Why must an anti-alias filter act on the analogue signal before it reaches the digitiser, and not on the digital samples afterwards?

Because the filter raises the Nyquist frequency of the digitiser, so a wider band of the signal can then be represented without error

Because digital filters have no poles near the Nyquist frequency and therefore cannot attenuate energy at any frequency above it

Because aliasing affects only random noise, so the filter is needed only to improve the signal-to-noise ratio of the final record

A periodic square wave of period T alternates between +A and −A, is odd about t = 0 and has half-wave symmetry, f(t + T/2) = −f(t). Its Fourier series contains:

only sine terms, at the odd harmonics of the fundamental

only cosine terms, at the odd harmonics of the fundamental

only sine terms, at every harmonic of the fundamental

both sine and cosine terms, at the odd harmonics only

A boxcar pulse of height 1 and duration T is replaced by a boxcar of height 1 and duration T/2. Compared with the spectrum of the original, the spectrum of the new pulse has:

a main lobe twice as wide and a zero-frequency value half as large

a main lobe half as wide and a zero-frequency value twice as large

a main lobe twice as wide and the same zero-frequency value

the same main-lobe width and a zero-frequency value half as large

a main lobe twice as wide and a zero-frequency value half as large

A 2 s record sampled at 500 Hz is padded with zeros to 8 s before the FFT is taken, and a rectangular window is used on the original record. Which statement is correct?

The bin spacing becomes 0.125 Hz, but tones 0.25 Hz apart stay unresolved because the data are still only 2 s long

The resolution improves to 0.125 Hz, so two tones 0.25 Hz apart now appear as two clearly separate peaks in the spectrum

The bin spacing becomes 0.125 Hz, but tones 0.25 Hz apart stay unresolved because the data are still only 2 s long

Aliasing is reduced because the Nyquist frequency rises in proportion to the length of the padded record

Spectral leakage disappears because the padded record now begins and ends with zeros, so no abrupt edges remain

Compared with a rectangular window of the same length, a Hann window gives:

a wider main lobe and much lower side lobes: weak peaks near strong ones show up, but close tones separate less well

a narrower main lobe and lower side lobes, so that resolution and leakage both improve together at no cost

the same main-lobe width but higher side lobes, because the taper removes energy from the ends of the record

a wider main lobe and higher side lobes, because the taper adds spectral energy at the ends of the record

a wider main lobe and much lower side lobes: weak peaks near strong ones show up, but close tones separate less well

Which of these causal systems, each driven by an input x[n], is stable, meaning that its output stays bounded for every bounded input?

y[n] = −0.9 y[n − 1] + x[n], whose single pole lies at z = −0.9

y[n] = 1.05 y[n − 1] + x[n], whose single pole lies at z = 1.05

y[n] = −0.9 y[n − 1] + x[n], whose single pole lies at z = −0.9

y[n] = y[n − 1] + x[n], whose single pole lies at z = 1 on the unit circle

y[n] = 0.5 y[n − 1] + 1.2 y[n − 2] + x[n], whose poles are the roots of z² − 0.5z − 1.2 = 0

With the Hilbert transform defined as convolution with 1/(πt), the Hilbert transform of cos(ω₀t), where ω₀ > 0, is:

cos(ω₀t) itself, because the transform has unit gain at every frequency

Which statement correctly contrasts the short-time Fourier transform (one fixed window) with the continuous wavelet transform?

The wavelet windows shorten as frequency rises, so Δf/f stays constant, whereas the fixed window has the same resolution at every frequency

The fixed window can reach arbitrarily fine time and frequency resolution together if its shape is chosen well, which the wavelet transform cannot

The wavelet transform uses a window of one fixed length at every frequency, so its time resolution is the same at all frequencies

The wavelet transform requires a stationary signal, whereas the short-time Fourier transform is meant for non-stationary signals such as seismic traces

Which of the following statements about convolution, correlation and deconvolution are correct?

Convolving a 6-sample trace with a 4-sample filter gives an output of 9 samples

The autocorrelation of a real series is symmetric about zero lag and has its maximum at zero lag

Convolving a 6-sample trace with a 4-sample filter gives an output of 9 samples

The autocorrelation of a real series is symmetric about zero lag and has its maximum at zero lag

In the frequency domain the cross-correlation of x and y is X·Y, exactly the same as their convolution

Deconvolution by plain spectral division is safe wherever the spectrum of the wavelet is very small

Which of the following statements about digital filters, their poles and zeros are correct?

