FE Electrical Signal Processing Practice Problems

Answer: C) \(-20\) dB

5000 Hz is exactly one decade above the 500 Hz corner. The Handbook's Bode table gives the single-pole term \((1 + j\omega/\omega_c)^{-1}\) a high-frequency magnitude of \(-20\log_{10}(\omega/\omega_c)\), so one decade out it is at \(-20\) dB - the slope p. 379 draws as \(-20\) dB/decade. The exact value from the first-order response is

$$|H| = \dfrac{1}{\sqrt{1 + (f/f_c)^2}} = \dfrac{1}{\sqrt{101}}, \qquad 20\log_{10}\dfrac{1}{\sqrt{101}} = -20.04\text{ dB}$$

The asymptote is accurate to a few hundredths of a decibel a decade out, which is why Bode's straight-line approximation is good enough for almost every hand calculation.

\(-40\) dB is a decade out on a second-order filter, two poles instead of one, and \(-60\) dB is the third-order figure. The slope per decade is \(-20\) dB for every pole the filter has, and a first-order filter has one.

\(-10\) dB applies the power formula, \(10\log_{10}\), to the voltage ratio of \(1/10\). Voltage ratios take \(20\log_{10}\); the \(-20\) dB and the power ratio of \(1/100\) under \(10\log_{10}\) agree.

Answer: C) 100 Hz

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Sampled Messages and Ideal-Impulse Sampling, p. 381 (PDF viewer: page 387). Sampling reproduces the message spectrum around every multiple of the sampling rate:

$$X_\delta(f) = f_s \sum_{k=-\infty}^{\infty} M(f - k f_s)$$

Step 1 – Confirm that aliasing occurs. The Handbook's criterion is \(f_s > 2W\), where \(W\) is the highest frequency present. Here \(2W = 2(900) = 1800\) Hz while \(f_s = 1000\) Hz, so the condition fails and the 900 Hz tone cannot be recovered.

Terminology, so the Handbook and the arithmetic agree: p. 381 calls \(2W = 1800\) Hz the Nyquist frequency — the minimum acceptable sampling rate. Half the sampling rate, \(f_s/2 = 500\) Hz, is the folding frequency; everything the sampled data can represent lies between 0 and 500 Hz, which is the shaded band in the figure.

Step 2 – Find the replica that lands in that band. Apply the summation with \(k = 1\):

$$f - k f_s = 900 - 1000 = -100 \text{ Hz}$$

A negative frequency is observed at its magnitude, so the sampled data contains a component at \(|-100| = 100\) Hz. Every other replica (\(900\), \(1900\), \(-1100\), …) lies outside 0 to 500 Hz.

Step 3 – Answer: The 900 Hz tone is recorded as a 100 Hz tone — the correct answer is (C).

Step 4 – Eliminate the distractors:

(A) 500 Hz is \(f_s/2\), the edge of the observable band. It is the limit the alias must fall below, not the alias itself.

(B) 900 Hz is what the samples would show only if \(f_s > 1800\) Hz had been used, satisfying the Handbook criterion.

(D) 1900 Hz is the upper replica, \(f + f_s\). It does appear in \(X_\delta(f)\), but it sits far outside the observable band, so no 1900 Hz tone can be read from the samples.

Answer: B) 2 Hz

The Handbook's Resonance entry relates the three quantities as \(BW = \omega_0/Q\), and its Band-Reject Filters entry defines that bandwidth as the 3-dB span \(\omega_U - \omega_L\). In hertz:

$$BW = \dfrac{f_0}{Q} = \dfrac{60}{30} = 2\text{ Hz}$$

The notch is only 2 Hz wide at its 3-dB points, from about 59 Hz to 61 Hz, which is the point of a high \(Q\): the interference is removed while signal content a few hertz away passes almost untouched. The trade-off is that the notch must be tuned accurately, since a line frequency 1 Hz off nominal already sits at the edge of the rejection band.

0.5 Hz is \(Q/f_0\), the ratio inverted.

30 Hz is the quality factor copied across as a bandwidth. \(Q\) is dimensionless; the bandwidth it implies depends on the centre frequency.

60 Hz is the centre frequency itself, which would correspond to \(Q = 1\).

Answer: A) approximately 0.743

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Digital Signal Processing, p. 376 (PDF viewer: page 382).

