Harmonic exponentials; CTFS conjugate symmetry

Representation with harmonically related exponentials

Take a periodic signal $x(t)$ with fundamental period $T$ and $\omega_0 = 2\pi/T$. The set of harmonically related complex exponentials

$$\phi_k(t) = e^{jk\omega_0 t}, \quad k = 0, \pm1, \pm2, \dots$$

are all periodic with period $T$, since each frequency is an integer multiple of $\omega_0$. The term for $k = 0$ is the DC part, $k = \pm1$ are the fundamental components, and $k = \pm N$ are the $N$th harmonics. A linear combination of them is periodic with period $T$:

$$x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t} \quad \text{(synthesis equation)}$$

The functions are orthogonal over one period: $\int_T e^{jk\omega_0 t}e^{-jn\omega_0 t}dt$ equals $T$ when $k = n$ and 0 otherwise. Multiplying the synthesis equation by $e^{-jn\omega_0 t}$ and integrating over $T$ therefore leaves only one term. This gives

$$a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt \quad \text{(analysis equation)}$$

The series converges to $x(t)$ when $x(t)$ meets the Dirichlet conditions. Example: $\cos\omega_0 t = \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t}$, so $a_{\pm1} = 1/2$.

Conjugation property

Statement: if $x(t) \xleftrightarrow{FS} a_k$, then

$$x^*(t) \xleftrightarrow{FS} a_{-k}^*$$

Proof: let $b_k$ be the coefficients of $x^*(t)$:

$$b_k = \frac{1}{T}\int_T x^*(t)e^{-jk\omega_0 t}dt = \left[\frac{1}{T}\int_T x(t)e^{jk\omega_0 t}dt\right]^* = \left[\frac{1}{T}\int_T x(t)e^{-j(-k)\omega_0 t}dt\right]^* = a_{-k}^*$$

Conjugate symmetry property

Statement: if $x(t)$ is real, its coefficients are conjugate symmetric:

$$a_{-k} = a_k^*$$

Proof: for a real signal $x(t) = x^*(t)$, so the two signals have the same coefficients: $a_k = b_k = a_{-k}^*$. Taking the conjugate of both sides gives $a_k^* = a_{-k}$.

Consequences for real $x(t)$:

• $|a_{-k}| = |a_k|$, so the magnitude spectrum is even.
• $\angle a_{-k} = -\angle a_k$, so the phase spectrum is odd.
• $\mathrm{Re}\{a_k\}$ is even and $\mathrm{Im}\{a_k\}$ is odd.
• Real and even $x(t)$ gives real, even $a_k$. Real and odd $x(t)$ gives purely imaginary, odd $a_k$.

For example, $\sin\omega_0 t$ has $a_1 = 1/(2j)$ and $a_{-1} = -1/(2j) = a_1^*$.