A causal FIR filter has no poles except at z = 0 and is therefore always stable

A zero placed on the unit circle at z = e^{iω₁} makes the gain vanish at frequency ω₁

A pole just inside the unit circle at angle ω₁ produces a sharp peak in the gain near ω₁

A causal FIR filter has no poles except at z = 0 and is therefore always stable

A causal filter is stable whenever all of its zeros lie inside the unit circle

A zero placed on the unit circle at z = e^{iω₁} makes the gain vanish at frequency ω₁

A pole just inside the unit circle at angle ω₁ produces a sharp peak in the gain near ω₁

Which of the following statements about the analytic signal and wavelets are correct?

The analytic signal x(t) + i·H{x}(t) contains no negative-frequency components

The envelope of a signal x(t) is the modulus |x(t) + i·H{x}(t)|

The analytic signal x(t) + i·H{x}(t) contains no negative-frequency components

The envelope of a signal x(t) is the modulus |x(t) + i·H{x}(t)|

The instantaneous frequency is defined as the time derivative of the envelope

The Ricker wavelet is a Gaussian multiplied by a complex sinusoid

Geophysical Signal Processing: Sampling, Fourier and Hilbert Transforms, Correlation and Convolution, Digital Filters and Time–Frequency Analysis

This chapter of Part B2 of the Geology and Geophysics (GG) paper answers the heading “Geophysical Signal Processing”: “Sampling theorem, Nyquist frequency, aliasing, Fourier series, periodic waveform, Fourier and Hilbert transform, Z-transform and wavelet transform; power spectrum, delta function, auto-correlation, cross-correlation, convolution, deconvolution, principles of digital filters, windows, poles, and zeros, Basics of Time-frequency analysis.” It teaches sampling and aliasing, Fourier series and the Fourier transform, the DFT, power spectrum and windows, the delta function with convolution, correlation and deconvolution, the Z-transform with poles, zeros and digital filters, the Hilbert transform, and the wavelet transform with time–frequency analysis. The numericals are the Nyquist and aliased frequency, a four-point DFT, discrete convolution, the gain of a one-pole filter, stability tests and the frequency resolution 1/T.

1. Sampling theorem, Nyquist frequency and aliasing

A seismic trace, a magnetometer record or a well log is continuous in nature but is stored as samples at an interval Δt. The sampling theorem says that a signal containing no frequency above f_max is recovered completely from its samples if the sampling frequency fs = 1/Δt exceeds 2 f_max. The limit fN = fs/2 = 1/(2Δt) is the Nyquist frequency, the highest frequency the samples can represent. A 2 ms seismic record has fs = 500 Hz and fN = 250 Hz; a 4 ms record has fN = 125 Hz.

Energy above fN is not lost: it is aliased (folded) back into the band 0 to fN as a false low frequency. A sinusoid of frequency f sampled at fs appears at f_a = |f − k·fs|, with the integer k chosen so that f_a lies between 0 and fs/2. Worked example: a 130 Hz signal sampled at 100 Hz (Δt = 10 ms, fN = 50 Hz) has k = 1, so f_a = 30 Hz; a 70 Hz tone at this rate also appears at 30 Hz. The samples of the 130 Hz and 30 Hz sinusoids are identical, so nothing in the data tells them apart.

⚠️ Aliasing cannot be undone afterwards

Once sampled, an alias is indistinguishable from a genuine low-frequency signal. The cure is applied before the digitiser: an analogue anti-alias filter that removes energy above fN. Sampling at exactly 2 f_max is only the theoretical limit; real filters roll off gradually, so practical rates are higher. The same law holds in space: receivers at spacing Δx represent wavenumbers only up to 1/(2Δx), and steeply dipping events with shorter apparent wavelengths are spatially aliased.

2. Fourier series, periodic waveforms and the Fourier transform

A periodic waveform of period T (fundamental f₀ = 1/T, ω₀ = 2π/T) is a sum of harmonics: f(t) = a₀/2 + Σ [aₙ cos nω₀t + bₙ sin nω₀t], with aₙ = (2/T)∫f cos nω₀t dt and bₙ = (2/T)∫f sin nω₀t dt over one period; the complex form is f(t) = Σ cₙ e^inω₀t. Symmetry saves work: an even waveform has only cosines, an odd one only sines, and half-wave symmetry, f(t + T/2) = −f(t), leaves only odd harmonics. A square wave of height ±A, odd about t = 0, has bₙ = 4A/(nπ) for odd n and zero otherwise, so its harmonics fall as 1, 1/3, 1/5 … of the fundamental. A periodic signal has a line spectrum at multiples of f₀, and Parseval’s relation gives its mean power as Σ|cₙ|².

