GATE EC · EE · IN

Sampling and reconstruction

Sampling turns a continuous-time signal into a sequence of numbers; the sampling theorem says when those numbers still carry the whole signal, and reconstruction says how an ideal lowpass filter rebuilds it from them. The standard calculations — the Nyquist rate of a sum, product or square of signals, the apparent frequency of an aliased tone, the allowed cutoff of the reconstruction filter, the output when a sampled signal is filtered, the minimum rate for a bandpass signal — all follow from one picture: sampling repeats the spectrum at every multiple of $f_{s}$ , scaled by $1/ T_{s}$ .

8 previous-year questions · 7 solved · 6 practice questions

Ideal sampling and the spectrum of the sampled signal

Conventions on this page. The Fourier transform is written in $f$ (Hz), symmetric form: $X (f) = \int_{- \infty}^{\infty} x (t) e^{- j 2 π f t} d t$ and $x (t) = \int_{- \infty}^{\infty} X (f) e^{j 2 π f t} df$ . In $ω$ (rad/s) the forward transform is $\int x (t) e^{- j ω t} d t$ and the inverse carries $1/2 π$ ; the $ω$ form of each result is given where you will meet it. $sinc (x) = \frac{sin ( π x )}{π x}$ (normalised: $sinc (0) = 1$ , zeros at every nonzero integer). $rect (x) = 1$ for $∣ x ∣ < \frac{1}{2}$ and $0$ for $∣ x ∣ > \frac{1}{2}$ ; $tri (x) = 1 - ∣ x ∣$ for $∣ x ∣ \leq 1$ and $0$ otherwise. $T_{s}$ is the sampling interval, $f_{s} = 1/ T_{s}$ the sampling rate in samples/s, $ω_{s} = 2 π f_{s}$ . Digital frequency $Ω$ is in rad/sample.

Ideal sampling multiplies $x (t)$ by a unit impulse train of period $T_{s}$ . Multiplication in time is convolution in frequency, and the transform of an impulse train is another impulse train — spacing $f_{s}$ , strength $f_{s}$ . Convolving $X (f)$ with it places a copy of $X (f)$ at every integer multiple of $f_{s}$ , each multiplied by $f_{s} = 1/ T_{s}$ .

So the sampled spectrum is periodic in $f$ with period $f_{s}$ . Sampling in time makes the spectrum periodic, just as a periodic time signal has a line spectrum.

IMPULSE TRAIN AND ITS TRANSFORM

n = - \infty \sum \infty δ (t - n T_{s}) \leftrightarrow f_{s} k = - \infty \sum \infty δ (f - k f_{s})

In $ω$ : the transform is $ω_{s} \sum_{k} δ (ω - k ω_{s})$ , with $ω_{s} = 2 π / T_{s}$ .

IDEALLY 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})

Uses $x (t) δ (t - t_{0}) = x (t_{0}) δ (t - t_{0})$ : only the sample values survive, as impulse strengths.

SPECTRUM OF THE IDEALLY SAMPLED SIGNAL

X_{s} (f) = \frac{1}{T _{s}} k = - \infty \sum \infty X (f - k f_{s}) = n = - \infty \sum \infty x (n T_{s}) e^{- j 2 π f n T_{s}}

Replicas of $X (f)$ at every multiple of $f_{s}$ , each scaled by $1/ T_{s}$ . In $ω$ : $X_{s} (ω) = \frac{1}{T _{s}} \sum_{k} X (ω - k ω_{s})$ — the scale factor is still $1/ T_{s}$ , not $ω_{s}$ .

EXAM TIP

The $1/ T_{s}$ scaling is why the ideal reconstruction filter is specified with passband gain $T_{s}$ : the two cancel and $x (t)$ comes back at its original amplitude. If a question's filter has gain 1 instead, the output is $f_{s} x (t)$ .

SUM OF THE SAMPLES

n = - \infty \sum \infty x (n T_{s}) = \frac{1}{T _{s}} k = - \infty \sum \infty X (k f_{s}) = \frac{X ( 0 )}{T _{s}} = \frac{1}{T _{s}} \int_{- \infty}^{\infty} x (t) d t (f_{m} < f_{s})

Put $f = 0$ in the spectrum formula above. The first equality holds generally; the second needs $X (k f_{s}) = 0$ for every $k \neq = 0$ , which a signal band-limited to $f_{m} < f_{s}$ guarantees — the rate does not have to reach the Nyquist rate for this. It turns an infinite sum of sinc-type samples into a single transform value.

WORKED EXAMPLE

Evaluate $n = - \infty \sum \infty sinc^{2} (\frac{n}{3})$ .

Solution

Read the terms as samples of $x (t) = sinc^{2} (t /3)$ taken at $t = n$ , so $T_{s} = 1$ s and $f_{s} = 1$ Hz.
$sinc^{2} (a t) \leftrightarrow \frac{1}{∣ a ∣} tri (\frac{f}{a})$ with $a = 1/3$ gives $X (f) = 3 tri (3 f)$ : band-limited to $f_{m} = 1/3$ Hz, with $X (0) = 3$ .
$f_{m} = 1/3 < f_{s} = 1$ , so $X (k) = 0$ for every $k \neq = 0$ and the sum is $X (0) / T_{s} = 3$ .
Direct check: $sin^{2} (π n /3) = 3/4$ unless 3 divides $n$ (then it is 0), so the sum is $1 + 2 \cdot \frac{9}{π ^{2}} \cdot \frac{3}{4} \sum_{n \geq 1, 3 ∤ n} \frac{1}{n ^{2}} = 1 + \frac{27}{2 π ^{2}} \cdot \frac{4 π ^{2}}{27} = 1 + 2 = 3$ , using $\sum_{3 ∤ n} 1/ n^{2} = \frac{π ^{2}}{6} \cdot \frac{8}{9}$ .

