Signal Processing

5% of exam

Sampling and aliasing, Nyquist, analog filters, digital filters and Z-transforms.

3 concepts

Exam areas

A. Sampling

Sampling and the Nyquist Theorem

Why sampling above twice the highest frequency is mandatory, how aliasing folds energy down, and how anti-alias filtering and reconstruction recover the original signal.

Every digital signal processing chain begins with a decision that cannot be undone: how fast to sample. Sample fast enough and the discrete sequence carries every bit of information the original analog signal held; sample too slowly and high-frequency content is permanently disguised as low-frequency content — aliasing — with no algorithm able to recover the truth afterward. The FE-ECE exam tests this as the gateway to the whole Signal Processing topic, and the handbook states the rule compactly in its Instrumentation 'Sampling' note: the sample rate must exceed twice the highest frequency in the signal,

f_s > 2 f_N

. Points are lost in the bookkeeping: confusing the signal's highest frequency with the Nyquist rate, or forgetting the folding that happens when the rule is broken.

Sampling rate and the sampling theorem

Sampling means recording the analog signal

x(t)

only at the instants

t = nT_s

, producing the sequence

x[n] = x(nT_s)

. The sample rate is

f_s = 1/T_s

in samples per second (Hz). The Nyquist–Shannon sampling theorem says a band-limited signal whose highest frequency component is

f_{max}

is completely determined by its samples provided you sample strictly faster than twice that highest frequency. The handbook denotes the highest signal frequency

f_N

and writes the requirement as

f_s > 2 f_N

f_s = \dfrac{1}{T_s}, \qquad f_s > 2 f_{max}

Nyquist rate vs. Nyquist frequency

Two terms sit one factor of two apart and the exam exploits the confusion. The Nyquist rate is the minimum acceptable sampling rate,

2 f_{max}

— a property of the signal you are about to sample. The folding frequency is

f_s/2

— a property of the sampler you have chosen, the highest signal frequency that a given

f_s

can represent without ambiguity. Read each problem for which one it is asking: 'minimum sample rate' means

2 f_{max}

; 'highest frequency that can be captured' means

f_s/2

. A terminology caution: the FE handbook loosely uses 'Nyquist frequency' two different ways — in Instrumentation it labels the highest signal frequency

f_N

that way (

f_s > 2 f_N

), and in 'Sampled Messages' it calls the rate

2W

the Nyquist frequency; this guide uses the standard convention in which the folding frequency is

f_s/2

f_{\text{Nyquist rate}} = 2 f_{max}, \qquad f_{\text{fold}} = \dfrac{f_s}{2}

Aliasing: how the disguise happens

Sampling at rate

f_s

makes the spectrum periodic: every real frequency component appears repeated at

\pm k f_s

for all integers

k

. When a component sits above the folding frequency

f_s/2

, one of those replicas lands inside the baseband

[0, f_s/2]

and is indistinguishable from a genuine low-frequency component. A signal at

f

therefore appears at the alias frequency obtained by subtracting the nearest integer multiple of

f_s

and taking the absolute value — energy 'folds' down across

f_s/2

. This is irreversible: once two frequencies map to the same sample sequence, no later processing can separate them.

f_{alias} = \left| f - k f_s \right|, \quad k = \text{round}\!\left(\dfrac{f}{f_s}\right)

Anti-alias filtering

Because aliasing cannot be undone, it must be prevented before the sampler. An anti-alias filter is an analog low-pass filter placed ahead of the analog-to-digital converter that attenuates everything above the folding frequency so those components are too small to corrupt the band of interest. The handbook's signal-conditioning note makes exactly this point: signal conditioning is 'often required to prevent alias frequencies from being measured.' In practice you set

f_s

comfortably above

2 f_{max}

to leave a transition (guard) band, because real filters cannot drop to zero instantly at the cutoff.

f_{c,\text{AAF}} \le \dfrac{f_s}{2}

Reconstruction and the ideal interpolator

Going back from samples to a continuous signal is reconstruction. The ideal reconstructor passes the sampled sequence through an ideal low-pass filter of cutoff

f_s/2

, which in the time domain interpolates the samples with shifted sinc functions — the Whittaker–Shannon formula. When the sampling theorem was satisfied the result is exactly

x(t)

