# Effect of sampling rate and duration on discrete parameters of sine (spectrum)?

The DFT of

$$\cos(2\pi f t + \phi)$$

peaks at $$k=\pm f$$ if $$t = \frac{1}{N}[0, 1, ..., N - 1]$$ (for integer $$f$$, & within Nyquist). What about other $$t$$? What if we double the sampling rate or duration?

How would one plug it into a sine DFT closed form solution? Given x = cos(2*pi*f*t + phi), would it be sine_dft(N, f*duration, phi), for example? Code validation is preferred but optional, and likewise for handling the case t[0] != 0.

Clarification: to be precise, I'm looking to "translate" physical changes to discrete, in terms of changes in the time vector. For example, for doubled duration, does DFT's peak location double, halve, or stay approximately same? Yet, I'm not asking about DFT's precise behavior - for fractional $$f$$, there's no longer one peak that can be tracked this way. It's about how the DFT "sees" input. In exact but somewhat circular terms, $$f$$ is the $$k$$ at which the DFT peaks, for integer $$f$$. Likewise for $$\phi$$, if anything's to be said about it.

In summary,

• Doubled duration: $$f$$ doubled
• Doubled sampling rate: $$f$$ unchanged
• Non-zero $$t[0]$$: $$\phi \rightarrow \phi + 2\pi f t[0]$$ (using old $$f$$)
• Duration, sampling rate: $$T = (t[1] - t[0])\cdot N$$, $$S = 1 / (t[1] - t[0])$$, $$S = N/T$$

In one equation: $$\texttt{sine_dft}\{N, fT, \phi + 2\pi f t[0]\}$$. In SR terms, $$fT \rightarrow f(N/S)$$.

Effects on power/energy are described in Relationship between energy, power, and sampling rate?

### Understanding SR vs Duration

Recall, the result's for

$$\texttt{DFT}\{\cos(2\pi f t + \phi)\},$$

where $$t = \frac{1}{N}[0, 1, ..., N - 1]$$. Let $$N = 4$$, so $$t = [0, 0.25, 0.5, 0.75]$$. We have:

• Doubled sampling rate: $$\rightarrow [0, 0.125, 0.25, 0.375, 0.5, 0.625, 0.75, 0.875]$$. So, $$N \rightarrow 2N$$ and $$f$$ unchanged, hence $$(f / N) \rightarrow (f / N) / 2$$. Note, despite the $$0.875$$, the duration hasn't changed; for $$N \rightarrow \infty$$, the last sample approaches $$1$$. This can be proven (without the $$\infty$$).
• Doubled duration: $$\rightarrow [0, 0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75]$$. So, $$N \rightarrow 2N$$, and now since there's twice the total oscillations, $$f \rightarrow 2f$$. Hence, $$f/N$$ is unchanged.

For spectral analysis, it can be more useful to reformulate these in terms of least possible change in DFT and signal, while staying equivalent in discrete terms. That would be:

• Doubled sampling rate: $$f \rightarrow f/2$$, unchanged $$N$$. Same DFT, but now the sine is halved in frequency.
• Doubled duration: $$N \rightarrow 2N$$, unchanged $$f/N$$. Same sine, but now the DFT has twice the samples.

For analyzing the effects on a reference signal, it's more useful to think of what actually happens with the parameters; that's in the first version. Expressed parametrically (follows first version),

\begin{alignat}{1} \text{Sampling rate:}\ \ & f_\text{new} &= f_\text{old}\quad & N_\text{new} = N_\text{old} (S_\text{new} / S_\text{old}) \\ \text{Duration:}\ \ & f_\text{new} &= f_\text{old} (T_\text{new} / T_\text{old}) \quad & N_\text{new} = N_\text{old} (T_\text{new} / T_\text{old}) \end{alignat}

Visually (follows first version),

Note, the DFT will not "see" the doubled duration plot, i.e. $$t \in [0, 2]$$; its domain always spans a unit interval, and everything else re-expresses itself around it to attain equivalence, hence the "spectral analysis" perspective. Below section extends this one.