Answer: 3

The sampling theorem and the Nyquist rate

Let $x (t)$ be band-limited: $X (f) = 0$ for $∣ f ∣ > f_{m}$ . The replica centred at $k f_{s}$ occupies $[k f_{s} - f_{m}, k f_{s} + f_{m}]$ . Neighbouring replicas do not overlap when $f_{s} - f_{m} \geq f_{m}$ , that is $f_{s} \geq 2 f_{m}$ , and then a lowpass filter can cut out the baseband copy intact.

The empty band $f_{s} - 2 f_{m}$ between the top of the baseband copy and the bottom of the first replica is the guard band. A real filter needs it to roll off from passband to stopband.

SAMPLING THEOREM

f_{s} \geq 2 f_{m} (T_{s} \leq \frac{1}{2 f _{m}})

For a signal band-limited to $f_{m}$ Hz. If $X (f)$ has impulses exactly at $\pm f_{m}$ (the highest component is a sinusoid), the inequality must be strict: $f_{s} > 2 f_{m}$ .

NYQUIST RATE AND NYQUIST INTERVAL

f_{N} = 2 f_{m}, T_{N} = \frac{1}{f _{N}} = \frac{1}{2 f _{m}}

Nyquist rate = twice the HIGHEST frequency present (equal to twice the bandwidth only for a lowpass signal). Nyquist interval = the largest allowed sampling interval. In rad/s: $ω_{N} = 2 ω_{m}$ .

BAND LIMITS OF THE SINC FORMS

\frac{sin ( 2 π B t )}{π t} = 2 B sinc (2 B t) \leftrightarrow rect (\frac{f}{2 B}), sinc^{2} (2 B t) \leftrightarrow \frac{1}{2 B} tri (\frac{f}{2 B})

$rect (f /2 B) = 1$ for $∣ f ∣ < B$ ; $tri (f /2 B) = 1 - ∣ f ∣/ (2 B)$ for $∣ f ∣ \leq 2 B$ . So $sin (2 π B t) / (π t)$ reaches $B$ Hz and its square reaches $2 B$ Hz. In $ω$ : $sin (W t) / (π t)$ is band-limited to $W$ rad/s $= W /2 π$ Hz, so its Nyquist rate is $W / π$ samples/s.

Most Nyquist-rate questions give a signal built from simpler ones. Find the highest frequency of the combination with the rules below, then double it. Let $x$ be band-limited to $B$ , $x_{1}$ to $B_{1}$ and $x_{2}$ to $B_{2}$ (all lowpass). The entries are the band edges in general; a special cancellation can only make them smaller.

Signal	Highest frequency	Nyquist rate
$x_{1} (t) + x_{2} (t)$	$max (B_{1}, B_{2})$	$2 max (B_{1}, B_{2})$
$x_{1} (t) x_{2} (t)$	$B_{1} + B_{2}$	$2 (B_{1} + B_{2})$
$x^{2} (t)$	$2 B$	$4 B$
$x^{n} (t)$	$n B$	$2 n B$
$x_{1} (t) * x_{2} (t)$ (convolution)	$min (B_{1}, B_{2})$	$2 min (B_{1}, B_{2})$
$x (a t)$	$∣ a ∣ B$	$2∣ a ∣ B$
$x (t - t_{0})$ , $\frac{d x}{d t}$ , $\int x d t$	$B$	$2 B$
$x (t) cos (2 π f_{0} t)$ (a bandpass signal when $f_{0} > B$ )	$f_{0} + B$	$2 (f_{0} + B)$
$x (t)$ after an ideal LPF of cutoff $f_{c} < B$	$f_{c}$	$2 f_{c}$

EXAM TIP

Multiplication in time adds band limits (the spectra convolve); convolution in time keeps the smaller one (the spectra multiply). Squaring doubles the band limit: $cos^{2} (2 π f_{0} t) = \frac{1}{2} + \frac{1}{2} cos (4 π f_{0} t)$ contains $2 f_{0}$ .

EXAM TIP

Convert to Hz before anything else. $cos (1000 π t)$ is a 500 Hz tone, not 1000 Hz. A missing factor of $2 π$ is the commonest slip in these questions.

EXAM TIP

A periodic signal with finitely many harmonics has Nyquist rate $2 \times$ its highest harmonic, not its fundamental. A time-limited signal (a rectangular or triangular pulse) is never band-limited: its Nyquist rate is infinite, and some aliasing is unavoidable unless it is filtered first.

WORKED EXAMPLE

Let $x_{1} (t) = \frac{sin ( 200 π t )}{π t}$ and $x_{2} (t) = [\frac{sin ( 300 π t )}{π t}]^{2}$ . Find the Nyquist rate of (a) $x_{1} (t) + x_{2} (t)$ , (b) $x_{1} (t) x_{2} (t)$ , (c) $x_{1} (t) cos (2000 π t)$ .

Solution

$sin (200 π t) / (π t) = sin (2 π \cdot 100 t) / (π t)$ is a rectangle in frequency reaching $B_{1} = 100$ Hz.
$sin (300 π t) / (π t)$ reaches 150 Hz. Squaring convolves its spectrum with itself, so $x_{2}$ reaches $B_{2} = 2 \times 150 = 300$ Hz.
(a) Sum: highest frequency $max (100, 300) = 300$ Hz, so the Nyquist rate is $600$ Hz.
(b) Product: highest frequency $100 + 300 = 400$ Hz, so the Nyquist rate is $800$ Hz.
(c) $cos (2000 π t)$ is a 1000 Hz carrier. The product occupies 900 to 1100 Hz; highest frequency 1100 Hz, Nyquist rate $2200$ Hz. (It is a bandpass signal of bandwidth 200 Hz, so bandpass sampling could go as low as $2 (1100) / ⌊ 1100/200 ⌋ = 440$ Hz — see the bandpass section.)

Answer: (a) 600 Hz, (b) 800 Hz, (c) 2200 Hz

Aliasing and the apparent frequency

If $f_{s} < 2 f_{m}$ , the replica centred at $f_{s}$ overlaps the baseband copy. A component at $f_{0}$ has images at $∣ f_{0} - k f_{s} ∣$ for every integer $k$ , and one of them lands in $[0, f_{s} /2]$ . After sampling, that image is indistinguishable from a genuine tone at that frequency — the two produce identical samples. This is aliasing.