The jump of a square wave also shows the Gibbs phenomenon: partial sums overshoot by about 9% of the jump however many terms are kept, and at the jump they converge to the mid-value. For a non-periodic signal let T → ∞: the lines merge into a continuum, F(ω) = ∫ f(t) e^−iωt dt and f(t) = (1/2π) ∫ F(ω) e^iωt dω. For a real f(t), |F| is even in ω and the phase is odd.

Fourier pairs and properties worth knowing
Time domain	Frequency domain
Delta function δ(t)	Constant: every frequency equally present
Constant (DC level)	A spike at zero frequency
cos ω₀t	Two spikes at +ω₀ and −ω₀
Boxcar of duration T	sinc shape, first zero at 1/T
Gaussian pulse	Gaussian, narrow in time means wide in frequency
Delay by t₀	Multiply by e^−iωt₀: same amplitude, linear phase
Derivative d/dt	Multiply by iω: boosts high frequencies
Convolution of two signals	Product of their transforms

Scaling follows: compressing a signal in time by a factor 2 widens its spectrum by 2 and halves its peak. The rms duration and rms bandwidth obey Δt·Δω ≥ 1/2, with equality for the Gaussian. A short sharp pulse therefore needs many frequencies, and sharp edges need high ones.

3. DFT, power spectrum, windows and frequency resolution

For N samples x[n], the discrete Fourier transform is X[k] = Σ x[n] e^−i2πkn/N, n and k from 0 to N − 1. Bin k corresponds to frequency k·fs/N, and for a real series the bins above N/2 are the mirror image of the negative frequencies. The spacing of bins, Δf = fs/N = 1/T, depends only on the record length T = NΔt: to separate two tones that differ by Δf the record must last at least about 1/Δf. The FFT computes the DFT in about N log₂N operations instead of N². Worked example for x = [1, 2, 3, 4], N = 4, with e^−i2π/4 = −i: X[0] = 10, X[1] = 1 − 2i − 3 + 4i = −2 + 2i, X[2] = 1 − 2 + 3 − 4 = −2, X[3] = −2 − 2i; so |X[1]|² = 8.

The power spectrum (periodogram) is |X|²/T: it keeps how much energy sits at each frequency and discards phase. Parseval’s theorem equates total energy in time and frequency, Σ|x|² = (1/N) Σ|X|², and the Wiener–Khinchin theorem says the power spectrum is the Fourier transform of the autocorrelation. A finite record is the signal multiplied by a window, so the spectrum is the true one convolved with the window’s own transform: its main lobe sets the resolution and its side lobes leak a strong peak onto weak neighbours.

Common windows for a record of length T
Window	Main-lobe width, null to null	Highest side lobe (about)
Rectangular (no taper)	2/T	−13 dB
Triangular (Bartlett)	4/T	−26 dB
Hann	4/T	−31 dB
Hamming	4/T	−43 dB
Blackman	6/T	−58 dB

⚠️ Zero-padding interpolates, it does not resolve

Appending zeros to a 2 s record before the FFT gives finer bin spacing, so the plotted spectrum looks smoother, but the true resolution is still about 1/2 s = 0.5 Hz because no new information was added. Tapering trades resolution for leakage: a smoother window has lower side lobes and a wider main lobe, and the rectangular window is the sharpest but leakiest.

4. Delta function, convolution, correlation and deconvolution

The delta function δ(t) is zero except at t = 0, has unit area, and sifts: ∫ f(t) δ(t − t₀) dt = f(t₀). Convolving with δ(t − t₀) only delays a signal by t₀. A system’s output for a unit impulse is its impulse response h. For any input the output is the convolution y[n] = Σ h[k] x[n − k], which has N + M − 1 samples for inputs of N and M samples, and in the frequency domain Y = H·X. The seismic convolutional model is trace = wavelet * reflectivity + noise. Worked example (flip, slide, multiply, add): [1, 2, 3] * [1, −1, 2] = [1, 1, 3, 1, 6]; as a check, the output sums to 12 = 6 × 2, the product of the input sums.

