Trigonometric form of Fourier series

For a real periodic signal, the Fourier series can be written using real cosines and sines instead of complex exponentials:

$x(t)=a_x[0]+{\sum }_{k=1}^{\infty }\left\{a_x[k]\cos (2\pi kt/T)+b_x[k]\sin (2\pi kt/T)\right\}.$

This is the trigonometric Fourier series. Sometimes more convenient than the complex form, especially for hand-computation and for visualizing what frequencies a real signal contains.

Derivation from the complex form

Start with the complex synthesis equation and group the $k=k_0$ term with $k=-k_0$, using conjugate symmetry $c_x[-k]=c_x^{\ast }[k]$:

$x(t)=c_x[0]+{\sum }_{k=1}^{\infty }(c_x[k]e^{j2\pi kt/T}+c_x^{\ast }[k]e^{-j2\pi kt/T}).$

Applying Euler’s formula and combining:

$=c_x[0]+{\sum }_{k=1}^{\infty }\left[(c_x[k]+c_x^{\ast }[k])\cos (2\pi kt/T)+j(c_x[k]-c_x^{\ast }[k])\sin (2\pi kt/T)\right].$

Define

$a_x[k]=c_x[k]+c_x^{\ast }[k]=2\mathrm{Re}{c}_x[k],b_x[k]=j(c_x[k]-c_x^{\ast }[k])=-2\mathrm{Im}{c}_x[k]$

for $k\ge 1$, with $a_x[0]=c_x[0]$ and $b_x[0]=0$. Both are real numbers.

Direct formulas

$a_x[k]=\frac{2}{T}{\int }_{T}x(t)\cos (2\pi kt/T)dt,b_x[k]=\frac{2}{T}{\int }_{T}x(t)\sin (2\pi kt/T)dt(k\ge 1),$

with

$a_x[0]=\frac{1}{T}{\int }_{T}x(t)dt\text{(DC, no factor of 2)}.$

Watch the DC term. The formula for $a_x[0]$ has prefactor $1/T$, not $2/T$. This asymmetry is one of the two most common errors with the trigonometric form — independent of the conversion-to-complex-form factor-of-$1/2$ mistake below. The reason: pairing $c_x[k]$ with $c_x[-k]$ produces a factor of 2 for $k\ge 1$, but the $k=0$ term has no pair, so it doesn’t double.

Converting between forms

$c_x[0]=a_x[0],c_x[k]=\frac{a_x[k]-j{b}_x[k]}{2},c_x[-k]=c_x^{\ast }[k]=\frac{a_x[k]+j{b}_x[k]}{2}(k\ge 1).$

Two things to internalize:

1. The factor of $1/2$. Forgetting this is the single most common error in converting forms. Write it carefully each time.
2. The negative sign on $b_x$ for $+k$. Notice $c_x[k]$ has $-j{b}_x[k]$, but $c_x[-k]$ has $+j{b}_x[k]$. The minus goes with positive $k$, the plus with negative $k$. (Mnemonic: the positive-$k$ coefficient corresponds to $e^{+j\omega t}=\cos +j\sin$, and the analysis side picks up the conjugate, flipping the sign on $\sin$.)

A quick sanity check

Take $x(t)=\cos (2\pi t/T)$. The trigonometric form should give $a_x[1]=1$, everything else zero. From the direct formula, $a_x[1]=(2/T){\int }_T{\cos }^2(2\pi t/T)dt=(2/T)(T/2)=1$. ✓

In complex form, this should give $c_x[1]=c_x[-1]=1/2$ (from the cosine pair $\cos \leftrightarrow (1/2)(\delta [k-1]+\delta [k+1])$). Converting: $c_x[1]=(a_x[1]-j{b}_x[1])/2=(1-0)/2=1/2$. ✓

Why use which form

Trigonometric is the natural form for real signals — coefficients are real, easy to plot, easy to interpret physically (each $a_x[k]$ is the amount of $\cos (2\pi kt/T)$ in the signal). Complex is the natural form for derivations involving the Fourier transform (which generalizes the complex form to a continuous integral) and for systems analysis (where complex exponentials are eigenfunctions of LTI systems).

Fluency requires switching between forms cleanly. The conversion formulas are on the sheet, but the factor of 2 and the sign on $b_x$ get checked often.