Sampling

Aliasing

Aliasing is the overlap of the shifted spectra of a sampled signal when the sampling frequency is less than twice the highest frequency ($f_s < 2f_m$). High-frequency components then appear as false low frequencies ("aliases"), and the original signal cannot be recovered. For example, a 7 kHz tone sampled at 10 kHz looks like a 3 kHz tone.

Nyquist rate of $x(t) = 1 + \cos(2000\pi t) + \sin(4000\pi t)$

Find the frequency of each term using $\omega = 2\pi f$:

Highest frequency: $f_m = 2000$ Hz.

$$\begin{aligned} \text{Nyquist rate } f_N &= 2f_m = 2 \times 2000 = 4000\ \text{Hz} \\ \text{Nyquist interval } T_N &= \frac{1}{f_N} = \frac{1}{4000} = 0.25\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 4000$ samples/s (4 kHz), i.e. $\omega_s = 8000\pi$ rad/s; Nyquist interval $= 0.25$ ms.

Sampling theorem

A continuous-time signal that is band-limited to $f_m$ Hz (no frequency components above $f_m$) is completely described by its samples taken at a uniform rate $f_s \ge 2f_m$. The signal can then be exactly recovered from the samples by passing them through an ideal low-pass filter with cutoff between $f_m$ and $f_s - f_m$.

Nyquist criterion

• The minimum sampling rate $f_N = 2f_m$ is the Nyquist rate; its reciprocal $T_N = \frac{1}{2f_m}$ is the Nyquist interval (maximum gap between samples).
• If $f_s < 2f_m$, the spectral copies overlap and aliasing occurs. In practice $f_s$ is kept above $2f_m$ (e.g. audio: $f_m = 20$ kHz, $f_s = 44.1$ kHz).

Nyquist rate of $x(t) = 20\sin(500\pi t) + 10\cos(300\pi t) - 50\cos(1000\pi t)$

Highest frequency $f_m = 500$ Hz (the amplitudes do not matter).

$$\begin{aligned} f_N &= 2f_m = 2\times500 = 1000\ \text{Hz} \\ T_N &= \frac{1}{1000} = 1\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 1000$ samples/s (1 kHz, or $2000\pi$ rad/s); Nyquist interval $= 1$ ms.

Aliasing is the distortion that occurs when a signal is sampled at a rate lower than twice its highest frequency ($f_s < 2f_m$). The shifted copies of the spectrum overlap, so high-frequency components fold back and appear as lower frequencies. After this, the original signal cannot be recovered by any filter.

Why it happens

Sampling with an impulse train of rate $f_s$ makes the spectrum periodic:

$$X_s(f) = f_s\sum_{k=-\infty}^{\infty}X(f - kf_s)$$

Each copy occupies $kf_s - f_m$ to $kf_s + f_m$. The copies are separate only if $f_s - f_m \ge f_m$, i.e. $f_s \ge 2f_m$.

 fs > 2fm (no aliasing):
   ___       ___       ___
  /   \     /   \     /   \
-+--+--+---+--+--+---+--+--+--> f
  -fm 0 fm    fs        2fs

 fs < 2fm (aliasing: tails overlap):
   ____ ____ ____
  /    X    X    \
-+---+----+----+---> f
     0    fs   2fs

Example: A 700 Hz tone sampled at 1000 Hz gives samples identical to a 300 Hz tone ($|700 - 1000| = 300$ Hz). The 700 Hz tone has been "aliased" to 300 Hz.

Effects

• False frequencies appear in the reconstructed signal.
• In images, aliasing causes jagged edges and moiré patterns; in video, wheels seem to turn backwards.

