Frequency domain to time domain conversion-(varying sample count)
Consider the DTFT of an N-sequence $x_n$:

$$
\begin{align*}
X(e^{j\omega}) &amp;= \sum_{n=0}^{N-1} x_n e^{-j\omega n} &amp; \text{(DTFT)} \\
&amp;= \sum_n \left ( \frac{1}{N} \sum_{k=0}^{N-1} X_k W_N^{-kn} \right ) e^{-j \omega n} &amp; \text{(subs. IDFT)}
\end{align*}
$$

Here $X_k = X(e^{j\frac{2\pi}{N}k}), \, W_N = e^{-j\frac{2\pi}{N}}$, and let $\omega_k = \frac{2\pi}{N} k$, then rearrange:

$$
\begin{align*}
X(e^{j\omega}) = \sum_{k=0}^{N-1} X_k \left ( \frac{1}{N} \sum_{n=0}^{N-1} e^{j(\omega_k - \omega) n} \right )
\end{align*}
$$

Hence the DTFT of an N-sequence is uniquely determined by N frequency samples uniformly spaced between $0$ and $2\pi$. The term in the parentheses is an interpolation function (it's actually the digital sinc).

Check whether your $x_n$ is periodic to 150 samples.

What happens when you zero pad the DFT to length $P &gt; N$ (add $P-N$ more $X_k = 0$ terms), and take the IDFT? You interpolate to a higher sampling rate (since $\omega_k = \frac{2\pi}{P}k$ is less than before).

Note that you're adding points after the first $N/2$ samples, explained here.