlap of copies has a direct time-domain meaning. Samples are only numbers. Two different sinusoids that take the same value at every $n T_{s}$ cannot be told apart from their samples. Adding $ω_{s}$ to a frequency changes the argument at the samples by a multiple of $2 π$ , which changes nothing.

TWO TONES WITH THE SAME SAMPLES

cos ((ω_{0} + k ω_{s}) n T_{s} + φ) = cos (ω_{0} n T_{s} + 2 π k n + φ) = cos (ω_{0} n T_{s} + φ)

Holds for every integer $k$ and $n$ , since $ω_{s} T_{s} = 2 π$ ; $ω_{0}$ in rad/s, $φ$ in rad. Cosine is even, so $- ω_{0}$ with phase $- φ$ gives the same samples. So a tone at $f$ matches tones at $f + k f_{s}$ and, phase reversed, at $- f + k f_{s}$ . Steps of $f_{s}$ bring $f$ into $[- f_{s} /2, f_{s} /2]$ ; a sign flip puts it in $[0, f_{s} /2]$ , its apparent frequency.

WORKED EXAMPLE

A 700 Hz cosine $cos (2 π 700 t)$ is sampled at 1000 Hz. Find the lowest-frequency cosine with the same samples.

Solution

$T_{s} = 1$ ms, so the samples are $x [n] = cos (2 π \cdot 700 \cdot n \cdot 0.001) = cos (2 π \cdot 0.7 n)$ : the hertz and the seconds cancel.
Write $0.7 = 1 - 0.3$ : $2 π \cdot 0.7 n = 2 π n - 2 π \cdot 0.3 n$ . The term $2 π n$ is a whole number of turns, so dropping it leaves the cosine unchanged: $x [n] = cos (- 2 π \cdot 0.3 n)$ .
Cosine is even, so $cos (- 2 π \cdot 0.3 n) = cos (2 π \cdot 0.3 n)$ ; the phase is zero, so there is no phase to reverse.
$0.3 \times 1000 = 300$ Hz lies in $[0, 500]$ Hz, so it is the apparent frequency.

Answer: 300 Hz.

Reconstruction and the sinc

Reconstruction means building a continuous signal from the samples: put a copy of one pulse shape $g (t)$ at each sample time, scaled by the sample value. In frequency, that is the sampled spectrum multiplied by $G (ω)$ . For perfect recovery, $G (ω)$ must keep the baseband copy and remove every other copy. That decides $g (t)$ . Below, $sinc (x) = sin (π x) / (π x)$ , with $sinc (0) = 1$ .

INTERPOLATION AS FILTERING

x_{r} (t) = n \sum x [n] g (t - n T_{s}) ⟺ X_{r} (ω) = G (ω) X_{s} (ω)

$x_{r} (t)$ is the reconstructed signal and $G (ω)$ the transform of $g (t)$ . Convolving $x [n] δ (t - n T_{s})$ with $g$ gives $x [n] g (t - n T_{s})$ , so $x_{r} = x_{s} * g$ and $X_{r} = G X_{s}$ . Take $G (ω) = T_{s}$ for $∣ ω ∣ < ω_{s} /2$ and $0$ elsewhere: it keeps only the $k = 0$ copy, and the gain $T_{s}$ cancels its $1/ T_{s}$ , so $X_{r} = X$ when $ω_{s} > 2 ω_{m}$ .

THE IDEAL KERNEL

g (t) = \frac{1}{2 π} \int_{- π / T_{s}}^{π / T_{s}} T_{s} e^{j ω t} d ω = \frac{T _{s}}{2 π} \cdot \frac{2 sin ( π t / T _{s} )}{t} = \frac{sin ( π t / T _{s} )}{π t / T _{s}} = sinc (\frac{t}{T _{s}})

Valid for $t \neq = 0$ ; at $t = 0$ the integral gives $1$ directly, matching $sinc (0) = 1$ . This is the inverse transform of $G (ω)$ , whose cut-off is $ω_{s} /2 = π / T_{s}$ rad/s; $t$ is in s. By Euler's formula, the integral of $e^{j ω t}$ over the band is $(e^{j π t / T_{s}} - e^{- j π t / T_{s}}) / (j t) = 2 sin (π t / T_{s}) / t$ .

