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Does anyone know how to represent the Discrete Fourier transform (DFT) coefficient, $X[k]$, with respect to the Continuous time-Fourier series (CT-FT) coefficient, $X_k$? I come to the conclusion as $X[k] = NX_k$. $\\\\$

As we know Fourier series is for a periodic signal. So DFT coefficient is just a sampled version of the Fourier series coefficient. Let $P$ be the period of CT signal $x(t)$, and $N$ be the sampled period of $x[n]$. We sampled $x(t)$ by $T=P/N$ as sample period. From $$x[n] = x(nP/N)\\=\sum_{k=-inf}^{inf}X_k e^{2\pi*k/P*(nP/N)}\\ =\sum_{k=-inf}^{inf}X_k e^{2\pi*kn/N}$$ Since for DFT, $$x[n] =1/N\sum_{k=-0}^{N-1}X[k] e^{2\pi*kn/N}$$, we can say that $X[k]=NX_k$ for $k=0,1,...,N-1$? $$\\\\$$Is my interpretation correct?

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No, it's not that simple. In general, sampling will introduce aliasing. The correct derivation is as follows. Let $T$ be the period of the continuous-time signal $x(t)$, and let $T_s$ be the sampling period satisfying $T=NT_s$ with integer $N$. It follows that the sampled sequence $x_d[n]=x(nT_s)$ is also periodic with period $N$.

We have

$$x(t)=\sum_{k=-\infty}^{\infty}c_ke^{j\frac{2\pi k}{T} t}\tag{1}$$

For the sampled sequence we obtain

$$x_d[n]=x(nT_s)=\sum_{k=-\infty}^{\infty}c_ke^{j\frac{2\pi k}{T} nT_s}=\sum_{k=-\infty}^{\infty}c_ke^{j\frac{2\pi k}{N} n}\tag{2}$$

where we've used $T_s=T/N$. Note that the term $e^{j\frac{2\pi k}{N} n}$ is $N$-periodic in $k$, so we can rewrite $(2)$ as

$$x_d[n]=\sum_{k=0}^{N-1}\sum_{l=-\infty}^{\infty}c_{k+lN}e^{j\frac{2\pi k}{N} n}\tag{3}$$

Comparing Eq. $(3)$ to the IDFT

$$x_d[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j\frac{2\pi kn}{N}}\tag{4}$$

yields

$$X[k]=N\sum_{l=-\infty}^{\infty}c_{k+lN}\tag{5}$$

Form Eq. $(5)$ it is obvious that the DFT coefficients $X[k]$ are a (scaled and) aliased version of the Fourier coefficients $c_k$ of the continuous-time function.

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  • $\begingroup$ Omg! You are amazing. I totally know where I got wrong. I feel that my derivation is a little bit wrong so I decide to ask people on this platform. Thank you for your help! $\endgroup$ – Yok Jye Tang Jan 31 at 17:06

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