Correlation slides without flipping. The autocorrelation r[τ] = Σ x[n] x[n + τ] is symmetric about zero lag and peaks there at the energy Σx²; a white (random) series has a spike, a resonant one a ringing autocorrelation. The cross-correlation of two records peaks at the lag by which one is delayed relative to the other: for x = [1, 2, 1, 0, 0] and y = [0, 0, 1, 2, 1] the peak is at lag +2. In frequency terms, cross-correlation is X*·Y (conjugate, so phases subtract), and convolution is X·Y. Correlating a Vibroseis record with the sweep compresses it into a zero-phase wavelet.

Deconvolution undoes a convolution. Dividing spectra, R = D/W, blows up where the wavelet spectrum is near zero, so a small stabiliser is added: R ≈ D·W*/(|W|² + ε). In the time domain the Wiener (least-squares) filter makes the wavelet as close as possible to a spike (spiking deconvolution) and needs only the autocorrelation, assuming white reflectivity and a minimum-phase wavelet. Predictive deconvolution with a prediction lag removes periodic reverberations and short-period multiples. The inverse of the two-term wavelet (1, −a) is 1 + a z⁻¹ + a² z⁻² + …, which converges only if |a| < 1.

🧠 Convolution flips, correlation does not

Correlation of x with y equals convolution of the time-reversed x with y. That one fact explains why autocorrelation is symmetric (the phase is lost, only |X|² remains), why cross-correlation keeps the delay, and why deconvolution amplifies noise wherever the signal spectrum is weak.

5. Z-transform, poles and zeros, and digital filters

The Z-transform of a sequence is X(z) = Σ x[n] z⁻ⁿ, so z⁻¹ is a one-sample delay. On the unit circle, z = e^iω with ω = 2πf/fs (Nyquist is ω = π), it reduces to the Fourier transform. A filter with y[n] = Σ bₖ x[n − k] − Σ aₖ y[n − k] has transfer function H(z) = B(z)/A(z) with Y = H·X. The zeros are the roots of B (H = 0) and the poles the roots of A (H → ∞). The gain at frequency ω is the product of the distances from e^iω to the zeros divided by the product of the distances to the poles: a zero on the unit circle kills that frequency, and a pole near the circle makes a sharp peak.

Stability. A causal filter is stable only if every pole lies inside the unit circle, |z| < 1; a pole on the circle gives a non-decaying response (the accumulator y[n] = y[n − 1] + x[n]), and a pole outside gives a growing one. An FIR filter has poles only at z = 0 and is always stable, and a symmetric FIR has linear phase. An IIR filter uses feedback, gives sharp responses cheaply, and can be unstable. Zeros inside the circle (minimum phase) concentrate energy early and give a stable inverse, but they do not decide stability.

Worked examples.
(i) y[n] = x[n] + x[n − 1] has H = 1 + z⁻¹, a zero at z = −1, and |H| = 2|cos(ω/2)|: gain 2 at DC, √2 at ω = π/2 and 0 at Nyquist, so it is a low-pass filter.
(ii) y[n] = 0.5 y[n − 1] + x[n] has H = 1/(1 − 0.5 z⁻¹), one pole at z = 0.5 (stable), and |H| = 1/|1 − 0.5 e^−iω|: gain 2 at DC, 1/√1.25 = 0.894 at ω = π/2 and 0.667 at Nyquist. Running a filter forwards and then backwards gives zero phase and a squared amplitude response.

⚠️ Stability is about poles, not zeros

A pole at z = −0.9 is stable (it lies inside the circle, and the sign only makes the response alternate), whereas a pole at z = 1.05 is not. Zeros anywhere, even outside the circle, never make a causal filter unstable; they only change the phase behaviour and whether the inverse filter is stable.

6. Hilbert transform and the analytic signal

The Hilbert transform is H{x}(t) = (1/π) P∫ x(τ)/(t − τ) dτ, the convolution of x with 1/(πt). In the frequency domain it multiplies the spectrum by −i·sgn(ω): every positive frequency is shifted by −90° and every negative one by +90°, with no change of amplitude. So H{cos ω₀t} = sin ω₀t, H{sin ω₀t} = −cos ω₀t, and applying it twice gives −x. For a causal signal the real and imaginary parts of its spectrum are a Hilbert pair (a Kramers–Kronig relation).

