# Represent DFT coefficients with respect to Continuous time-Fourier series coefficients

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?

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.

• 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! – Yok Jye Tang Jan 31 at 17:06