Huiswerkopgaven 5a - numerical problems
12 important questions on Huiswerkopgaven 5a - numerical problems
Nyquist Rate from the Frequency Domain. Figure 1 shows the one-sided magnitude spectra of two signals, GA(f ) and GB (f ). From the spectra determine:
(a) The Nyquist rate and sampling interval for gA(t).
(b) The Nyquist rate and sampling interval for gB (t).![]()
Let fA = 4 kHz and fB = 6 kHz be the respective bandwidths.
(a) fN = 2fA = 8 kHz; TN = 125 μs.
(b) fN = 2fB = 12 kHz; TN = 83.3 μs.
(c) The Nyquist rate and sampling interval for gA(t) · gB (t).
(d) The Nyquist rate and sampling interval for g_A^2(t).
(c) Multiplication ⇔ convolution of spectra; bandwidth of product = fA + fB = 10 kHz. fN = 20 kHz; TN = 50 μs.
(d) Squaring doubles the bandwidth: 2fA = 8 kHz. fN = 16 kHz; TN = 62.5 μs
Sketching the Sampled Spectrum. A signal has the triangular Fourier spectrum with B = 5 kHz. For each of the sampling frequencies below, sketch |Gs(f )| (the spectrum of the sampled signal) over the range −20 kHz to 20 kHz, and state whether g(t) can be perfectly recovered and what the output of a 5 kHz low-pass filter would be.
(a) fs = 20 kHz (well above Nyquist)![]()
fs = 20 kHz: copies centred at 0, ±20, ±40, . . . kHz – well separated with a 10 kHz gap between adjacent lobes.
Signal can be recovered. The 5 kHz LPF (cut-off fc) passes only the central copy; output = G(f )/fs.
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Sketching the Sampled Spectrum. A signal has the triangular Fourier spectrum
![]()
with B = 5 kHz. For each of the sampling frequencies below, sketch |Gs(f )| (the spec-
trum of the sampled signal) over the range −20 kHz to 20 kHz, and state whether g(t)
can be perfectly recovered and what the output of a 5 kHz low-pass filter would b
(b) fs = 10 kHz (exactly at Nyquist
Adjacent triangles touch at ±5 kHz (zero amplitude) but do not overlap. Signal can just be recovered with an ideal 5 kHz LPF; a real filter would need a guard band.
Sketching the Sampled Spectrum. A signal has the triangular Fourier spectrum
with B = 5 kHz. For each of the sampling frequencies below, sketch |Gs(f )| (the spectrum of the sampled signal) over the range −20 kHz to 20 kHz, and state whether g(t) can be perfectly recovered and what the output of a 5 kHz low-pass filter would be.![]()
(c) fs = 7 kHz (below Nyquist – aliasing occurs
fs = 7 kHz: copies centred at 0, ±7, ±14, . . . kHz. The copy at +7 kHz spans [2, 12] kHz, overlapping the central copy over [2, 5] kHz (symmetrically on the negative side). In the overlap region the two triangles sum to a constant:
Aliasing occurs (shaded regions); the corrupted content cannot be separated from the genuine baseband by any linear filter. The 5 kHz LPF output contains the distorted central lobe – not the original spectrum.
Alias Frequency Calculation. A data-acquisition system samples at fs = 8 kHz. No anti-aliasing filter is used. Determine the frequency at which each of the following input tones will appear in the reconstructed (baseband) output. For each case state whether aliasing has occurred.
(a) f1 = 3 kHz
Alias Frequency Calculation. A data-acquisition system samples at fs = 8 kHz. No anti-aliasing filter is used. Determine the frequency at which each of the following input tones will appear in the reconstructed (baseband) output. For each case state whether aliasing has occurred.
(b) f2 = 4 kHz (exactly Nyquist).
f2 = 4 kHz = fs/2: exactly the Nyquist frequency; strictly does not alias, so it appears at 4 kHz. However, reconstruction requires an ideal brick-wall filter; in practice this boundary frequency is unreliable and is typically avoided by choosing fs slightly above 2fm.
Alias Frequency Calculation. A data-acquisition system samples at fs = 8 kHz. No anti-aliasing filter is used. Determine the frequency at which each of the following input tones will appear in the reconstructed (baseband) output. For each case state whether
aliasing has occurred.
(c) f3 = 6 kHz
Alias Frequency Calculation. A data-acquisition system samples at fs = 8 kHz. No anti-aliasing filter is used. Determine the frequency at which each of the following input tones will appear in the reconstructed (baseband) output. For each case state whether
aliasing has occurred.
(d) f4 = 11 kHz
(d) f4 = 11 kHz: f4/fs = 11/8 ≈ 1.375, nearest integer is 1. fa = |11 − 8| = 3 kHz.
Aliases to 3 kHz. (Aliasing occurs.)
Alias Frequency Calculation. A data-acquisition system samples at fs = 8 kHz. No anti-aliasing filter is used. Determine the frequency at which each of the following input tones will appear in the reconstructed (baseband) output. For each case state whether
aliasing has occurred.
(e) f5 = 20 kHz
(e) f5 = 20 kHz: f5/fs = 20/8 = 2.5, nearest integer is 2 or 3. Using k = 2:
fa = |20 − 16| = 4 kHz; using k = 3: fa = |20 − 24| = 4 kHz. Aliases to 4 kHz.
(Aliasing occurs.)
Anti-Aliasing Filter Design An engineer is digitising a sensor signal with an ADC. The ADC samples at fs = 8 kHz. The sensor output contains the following sinusoidal components:
x(t) = 0.8 cos(2π × 1200 t) + 0.3 cos(2π × 3500 t) + 0.05 cos(2π × 9000 t).
a) What is the Nyquist frequency of the ADC (i.e. the highest frequency that can be represented without aliasing at this sampling rate)
(a) The Nyquist frequency is fs/2 = 4 kHz. This is the highest frequency the system can faithfully represent at the chosen sampling rate – not to be confused with the Nyquist rate that would be required to capture the full signal (2 × 9 kHz = 18 kHz), which this ADC does not meet.
Anti-Aliasing Filter Design An engineer is digitising a sensor signal with an ADC. The ADC samples at fs = 8 kHz. The sensor output contains the following sinusoidal
components:
x(t) = 0.8 cos(2π × 1200 t) + 0.3 cos(2π × 3500 t) + 0.05 cos(2π × 9000 t).
(b) Identify which components will alias if no anti-aliasing filter is used. For each aliased
component, calculate the alias frequency that will appear in the output.
(b) Only components above 4 kHz will alias. The 1200 Hz and 3500 Hz components are both below the Nyquist frequency and pass through unaffected. The 9000 Hz tone exceeds the Nyquist frequency; its alias frequency is:
fa = |f − kfs| = |9000 − 1 × 8000| = 1000 Hz.
Without an anti-aliasing filter, a spurious tone at 1 kHz with amplitude 0.05 will appear in the sampled output
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