Physical Meaning of 1-D FT

Here we only consider the Fourier expansion of continuous and periodic signals. The result here can be easily generalized to all other forms of the Fourier transform. The Fourier expansion of a 1D periodic signal

$$x_T(t)=\sum_{n=-\infty}^{\infty} X[n]e^{j2\pi nf_0t}$$

represents the signal as a weighted sum of complex exponential functions. Here $X[n]$ is the weight of the nth term (the Fourier coefficient) which is in general a complex number

$$X[n]=X_r[n]+jX_j[n]=\vert X[n]\vert e^{j\angle(X[n])}=\frac{1}{T}\int_T x_T(t)e^{-j2\pi nf_0t} dt$$

where $\vert X[n]\vert$ and $\angle X[n]$ are respectively the magnitude and phase of the nth coefficient $X[n]$:

$$\left\{ \begin{array}{l} \vert X[n]\vert=\sqrt{X_r[n]^2+X_j[... ...n]}=\tan^{-1} \left[ X_j[n]/X_r[n] \right] \end{array} \right.$$

In particular, when $n=0$, $X[0]$ is the average or DC component of the signal:

$$X[0]=\frac{1}{T}\int_T x(t) dt$$

The Fourier coefficient can be further written as:

$$X[n] = \frac{1}{T}\int_T x(t)e^{-j2\pi f_0nt}\;dt=\frac{1}{T}\int_T [x_r(t)+jx_i(t)]\; \left[\cos(2\pi f_0nt)-j\;sin(2\pi f_0nf)\right] dt$$

$$= \frac{1}{T}\int_T \left[x_r(t)\cos(2\pi f_0nt)+x_i(t)\sin(2\pi f_... ...\frac{j}{T}\int_T \left[x_i(t)\cos(2\pi f_0nt)-x_r(t)\sin(2\pi f_0n)\right] dt$$

$$= X_r[n]+jX_i[n]$$

where $X_r[n]$ and $X_i[n]$ are respectively the real and imaginary part of the spectrum:

$$X_r[n]=Re[X[n]]=\frac{1}{T}\int_T \left[x_r(t)\cos(2\pi f_0nt)+x_i(t)\sin(2\pi f_0nt)\right] dt$$

and

$$X_i[n]=Im[X[n]]=\frac{1}{T}\int_T \left[x_i(t)\cos(2\pi f_0nt)-x_r(t)\sin(2\pi f_0n)\right] dt$$

In particular, if $x_i(t)=0$, i.e., the signal $x(t)=x_r(t)$ is real, and the real and imaginary parts of $X[n]$ become even and odd functions of $n$, respectively:

$$X_r[n]=\frac{1}{T}\int_T x_r(t)\cos(2\pi f_0nt) dt=X_r[-n]... ...; X_i[n]=\frac{1}{T}\int_T -x_r(t)\sin(2\pi f_0n) dt=-X_i[-n]$$

and we have

$$X[-n]=X_r[-n]+jX_i[-n]=X_r[n]-jX_i[n]=X^*[n]$$

Now the Fourier expansion of $x(t)$ can be written as

$$x(t) = \sum_{n=-\infty}^{\infty} X[n]e^{j2\pi nf_0t} = X[0]+\sum_{n=1}^{\infty}\left[ X[n]e^{j2\pi nf_0t}+X[-n]e^{-j2\pi nf_0t}\right]$$

$$= X[0]+\sum_{n=1}^{\infty} \left[ X[n]e^{j2\pi nf_0t}+X^*[n]e^{-j2\... ...e X[n]}e^{j2\pi nf_0t}+\vert X[n]\vert e^{-j\angle X[n]}e^{-j2\pi nf_0t}\right]$$

$$= X[0]+\sum_{n=1}^{\infty} \vert X[n]\vert \left[e^{j(2\pi nf_0+\an... ...ht] = X[0]+\sum_{n=1}^{\infty} 2\vert X[n]\vert\;\cos(2\pi nf_0t+\angle X[n] )$$

Now we see that any real signal $x(t)$ can be represented as a weighted sum of an infinite number of sinusoids of frequencies $f_n=nf_0$ with amplitudes $2 \vert X[n]\vert$ and phases $\angle X[n]$.