Prevention

1. Sample above the Nyquist rate: choose $f_s > 2f_m$, usually with a margin (e.g. CD audio uses 44.1 kHz for a 20 kHz band).
2. Anti-aliasing (pre-)filter: pass the analog signal through a low-pass filter with cutoff $f_s/2$ before the sampler. Real signals are never perfectly band-limited, and noise has high frequencies; the filter removes these before they can fold back.
3. Guard band: since practical filters are not ideal, keep $f_s$ comfortably above $2f_m$ so that the filter's transition band fits between $f_m$ and $f_s - f_m$.
4. Oversampling: sample at a much higher rate, then filter and decimate digitally; this relaxes the analog filter requirement.

x(t) -> [Anti-aliasing LPF, fc = fs/2] -> [Sampler fs] -> [ADC]

Sampling theorem

A band-limited signal with no frequency components above $f_m$ Hz can be completely represented by, and exactly recovered from, its uniformly spaced samples if the sampling frequency satisfies $f_s \ge 2f_m$. The minimum rate $2f_m$ is the Nyquist rate; recovery is done with an ideal low-pass filter.

Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(4000\pi t)\cos(1000\pi t)$

Step 1: convert the product to a sum using $\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\left[\cos(5000\pi t) + \cos(3000\pi t)\right] \\ &= \frac{1}{4\pi}\cos(5000\pi t) + \frac{1}{4\pi}\cos(3000\pi t) \end{aligned}$$

Step 2: frequencies.

Highest frequency $f_m = 2500$ Hz.

Step 3: Nyquist rate and interval.

$$\begin{aligned} f_N &= 2f_m = 2\times2500 = 5000\ \text{Hz} \\ T_N &= \frac{1}{f_N} = \frac{1}{5000} = 0.2\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 5000$ samples/s (5 kHz, $\omega_N = 10000\pi$ rad/s); Nyquist interval $= 0.2$ ms $= 200\ \mu$s.

Note: the highest frequency is not 2000 Hz (from $\cos 4000\pi t$ alone); multiplication creates the sum frequency 2500 Hz.

Sampling theorem

Statement: A signal $x(t)$ band-limited to $\omega_m$ ($X(j\omega) = 0$ for $|\omega| > \omega_m$) is uniquely determined by its samples $x(nT)$ if the sampling frequency $\omega_s = \frac{2\pi}{T} \ge 2\omega_m$.

Proof: Model sampling as multiplication by an impulse train:

$$x_p(t) = x(t)\,p(t), \qquad p(t) = \sum_{n=-\infty}^{\infty}\delta(t - nT)$$

The FT of the periodic impulse train is $P(j\omega) = \frac{2\pi}{T}\sum_k\delta(\omega - k\omega_s)$. Multiplication in time is convolution in frequency:

$$\begin{aligned} X_p(j\omega) &= \frac{1}{2\pi}X(j\omega) * P(j\omega) \\ &= \frac{1}{T}\sum_{k=-\infty}^{\infty}X(j(\omega - k\omega_s)) \end{aligned}$$

So the sampled spectrum is the original spectrum repeated every $\omega_s$. The copy at $k=0$ spans $-\omega_m$ to $\omega_m$, and the next starts at $\omega_s - \omega_m$. They do not overlap if $\omega_s - \omega_m \ge \omega_m$, i.e. $\omega_s \ge 2\omega_m$. Then an ideal low-pass filter with gain $T$ and cutoff $\omega_m \le \omega_c \le \omega_s - \omega_m$ passes only the $k=0$ copy, giving back $X(j\omega)$ and hence $x(t)$ exactly. If $\omega_s < 2\omega_m$, the copies overlap (aliasing) and recovery is impossible. Hence proved.

Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(4000\pi t)\cos(1000\pi t)$

$$x(t) = \frac{1}{4\pi}\left[\cos(5000\pi t) + \cos(3000\pi t)\right]$$

Frequencies: $\frac{5000\pi}{2\pi} = 2500$ Hz and $\frac{3000\pi}{2\pi} = 1500$ Hz, so $f_m = 2500$ Hz.

$$f_N = 2f_m = 5000\ \text{Hz}, \qquad T_N = \frac{1}{5000} = 0.2\ \text{ms}$$

Answer: Nyquist rate $= 5000$ samples/s (5 kHz); Nyquist interval $= 0.2$ ms.