$f_{s} /2$ is the folding frequency: the part of the spectrum above it is reflected back into $[0, f_{s} /2]$ , like a sheet folded at $f_{s} /2$ , then at $f_{s}$ , and so on.

APPARENT (ALIASED) FREQUENCY

f_{a} = ∣ f_{0} - k f_{s} ∣, k = integer nearest to \frac{f _{0}}{f _{s}}, 0 \leq f_{a} \leq \frac{f _{s}}{2}

The frequency an ideal reconstruction filter of cutoff $f_{s} /2$ outputs for a tone at $f_{0}$ sampled at $f_{s}$ . If $f_{0} < f_{s} /2$ , $k = 0$ and $f_{a} = f_{0}$ — no aliasing.

Finding the output tone, with its phase, when the samples pass through an ideal lowpass filter of cutoff $f_{s} /2$ and passband gain $T_{s}$ (so amplitudes carry over unchanged):

Write the tone in Hz: $A cos (2 π f_{0} t + θ)$ .
Reduce modulo $f_{s}$ : $f^{'} = f_{0} - m f_{s}$ with $m$ chosen so that $0 \leq f^{'} < f_{s}$ .
If $f^{'} < f_{s} /2$ : the output is $A cos (2 π f^{'} t + θ)$ — same phase. (If $f^{'} = 0$ this is the DC value $A cos θ$ .)
If $f_{s} /2 < f^{'} < f_{s}$ : the tone folds. The output frequency is $f_{a} = f_{s} - f^{'}$ and the phase changes sign: $A cos (2 π f_{a} t - θ)$ . A sine also flips sign: $A sin (2 π f^{'} t + θ)$ becomes $- A sin (2 π f_{a} t - θ)$ .
If $f^{'} = f_{s} /2$ exactly: the samples are $A (- 1)^{n} cos θ$ and the phase is lost (next formula).
Tones that land on the same frequency add as phasors. If the filter's passband gain is 1 instead of $T_{s}$ , multiply the result by $f_{s}$ .

SINUSOID SAMPLED AT EXACTLY TWICE ITS FREQUENCY

f_{s} = 2 f_{0} : x (n T_{s}) = A cos (π n + θ) = A (- 1)^{n} cos θ

The samples depend only on $A cos θ$ : a sine ( $θ = - π /2$ ) gives all-zero samples, and different $(A, θ)$ pairs give identical samples. A sinusoid therefore needs $f_{s} > 2 f_{0}$ strictly. In the spectrum, the impulses at $\pm f_{0}$ fall on each other's replicas at $\pm f_{s} /2$ .

WORKED EXAMPLE

$x (t) = 3 sin (1400 π t) + 2 sin (2600 π t)$ is sampled by an ideal impulse train at $f_{s} = 1$ kHz, and the samples are passed through an ideal lowpass filter with cutoff 500 Hz and passband gain $T_{s}$ . Find the output $y (t)$ .

Solution

In Hz the tones are 700 Hz and 1300 Hz. Both exceed $f_{s} /2 = 500$ Hz, so both alias.
700 Hz: $f^{'} = 700 > 500$ , so it folds to $f_{a} = 1000 - 700 = 300$ Hz and the sine changes sign: $3 sin (1400 π t) \to - 3 sin (600 π t)$ .
1300 Hz: $f^{'} = 1300 - 1000 = 300 < 500$ , so no fold: $2 sin (2600 π t) \to 2 sin (600 π t)$ .
Check on the samples ( $t = n /1000$ ): $3 sin (1.4 π n) = 3 sin (1.4 π n - 2 π n) = - 3 sin (0.6 π n)$ and $2 sin (2.6 π n) = 2 sin (0.6 π n)$ .
Both land on 300 Hz and add: $- 3 + 2 = - 1$ . The gain $T_{s}$ cancels the $1/ T_{s}$ from sampling.

Answer: $y (t) = - sin (600 π t)$ — a 300 Hz tone of amplitude 1; the two inputs partly cancel.

EXAM TIP

Never add the amplitudes of tones that alias onto the same frequency without checking which of them folded: a folded tone has its phase reversed, so they can reinforce or cancel.

EXAM TIP

In rad/s the same rule reads $ω_{a} = ∣ ω_{0} - k ω_{s} ∣$ with $ω_{s} = 2 π / T_{s}$ . Pick one unit and stay in it.

Reconstruction with an ideal lowpass filter

When $f_{s} > 2 f_{m}$ , an ideal lowpass filter with passband gain $T_{s}$ keeps the baseband copy and rejects every replica. Its cutoff only has to lie in the empty band between the top of the baseband copy ( $f_{m}$ ) and the bottom of the first replica ( $f_{s} - f_{m}$ ).

In time, the filter's impulse response is a sinc, so the output is a sum of sinc pulses, one centred on each sampling instant and weighted by that sample — the sinc interpolation formula. With $f_{c} = f_{s} /2$ each sinc is zero at all the other sampling instants, so the rebuilt signal passes exactly through the samples.

RECONSTRUCTION FILTER AND ALLOWED CUTOFF

H_{r} (f) = T_{s} rect (\frac{f}{2 f _{c}}), f_{m} < f_{c} < f_{s} - f_{m}

Passband gain $T_{s}$ restores the amplitude. The range exists only when $f_{s} > 2 f_{m}$ ; at $f_{s} = 2 f_{m}$ it shrinks to the single value $f_{c} = f_{s} /2$ . The endpoints are also acceptable when $X (f)$ has no impulses at $\pm f_{m}$ .

IMPULSE RESPONSE OF THE RECONSTRUCTION FILTER

h_{r} (t) = 2 f_{c} T_{s} sinc (2 f_{c} t) f_{c} = f_{s} /2 h_{r} (t) = sinc (\frac{t}{T _{s}})

With $f_{c} = f_{s} /2$ : $h_{r} (0) = 1$ and $h_{r} (n T_{s}) = 0$ for every $n \neq = 0$ .