The analytic signal z(t) = x(t) + i·H{x}(t) has no negative-frequency content. Its modulus is the envelope A(t) = √(x² + H²) (reflection strength in seismics), its angle the instantaneous phase φ(t) = tan⁻¹(H/x), and (1/2π) dφ/dt the instantaneous frequency. Worked example: x = 3 cos ω₀t + 4 sin ω₀t has H{x} = 3 sin ω₀t − 4 cos ω₀t, so A² = 9 + 16 = 25 and A = 5 at all times: the signal is 5 cos(ω₀t − φ₀) with tan φ₀ = 4/3.

7. Wavelet transform and basics of time–frequency analysis

The Fourier transform says which frequencies are present, not when, and it assumes the signal is stationary. A seismic trace is not: attenuation strips high frequencies from late arrivals. The short-time Fourier transform (STFT) slides a window w(t − τ) along the trace and transforms each piece; the squared modulus is the spectrogram. One window length fixes both resolutions, time about T_w and frequency about 1/T_w, and the Gabor–Heisenberg limit Δt·Δf ≥ 1/(4π) forbids both being small. A chirp, such as a Vibroseis sweep, shows a ridge in the spectrogram although its ordinary spectrum is essentially flat within its band.

The continuous wavelet transform is W(a, b) = (1/√a) ∫ x(t) ψ*((t − b)/a) dt, where the mother wavelet ψ has zero mean and the scale a stretches it (low frequency, f ≈ f_c/a) or compresses it (high frequency). The windows are short at high frequency and long at low frequency, so the relative bandwidth Δf/f stays constant: good time resolution for sharp transients and good frequency resolution for slow trends. Common choices are the Morlet (a Gaussian times a complex sinusoid), the Ricker or Mexican hat (second derivative of a Gaussian, real and zero-phase, the standard model seismic wavelet with peak frequency f_p) and the Haar (the simplest orthogonal wavelet).

The discrete wavelet transform uses dyadic scales 2ʲ and a filter bank that splits the signal into a low-pass (approximation) and a high-pass (detail) half, each down-sampled by 2, so N samples give N coefficients. Thresholding small detail coefficients de-noises and compresses data. In exploration, wavelet and spectral-decomposition methods map thin-bed tuning frequencies and frequency-dependent attenuation.

🎯 Why wavelets suit seismic data

Reflections at depth are low-frequency and long, while shallow ones are sharp. A fixed STFT window is a compromise for both, whereas the constant-Q tiling of a wavelet analyses each in a window matched to its own frequency.

