FE Electrical Signal Processing Practice Problems

Q: How many signal processing questions are on the FE Electrical exam?

NCEES specifies 5-8 questions out of 110, about 5 to 7 percent of the exam.

Q: What is the Nyquist rate?

Twice the highest frequency in the signal: the minimum sampling rate that avoids aliasing. Practical systems sample well above it to leave room for a realizable anti-aliasing filter.

Q: How much digital filter theory is tested?

Difference equations, the z-transform of a delay, the transfer function of a short filter, stability from the pole locations and the frequency response at DC and at half the sampling rate. Filter design procedures are not tested.

Q: Where does quantization fit?

An n-bit converter has a step of the full-scale range over two to the n, and a full-scale sinusoid gives a signal-to-quantization-noise ratio of 6.02n plus 1.76 decibels. Both appear in this area and in Communications.

Signal Processing is 5-8 questions on the FE Electrical exam and one of the areas that most distinguishes it from the other FE disciplines. The three parts are sampling, analog filters and digital filters.

Sampling questions are arithmetic once the rule is clear; filter questions are about reading a transfer function or a difference equation and stating what it does.

Exam weight: NCEES lists Signal Processing at 5-8 questions (5-7%) of the 110-question FE Electrical and Computer exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.

What NCEES Tests in Signal Processing

The specification lists sampling (aliasing, Nyquist theorem), analog filters and digital filters (difference equations, Z-transforms).

The specification names sampling (aliasing, Nyquist theorem), analog filters and digital filters (difference equations, z-transforms). Expect the aliased frequency of a component sampled too slowly, the minimum sampling rate for a band-limited signal, the cutoff frequency and roll-off of an RC filter, the response of a second-order filter at resonance, and the transfer function, stability or frequency response of a short difference equation.

The Handbook calls the rate 2W the Nyquist frequency, while many textbooks call it the Nyquist rate and reserve Nyquist frequency for half the sampling rate. Read the question's definition before answering.

5 Free Signal Processing Practice Problems

Each problem below comes from the PECivilClick FE Electrical question bank, with a worked solution that cites its FE Reference Handbook page, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.

Problem 1 — A. Sampling (aliasing, Nyquist theorem)

A continuous-time signal contains no frequency components above 4 kHz. What is the minimum sampling rate that allows it to be reconstructed without aliasing?

Answer: C) 8 kHz

The Handbook states the sampling theorem for a low-pass message whose spectrum is zero above $W$: exact reconstruction is possible from samples taken at $f_s > 2W$. With $W = 4$ kHz,

$$f_s = 2W = 2 \times 4\text{ kHz} = 8\text{ kHz}$$

is the threshold - the Nyquist rate, which the Handbook calls the Nyquist frequency. Strictly the inequality is $f_s > 2W$, so 8 kHz is the limiting value rather than a rate at which everything works perfectly; in practice the rate is set comfortably above it to leave the reconstruction filter a transition band. The reason for the factor of 2 is in the sampled spectrum on the same page: sampling replicates $M(f)$ at every multiple of $f_s$, and the copies stay clear of each other only if $f_s - W > W$.

4 kHz is the highest signal frequency itself. Sampling at the signal frequency would put the spectral copies exactly on top of the original.

2 kHz is half of that, the factor of 2 applied in the wrong direction.

16 kHz would work, but it is twice the minimum asked for.

Problem 2 — B. Analog filters

A first-order low-pass filter has the transfer function $H(s) = \dfrac{1000}{s + 1000}$. What is its corner (cutoff) frequency in hertz?

Answer: A) 159.2 Hz

Compare with the Handbook's first-order low-pass form, whose denominator is $s + \omega_c$ with $\omega_c$ in radians per second. Here $\omega_c = 1000$ rad/s, and hertz are cycles rather than radians:

$$f_c = \dfrac{\omega_c}{2\pi} = \dfrac{1000}{2\pi} = 159.2\text{ Hz}$$

At that frequency $|H(j\omega_c)| = 1000/|j1000 + 1000| = 1/\sqrt{2}$, which is the Handbook's defining property of the corner: the gain is down to $1/\sqrt{2}$ of its dc value, $-3$ dB. The numerator 1000 makes the dc gain exactly 1.

1000 Hz reads the radian frequency off the transfer function as if it were already in hertz. Transfer functions in $s$ are always in radians per second; the $2\pi$ is where most slips on this item happen.

6283 Hz multiplies by $2\pi$ instead of dividing.

500 Hz halves the 1000 for no reason the transfer function supplies.

Problem 3 — C. Digital filters (difference equations, Z-transforms)

What is the z-transform of $x[n] = (0.5)^n u[n]$?