THE RECONSTRUCTION FORMULA

x (t) = n = - \infty \sum \infty x [n] sinc (\frac{t - n T _{s}}{T _{s}}) = n = - \infty \sum \infty x [n] \frac{sin ( π ( t - n T _{s} ) / T _{s} )}{π ( t - n T _{s} ) / T _{s}}

Holds when $X (ω) = 0$ for $∣ ω ∣ > ω_{m}$ and $ω_{s} > 2 ω_{m}$ . At a sample time $t = k T_{s}$ ( $k$ an integer) the term $n = k$ is $x [k]$ , and every other term is $0$ , because its sine is $sin (π (k - n)) = 0$ . So the curve passes through every sample, and between samples the neighbouring sincs fill in the values.

The sinc is ideal in the sense that its spectrum is flat over the baseband and exactly zero outside. It is also not buildable: it extends over all $t$ , so it needs samples that have not arrived yet (it is non-causal). Real converters use shorter pulses and accept some error.

WORKED EXAMPLE

Samples at $T_{s} = 1$ ms are $x [0] = 2$ , $x [1] = 1$ and zero elsewhere. Find the ideal reconstruction at $t = 0.5$ ms.

Solution

Only two samples are non-zero, so the sum has two terms: $2 sinc (t / T_{s}) + 1 \cdot sinc ((t - T_{s}) / T_{s})$ .
At $t = 0.5$ ms, $t / T_{s} = 0.5$ and $(t - T_{s}) / T_{s} = - 0.5$ : the milliseconds cancel.
$sinc (\pm 0.5) = sin (\pm π /2) / (\pm π /2) = 2/ π \approx 0.6366$ , since the sign cancels.
Total: $(2 + 1) \times 0.6366$ .

Answer: $x_{r} (0.5 ms) \approx 1.91$ , in the same units as the samples.

From analog to digital frequency

The sampled signal has a second description: the list $x [n] = x (n T_{s})$ and its DTFT $X (e^{j Ω}) = \sum_{n} x [n] e^{- j Ω n}$ , where $Ω$ is in rad/sample and has period $2 π$ . (The DTFT notes write lower-case $ω$ for this same $Ω$ .) $X (e^{j Ω})$ and $X (ω)$ are different functions: the first is the DTFT of $x [n]$ , the second the transform of $x (t)$ , and the argument shows which is meant. Take the transform of $x_{s} (t)$ term by term and the two descriptions meet.

DIGITAL FREQUENCY FROM ANALOG FREQUENCY

X_{s} (ω) = n \sum x [n] e^{- j ω n T_{s}} = X (e^{j Ω})_{Ω = ω T_{s}}, Ω = ω T_{s} = \frac{2 π f}{f _{s}}

Uses the transform of $δ (t - n T_{s})$ , which is $e^{- j ω n T_{s}}$ . $Ω$ is in rad/sample, $ω$ in rad/s, $T_{s}$ in s, $f$ and $f_{s}$ in Hz. $f = f_{s} /2$ gives $Ω = π$ , where $cos (π n) = (- 1)^{n}$ changes sign fastest. $f = f_{s}$ gives $Ω = 2 π$ : one period $ω_{s}$ of $X_{s} (ω)$ maps onto one period $2 π$ of $X (e^{j Ω})$ .

THE DTFT FROM THE ANALOG SPECTRUM

X (e^{j Ω}) = \frac{1}{T _{s}} k = - \infty \sum \infty X (\frac{Ω - 2 π k}{T _{s}})

Put $ω = Ω/ T_{s}$ in the sampled spectrum: since $k ω_{s} T_{s} = 2 π k$ , $Ω/ T_{s} - k ω_{s} = (Ω - 2 π k) / T_{s}$ . It holds wherever that spectrum holds. $X (\cdot)$ on the right is the transform of $x (t)$ , its argument in rad/s. If $X (ω) = 0$ for $∣ ω ∣ > ω_{m}$ , the copies are separate when $ω_{m} T_{s} < π$ , which is $ω_{s} > 2 ω_{m}$ again.