Key takeaways

Sampling at Δt gives Nyquist frequency fN = 1/(2Δt); a frequency f above it aliases to |f − k·fs| in the band 0 to fN and cannot be recognised afterwards, so an anti-alias filter must precede digitising.
A periodic waveform is a sum of harmonics with a line spectrum (odd square wave: bₙ = 4A/nπ for odd n; Gibbs overshoot about 9%); the Fourier transform extends this to non-periodic signals, and a signal narrow in time is wide in frequency.
DFT bin spacing and resolution are 1/T; zero-padding only interpolates. Windows trade main-lobe width against side-lobe level (rectangular 2/T and −13 dB, Hann 4/T and −31 dB, Blackman 6/T and −58 dB), and the power spectrum is the transform of the autocorrelation.
Convolution (N + M − 1 samples, Y = H·X) builds the trace from wavelet and reflectivity; correlation does not flip, so autocorrelation is symmetric and cross-correlation peaks at the delay; deconvolution needs a stabiliser because it amplifies noise where the spectrum is weak.
In H(z) = B/A the zeros are roots of B and the poles roots of A; the gain is the product of distances to zeros over the product of distances to poles. A causal filter is stable only when all poles lie inside the unit circle, and FIR filters are always stable.
The Hilbert transform shifts phase by 90° and builds the analytic signal, whose modulus is the envelope; the STFT has one fixed resolution tile, while the wavelet transform has constant-Q tiles that are short at high frequency and long at low frequency.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A seismic recorder samples a signal every 4 ms. What is the Nyquist frequency of the recorded data?
1. 62.5 Hz
2. 125 Hz
3. 250 Hz
4. 500 Hz
Show answer
Answer: B — 125 Hz
The sampling frequency is fs = 1/Δt = 1/0.004 s = 250 Hz, and the Nyquist frequency is half of it, fs/2 = 125 Hz. The value 250 Hz is the tempting answer because it is the sampling frequency itself, not the highest frequency that the samples can represent.
Why must an anti-alias filter act on the analogue signal before it reaches the digitiser, and not on the digital samples afterwards?
1. Because the filter raises the Nyquist frequency of the digitiser, so a wider band of the signal can then be represented without error
2. Because digital filters have no poles near the Nyquist frequency and therefore cannot attenuate energy at any frequency above it
3. Because once sampled, an aliased component is indistinguishable from a genuine low-frequency component and can no longer be separated
4. Because aliasing affects only random noise, so the filter is needed only to improve the signal-to-noise ratio of the final record
Show answer
Answer: C — Because once sampled, an aliased component is indistinguishable from a genuine low-frequency component and can no longer be separated
Sampling folds all energy above fN into the band 0 to fN, where it sits exactly on top of true low-frequency signal; the samples of an aliased 130 Hz tone and of a real 30 Hz tone (at fs = 100 Hz) are identical, so no later digital processing can tell them apart. The filter does not change the Nyquist frequency, which is fixed by Δt, and aliasing affects signal as much as noise.
A periodic square wave of period T alternates between +A and −A, is odd about t = 0 and has half-wave symmetry, f(t + T/2) = −f(t). Its Fourier series contains:
1. only sine terms, at the odd harmonics of the fundamental
2. only cosine terms, at the odd harmonics of the fundamental
3. only sine terms, at every harmonic of the fundamental
4. both sine and cosine terms, at the odd harmonics only
Show answer
Answer: A — only sine terms, at the odd harmonics of the fundamental
Oddness about t = 0 removes every cosine term, and half-wave symmetry removes every even harmonic, leaving bₙ = 4A/(nπ) for odd n only. Sine terms at all harmonics would describe an odd wave without half-wave symmetry, such as a sawtooth.
A boxcar pulse of height 1 and duration T is replaced by a boxcar of height 1 and duration T/2. Compared with the spectrum of the original, the spectrum of the new pulse has:
1. a main lobe half as wide and a zero-frequency value twice as large
2. a main lobe twice as wide and the same zero-frequency value
3. the same main-lobe width and a zero-frequency value half as large
4. a main lobe twice as wide and a zero-frequency value half as large
Show answer
Answer: D — a main lobe twice as wide and a zero-frequency value half as large
The first zero of the spectrum sits at 1/duration, so halving the duration moves it from 1/T to 2/T, doubling the width. The value at zero frequency equals the area under the pulse, which halves from T to T/2. The reverse pairing (narrower and taller) is the tempting error, but it is what a longer pulse gives.
A 2 s record sampled at 500 Hz is padded with zeros to 8 s before the FFT is taken, and a rectangular window is used on the original record. Which statement is correct?
1. The resolution improves to 0.125 Hz, so two tones 0.25 Hz apart now appear as two clearly separate peaks in the spectrum