Nyquist criterion

For a signal band-limited to $f_m$ Hz, the sampling frequency must satisfy

$$f_s \ge 2f_m$$

The minimum value $f_N = 2f_m$ is the Nyquist rate and $T_N = 1/(2f_m)$ is the Nyquist interval. When this holds, the spectral copies produced by sampling, $X_s(f) = f_s\sum_k X(f - kf_s)$, do not overlap, and an ideal low-pass filter of cutoff $f_m$ to $f_s - f_m$ recovers the original signal exactly.

Aliasing effect

If $f_s < 2f_m$, the copies at $0$ and $\pm f_s$ overlap. A component of frequency $f > f_s/2$ then appears at the alias frequency $|f - kf_s|$ (the nearest one in $0$ to $f_s/2$). Example: 900 Hz sampled at 1200 Hz looks like 300 Hz. The overlap cannot be undone after sampling. It is prevented by an anti-aliasing low-pass filter before the sampler and by choosing $f_s > 2f_m$.

 fs >= 2fm:  /\     /\     /\        separate copies
           -+--+---+--+---+--+--> f
              0       fs     2fs
 fs < 2fm:   /\/\/\/\/\              copies overlap
           -+----+----+---> f
            0    fs   2fs

Nyquist rate of $x(t) = 30\sin(100\pi t) + 40\cos(400\pi t)\sin(300\pi t)$

Convert the product using $\sin A\cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$ with $A = 300\pi t$, $B = 400\pi t$:

$$\begin{aligned} 40\cos(400\pi t)\sin(300\pi t) &= 20\left[\sin(700\pi t) + \sin(-100\pi t)\right] \\ &= 20\sin(700\pi t) - 20\sin(100\pi t) \end{aligned}$$

So

$$x(t) = 10\sin(100\pi t) + 20\sin(700\pi t)$$

Highest frequency $f_m = 350$ Hz.

$$f_N = 2f_m = 700\ \text{Hz}, \qquad T_N = \frac{1}{700} = 1.43\ \text{ms}$$

Answer: Nyquist rate $= 700$ samples/s ($\omega_N = 1400\pi$ rad/s); Nyquist interval $\approx 1.43$ ms.

Frequencies in $x(t)$ ($\omega = 2\pi f$):

(i) Minimum sampling rate

$f_m = 250$ Hz, so

$$F_{s,\min} = 2f_m = 2\times250 = 500\ \text{Hz}$$

Answer: 500 samples/s (Nyquist rate).

(ii) Discrete-time signal for $F_s = 400$ Hz

Put $t = nT = n/F_s = n/400$:

$$\begin{aligned} x[n] &= 3\cos\left(\frac{200\pi n}{400}\right) + 2\sin\left(\frac{300\pi n}{400}\right) - 3\cos\left(\frac{500\pi n}{400}\right) \\ &= 3\cos\left(\frac{\pi n}{2}\right) + 2\sin\left(\frac{3\pi n}{4}\right) - 3\cos\left(\frac{5\pi n}{4}\right) \end{aligned}$$

The last term has digital frequency $5\pi/4 > \pi$, so it is aliased. Since $\cos\left(\frac{5\pi n}{4}\right) = \cos\left(2\pi n - \frac{3\pi n}{4}\right) = \cos\left(\frac{3\pi n}{4}\right)$:

$$x[n] = 3\cos\left(\frac{\pi n}{2}\right) + 2\sin\left(\frac{3\pi n}{4}\right) - 3\cos\left(\frac{3\pi n}{4}\right)$$

In normalized frequency: $\frac{100}{400} = \frac14$, $\frac{150}{400} = \frac38$, and $\frac{250}{400} = \frac58$, which aliases to $1 - \frac58 = \frac38$ (i.e. 150 Hz).

(iii) Reconstructed signal

An ideal reconstruction filter passes only $|f| \le F_s/2 = 200$ Hz, so each digital frequency $\omega$ in $[0, \pi]$ maps back to $f = \frac{\omega F_s}{2\pi}$:

$$x_r(t) = 3\cos(200\pi t) + 2\sin(300\pi t) - 3\cos(300\pi t)$$

Answer: $x_r(t) = 3\cos 200\pi t + 2\sin 300\pi t - 3\cos 300\pi t$.