SINC INTERPOLATION FORMULA

x (t) = n = - \infty \sum \infty x (n T_{s}) sinc (\frac{t - n T _{s}}{T _{s}})

Exact for $x (t)$ band-limited to $f_{m} < f_{s} /2$ , using $f_{c} = f_{s} /2$ . For any other allowed cutoff: $x (t) = 2 f_{c} T_{s} \sum_{n} x (n T_{s}) sinc (2 f_{c} (t - n T_{s}))$ .

ENERGY FROM THE SAMPLES

\int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t = T_{s} n = - \infty \sum \infty ∣ x (n T_{s}) ∣^{2}

For an energy signal band-limited to $f_{m}$ with $f_{s} \geq 2 f_{m}$ . It follows from the sinc interpolation formula: the pulses $sinc ((t - n T_{s}) / T_{s})$ are mutually orthogonal and each has energy $T_{s}$ .

Below the Nyquist rate no cutoff works. The overlapping parts of adjacent replicas are already added together inside $X_{s} (f)$ , and no filter can separate a sum. The ideal LPF with cutoff $f_{s} /2$ then outputs the original components below $f_{s} /2$ plus the aliases found in the previous section.

EXAM TIP

Check a given cutoff against both ends of the gap. A cutoff above $f_{s} - f_{m}$ lets in the bottom of the first replica even though $f_{s} > 2 f_{m}$ ; for a tone at $f_{0}$ sampled at $f_{s}$ , the first replica tone sits at $f_{s} - f_{0}$ , and the filter must stop it.

Practical sampling: natural, flat-top, aperture effect and anti-aliasing

Real samplers use pulses of finite width $τ$ , not impulses. In natural sampling a switch passes $x (t)$ for $τ$ seconds in every $T_{s}$ , so the pulse tops follow the signal. In flat-top sampling (sample-and-hold) the value at the sampling instant is held constant for $τ$ , so the pulse tops are flat.

The difference shows up in the spectrum: natural sampling weights each replica by a constant, while flat-top sampling multiplies the whole spectrum, baseband copy included, by $sinc (f τ)$ .

NATURAL SAMPLING

X_{n s} (f) = \frac{τ}{T _{s}} k = - \infty \sum \infty sinc (\frac{k τ}{T _{s}}) X (f - k f_{s})

$x (t)$ times a unit-height rectangular pulse train (width $τ$ , period $T_{s}$ , pulses centred on $n T_{s}$ ). $c_{k} = (τ / T_{s}) sinc (k τ / T_{s})$ are the pulse train's Fourier-series coefficients — one constant per replica. The baseband copy is $(τ / T_{s}) X (f)$ , undistorted; an ideal LPF with gain $T_{s} / τ$ recovers $x (t)$ .

FLAT-TOP (SAMPLE-AND-HOLD) SAMPLING

X_{f t} (f) = \frac{τ}{T _{s}} sinc (f τ) e^{- j π f τ} k = - \infty \sum \infty X (f - k f_{s})

Impulse samples convolved with the hold pulse $rect ((t - τ /2) / τ)$ (width $τ$ , starting at each sampling instant), whose transform is $τ sinc (f τ) e^{- j π f τ}$ . The factor varies with $f$ , so the baseband copy itself is distorted — the aperture effect — and delayed by $τ /2$ .

APERTURE-EFFECT ATTENUATION AND EQUALISER

A (f) = ∣ sinc (f τ) ∣ = \frac{sin ( π f τ )}{π f τ}, H_{e q} (f) = \frac{1}{sinc ( f τ )}, ∣ f ∣ \leq f_{m}

Attenuation relative to DC, worst at the top of the band $f = f_{m}$ . The equaliser, placed with the reconstruction filter, undoes the magnitude distortion; the delay $τ /2$ is simply left in. The effect is small when $τ ≪ T_{s}$ : with $τ / T_{s} = 0.1$ the loss even at $f_{s} /2$ is under 0.04 dB.

ZERO-ORDER HOLD (DAC) RECONSTRUCTION

H_{0} (f) = T_{s} sinc (f T_{s}) e^{- j π f T_{s}}, \frac{H _{0} ( f _{s} /2 )}{H _{0} ( 0 )} = \frac{2}{π} \approx 0.637 (- 3.92 dB)

Flat-top with $τ = T_{s}$ : each sample is held until the next, giving a staircase. It droops across the band (needs a $1/ sinc$ equaliser) and passes attenuated images (needs a smoothing lowpass filter).

Property	Ideal (impulse)	Natural	Flat-top
Pulse top	impulse of strength $x (n T_{s})$	follows $x (t)$ for $τ$	held at $x (n T_{s})$ for $τ$
Weight on replica $k$	$1/ T_{s}$	$(τ / T_{s}) sinc (k τ / T_{s})$ , a constant	$(τ / T_{s}) sinc (f τ) e^{- j π f τ}$ , varies with $f$
Baseband copy	exact, scaled by $1/ T_{s}$	exact, scaled by $τ / T_{s}$	distorted by $sinc (f τ)$ , delayed by $τ /2$
Recovery	ideal LPF, gain $T_{s}$	ideal LPF, gain $T_{s} / τ$	ideal LPF, gain $T_{s} / τ$ , plus $1/ sinc (f τ)$ equaliser

Anti-aliasing filter. An analog lowpass filter placed before the sampler, removing signal and noise above the wanted band so that nothing can fold into it. It cannot work after the sampler: once a component has folded it sits at a legitimate in-band frequency.

Its transition band may run from $f_{m}$ up to $f_{s} - f_{m}$ . A component at $f$ between $f_{s} /2$ and $f_{s} - f_{m}$ folds to $f_{s} - f$ , which lies between $f_{m}$ and $f_{s} /2$ — outside the wanted band $[0, f_{m}]$ , so a later filter that keeps only $[0, f_{m}]$ removes it. A component in $[f_{s} - f_{m}, f_{s} + f_{m}]$ (and likewise near $2 f_{s}$ , $3 f_{s}$ , …) folds into $[0, f_{m}]$ , so the filter must stop everything from $f_{s} - f_{m}$ upward. This is why practical systems sample faster than $2 f_{m}$ : the guard band is room for real filters to roll off.