### Effects on parameters

We could express a sine DFT solution purely in terms of sampling rate and duration, but it suffices, and is useful and much simpler, to simply understand how $$f, N, \phi$$ change with changes in physical characteristics. Of course, here, $$t$$ must be uniform (all $$t[n] - t[n - 1]$$ same).

Let $$S = \text{sampling rate}$$, $$T = \text{duration}$$, hence $$t = [t_0, t_0 + 1/S, ..., t_0 + 1/S\cdot(N - 1)]$$ or (same) $$t = [t_0, t_0 + T/N, ..., t_0 + T/N\cdot(N - 1)]$$. Some info repeats the previous section.

• $$f \rightarrow f\cdot T$$: doubled duration means doubled the wiggles - DFT's $$f$$ counts the wiggles
• $$f \rightarrow f (N/S)$$ per previous, since $$T = N/S$$
• $$\phi \rightarrow \phi + 2\pi ft_0$$: starting at $$t_0$$ is same as time-shifting to $$t_0$$: $$\cos(2\pi f t + \phi)=$$ $$\cos(2\pi f (t' + t_0) + \phi)$$, where $$t' = [0, 1/S, ..., (N-1)/S]$$. Expanding second's arg-ument, $$2\pi f t' + 2\pi ft_0 + \phi$$; this matches the original solution's formulation, with the sub.
• $$T \neq t[-1] - t[0]$$, and $$S \neq N/(t[-1] - t[0])$$. Common pitfalls. The duration is tricky - see below, also explained here.

### Code validation & illustration

Illustrates above points, and shows equivalence of different constructions of $$t$$. sine_dft is found here.

import numpy as np
from numpy.fft import fft

# configure
N, f, phi, t0, S = (128, 3.5, 1.1, 3.5, 12.9)

# make time vector; show from all major perspectives
t = np.zeros(N)
t[0] = t0
for n in range(1, N):
t[n] = t[n - 1] + 1/S
assert np.allclose(t, np.linspace(t0, t0 + N/S, N, endpoint=False))
assert np.allclose(t, np.arange(start=t0, stop=t0 + N/S, step=1/S))

# make sine
x = np.cos(2*np.pi * f * t + phi)

# show how we retrieve parameters
S = 1 / (t[1] - t[0])
T = N / S
phi0 = phi + 2*np.pi*f*t[0]

# assert equality

• Increased sampling rate: "refines" the existing spectrum by reducing leakage / aliasing, without necessarily changing peaks' location. If original SR was below Nyquist (exact or "effective"), then peak locations$${}^1$$ may change. For any amount of aliasing, the "effective" peak location (e.g. as mean of a few peaks) will almost always change, but negligibly if we're sufficiently above "effective Nyquist" - this happens very quickly for sines.
• Underlying function is sampled over a greater interval: anything could happen, new contents are introduced (almost anything). If "new" exactly repeats "original", that's periodicity and inserts zeros between existing DFT bins. If "new" samples a function that repeats, e.g. sine, then peak locations are moved up by approximately $$T_\text{new}/T_\text{orig}$$ (only the newly-sampled interval must "repeat").
1: for sines, SR still doesn't change $$f$$. In short, $$fT$$ remains valid for $$\texttt{sine_dft}$$, even if below Nyquist, and $$k' = \pm f_d$$ remains true, where $$k'$$ is continuous and maps to the underlying function that $$k$$ is sampling (the sine solution), and $$f_d$$ is $$fT$$ after accounting for $$N/2$$-modularity ("wrap"). SR changes $$f_d$$ if going from below to above Nyquist. $$f_d$$ is always recoverable exactly (see sine solution Q&A), and it matches $$fT$$ if above Nyquist, else it can be matched with more info.