Spectral leakage when performing fft

In this entire discussion, let $k,r,N,N_0,m,n \in \mathbb{Z}$.

Derivation of DFT for a cosine signal sampled exactly $p$ periods

The DFT of a discrete time signal $x[k]$ with length $N_0$ is defined as

$$X[r] = \sum_{k=0}^{N_0-1}x[k]e^{-jr\Omega_0 k}, \; \; \; \Omega_0 = \frac{2\pi}{N_0} \tag1$$

where we only need to perform calculations for $r \in [0, N_0-1]$ because the DFT is $N_0$-periodic in $r$ ($2\pi$-periodic in $\Omega$).

We know from continuous time that

$$\int_{T} \cos(n\omega_0 t)\cos(m\omega_0 t) \: \text{d}t = \begin{cases} 0 \: \: \: n \neq m \\\\ \frac{T}{2} \: \: \: n = m\end{cases} \tag2$$ $$\int_{T} \cos(n\omega_0 t)\sin(m\omega_0 t) \: \text{d}t = 0 \: \: \: \text{for all n and m} \tag3$$

where $T$ denotes one period. The discretized versions are

$$\sum_{k=0}^{N-1} \cos(n\Omega k)\cos(m\Omega k) = \begin{cases} 0 \: \: \: n \neq m \\\\ \frac{N}{2} \: \: \: n = m, \; \; \; \Omega \neq \pm \pi \\\\ N \: \: \: n = m, \; \, \; \Omega = \pm \pi \end{cases} \tag4$$ $$\sum_{k=0}^{N-1} \cos(n\Omega k)\sin(m\Omega k) = 0 \: \: \: \text{for all n and m} \tag5$$

where $N$ denotes one period.

If the signal we are performing the DFT on is a cosine signal $x[k] = \cos(\Omega k), \Omega \in [-\pi, \pi)$, then

$$\begin{align*} X[r] &= \sum_{k=0}^{N_0-1}\cos(\Omega k)e^{-jr \Omega_0 k} \\\\ \tag6 X[r] &= \sum_{k=0}^{N_0-1}\cos(\Omega k)\cos(r\Omega_0 k) - i \sum_{k=0}^{N_0-1}\cos(\Omega k)\sin(r\Omega_0 k) \end{align*} $$

Note that $x[k]$ has period $N = \frac{2\pi}{\Omega}$. If we have sampled exactly $p$ periods of $x[k]$, that is, $N_0 = pN$, then $\Omega_0 = \frac{2\pi}{pN}$ and equation $(6)$ becomes

$$X[r] = p\sum_{k=0}^{N-1}\cos(\Omega k)\cos(r\Omega_0 k) - ip \sum_{k=0}^{N-1}\cos(\Omega k)\sin(r\Omega_0 k) \tag7$$

Applying $(5)$ on $(7)$ removes the second sum

$$X[r] = p\sum_{k=0}^{N-1}\cos(\Omega k)\cos(r\Omega_0 k) \tag8$$

Substituting $\Omega = \frac{2\pi}{N}$ and $ \Omega_0 = \frac{2\pi}{pN} $ we are left with

$$X[r] = p\sum_{k=0}^{N_0-1}\cos\bigg(\frac{2\pi}{N} k \bigg)\cos \bigg(\frac{r}{p} \frac{2\pi}{N} k \bigg) \tag8$$

Here, we need to consider two cases: When $\Omega \neq \pm \pi$, that is, $N \neq 2, $or $\Omega = \pm \pi$, that is, $N = 2$.

In the special case where $\Omega = \pi$ we have only one value of $r$ with a non-zero result.

$$X[r] = p\sum_{k=0}^{1}\cos(\pi k)\cos(\frac{r}{p}\pi k) = \begin{cases}0 \; \; \; r \neq p \\\\ pN \; \; \; r = p \end{cases} \tag9$$

In the general case we have two values of $r$ yielding non-zero results. Recall that $\cos(\Omega k) = \cos( (-\Omega + 2\pi) k)$, which is equivalent to $\cos\bigg( (-\frac{2\pi}{N} + 2\pi) k \bigg) =\cos\bigg( (N-1)\frac{2\pi}{N}k \bigg)$.