WORKED EXAMPLE

A signal band-limited to 4 kHz is sampled at 10 kHz. Find the aperture-effect attenuation at 4 kHz, relative to DC, in dB, for (a) flat-top pulses of width $τ = 25 μ s$ and (b) a zero-order hold ( $τ = T_{s}$ ).

Solution

$T_{s} = 1/ (10 kHz) = 100 μ s$ . The attenuation factor at $f = 4$ kHz is $∣ sinc (f τ) ∣$ .
(a) $f τ = 4000 \times 25 \times 1 0^{- 6} = 0.1$ , so $sinc (0.1) = \frac{sin ( 0.1 π )}{0.1 π} = \frac{0.3090}{0.3142} = 0.9836$ , and $20 lo g_{10} 0.9836 = - 0.14$ dB.
(b) $f τ = 4000 \times 100 \times 1 0^{- 6} = 0.4$ , so $sinc (0.4) = \frac{sin ( 0.4 π )}{0.4 π} = \frac{0.9511}{1.2566} = 0.7568$ , and $20 lo g_{10} 0.7568 = - 2.42$ dB.
Narrow pulses make the aperture effect negligible. The full-interval hold needs an equaliser gain of $1/0.7568 \approx 1.32$ at 4 kHz.

Answer: (a) about $- 0.14$ dB; (b) about $- 2.42$ dB

Bandpass sampling

A bandpass signal occupies only $f_{L} \leq ∣ f ∣ \leq f_{H}$ , with bandwidth $B = f_{H} - f_{L}$ . Sampling at $2 f_{H}$ always works, but much lower rates can too: the shifted copies of the positive and negative bands only have to avoid the original band — they may sit in the empty space between $- f_{L}$ and $f_{L}$ or above $f_{H}$ . The minimum depends on how many whole bandwidths fit under $f_{H}$ .

MINIMUM BANDPASS SAMPLING RATE

f_{s, m i n} = \frac{2 f _{H}}{⌊ f _{H} / B ⌋}, B = f_{H} - f_{L}

$⌊ \cdot ⌋$ is the greatest-integer function. If $f_{H}$ is an integer multiple of $B$ , $f_{s, m i n} = 2 B$ . In general $2 B \leq f_{s, m i n} < 4 B$ .

ALL ALIAS-FREE UNIFORM SAMPLING RATES

\frac{2 f _{H}}{k} \leq f_{s} \leq \frac{2 f _{L}}{k - 1}, k = 1, 2, \dots, ⌊ \frac{f _{H}}{B} ⌋

For $k = 1$ the upper limit is $\infty$ : ordinary sampling at $f_{s} \geq 2 f_{H}$ . Rates in the gaps between these windows alias, so a rate above the minimum is not automatically safe.

WORKED EXAMPLE

A bandpass signal occupies 18 kHz to 25 kHz. Find the minimum sampling rate that avoids aliasing, and decide whether $f_{s} = 20$ kHz is acceptable.

Solution

$B = 25 - 18 = 7$ kHz and $f_{H} / B = 25/7 = 3.57$ , so $⌊ f_{H} / B ⌋ = 3$ .
$f_{s, m i n} = 2 (25) /3 = 16.67$ kHz.
Alias-free windows: $k = 3$ : $16.67 \leq f_{s} \leq 2 (18) /2 = 18$ kHz; $k = 2$ : $25 \leq f_{s} \leq 36$ kHz; $k = 1$ : $f_{s} \geq 50$ kHz.
20 kHz lies in the gap between 18 and 25 kHz, so it aliases. Check: the negative band $[- 25, - 18]$ kHz shifted by $2 f_{s} = 40$ kHz lands on $[15, 22]$ kHz, overlapping the original $[18, 25]$ kHz.

Answer: $f_{s, m i n} \approx 16.67$ kHz; 20 kHz is not acceptable (it causes aliasing).

EXAM TIP

The Nyquist rate is $2 f_{m a x}$ by definition. The bandpass formula answers a different question — the lowest rate at which a bandpass signal can be sampled without aliasing. For $m (t) cos (2 π f_{c} t)$ with $m$ band-limited to $f_{m}$ , the Nyquist rate is $2 (f_{c} + f_{m})$ , while the bandpass minimum uses $f_{H} = f_{c} + f_{m}$ and $B = 2 f_{m}$ . Read which one is asked.

From continuous to discrete frequency, and sampling followed by filtering

The samples $x [n] = x (n T_{s})$ form a discrete-time signal. Its DTFT is the sampled spectrum with the frequency axis relabelled: a continuous-time frequency $f$ (Hz) or $ω$ (rad/s) appears at the digital frequency $Ω = ω T_{s} = 2 π f / f_{s}$ rad/sample. The folding frequency $f_{s} /2$ maps to $Ω = π$ , and the $f_{s}$ -periodicity of $X_{s} (f)$ becomes the $2 π$ -periodicity of the DTFT.

CT TO DT FREQUENCY MAPPING

Ω = ω T_{s} = \frac{2 π f}{f _{s}} rad/sample, f = \frac{Ω f _{s}}{2 π}

One-to-one only on the baseband, $∣Ω∣ \leq π \leftrightarrow ∣ f ∣ \leq f_{s} /2$ . $cos (2 π f_{0} t)$ sampled gives $cos (Ω_{0} n)$ with $Ω_{0} = 2 π f_{0} / f_{s}$ ; reduce $Ω_{0}$ modulo $2 π$ into $(- π, π]$ to read off the alias.