Answer: D) $\dfrac{z}{z - 0.5}$

This is the Handbook's pair $\beta^k \leftrightarrow 1/(1 - \beta z^{-1})$ with $\beta = 0.5$:

$$X(z) = \dfrac{1}{1 - 0.5z^{-1}} = \dfrac{z}{z - 0.5}, \qquad |z| > 0.5$$

The derivation is the same geometric series as for the step, $\sum (0.5/z)^n$, which converges when $|0.5/z| < 1$. The pole sits at $z = 0.5$, inside the unit circle, and that is what a decaying sequence looks like in the $z$-domain: the closer the pole to the origin, the faster the decay. This single pair is the workhorse of the whole topic, because every first-order IIR filter has an impulse response of exactly this shape.

$1/(z-0.5)$ has the pole right and the numerator wrong; it is $z^{-1}$ times the answer, the transform of $(0.5)^{n-1}u[n-1]$.

$z/(z+0.5)$ is the transform of $(-0.5)^n u[n]$, a sequence that alternates in sign.

$0.5z/(z-1)$ is half a unit step, with the 0.5 in the wrong place altogether.

Problem 4 — A. Sampling (aliasing, Nyquist theorem)

The signal $x(t) = \cos(2\pi \cdot 500t) + \cos(2\pi \cdot 1200t)$ is sampled at 3000 Hz. Does aliasing occur?

Answer: D) No, because $f_s > 2 \times 1200$ Hz

Only the highest frequency present matters. The Handbook's condition is $f_s > 2W$ with $W$ the bandwidth of the message, and here $W = 1200$ Hz:

$$2W = 2400\text{ Hz} < 3000\text{ Hz} = f_s$$

so the sampling rate exceeds the Nyquist rate and neither tone aliases. The 500 Hz tone is irrelevant to the test: a lower tone is always safe once the highest one is. What the sum $500 + 1200$ would describe is a product of the two tones, which does not exist here - the signal is a sum, and sampling acts on each component separately.

The two options that compare $f_s$ with $500 + 1200 = 1700$ Hz apply a criterion that has no basis in the sampling theorem. One of them happens to reach the right verdict, since 3000 does exceed 1700, but for the wrong reason: a signal with tones at 500 and 1200 Hz sampled at 2000 Hz would pass that test and still alias badly.

The option claiming $f_s < 2 \times 1200$ has the inequality backwards; 3000 is greater than 2400, not less.

Problem 5 — A. Sampling (aliasing, Nyquist theorem)

A signal is sampled at $f_s = 10$ kHz, and the sampled spectrum shows a component at 3.5 kHz that was not in the original baseband. The original component is known to lie between $f_s/2$ and $f_s$. What was its frequency?

Answer: B) 6.5 kHz

Sampling copies the spectrum to every multiple of $f_s$, so a component at $f$ shows up at $|f - kf_s|$. An alias at 3.5 kHz therefore came from some $f = kf_s \pm 3.5$ kHz: 6.5 or 13.5 kHz for $k = 1$, 16.5 or 23.5 kHz for $k = 2$, and so on. That is the essence of aliasing - the sampled data cannot distinguish between them, which is why the question has to say where the original lay. Between $f_s/2 = 5$ kHz and $f_s = 10$ kHz there is exactly one candidate:

$$f = f_s - 3.5 = 10 - 3.5 = 6.5\text{ kHz}$$

Equivalently, a component above $f_s/2$ is reflected about $f_s/2$: 6.5 kHz sits 1.5 kHz above the fold at 5 kHz and lands 1.5 kHz below it, at 3.5 kHz.

13.5 kHz also aliases to 3.5 kHz - it is the $k = 1$ solution on the other side - but it lies above $f_s$, outside the range the question specifies. Without that restriction the item would have two correct answers.

8.5 kHz adds 3.5 kHz to the fold frequency instead of reflecting across it; 8.5 kHz would alias to $10 - 8.5 = 1.5$ kHz.

3.5 kHz assumes no aliasing at all, which the question rules out by saying the component was absent from the original baseband.

Using the FE Reference Handbook for Signal Processing

Sampled messages and ideal-impulse sampling are on p. 377 of the Handbook, the analog-to-digital conversion formulas with the sampling condition on p. 224, first-order low-pass filters on p. 380, decibels and Bode plots on p. 374, and difference equations, z-transforms and convolution on pp. 370-371. Butterworth and other named filter responses are not in the Handbook, so a question about them gives the magnitude function.

Four Mistakes That Cost Points

Folding the aliased frequency incorrectly. A component above half the sampling rate appears at the distance to the nearest multiple of the sampling rate. A 7 kHz tone sampled at 10 kHz appears at 3 kHz, not at 2 kHz or 7 kHz.
Filtering after sampling to remove aliasing. Once a component has aliased it is indistinguishable from a true low-frequency component. The anti-aliasing filter must precede the sampler.
Reading the wrong side of a difference equation as the poles. The coefficients of the past output terms form the denominator and set the poles; the input terms form the numerator. Swapping them inverts the transfer function.
Judging discrete-time stability with the s-plane rule. A digital filter is stable when its poles lie inside the unit circle, not in a left half-plane. A pole at 1.1 is unstable even though it is real and positive.

Frequently Asked Questions

How many signal processing questions are on the FE Electrical exam?