; when it was violated the reconstructor faithfully rebuilds the aliased (wrong) signal instead. Real converters approximate this with a sample-and-hold (zero-order hold) followed by an analog reconstruction (smoothing) low-pass filter.

x(t) = \sum_{n=-\infty}^{\infty} x[n]\,\operatorname{sinc}\!\left(\dfrac{t - nT_s}{T_s}\right)

Exam strategy

Decide first which quantity is asked. 'Minimum sampling rate' or 'Nyquist rate' is

2 f_{max}

; 'maximum signal frequency for this rate' or 'folding frequency' is

f_s/2

. For an alias question, compute

f_{alias} = |f - k f_s|

with

k

the nearest integer to

f/f_s

— a component just below

f_s

aliases to just above

0

, and one at exactly

f_s/2

sits on the fold. Remember

f_s > 2 f_{max}

is strict: a 1 kHz tone sampled at exactly 2 kHz is the textbook failure (you can land on the zero crossings). When a problem mentions an anti-alias filter, its cutoff is at or below

f_s/2

, not at

f_s

Key equations

Sampling interval and rate

f_s = \dfrac{1}{T_s}

$T_s$ = time between samples (s); $f_s$ = sample rate (samples/s = Hz). The fundamental link for every sampling problem.

Sampling theorem (Nyquist criterion)

f_s > 2 f_{max}

Sample strictly faster than twice the highest frequency $f_{max}$ present in the band-limited signal to avoid aliasing. Handbook 'Sampling' note: $f_s > 2 f_N$ .

Nyquist rate

f_{NR} = 2 f_{max}

Minimum acceptable sample rate (Hz). A property of the signal. Used when a problem asks for the smallest $f_s$ .

Folding (Nyquist) frequency

f_{fold} = \dfrac{f_s}{2}

Highest frequency a chosen $f_s$ can represent without ambiguity (Hz). A property of the sampler. Frequencies above it alias.

Alias frequency

f_{alias} = \left| f - k f_s \right|, \;\; k = \operatorname{round}\!\left(\tfrac{f}{f_s}\right)

Apparent (baseband) frequency that a component $f > f_s/2$ takes after sampling. $k$ is the nearest integer multiple of $f_s$ to $f$ .

First-zone alias (folding)

f_{alias} = f_s - f \quad (f_s/2 < f < f_s)

Quick form when $f$ lies in the first zone above the fold: the apparent frequency is the mirror image about $f_s/2$ .

Anti-alias filter cutoff

f_{c} \le \dfrac{f_s}{2}

Analog low-pass cutoff placed before the ADC; attenuates content above the folding frequency so it cannot alias into the band of interest.

Ideal reconstruction (sinc interpolation)

x(t) = \sum_{n} x[n]\,\operatorname{sinc}\!\left(\dfrac{t-nT_s}{T_s}\right)

Whittaker–Shannon formula; exactly recovers a band-limited $x(t)$ from its samples when the sampling theorem was satisfied.

ADC voltage resolution

\Delta V = \dfrac{V_H - V_L}{2^{n}}

Quantization step of an $n$ -bit converter over range $[V_L, V_H]$ (V). Sampling sets the time axis of the digitized signal; this companion handbook formula sets the amplitude axis.

Worked examples

Alias of a component above the fold

Problem. A signal containing tones at $1.5\,\text{kHz}$ and $3.5\,\text{kHz}$ is sampled at $f_s = 5\,\text{kHz}$ . Which tone is captured correctly, and at what frequency does the other one appear?

Solution. Folding frequency:

f_s/2 = 2.5\,\text{kHz}

. The

1.5\,\text{kHz}

tone is below the fold, so it is captured correctly at

1.5\,\text{kHz}

. The

3.5\,\text{kHz}

tone is above the fold and aliases. Nearest multiple of

f_s

k = \operatorname{round}(3.5/5) = 1

, so

f_{alias} = |3.5 - 1\cdot 5| = 1.5\,\text{kHz}

. The

3.5\,\text{kHz}

tone masquerades as

1.50\,\text{kHz}

— it lands exactly on top of the real

1.5\,\text{kHz}

tone. Sanity check: a tone at

f_s - f = 5 - 3.5 = 1.5\,\text{kHz}

, consistent.

f_{alias} = |3.5 - (1)(5)| = 1.5\ \text{kHz}

Choosing a sample rate with a guard band

Problem. An instrument signal has no energy above $18\,\text{kHz}$ . (a) What is the minimum (Nyquist) sample rate? (b) If you sample at $44\,\text{kHz}$ , what is the folding frequency, and is the signal safe from aliasing?