PERIODICITY OF A SAMPLED SINUSOID

x [n] = A cos (2 π \frac{f _{0}}{f _{s}} n + θ) is periodic only if \frac{f _{0}}{f _{s}} = \frac{p}{q} (integers, lowest terms, q > 0); then N = q

A continuous-time sinusoid is always periodic; its samples need not be — $cos (n)$ , from $f_{0} / f_{s} = 1/2 π$ , never repeats. Examples: 1 kHz sampled at 3 kHz gives $p / q = 1/3$ , $N = 3$ ; 700 Hz sampled at 1 kHz gives $7/10$ , $N = 10$ . A sum of such sequences with periods $N_{1}$ and $N_{2}$ repeats every $lcm (N_{1}, N_{2})$ samples.

DTFT OF THE SAMPLES

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

$X_{c} (ω)$ is the continuous-time transform in $ω$ (rad/s). This is $X_{s} (ω)$ evaluated at $ω = Ω/ T_{s}$ .

DISCRETE-TIME PROCESSING OF A CT SIGNAL

H_{eff} (ω) = {H (e^{j ω T_{s}}), 0, ∣ ω ∣ < ω_{s} /2 ∣ ω ∣ > ω_{s} /2

Sampler at $f_{s}$ , then an LTI discrete-time system $H (e^{j Ω})$ , then ideal reconstruction (cutoff $f_{s} /2$ , gain $T_{s}$ ). Valid when $x (t)$ is band-limited below $f_{s} /2$ . A DT cutoff $Ω_{c}$ acts as a CT cutoff $f_{c} = Ω_{c} f_{s} / (2 π)$ : at $f_{s} = 8$ kHz, $Ω_{c} = π /4$ gives $f_{c} = 1$ kHz.

Sampling followed by a bandpass filter. The impulse-sampled signal contains every replica, so a bandpass filter after the sampler can select a replica instead of the baseband copy — sampling used as frequency translation. The output can sit at a frequency the input never contained. That is not aliasing; aliasing is a replica landing inside the band the reconstruction filter keeps. Work line by line.

Procedure: sample, then filter.

Write $x (t)$ as spectral lines: $A cos (2 π f_{0} t + θ)$ has $\frac{A}{2} e^{j θ}$ at $+ f_{0}$ and $\frac{A}{2} e^{- j θ}$ at $- f_{0}$ .
Impulse sampling copies every line to $k f_{s} \pm f_{0}$ for all integers $k$ , with strength multiplied by $f_{s}$ . (For natural sampling, multiply by that replica's weight $c_{k}$ instead.)
Keep the lines inside the filter's passband — a real filter passes the $+$ and $-$ bands together — and multiply by its gain.
Pair each line at $+ f$ with its mirror at $- f$ to form a real cosine. A line at $k f_{s} + f_{0}$ keeps phase $θ$ ; a line at $k f_{s} - f_{0}$ (with $k \geq 1$ ) came from the $- f_{0}$ line and carries $- θ$ .
Add everything that lands on the same frequency as phasors.

WORKED EXAMPLE

$x (t) = 2 cos (2000 π t + π /6)$ is sampled by an ideal impulse train at $f_{s} = 3$ kHz. The sampled signal passes through an ideal bandpass filter with gain $T_{s}$ for $3.5 kHz \leq ∣ f ∣ \leq 5.5 kHz$ and zero elsewhere. Find the output $y (t)$ .

Solution

$x (t)$ is a 1 kHz tone: lines $e^{j π /6}$ at $+ 1$ kHz and $e^{- j π /6}$ at $- 1$ kHz.
The sampled spectrum has lines at $3 k \pm 1$ kHz — at 1, 2, 4, 5, 7, 8, … kHz and their negatives — each multiplied by $f_{s}$ .
The passband keeps 4 kHz ( $= 3 + 1$ , from the $+ 1$ kHz line, phase $+ π /6$ ) and 5 kHz ( $= 6 - 1$ , from the $- 1$ kHz line, phase $- π /6$ ). The gain $T_{s}$ cancels the factor $f_{s}$ .
Pairing with the mirror lines at $- 4$ and $- 5$ kHz gives $2 cos (8000 π t + π /6)$ and $2 cos (10000 π t - π /6)$ .

Answer: $y (t) = 2 cos (8000 π t + π /6) + 2 cos (10000 π t - π /6)$

EXAM TIP

A DT sinusoid $cos (Ω_{0} n)$ is unchanged when $Ω_{0}$ changes by $2 π$ , and $cos ((2 π - Ω_{0}) n) = cos (Ω_{0} n)$ . These two identities are aliasing and folding seen from the discrete-time side; use them to check any apparent-frequency answer.

PREVIOUS-YEAR QUESTIONS

8 questions from the EC, EE and IN papers

Try each question first, then open the answer. One shows the official answer only. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2018 · GATE EE 2023 · GATE IN 2024 · GATE IN 2020

GATE 2026 EC Q.15

GATE 2026 · ECQ.15 · 1 mark · MCQ

Two analog signals $x_{1} (t)$ and $x_{2} (t)$ ( $t$ in second), are sampled at a rate $F_{s} = 40 Hz$ , where

$x_{1} (t) = cos (20 π t), t \geq 0$ , and $x_{2} (t) = cos (100 π t), t \geq 0$ .

The first ten samples (starting from $t = 0$ ) are considered for the analysis.

Which of the following statements is TRUE?

(A)
All of the first three samples of $x_{1} (t)$ are greater than the corresponding samples of $x_{2} (t)$ .
(B)
All of the last three samples of $x_{1} (t)$ are greater than the corresponding samples of $x_{2} (t)$ .
(C)
All of the samples of $x_{2} (t)$ are greater than the corresponding samples of $x_{1} (t)$ .
(D)
All of the fourth to seventh samples of $x_{1} (t)$ are equal to the corresponding samples of $x_{2} (t)$ .

Answer and solution

GATE 2026 EC Q.15: answer and solution

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GATE 2025 EC Q.38

GATE 2025 · ECQ.38 · 2 marks · MCQ

Consider a continuous-time finite-energy signal $f (t)$ whose Fourier transform vanishes outside the frequency interval $[- ω_{c}, ω_{c}]$ , where $ω_{c}$ is in rad/sec.