Because $F_s = 400$ Hz is below the Nyquist rate (500 Hz), the 250 Hz component has been aliased to 150 Hz, so $x_r(t) \ne x(t)$.

Nyquist sampling theorem

A continuous-time signal band-limited to $f_m$ Hz can be exactly recovered from its uniformly spaced samples if the sampling rate is at least twice the highest frequency, $f_s \ge 2f_m$. The minimum rate $2f_m$ is the Nyquist rate, and $1/(2f_m)$ is the Nyquist interval. Sampling below this rate causes aliasing.

Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(50\pi t)\cos(500\pi t)$

Using $\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\left[\cos(550\pi t) + \cos(450\pi t)\right] \\ &= \frac{1}{4\pi}\cos(550\pi t) + \frac{1}{4\pi}\cos(450\pi t) \end{aligned}$$

Highest frequency $f_m = 275$ Hz.

$$\begin{aligned} f_N &= 2f_m = 2\times275 = 550\ \text{Hz} \\ T_N &= \frac{1}{550} = 1.82\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 550$ samples/s ($\omega_N = 1100\pi$ rad/s); Nyquist interval $\approx 1.82$ ms.

Impulse train sampling

Impulse train (ideal) sampling multiplies the continuous signal $x(t)$ by a periodic train of unit impulses spaced $T$ seconds apart. The result is a train of impulses whose strengths are the sample values:

$$x_p(t) = x(t)\sum_{n=-\infty}^{\infty}\delta(t - nT) = \sum_{n=-\infty}^{\infty}x(nT)\,\delta(t - nT)$$

$T$ is the sampling period and $\omega_s = 2\pi/T$ the sampling frequency.

Spectrum of the sampled signal

$p(t) = \sum\delta(t - nT) \leftrightarrow P(j\omega) = \frac{2\pi}{T}\sum_k\delta(\omega - k\omega_s)$. Multiplication in time is convolution in frequency:

$$X_p(j\omega) = \frac{1}{2\pi}X(j\omega) * P(j\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty}X\big(j(\omega - k\omega_s)\big)$$

So $X_p(j\omega)$ is $X(j\omega)$ repeated at every multiple of $\omega_s$, scaled by $1/T$.

 X(jw)            Xp(jw), ws > 2wm
   /\               /\      /\      /\
  /  \             /  \    /  \    /  \
-+-+--+-->  w   --+--+--+-+--+--+-+--+--> w
 -wm 0 wm          -ws    0  wm   ws

Reconstruction from samples

1. Condition: $x(t)$ must be band-limited to $\omega_m$, and $\omega_s > 2\omega_m$ (Nyquist). Then the copies do not overlap.
2. Ideal low-pass filter: pass $x_p(t)$ through an LPF with gain $T$ and cutoff $\omega_c$, where $\omega_m < \omega_c < \omega_s - \omega_m$ (usually $\omega_c = \omega_s/2$):

$$H(j\omega) = \begin{cases} T, & |\omega| < \omega_c \\ 0, & \text{otherwise}\end{cases} \quad\Rightarrow\quad X_r(j\omega) = X_p(j\omega)H(j\omega) = X(j\omega)$$

3. Time-domain view (interpolation): the filter's impulse response is $h(t) = \frac{T\sin(\omega_c t)}{\pi t}$. Convolving it with the impulse train gives

$$x_r(t) = \sum_{n=-\infty}^{\infty}x(nT)\,\frac{T\omega_c}{\pi}\,\frac{\sin\big(\omega_c(t - nT)\big)}{\omega_c(t - nT)}$$

With $\omega_c = \omega_s/2$ this becomes $x_r(t) = \sum x(nT)\,\text{sinc}\left(\frac{t - nT}{T}\right)$: each sample is replaced by a sinc pulse, and the sum of these pulses fills in the values between samples exactly.

 x(t) --(x)--> xp(t) --> [ LPF: gain T, cutoff wc ] --> xr(t) = x(t)
         ^
     p(t) = sum delta(t - nT)

Practical note: an ideal LPF is non-causal, so real systems use a zero-order hold (D/A converter) followed by a smoothing filter, and sample a little above the Nyquist rate to leave a guard band.

