Question bank — scipy.signal — filtering, convolution, FFT-based analysis

False. Correlation is convolution without flipping one kernel; the two agree only when the kernel is symmetric ($h[n]=h[-n]$), because then flipping changes nothing.

False. By the convolution theorem, time convolution equals frequency multiplication exactly; fftconvolve returns the same result to floating-point rounding, just in $O(N\log N)$ instead of $O(N^2)$.

False. The highest honest frequency is Nyquist $f_s/2$, fixed by the sampling rate. More samples only shrink the bin spacing $\Delta f=f_s/N$ — finer resolution, not higher reach. See Aliasing and Sampling Theorem.

False. lfilter is causal and delays the signal; filtfilt runs the filter forward then backward so the two delays cancel, giving zero phase and roughly the squared magnitude response (filter applied twice).

False. IIR filters use feedback from past outputs and can be numerically unstable, especially at high order or extreme cutoffs; FIR filters (firwin) are always stable because they only sum past inputs.

True in principle, but you must use enough FFT length (zero-padding to $\ge M+N-1$) or the wrap-around gives circular convolution, not the linear one you wanted.

True. $\Delta f=f_s/N$; keeping $f_s$ fixed and doubling $N$ (a longer recording) halves $\Delta f$, so two close peaks become resolvable.

False. It has length $M+N-1$: the first sample is one-term overlap and the last is one-term overlap, so you gain $M+N-1$ distinct slide positions, not $M+N$.

Error: without fs=, the cutoff is a fraction of Nyquist, so it must lie in $(0,1)$. Passing 80 is illegal. Fix: butter(4, 80, fs=1000) or use $W_n=80/(1000/2)=0.16$.

Error: the raw magnitude is unscaled. For true one-sided amplitudes divide by $N$ and multiply the interior bins by 2 (DC and Nyquist are not doubled, since they have no mirror partner).

Error: the shift is the causal filter's own phase delay, not a real change in the data. For offline analysis use filtfilt (zero phase) so features stay put.

Error: raw correlation grows with signal energy, so a loud noisy region can outscore the true match. Normalise (e.g. divide by local norms) for a fair template score — see Cross-correlation and Template Matching.

Error: 'same' returns output exactly the length of x, centered. If you saw something shorter you likely used 'valid', which returns only fully-overlapping positions.

Error: $600>f_s/2=500$. There is no representable frequency there, so scipy raises an error; a low-pass cutoff must satisfy $0<f_{cut}<f_s/2$.

Error: a real signal's spectrum is conjugate-symmetric, so the upper half mirrors the lower. Use rfft/rfftfreq to get the meaningful positive half only.

Because any input is a sum of scaled, shifted spikes; linearity turns "response to a sum" into "sum of responses", and time-invariance makes each shifted spike give the same $h$ just shifted. Adding those up is exactly the convolution sum. See Linear Time-Invariant Systems.

The sum is an inner product between the signal and that pure wave; it is large exactly when the signal resembles (aligns with) the wave, so $|X[k]|$ measures overlap with frequency $f_k$. See Fourier Transform.

Running the filter twice (forward + reverse) cancels the phase (delays subtract) but multiplies the magnitude response by itself — so the effective attenuation is roughly squared. Design cutoffs with this doubling in mind.

FFT is $O(N\log N)$ vs direct $O(N^2)$, but FFTs carry overhead (padding, transforms). For small $N$ the constant factors dominate and direct convolution can win; the crossover is around a few dozen to a few hundred samples.

If the recording is too short, $\Delta f=f_s/N$ may exceed 1 Hz, so both fall in (or spill across) the same bin. Record longer to shrink $\Delta f$ and split them.

The FFT assumes the signal is periodic, so without padding the tail wraps around and contaminates the start (circular convolution). Padding to $\ge M+N-1$ gives room for the full linear overlap.

FIR output is a finite weighted sum of inputs only — bounded input gives bounded output automatically. IIR feeds past outputs back in, so poor coefficients can let the output grow without bound.

Nothing — it returns the signal unchanged. The impulse is the identity of convolution, matching the idea that $h[n]$ is the system's reaction to a lone spike.

$X[0]={\sum }_{n}x[n]$, the (unnormalised) average/offset of the signal — the "zero-frequency" component. It has no mirror twin, so it is never doubled in one-sided scaling.

It is a real, single (unpaired) value like DC — so, like DC, it is not multiplied by 2 in one-sided amplitude scaling. Forgetting this over-counts its amplitude.

All the energy sits in the DC bin $X[0]$ and every other bin is (ideally) zero — a flat line has no frequency content besides its offset.

Yes. Convolution is symmetric, $x\ast h=h\ast x$; the roles of "signal" and "kernel" are interchangeable and the output length is still $M+N-1$.

The low-pass leaves them essentially untouched (passband gain $\approx 1$); you mostly see only edge/transient effects at the very start and end from the filter's startup transient.

It peaks at zero lag, because a signal aligns best with an unshifted copy of itself; the peak height equals the signal's energy ${\sum }_{n}x[n{]}^2$.

An aliased ghost. Since $120>f_s/2=100$, it folds to $200-120=80$ Hz. Nyquist is violated, so the true frequency is lost — the core warning of Aliasing and Sampling Theorem.