The signal $f (t)$ is uniformly sampled to obtain $y (t) = f (t) p (t)$ . Here,
$p (t) = n = - \infty \sum \infty δ (t - τ - n T_{s}),$
with $δ (t)$ being the Dirac impulse, $T_{s} > 0$ , and $τ > 0$ . The sampled signal $y (t)$ is passed through an ideal lowpass filter $h (t) = ω_{c} T_{s} \frac{sin ( ω _{c} t )}{π ω _{c} t}$ with cutoff frequency $ω_{c}$ and passband gain $T_{s}$ .

The output of the filter is given by ________.

(A)
$f (t)$ if $T_{s} < π / ω_{c}$
(B)
$f (t - τ)$ if $T_{s} < π / ω_{c}$
(C)
$f (t - τ)$ if $T_{s} < 2 π / ω_{c}$
(D)
$T_{s} f (t)$ if $T_{s} < 2 π / ω_{c}$

Answer and solution

GATE 2025 EC Q.38: answer and solution

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GATE 2024 EC Q.48

GATE 2024 · ECQ.48 · 2 marks · MCQ

A continuous time signal $x (t) = 2 cos (8 π t + π /3)$ is sampled at a rate of 15 Hz. The sampled signal $x_{s} (t)$ when passed through an LTI system with impulse response

h (t) = (\frac{sin 2 π t}{π t}) cos (38 π t - π /2)

produces an output $x_{o} (t)$ . The expression for $x_{o} (t)$ is ______.

(A)
$15 sin (38 π t + π /3)$
(B)
$15 sin (38 π t - π /3)$
(C)
$15 cos (38 π t - π /6)$
(D)
$15 cos (38 π t + π /6)$

Answer and solution

GATE 2024 EC Q.48: answer and solution

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GATE 2024 IN Q.27

GATE 2024 · INQ.27 · 1 mark · NAT

The Nyquist sampling frequency for $x (t) = 10 sin^{2} (200 π t)$ is ________ Hz (rounded off to nearest integer).

Answer and solution

GATE 2024 IN Q.27: answer and solution

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GATE 2023 EC Q.50

GATE 2023 · ECQ.50 · 2 marks · MCQ

Let $x_{1} (t)$ and $x_{2} (t)$ be two band-limited signals having bandwidth $B = 4 π \times 1 0^{3}$ rad/s each. In the figure below, the Nyquist sampling frequency, in rad/s, required to sample $y (t)$ , is

GATE 2023 EC question 50: Block diagram with two branches. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
$20 π \times 1 0^{3}$
(B)
$40 π \times 1 0^{3}$
(C)
$8 π \times 1 0^{3}$
(D)
$32 π \times 1 0^{3}$

Answer and solution

GATE 2023 EC Q.50: answer and solution

GATE 2023 EE Q.63

GATE 2023 · EEQ.63 · 2 marks · NAT

A signal $x (t) = 2 cos (180 π t) cos (60 π t)$ is sampled at 200 Hz and then passed through an ideal low pass filter having cut-off frequency of 100 Hz.

The maximum frequency present in the filtered signal in Hz is ________ (Round off to the nearest integer).

Answer and solution

GATE 2023 EE Q.63: answer and solution

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GATE 2020 IN Q.22

GATE 2020 · INQ.22 · 1 mark · NAT

A sinusoid of 10 kHz is sampled at 15 k samples/s. The resulting signal is passed through an ideal low pass filter (LPF) with cut-off frequency of 25 kHz. The maximum frequency component at the output of the LPF (in kHz) is ____

Official answer

Official answer: 25Source: Official answer key, GATE 2020 IN (IIT Delhi)

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GATE 2018 EC Q.54

GATE 2018 · ECQ.54 · 2 marks · NAT

A band limited low-pass signal $x (t)$ of bandwidth 5 kHz is sampled at a sampling rate $f_{s}$ . The signal $x (t)$ is reconstructed using the reconstruction filter $H (f)$ whose magnitude response is shown below:

[Figure]

The minimum sampling rate $f_{s}$ (in kHz) for perfect reconstruction of $x (t)$ is ______.

GATE 2018 EC question 54: Plot of |H(f)| (vertical axis) against f (kHz) (horizontal axis). — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

GATE 2018 EC Q.54: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

Let $x (t) = \frac{sin ( 100 π t )}{π t} + cos (240 π t)$ and $y (t) = x^{2} (t)$ . The Nyquist rate of $y (t)$ (twice its highest frequency) is

(A)
240 Hz
(B)
340 Hz
(C)
480 Hz
(D)
960 Hz

Answer and solution

Answer: (C)

$sin (100 π t) / (π t) = sin (2 π \cdot 50 t) / (π t)$ is band-limited to 50 Hz.
$cos (240 π t)$ is a 120 Hz tone, so $x (t)$ reaches 120 Hz.
Squaring doubles the band limit: $y (t)$ reaches 240 Hz (the term $\frac{1}{2} cos (480 π t)$ from $cos^{2} (240 π t)$ is present; the cross term reaches only $120 + 50 = 170$ Hz and the squared sinc only 100 Hz).
Nyquist rate $= 2 \times 240 = 480$ Hz.

PRACTICE 22 marks · MSQ

A real bandpass signal has its spectrum confined to $30 kHz \leq ∣ f ∣ \leq 38 kHz$ . It is sampled uniformly by an ideal impulse train. Which of the following sampling rates avoid aliasing, so that the signal can be recovered exactly by an ideal bandpass filter?