Practical sampling

Ideal impulses cannot be generated, so in practice sampling is done with pulses of finite width using an electronic switch (natural sampling) or a sample-and-hold (S/H) circuit (flat-top sampling).

Sample-and-hold circuit:

           S (FET switch)          buffer
 x(t) ---[ buffer ]---o/ o----+----[ >  ]----> xs(t)
                      ^       |
       sampling pulse |      === C (hold
       (period T)            |   capacitor)
                            GND

Working:

1. Sample mode: a short sampling pulse closes the switch S (a FET). The capacitor $C$ charges quickly to the input value $x(nT)$ through the low output resistance of the input buffer.
2. Hold mode: the switch opens. The capacitor holds the voltage for the rest of the period $T$ because the output buffer has very high input impedance. The output is a staircase (flat-top samples), which the ADC then converts.
3. Mathematically, flat-top sampling = ideal sampling followed by a hold filter $h(t)$ (a pulse of width $\tau$). Its spectrum is $X_s(j\omega) = \frac{1}{T}H(j\omega)\sum_k X(j(\omega - k\omega_s))$, with $H(j\omega) = \tau\,\text{sinc}$-shaped. This causes the aperture effect (slight high-frequency loss), corrected with an equalizer $1/H(j\omega)$ at reconstruction.

With natural sampling (switch only, no capacitor), the pulse tops follow the signal and the spectrum copies are only weighted by constants, so no aperture distortion occurs.

Band-pass sampling theorem

For a band-pass signal occupying $f_L$ to $f_H$ with bandwidth $B = f_H - f_L$, sampling at $2f_H$ is not necessary. The signal can be recovered from samples taken at a rate as low as $2B$, provided

$$\frac{2f_H}{k} \le f_s \le \frac{2f_L}{k-1}, \qquad k = 1, 2, \dots, \left\lfloor\frac{f_H}{B}\right\rfloor$$

In particular, if $f_H$ is an integer multiple of $B$, then $f_s = 2B$ works. Example: a signal from 20 to 25 kHz ($B = 5$ kHz, $f_H = 5B$) can be sampled at 10 kHz instead of 50 kHz.

The Nyquist criterion says a signal band-limited to $f_m$ must be sampled at $f_s \ge 2f_m$. It prevents aliasing because it keeps the spectral copies created by sampling from overlapping.

How it works: Sampling at rate $f_s$ makes the spectrum periodic:

$$X_s(f) = f_s\sum_{k=-\infty}^{\infty}X(f - kf_s)$$

The original band occupies $-f_m$ to $f_m$. The first copy starts at $f_s - f_m$. There is no overlap when

$$f_s - f_m \ge f_m \;\Rightarrow\; f_s \ge 2f_m$$

 (a) fs > 2fm : copies separate, guard band
     ___         ___         ___
    /   \       /   \       /   \
 --+--+--+-----+--+--+-----+--+--+--> f
  -fm 0 fm  fs-fm fs      2fs
            |<->| guard band
     [ LPF cutoff fs/2 recovers centre copy ]

 (b) fs = 2fm : copies just touch (ideal LPF needed)
     ___ ___ ___
    /   V   V   \
 --+---+---+---+--> f
       0   fs  2fs

 (c) fs < 2fm : copies overlap -> aliasing
     ____ ____
    /    X    \
 --+---+---+----> f
       0   fs

Explanation:

• In (a), the original spectrum stays separate, so an ideal low-pass filter with cutoff $f_s/2$ removes the copies and gives back $x(t)$ exactly.
• In (b), recovery is possible only with an ideal brick-wall filter.
• In (c), the tail of the copy at $f_s$ folds into the band $0$ to $f_s/2$. Those components appear at false frequencies $|f - f_s|$ and cannot be separated by any filter. This is aliasing.

Example: A 3 kHz tone needs $f_s \ge 6$ kHz. At $f_s = 8$ kHz it is recovered correctly; at $f_s = 4$ kHz it appears as $|3 - 4| = 1$ kHz.