2. The bin spacing becomes 0.125 Hz, but tones 0.25 Hz apart stay unresolved because the data are still only 2 s long
3. Aliasing is reduced because the Nyquist frequency rises in proportion to the length of the padded record
4. Spectral leakage disappears because the padded record now begins and ends with zeros, so no abrupt edges remain
Show answer
Answer: B — The bin spacing becomes 0.125 Hz, but tones 0.25 Hz apart stay unresolved because the data are still only 2 s long
Padding to 8 s makes the bin spacing 1/8 s = 0.125 Hz, which only interpolates the spectrum of the 2 s record. The true resolution stays about 1/2 s = 0.5 Hz, so two tones 0.25 Hz apart lie inside one main lobe and are not separated. The Nyquist frequency depends only on the sampling interval (250 Hz here), and the abrupt cut-off of the original record still causes leakage whatever follows it.
Compared with a rectangular window of the same length, a Hann window gives:
1. a narrower main lobe and lower side lobes, so that resolution and leakage both improve together at no cost
2. the same main-lobe width but higher side lobes, because the taper removes energy from the ends of the record
3. a wider main lobe and higher side lobes, because the taper adds spectral energy at the ends of the record
4. a wider main lobe and much lower side lobes: weak peaks near strong ones show up, but close tones separate less well
Show answer
Answer: D — a wider main lobe and much lower side lobes: weak peaks near strong ones show up, but close tones separate less well
The Hann taper has a null-to-null main lobe of 4/T against 2/T for the rectangular window, and its highest side lobe is about −31 dB against −13 dB. Smoother tapering always trades resolution for lower leakage; no window improves both, which rules out the claim that both improve.
Which of these causal systems, each driven by an input x[n], is stable, meaning that its output stays bounded for every bounded input?
1. y[n] = 1.05 y[n − 1] + x[n], whose single pole lies at z = 1.05
2. y[n] = −0.9 y[n − 1] + x[n], whose single pole lies at z = −0.9
3. y[n] = y[n − 1] + x[n], whose single pole lies at z = 1 on the unit circle
4. y[n] = 0.5 y[n − 1] + 1.2 y[n − 2] + x[n], whose poles are the roots of z² − 0.5z − 1.2 = 0
Show answer
Answer: B — y[n] = −0.9 y[n − 1] + x[n], whose single pole lies at z = −0.9
Stability of a causal filter needs every pole strictly inside |z| < 1. The pole at −0.9 is inside, and its negative sign only makes the impulse response alternate in sign while decaying. The pole at 1.05 is outside (growing response), the pole at 1 is on the circle (the accumulator never decays, so a constant input grows without bound), and z² − 0.5z − 1.2 has roots near 1.37 and −0.87, one of which is outside the circle.
With the Hilbert transform defined as convolution with 1/(πt), the Hilbert transform of cos(ω₀t), where ω₀ > 0, is:
1. −sin(ω₀t)
2. −cos(ω₀t)
3. sin(ω₀t)
4. cos(ω₀t) itself, because the transform has unit gain at every frequency
Show answer
Answer: C — sin(ω₀t)
The transform multiplies positive frequencies by −i, a phase shift of −90°, and a cosine delayed by 90° is a sine: H{cos ω₀t} = sin ω₀t. The result −sin(ω₀t) corresponds to the opposite sign convention (+90° for positive frequencies), and −cos(ω₀t) is what two applications give. Unit gain means the amplitude is unchanged, not that the waveform is.
Which statement correctly contrasts the short-time Fourier transform (one fixed window) with the continuous wavelet transform?
1. The wavelet windows shorten as frequency rises, so Δf/f stays constant, whereas the fixed window has the same resolution at every frequency
2. The fixed window can reach arbitrarily fine time and frequency resolution together if its shape is chosen well, which the wavelet transform cannot
3. The wavelet transform uses a window of one fixed length at every frequency, so its time resolution is the same at all frequencies
4. The wavelet transform requires a stationary signal, whereas the short-time Fourier transform is meant for non-stationary signals such as seismic traces
Show answer
Answer: A — The wavelet windows shorten as frequency rises, so Δf/f stays constant, whereas the fixed window has the same resolution at every frequency
A wavelet is stretched for low frequencies and compressed for high ones, giving short windows (good timing) at high frequency and long windows (good frequency discrimination) at low frequency, a constant-Q tiling. The fixed STFT window has one tile shape everywhere, and no window can beat the Gabor–Heisenberg limit Δt·Δf ≥ 1/(4π). Both transforms were introduced for non-stationary signals, so the claim about stationarity is reversed.
Which of the following statements about convolution, correlation and deconvolution are correct?
1. Convolving a 6-sample trace with a 4-sample filter gives an output of 9 samples
2. The autocorrelation of a real series is symmetric about zero lag and has its maximum at zero lag
3. In the frequency domain the cross-correlation of x and y is X·Y, exactly the same as their convolution
