The DFT represents a finite-duration signal by a discrete set of spectral samples, obtained from a finite sum over the available time samples. When you have N time samples x[0] through x[N−1], the DFT computes N complex values X[k] = sum_{n=0}^{N−1} x[n] e^{-j 2π kn / N}, which correspond to N equally spaced frequency bins ω_k = 2π k / N. This means you are sampling the continuous spectrum of the signal (which would exist for all frequencies if you looked at the full Fourier transform) at N discrete points, not producing a continuous spectrum. The input need not be strictly periodic in time; the finite sequence is treated as one period of a periodic extension to justify the discrete spectrum, and the result is a finite, discrete set of spectral components. That’s why the statement about using a finite sum over N samples to approximate the spectrum is the accurate description.

In decibels, the SNR is defined as 10 times the base-10 logarithm of the power ratio because decibels are a logarithmic scale for power. Since SNR compares signal power to noise power, the appropriate form is 10 log10(P_signal / P_noise). If you were instead dealing with voltages or amplitudes, you’d use 20 log10(V_s / V_n) because power is proportional to the square of the voltage (P → V^2); the extra factor of 2 inside the log translates to the 20 in front when you convert a voltage ratio to decibels. The other options don’t match this power-based definition: 20 log10(P_s / P_n) would misalign with the standard power-to-decibel relation, log10(P_s / P_n) lacks the 10 to place it in decibels, and 40 log10(P_s / P_n) isn’t a recognized form for SNR.

The cutoff frequency is the point where the magnitude of the frequency response has fallen to 1/√2 of its maximum value. This is the -3 dB point, tied to the fact that power → |H(jω)|^2, so reducing power by half corresponds to reducing the magnitude by 1/√2. If the passband gain is normalized to 1, then the cutoff occurs at |H(jωc)| = 1/√2. The other options don’t match this standard definition: maximum magnitude would mean no attenuation, zero would imply complete attenuation, and half of the maximum does not correspond to the half-power point.

Sampling a signal in time causes its spectrum to repeat at intervals of the sampling frequency. This happens because sampling is multiplication of the signal by a periodic impulse train in time, and multiplication in time corresponds to convolution in frequency with that train’s spectrum. The spectrum of the impulse train is another train of impulses spaced by Fs, so the original spectrum is copied and shifted to every k·Fs. Each replica has the same shape as the original (and, under typical conventions, the same amplitude), just located at multiples of Fs. This is why you see multiple copies in the spectrum of a sampled signal. If the sampling rate is high enough that these copies don’t overlap, you can perfectly reconstruct the original signal; if not, they overlap and cause aliasing.