In practice, an anti-aliasing low-pass filter (cutoff $f_s/2$) is placed before the sampler to make sure the signal really is band-limited.

Step 1: convert the product to a sum using $2\cos A\cos B = \cos(A+B) + \cos(A-B)$:

$$\begin{aligned} x(t) &= 2\cos(100\pi t)\cos(300\pi t) \\ &= \cos(400\pi t) + \cos(200\pi t) \end{aligned}$$

Step 2: frequencies ($f = \omega/2\pi$):

Highest frequency $f_m = 200$ Hz.

Step 3: Nyquist rate and interval.

$$\begin{aligned} f_N &= 2f_m = 2\times200 = 400\ \text{Hz} \\ T_N &= \frac{1}{f_N} = \frac{1}{400} = 2.5\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 400$ samples/s ($\omega_N = 800\pi$ rad/s); Nyquist interval $= 2.5$ ms.

Note: the highest frequency is 200 Hz, not 150 Hz; multiplying two sinusoids creates the sum frequency.

Sampling theorem

Sampling theorem: A continuous-time signal $x(t)$ that is band-limited to $\omega_M$ (that is, $X(j\omega)=0$ for $|\omega|>\omega_M$) is uniquely determined by its samples $x(nT)$ if the sampling frequency satisfies

$$\omega_s = \frac{2\pi}{T} > 2\omega_M \quad\text{or}\quad f_s > 2f_M$$

The minimum rate $2f_M$ is called the Nyquist rate, and $T_N = 1/(2f_M)$ is the Nyquist interval.

The original signal is recovered by passing the samples through an ideal low-pass filter of cutoff between $\omega_M$ and $\omega_s-\omega_M$ and gain $T$.

Nyquist rate of $x(t) = 1 + \cos(2000\pi t) + \sin(4000\pi t)$

Write each term as $\cos(2\pi f t)$ or $\sin(2\pi f t)$ and read its frequency:

The highest frequency present is $f_M = 2000$ Hz, so

$$\begin{aligned} f_N &= 2 f_M = 2 \times 2000 = 4000\ \text{Hz} \\ \omega_N &= 2\pi f_N = 8000\pi\ \text{rad/s} \end{aligned}$$

Answer: Nyquist rate = 4000 Hz (4000 samples/s, or $8000\pi$ rad/s).

Aliasing

Aliasing is the distortion that occurs when a signal is sampled at a rate lower than its Nyquist rate ($f_s < 2f_M$). The shifted copies of the spectrum in the sampled signal overlap, so a high-frequency component "folds back" and appears as a lower frequency (its alias). After overlap the original spectrum cannot be separated by any filter, so the signal cannot be recovered exactly.

Example: a 700 Hz tone sampled at $f_s = 1000$ Hz gives the same samples as a $1000-700 = 300$ Hz tone.

   overlap (aliasing) when ws < 2wM
     ___       ___       ___
    /   \     /   \     /   \
   /     \   /     \   /     \
--/-------\-/-------\-/-------\--> w
       -ws  \ /  0   \ /  ws
             X        X   <- overlapped region

Nyquist rate of $x(t) = 5\cos(1500\pi t) - 2\sin(2000\pi t) + 3\sin(1850\pi t)$

The maximum frequency is $f_M = 1000$ Hz.

$$f_N = 2 f_M = 2 \times 1000 = 2000\ \text{Hz}$$

Answer: Nyquist rate = 2000 Hz (2000 samples/s, i.e. $\omega_N = 4000\pi$ rad/s).

Nyquist sampling theorem: a signal band-limited to $\omega_M$ is completely determined by its samples $x(nT)$ if $\omega_s = 2\pi/T > 2\omega_M$. The minimum rate $2\omega_M$ is the Nyquist rate.