Solution. (a) Nyquist rate

= 2 f_{max} = 2(18) = 36\,\text{kHz}

. Any

f_s

strictly above this works. (b) Folding frequency

= f_s/2 = 44/2 = 22.0\,\text{kHz}

. Since

f_{max} = 18\,\text{kHz} < 22\,\text{kHz}

, all signal energy is below the fold, so nothing aliases — the

4\,\text{kHz}

guard band (

22 - 18

) gives the anti-alias filter room to roll off. Sanity check:

44\,\text{kHz} > 36\,\text{kHz}

Nyquist rate, so the theorem is satisfied.

Out-of-band interference folding into audio

Problem. An audio ADC samples at $f_s = 44.1\,\text{kHz}$ . A $30\,\text{kHz}$ ultrasonic interferer leaks past a weak anti-alias filter. Where does it appear in the recorded baseband?

Solution. Folding frequency:

44.1/2 = 22.05\,\text{kHz}

; the

30\,\text{kHz}

interferer is above it, so it aliases.

k = \operatorname{round}(30/44.1) = 1

, so

f_{alias} = |30 - 1\cdot 44.1| = 14.1\,\text{kHz}

. The ultrasonic tone folds down to

14.1\,\text{kHz}

— squarely in the audible band, which is exactly why a sharp anti-alias filter is required. Sanity check:

f_s - f = 44.1 - 30 = 14.1\,\text{kHz}

, matches.

Common pitfalls

•Confusing Nyquist rate ( $2f_{max}$ , a property of the signal) with the folding frequency ( $f_s/2$ , a property of the sampler) — they differ by exactly a factor of two and the wrong one is a stock distractor.
•Treating $f_s = 2 f_{max}$ as adequate. The theorem requires a STRICT inequality $f_s > 2f_{max}$ ; sampling a sinusoid at exactly twice its frequency can land on zero crossings and lose it entirely.
•Computing the alias as $f - f_s$ without the absolute value or the nearest-multiple rounding — for $f$ between $f_s/2$ and $f_s$ the alias is $f_s - f$ , a positive baseband frequency.
•Putting the anti-alias filter cutoff at $f_s$ instead of $\le f_s/2$ . It must stop content above the FOLDING frequency, not above the sample rate.
•Believing aliasing can be filtered out after sampling. Once two frequencies share the same samples they are inseparable — anti-aliasing must happen in the analog domain before the ADC.
•Forgetting that real signals are not perfectly band-limited, so practical designs sample well above $2f_{max}$ to leave a guard band for the filter transition region.
•Tripping over the word 'Nyquist frequency': the handbook uses it loosely (for $f_N$ and for $2W$ ), so anchor on the unambiguous quantities — Nyquist rate $=2f_{max}$ and folding frequency $=f_s/2$ — rather than the name.

References

NCEES FE Reference Handbook — Instrumentation, Measurement, and Control: Sampling — States $f_s > 2 f_N$ and the alias-frequency concept; also gives $\Delta V=(V_H-V_L)/2^n$ for ADC resolution.
NCEES FE Reference Handbook — Electrical and Computer Engineering: Sampled Messages — Reconstruction from samples taken at $f_s > 2W$ ; labels $2W$ the Nyquist frequency.
Oppenheim & Schafer, Discrete-Time Signal Processing — Sampling theorem and ideal reconstruction (sinc interpolation).

B. Analog filters

Analog Filters

Read and design first- and second-order RC and RLC filters: transfer functions, cutoff and resonant frequencies, roll-off rate, and the asymptotic Bode magnitude sketch the exam expects.

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C. Digital filters

Digital Filters and the Z-Transform

Move from difference equations to H(z), classify FIR vs IIR, locate poles and zeros, and apply the |z|<1 stability test that the FE-ECE digital-filter questions hinge on.

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