Step 1 – Move onto the unit circle. The frequency response of a discrete-time system is \(H(z)\) evaluated at \(z = e^{j\omega}\). As the figure shows, \(\omega = \pi/2\) puts the evaluation point at the top of the unit circle:

$$z = e^{j\pi/2} = j$$

Step 2 – Substitute.

$$H(e^{j\pi/2}) = \frac{j}{j - 0.9}$$

Step 3 – Take magnitudes. The magnitude of a ratio is the ratio of the magnitudes:

$$|H| = \frac{|j|}{|j - 0.9|} = \frac{1}{\sqrt{(-0.9)^2 + 1^2}} = \frac{1}{\sqrt{1.81}} = \frac{1}{1.345} = 0.743$$

Geometrically these are distances in the figure: the numerator is the distance from the zero at \(z = 0\) to the point \(j\), which is 1, and the denominator is the distance from the pole at \(z = 0.9\) to the same point.

Step 4 – Answer: (A) approximately 0.743.

Step 5 – Eliminate the distractors. Three of the four options are genuine quantities of this filter, so the one you land on identifies the slip:

(B) 10.0 is \(|H|\) at \(\omega = 0\). At DC the evaluation point is \(z = 1\), only \(0.1\) away from the pole, giving \(1/0.1 = 10\). This is a single-pole low-pass response, and the whole point of the question is that \(\pi/2\) is far from DC.

(C) 0.526 is \(|H|\) at \(\omega = \pi\), the other end of the band, where \(z = -1\) and the pole distance is \(1.9\). It is also what you get at \(\pi/2\) by adding magnitudes as if they were collinear, \(1/(1 + 0.9)\), instead of combining the real and imaginary parts with Pythagoras.

(D) 1.345 is \(\sqrt{1.81}\), the denominator distance itself — the fraction was inverted. A stable single-pole filter evaluated away from its pole cannot have gain above its DC value, so an answer larger than 1 here should prompt a second look.

Answer: B) \(3\cos(2\pi \cdot 300t)\)

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Sampled Messages and Ideal-Impulse Sampling, p. 381 (PDF viewer: page 387). The Handbook notes that the message is recovered with an ideal low-pass filter, which passes everything below \(f_s/2\) and nothing above.

Step 1 – Folding frequency. \(f_s = 800\) Hz, so the reconstruction filter passes 0 to \(f_s/2 = 400\) Hz.

Step 2 – The 300 Hz component. \(300 < 400\) Hz, so it survives at 300 Hz with amplitude 2.

Step 3 – The 1100 Hz component. \(1100 > 400\) Hz, so it aliases. Testing values of \(k\) until the result falls inside 0 to 400 Hz:

\(k = 1\): \(|1100 - 800| = 300\) Hz — inside the band.

\(k = 2\): \(|1100 - 1600| = 500\) Hz — outside.

So it returns at 300 Hz, with its amplitude of 1 intact.

Step 4 – The two land on top of each other. The alias falls exactly on the frequency of the genuine component, and it arrives in phase: sampling \(\cos(2\pi \cdot 1100 nT)\) with \(T = 1/800\) gives \(\cos(2\pi \cdot 1.375 n) = \cos(2\pi \cdot 0.375 n)\), which is the same sequence as \(\cos(2\pi \cdot 300 nT)\) — same sign, no inversion. The two therefore add:

$$2\cos(2\pi \cdot 300t) + 1\cos(2\pi \cdot 300t) = 3\cos(2\pi \cdot 300t)$$

Step 5 – Answer: (B) \(3\cos(2\pi \cdot 300t)\).

Why this case matters. The output is a single 300 Hz tone of amplitude 3. Nothing in the sampled data distinguishes the 2 V of real signal from the 1 V of aliased interference — they are the same sequence of numbers. No amount of later filtering can separate them, which is precisely why the anti-aliasing filter has to come before the sampler.

Step 6 – Eliminate the distractors:

(A) assumes the 1100 Hz component is simply lost. Sampling relocates energy in frequency; it does not remove it. Without an anti-aliasing filter that energy is still in the output, now disguised as 300 Hz content.

(C) is the original signal, which would require \(f_s > 2200\) Hz.

(D) uses \(k = 2\), giving 500 Hz. That replica exists in \(X_\delta(f)\), but 500 Hz is above \(f_s/2 = 400\) Hz, so the reconstruction filter removes it. Only the replica that lands inside the passband appears at the output.