(A)
19.5 kHz
(B)
22 kHz
(C)
28 kHz
(D)
70 kHz

Answer and solution

Answer: (A), (C)

$B = 38 - 30 = 8$ kHz and $⌊ f_{H} / B ⌋ = ⌊ 4.75 ⌋ = 4$ .
Alias-free windows $2 f_{H} / k \leq f_{s} \leq 2 f_{L} / (k - 1)$ , in kHz: $k = 4$ : $19 \leq f_{s} \leq 20$ ; $k = 3$ : $25.33 \leq f_{s} \leq 30$ ; $k = 2$ : $38 \leq f_{s} \leq 60$ ; $k = 1$ : $f_{s} \geq 76$ .
19.5 kHz lies in the $k = 4$ window and 28 kHz in the $k = 3$ window, so both are valid.
22 kHz (between 20 and 25.33) and 70 kHz (between 60 and 76) lie in gaps. Check 70 kHz: the band $[- 38, - 30]$ kHz shifted by $+ 70$ kHz lands on $[32, 40]$ kHz, overlapping $[30, 38]$ kHz.

PRACTICE 31 mark · NAT

The value of $n = - \infty \sum \infty \frac{sin ( π n /4 )}{π n}$ , with the $n = 0$ term taken as its limiting value $1/4$ , is ________.

Answer and solution

Answer: $1$

The terms are the samples at $t = n$ ( $T_{s} = 1$ s, $f_{s} = 1$ Hz) of $x (t) = \frac{sin ( π t /4 )}{π t} = \frac{sin ( 2 π \cdot \frac{1}{8} t )}{π t}$ .
By the sinc-form pair with $B = 1/8$ Hz, $X (f) = rect (4 f)$ : band-limited to $1/8$ Hz, with $X (0) = 1$ .
$f_{m} = 1/8 < f_{s} = 1$ Hz, so $X (k) = 0$ for every $k \neq = 0$ and the sum equals $X (0) / T_{s} = 1$ .
Check: $\sum_{n \geq 1} sin (n α) / n = (π - α) /2$ for $0 < α < 2 π$ ; with $α = π /4$ the full sum is $\frac{1}{4} + \frac{2}{π} \cdot \frac{3 π}{8} = \frac{1}{4} + \frac{3}{4} = 1$ .

PRACTICE 42 marks · NAT

The signal $x (t) = 4 cos (2000 π t)$ is multiplied by a periodic rectangular pulse train of unit height, pulse width $τ = 50 μ s$ and period $T_{s} = 200 μ s$ , with the pulses centred at $t = n T_{s}$ (natural sampling). The product passes through an ideal bandpass filter of unit gain for $3.5 kHz \leq ∣ f ∣ \leq 4.5 kHz$ and zero elsewhere. The amplitude of the output sinusoid is ________ (round off to two decimal places).

Answer and solution

Answer: $0.9$ (accepted range $\pm 0.01$ )

$f_{s} = 1/ T_{s} = 5$ kHz and the duty cycle is $d = τ / T_{s} = 0.25$ . The pulse train's Fourier-series coefficients are $c_{k} = d sinc (k d)$ , with $sinc (x) = sin (π x) / (π x)$ .
$x (t)$ has lines of strength 2 at $\pm 1$ kHz. The product has lines at $5 k \pm 1$ kHz: 1, 4, 6, 9, … kHz and their negatives.
Only the 4 kHz pair is in the passband: $+ 4 = 5 - 1$ (from $k = 1$ ) and $- 4 = - 5 + 1$ (from $k = - 1$ ), each of strength $2 c_{1}$ since $c_{- 1} = c_{1}$ .
$c_{1} = 0.25 sinc (0.25) = 0.25 \cdot \frac{sin ( π /4 )}{π /4} = 0.25 \times 0.9003 = 0.2251$ .
The output is $2 \times (2 c_{1}) cos (8000 π t) = 4 c_{1} cos (8000 π t)$ , amplitude $4 \times 0.2251 = 0.90$ (exactly $22 / π$ ).

PRACTICE 51 mark · NAT

The signal $x (t) = 5 cos (5200 π t + π /4)$ is sampled by an ideal impulse train at 1000 samples/s, and the samples are passed through an ideal lowpass filter with cutoff 500 Hz and passband gain $T_{s}$ . The output is a single sinusoid. Its frequency, in Hz, is ________.

Answer and solution

Answer: $400$

$5200 π = 2 π \times 2600$ , so $f_{0} = 2600$ Hz.
Reduce modulo $f_{s}$ : $f^{'} = 2600 - 2 (1000) = 600$ Hz.
$600 > f_{s} /2 = 500$ , so the tone folds to $f_{a} = 1000 - 600 = 400$ Hz, with its phase reversed: the output is $5 cos (800 π t - π /4)$ .

PRACTICE 61 mark · MSQ

Which of the following statements are TRUE?

(A)
In natural sampling with a rectangular pulse train, the replica centred at $k f_{s}$ is $X (f - k f_{s})$ multiplied by a constant that depends on $k$ but not on $f$ .
(B)
In flat-top sampling with pulse width $τ$ , the baseband part of the sampled spectrum is $X (f)$ multiplied by a factor whose magnitude is proportional to $∣ sinc (f τ) ∣$ .
(C)
Sampling $x (t) = cos (2 π f_{0} t)$ at exactly $f_{s} = 2 f_{0}$ , with samples at $t = n T_{s}$ , gives all-zero samples.
(D)
An anti-aliasing filter placed after the sampler removes aliasing that has already occurred.

Answer and solution

Answer: (A), (B)

A — true: natural sampling multiplies $x (t)$ by a pulse train with Fourier-series coefficients $c_{k} = (τ / T_{s}) sinc (k τ / T_{s})$ , giving $\sum_{k} c_{k} X (f - k f_{s})$ ; each $c_{k}$ depends on $k$ only.
B — true: flat-top sampling convolves the impulse samples with a hold pulse whose transform is $τ sinc (f τ) e^{- j π f τ}$ , which multiplies the baseband copy (aperture effect).
C — false: $cos (2 π f_{0} \cdot n / (2 f_{0})) = cos (π n) = (- 1)^{n}$ . It is a sine that gives all-zero samples.
D — false: a folded component occupies an in-band frequency and cannot be filtered out afterwards; the anti-aliasing filter must come before the sampler.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

Nyquist sampling theorem, sampling and reconstruction

GATE EE

sampling theorem

GATE IN

sampling

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