This means that $(8)$ has two values of $r$ in $[0, N_0-1]$ that yields a non-zero result: $r=p$ and $\frac{r}{p}=N-1 \Rightarrow r = (N-1)p$.

Conclusion

The DFT of a time-discrete cosine signal $x[k] = \cos(\Omega k)$ with exactly $p$-periods of samples and length $N_0$ has one non-zero value if $\Omega = \pm \pi$ and two non-zeros values for the general case, when $r\in[0,N_0-1]$. More specifically $$X[r] = \sum_{k=0}^{N_0-1}x[k]e^{-jr\Omega_0 k} = \begin{cases} pN \; \; \; \Omega = \pm \pi , \; \; \; r = p \\\\ p\frac{N}{2} \; \; \; \Omega \neq \pm \pi, \; \; \; r = p \; \; \; \text{or} \; \; \; r = (N-1)p \\\\ 0 \; \; \; \text{otherwise}\end{cases}$$

This shows why the DFT of 500 Hz signal in the question has only two places where it is different from zero.

The DFT of a finite duration sine/cosine is a sinc pulse with zero crossing at multiples of $1/T$Hz. The FFT is just sampled version of DFT, with samples at $f_s/N$ = $2$Hz in this case.

In the first case, the zero crossings were at multiples of $1/T$=$1/0.5$=$2$Hz, and coincidentally where the FFT samples the DFT waveform. So you didn't see any magnitude at those points. In the second case, the waveform shifted away from the sampling points of FFT. The waveform zero crossing are $499 \pm 2n$ Hz, but FFT sampling points are at multiples of $2$Hz. So you will see non-zero magnitude in FFT.

You can model your discrete signal $y[n]$ as an infinite sinusoid $x[n]=cos(2\pi f_0 n), n=[-\infty,\infty]$ multiplied by a (boxcar/rectangular) window function of 0.5 sec

$$w[n] = \begin{cases} 1 \qquad & \text{for } \ 0<n<f_s/2, \\ 0 \qquad & \text{otherwise} \\ \end{cases}$$

So we have

$$y[n] = x[n]w[n]=\cos(2\pi f_0n)w[n]$$

We can perform a Fourier Transform to get (the Discrete Time Fourier Transform, DTFT): $$Y(f) = X(f)*W_n(f)$$ Since y[n] was discrete, you have replicas of the baseband spectrum every $f_s$, but let's focus only on the baseband. An infinite length sinusoid has a Fourier Transform that is a delta function around its frequency, and a rectangular/boxcar window has a Fourier Transform called a Dirichlet Kernel (similar to the continuous sinc function) $$Y(f) = 0.5(\delta(f-f_0)+\delta(f+f_0))*W_n(f)=0.5W_n(f-f_0) + 0.5W_n(f+f_0)$$ where $W_n(f)$ is the Dirichlet kernel for $n=0.5f_s=2500$. So what you get is the Dirichlet kernel centered around the frequency $f_0$ (and $-f_0$) rather than 0. This derivation is for any $f_0$. We can think of the DFT as sampling the DTFT (which is continuous in the frequency domain) in the dft bins, i.e. $f=\frac{2\pi k}{N}$. So, even in the first case you showed, there is a Dirichlet kernel "hiding" there, you just don't see it. Why? because it happens that all all the other dft bins fall on the zeros of the dirichlet kernel, which are the points $x_k=\frac{2\pi k}{2n+1}, 𝑘∈ℤ$. To see that even in your first example there is still a Dirichlet kernel (i.e., the Fourier Transform of your first signal is not a pure delta function), you can pad the time domain signal with zeros and see the result. As for the second case, in this case you centered the dirichlet kernel to a place where you don't sample it only when it is equal to zero (with the dft bins of nfft=n)