CTFS time reversal, scaling, frequency shift

Let $x(t)$ be periodic with period $T$, $\omega_0 = 2\pi/T$, and $x(t) \xleftrightarrow{FS} a_k$.

Time reversal

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

Put $k = -m$ in the synthesis equation:

$$x(-t) = \sum_k a_k e^{-jk\omega_0 t} = \sum_m a_{-m}e^{jm\omega_0 t}$$

Reversing the signal in time reverses its coefficient sequence. If $x(t)$ is even, $a_{-k} = a_k$. If it is odd, $a_{-k} = -a_k$.

Time scaling

$$x(\alpha t),\ \alpha > 0: \quad x(\alpha t) = \sum_k a_k e^{jk(\alpha\omega_0)t}$$

$x(\alpha t)$ has period $T/\alpha$ and fundamental frequency $\alpha\omega_0$. The coefficients $a_k$ stay the same, and only the harmonic spacing changes from $\omega_0$ to $\alpha\omega_0$. Compressing the signal ($\alpha > 1$) spreads the spectral lines apart.

Frequency shifting

$$e^{jM\omega_0 t}x(t) \xleftrightarrow{FS} a_{k-M}$$

The coefficients of the product are

$$\frac{1}{T}\int_T e^{jM\omega_0 t}x(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T x(t)e^{-j(k-M)\omega_0 t}dt = a_{k-M}$$

Multiplying by a harmonic exponential moves the whole line spectrum $M$ places to the right. This is the dual of the time-shift property.