Frequency domain to time domain conversion-(varying sample count)

Question

I need to convert 150 Frequency domain samples to the time domain. The problem is, I have been provided with 250 time domain samples for comparing my results. How am I supposed to relate the 150 samples to 250 samples? I can't find any relationship like interpolation/decimation between the two.

Does this have anything to do with windowing (hamming,hanning,etc), sampling frequency, min and max frequency in the original signal?

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@Jim: I have 150 frequency domain samples and i need to get 250 time domain samples using them. If I do a normal iFFT, I would get only 150 TD samples. So I tried to append 100 zeros to the FD samples and then did the iFFT. But then, my samples are different from the original answers.

The input set I have is : 150 TD samples, sampling frequency=10GHZ, Min and Max frequency of the original signal = 500MHZ and 3050 MHZ, Window= Kaiser (beta=6)

I believe that I have been provided with either insufficient or incorrect data. Am i right?

Answer 1

Consider the DTFT of an N-sequence $x_n$:

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

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 $P > N$ (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.