Frequency-domain analysis of impulse-train sampling

Let $x(t)$ be band-limited to $\omega_M$. Impulse-train sampling multiplies it by a periodic impulse train of period $T$:

$$p(t) = \sum_{n=-\infty}^{\infty} \delta(t - nT), \qquad x_p(t) = x(t)p(t) = \sum_{n=-\infty}^{\infty} x(nT)\,\delta(t-nT)$$

The Fourier transform of the impulse train is also an impulse train, with spacing $\omega_s = 2\pi/T$:

$$P(j\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta(\omega - k\omega_s)$$

Multiplication in time is convolution in frequency:

$$\begin{aligned} X_p(j\omega) &= \frac{1}{2\pi}\, X(j\omega) * P(j\omega) \\ &= \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big) \end{aligned}$$

So the spectrum of the sampled signal is the original spectrum, scaled by $1/T$, repeated at every multiple of $\omega_s$.

X(jw):            /\
                 /  \
           -----/----\------> w
              -wM    wM

Xp(jw), ws > 2wM (no overlap):
    /\        /\        /\
   /  \      /  \      /  \
--/----\----/----\----/----\--> w
    -ws        0        ws
       gap between copies = ws - 2wM

Conclusion from the spectrum

• The copy centred at $k=0$ occupies $-\omega_M$ to $\omega_M$; the next copy starts at $\omega_s - \omega_M$.
• The copies do not overlap only if $\omega_s - \omega_M > \omega_M$, i.e. $\omega_s > 2\omega_M$. This is the Nyquist condition.
• When it holds, an ideal low-pass filter with gain $T$ and cutoff $\omega_c$ such that $\omega_M < \omega_c < \omega_s - \omega_M$ passes only the $k=0$ copy, giving back $X(j\omega)$ exactly. Hence $x(t)$ is fully recovered.
• If $\omega_s < 2\omega_M$, the copies overlap (aliasing) and $x(t)$ cannot be recovered.

Sampling and the required sampling frequency

Sampling is the process of converting a continuous-time signal $x(t)$ into a discrete-time signal $x[n] = x(nT)$ by taking its values at equally spaced instants $t = nT$. $T$ is the sampling interval and $f_s = 1/T$ the sampling frequency.

Requirement: for a signal band-limited to $f_M$, the sampling frequency must be more than twice the highest frequency:

$$f_s > 2 f_M$$

$2f_M$ is the Nyquist rate. If this holds, $x(t)$ can be reconstructed exactly from its samples by an ideal low-pass filter; if not, aliasing occurs.

Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(100\pi t)\cos(300\pi t)$

The product must first be written as a sum, using $\cos A\cos B = \tfrac12[\cos(A-B) + \cos(A+B)]$:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\big[\cos(300\pi t - 100\pi t) + \cos(300\pi t + 100\pi t)\big] \\ &= \frac{1}{4\pi}\big[\cos(200\pi t) + \cos(400\pi t)\big] \end{aligned}$$

$f_M = 200$ Hz, so

$$f_N = 2 f_M = 400\ \text{Hz}$$

Answer: Nyquist rate = 400 Hz (400 samples/s, $\omega_N = 800\pi$ rad/s).

Aliasing effect

Aliasing is the distortion that occurs when a signal is sampled at a rate lower than its Nyquist rate ($f_s < 2f_M$). The shifted copies of the spectrum in the sampled signal overlap, so a high-frequency component "folds back" and appears as a lower frequency (its alias). After overlap the original spectrum cannot be separated by any filter, so the signal cannot be recovered exactly.

Example: a 700 Hz tone sampled at $f_s = 1000$ Hz gives the same samples as a $1000-700 = 300$ Hz tone.

How to overcome aliasing:

1. Sample fast enough: choose $f_s > 2f_M$ (in practice $f_s$ is kept 2.5 to 5 times $f_M$ to leave a guard band).
2. Anti-aliasing filter: pass the signal through an analog low-pass filter with cutoff $f_s/2$ before sampling, so that no component above $f_s/2$ reaches the sampler. Real signals and noise are never strictly band-limited, so this filter is always used in practice.
3. Use a reconstruction filter with a sharp cutoff so that neighbouring spectral copies are rejected.

