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

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?

• When you say that you have 250 frequency domain samples, how exactly did you get them? Were they time domain samples that you did an FFT on? Do you know the sample frequency of both sample sets? Nov 10 '13 at 3:42

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.