4. Deconvolution by plain spectral division is safe wherever the spectrum of the wavelet is very small
Show answer
Answer: A — Convolving a 6-sample trace with a 4-sample filter gives an output of 9 samples; B — The autocorrelation of a real series is symmetric about zero lag and has its maximum at zero lag
The length of a convolution is N + M − 1 = 6 + 4 − 1 = 9, and an autocorrelation is symmetric with its peak at the energy value at zero lag. Cross-correlation is X*·Y with a conjugate, so phases subtract and the delay is preserved, which differs from convolution. Dividing by a very small wavelet spectrum multiplies noise enormously, which is why a stabiliser is added.
Which of the following statements about digital filters, their poles and zeros are correct?
1. A causal FIR filter has no poles except at z = 0 and is therefore always stable
2. A causal filter is stable whenever all of its zeros lie inside the unit circle
3. A zero placed on the unit circle at z = e^iω₁ makes the gain vanish at frequency ω₁
4. A pole just inside the unit circle at angle ω₁ produces a sharp peak in the gain near ω₁
Show answer
Answer: A — A causal FIR filter has no poles except at z = 0 and is therefore always stable; C — A zero placed on the unit circle at z = e^{iω₁} makes the gain vanish at frequency ω₁; D — A pole just inside the unit circle at angle ω₁ produces a sharp peak in the gain near ω₁
An FIR filter’s only poles sit at the origin, so it cannot be unstable; a zero on the circle gives zero gain at its angle; and a pole close to the circle makes the distance to it small there, so the gain (zero distances over pole distances) peaks. Stability is decided by poles alone, so zeros inside the circle do not guarantee it.
Which of the following statements about the analytic signal and wavelets are correct?
1. The analytic signal x(t) + i·H{x}(t) contains no negative-frequency components
2. The envelope of a signal x(t) is the modulus |x(t) + i·H{x}(t)|
3. The instantaneous frequency is defined as the time derivative of the envelope
4. The Ricker wavelet is a Gaussian multiplied by a complex sinusoid
Show answer
Answer: A — The analytic signal x(t) + i·H{x}(t) contains no negative-frequency components; B — The envelope of a signal x(t) is the modulus |x(t) + i·H{x}(t)|
Adding i·H{x} cancels the negative half of the spectrum, and its modulus is the envelope. Instantaneous frequency is the derivative of the instantaneous phase divided by 2π, not of the envelope. A Gaussian times a complex sinusoid is the Morlet wavelet; the Ricker wavelet is the real second derivative of a Gaussian.
A 130 Hz sinusoid is sampled at 100 samples per second with no anti-alias filter. The frequency, in Hz, at which it appears in the sampled record is ____.
Numerical answer — type the value.
Show answer
Answer: 30
The Nyquist frequency is 100/2 = 50 Hz, so 130 Hz is aliased. It appears at |f − k·fs| with k chosen to land between 0 and 50 Hz: for k = 1, |130 − 100| = 30 Hz, which lies in the band, whereas k = 0 gives 130 Hz and k = 2 gives 70 Hz, both outside it. The answer is 30 Hz.
A four-point sequence is x = [2, 0, 1, 3], and its DFT is X[k] = Σ x[n] e^−i2πkn/4, with n and k running from 0 to 3. The value of |X[1]|² is ____.
Numerical answer — type the value.
Show answer
Answer: 10
For k = 1 the factors e^−iπn/2 are 1, −i, −1 and +i for n = 0, 1, 2, 3. So X[1] = 2·1 + 0·(−i) + 1·(−1) + 3·(i) = 1 + 3i. Then |X[1]|² = 1² + 3² = 10. Using the opposite sign convention, e^+iπn/2, gives the complex conjugate 1 − 3i with the same modulus, so |X[1]|² does not depend on that convention.
The sequence a = [2, −1, 1] is convolved with b = [1, 3] to give the full output y (all N + M − 1 samples). The sum of the squares of all the samples of y is ____.
Numerical answer — type the value.
Show answer
Answer: 42
The output has 3 + 2 − 1 = 4 samples: y₀ = 2·1 = 2; y₁ = 2·3 + (−1)·1 = 5; y₂ = (−1)·3 + 1·1 = −2; y₃ = 1·3 = 3. As a check, the samples sum to 8 = (2 − 1 + 1)(1 + 3). The sum of squares is 4 + 25 + 4 + 9 = 42.
A digital filter obeys y[n] = 0.8 y[n − 1] + x[n]. Its gain |H| at a frequency equal to one quarter of the sampling frequency (ω = 2πf/fs = π/2) is ____ to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 0.78
Taking Z-transforms gives H(z) = 1/(1 − 0.8 z⁻¹), with a single pole at z = 0.8 (stable). On the unit circle at ω = π/2, z⁻¹ = e^−iπ/2 = −i, so H = 1/(1 + 0.8i) and |H| = 1/√(1 + 0.64) = 1/√1.64 = 0.781, which is 0.78 to two decimals. The DC gain 1/(1 − 0.8) = 5 is the tempting wrong frequency to evaluate.
Two stationary sinusoids of frequencies 30.0 Hz and 30.4 Hz must be resolved as separate peaks in a DFT of a single record taken with a rectangular window. Using the rule that the record length must be at least 1/Δf, where Δf is the frequency separation, the minimum record length is ____ seconds, to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 2.5
The separation is Δf = 30.4 − 30.0 = 0.4 Hz, and the DFT resolution is 1/T, so T ≥ 1/0.4 = 2.5 s. The sampling rate does not enter, since adding samples within the same duration (or zeros) gives no extra resolution.