Nyquist rate of $x(t) = 2\sin 40\pi t + 10\cos 200\pi t + 5\cos 100\pi t$

$f_M = 100$ Hz.

$$f_N = 2 f_M = 2\times 100 = 200\ \text{Hz}$$

Answer: Nyquist rate = 200 Hz (200 samples/s, $\omega_N = 400\pi$ rad/s).

Sampling

Sampling is the conversion of a continuous-time signal $x(t)$ into a discrete-time sequence $x[n] = x(nT)$ by taking its values at regular intervals $T$ (sampling period). The sampling frequency is $f_s = 1/T$. By the sampling theorem, a band-limited signal can be recovered from its samples if $f_s > 2f_M$, where $f_M$ is its highest frequency. Sampling is the first step of analog-to-digital conversion.

Nyquist rate of $x(t) = 1 + \cos(200\pi t) + \sin(4000\pi t)$

$f_M = 2000$ Hz.

$$f_N = 2 f_M = 4000\ \text{Hz}$$

Answer: Nyquist rate = 4000 Hz (4000 samples/s).

The Nyquist (sampling) theorem states that a continuous-time signal band-limited to a highest frequency $f_M$ can be exactly recovered from its samples if it is sampled at a rate

$$f_s > 2 f_M$$

• $2f_M$ is the Nyquist rate; $T_N = 1/(2f_M)$ is the Nyquist interval.
• In the frequency domain, sampling repeats the spectrum $X(j\omega)$ at every multiple of $\omega_s$. With $\omega_s > 2\omega_M$ the copies do not overlap.
• Recovery is done with an ideal low-pass filter (gain $T$, cutoff between $\omega_M$ and $\omega_s - \omega_M$).
• If $f_s < 2f_M$, copies overlap and aliasing occurs; high frequencies appear as low ones and the signal is lost.
• In practice an anti-aliasing low-pass filter is placed before the sampler and $f_s$ is kept above $2f_M$ (e.g. audio with $f_M = 20$ kHz is sampled at 44.1 kHz).

Sampling is the process of converting a continuous-time signal $x(t)$ into a discrete-time signal by measuring its value at equally spaced instants $t = nT$:

$$x[n] = x(nT), \qquad n = 0, \pm1, \pm2, \dots$$

Here $T$ is the sampling period and $f_s = 1/T$ is the sampling rate. Mathematically it is modelled as multiplying $x(t)$ by an impulse train $p(t)=\sum_n \delta(t-nT)$. Sampling is the first step in analog-to-digital conversion and lets continuous signals be processed by digital systems. A band-limited signal can be rebuilt from its samples only if $f_s > 2f_M$ (sampling theorem); otherwise aliasing occurs.

The analog signal is not printed in the question. This answer assumes the standard textbook signal used with $F_s = 300$ Hz (Proakis):

$$x_a(t) = 3\cos 50\pi t + 10\sin 300\pi t - \cos 100\pi t$$

The same method works for any given $x_a(t)$.

Method

Sampling at $t = nT = n/F_s$ turns an analog frequency $F$ into a discrete-time frequency

$$f = \frac{F}{F_s}\ \text{cycles/sample}, \qquad \omega = 2\pi\frac{F}{F_s}\ \text{rad/sample}$$

Frequencies of each term

Discrete-time signal

$$\begin{aligned} x[n] &= x_a\!\left(\frac{n}{300}\right) \\ &= 3\cos\frac{\pi n}{6} + 10\sin(\pi n) - \cos\frac{\pi n}{3} \end{aligned}$$

Since $\sin(\pi n) = 0$ for every integer $n$, the 150 Hz term gives all-zero samples:

$$x[n] = 3\cos\frac{\pi n}{6} - \cos\frac{\pi n}{3}$$

Answer: the discrete-time frequencies are $f = 1/12,\ 1/2,\ 1/6$ cycles/sample ($\omega = \pi/6,\ \pi,\ \pi/3$ rad/sample). The highest frequency is 150 Hz, so the Nyquist rate is 300 Hz; sampling at exactly 300 Hz loses the sine term, which shows why $F_s$ must be strictly greater than $2F_{max}$.