WORKED EXAMPLE

A digital filter has cut-off $Ω_{c} = 0.4 π$ rad/sample and runs on data sampled at $20$ kHz. What analog cut-off does it have? A tone $cos (2 π \times 15000 t)$ reaches the sampler with no filter in front of it: at what frequency does it appear?

Solution

From $Ω = 2 π f / f_{s}$ , $f = Ω f_{s} /2 π$ , so $f_{c} = 0.4 π \times 20000/ (2 π)$ .
$f_{c} = 4000$ Hz.
The tone has $Ω = 2 π \times 15/20 = 1.5 π$ rad/sample; the kilohertz cancel.
$Ω$ repeats every $2 π$ , so $1.5 π \equiv - 0.5 π$ . Cosine is even, so $cos (- 0.5 π n) = cos (0.5 π n)$ : the apparent value is $0.5 π$ .
That is $0.5 π \times 20000/ (2 π) = 5000$ Hz, so the tone is aliased to $5$ kHz.

Answer: The cut-off is $4$ kHz, and the $15$ kHz tone appears as $5$ kHz.

WATCH OUT FOR

$ω$ and $Ω$ are different quantities. Test: check units. $ω$ is in rad/s and belongs to the signal; $Ω = ω T_{s}$ is in rad/sample, so the same tone gets a new $Ω$ when $T_{s}$ changes, and $Ω$ always has period $2 π$ .

Check yourself

CHECK YOURSELF 1

The signal $cos (2 π 5000 t)$ is sampled at $f_{s} = 8$ kHz. Which zero-phase cosine of lower frequency gives the same samples?

(A)
5 kHz is already the lowest
(B)
3 kHz
(C)
13 kHz
(D)
1 kHz

Answer and solution

Answer: (B)

Zero-phase tones at $\pm f + k f_{s}$ share samples.
$- 5 + 8 = 3$ kHz, and $3$ lies in $[0, 4]$ kHz.
So the apparent frequency is $3$ kHz.

CHECK YOURSELF 2

A signal is sampled at $f_{s} = 12$ kHz. What is the digital frequency $Ω$ , in rad/sample, of a $3$ kHz tone?

Answer and solution

Answer: $1.5708$ (accepted range $\pm 0.01$ )

$Ω = 2 π f / f_{s}$ .
$= 2 π \times 3/12 = π /2$ .
$π /2 \approx 1.5708$ rad/sample.

CHECK YOURSELF 3

Samples are taken at $f_{s} = 10$ kHz. An ideal reconstruction filter must recover every signal band-limited below $f_{s} /2$ . What cut-off frequency must it have?

(A)
5 kHz
(B)
10 kHz
(C)
2.5 kHz
(D)
20 kHz

Answer and solution

Answer: (A)

A signal may have content just below $f_{s} /2$ , so a cut-off below $f_{s} /2$ would remove part of it.
For such a signal, the copy around $f_{s}$ starts just above $f_{s} - f_{s} /2 = f_{s} /2$ , so a cut-off above $f_{s} /2$ would pass part of that copy.
Only $f_{s} /2 = 5$ kHz meets both.

CHECK YOURSELF 4One or more options are correct

Every signal with $X (ω) = 0$ for $∣ ω ∣ > 2 π \times 3000$ rad/s must be recovered exactly from its samples by ideal sinc interpolation. Which sampling rates $f_{s}$ achieve this?

(A)
4 kHz
(B)
6 kHz
(C)
6.5 kHz
(D)
10 kHz

Answer and solution

Answer: (C), (D)

The copies stay apart for every such signal only when $f_{s} > 2 f_{m} = 6$ kHz.
At $4$ kHz, the copy around $4$ kHz covers $1$ to $7$ kHz, so it overlaps any baseband content above $1$ kHz. At exactly $6$ kHz a $3$ kHz sine gives $sin (π n) = 0$ at every sample, so it is lost.
$6.5$ kHz and $10$ kHz both exceed $6$ kHz, so C and D.

Sampling from first principles

What you will be able to do

Before you start

Sampling copies the spectrum

Aliasing seen in the time domain

Reconstruction and the sinc

From analog to digital frequency

Check yourself

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