Time-scaling property of DTFS

In discrete time, $x[an]$ is not defined for non-integer $an$, so time scaling is defined through the time-expanded signal (inserting zeros).

Statement

Let $x[n]$ be periodic with period $N$ and DTFS coefficients $a_k$. For a positive integer $m$ define

$$x_{(m)}[n] = \begin{cases} x[n/m], & n \text{ a multiple of } m \\ 0, & \text{otherwise} \end{cases}$$

($m - 1$ zeros are inserted between samples). Then $x_{(m)}[n]$ is periodic with period $mN$ and

$$x_{(m)}[n] \overset{FS}{\longleftrightarrow} \frac{1}{m}a_k$$

(with $a_k$ viewed as periodic with period $N$, so the $mN$ coefficients are $a_k/m$ for $k = 0, 1, \dots, mN-1$).

Proof

The period of $x_{(m)}[n]$ is $mN$, so its coefficients are

$$b_k = \frac{1}{mN}\sum_{n=0}^{mN-1}x_{(m)}[n]e^{-jk\frac{2\pi}{mN}n}$$

Only $n = mr$ ($r = 0, 1, \dots, N-1$) give non-zero terms, where $x_{(m)}[mr] = x[r]$:

$$\begin{aligned} b_k &= \frac{1}{mN}\sum_{r=0}^{N-1}x[r]\,e^{-jk\frac{2\pi}{mN}mr} \\ &= \frac{1}{m}\left[\frac{1}{N}\sum_{r=0}^{N-1}x[r]e^{-jk\frac{2\pi}{N}r}\right] \\ &= \frac{1}{m}a_k \end{aligned}$$

Interpretation

• Expanding in time by $m$ multiplies the period by $m$, so the fundamental frequency becomes $\omega_0/m = 2\pi/(mN)$ and the spectrum is packed $m$ times more densely.
• The coefficient values are divided by $m$, because the same $N$ non-zero samples of one period are now averaged over $mN$ samples instead of $N$.
• The coefficient pattern of $x[n]$ is repeated $m$ times within one period of $b_k$.

Example: $x[n] = \{1, 1\}$ with $N = 2$ has $a_0 = 1$, $a_1 = 0$. With $m = 2$, $x_{(2)}[n] = \{1, 0, 1, 0\}$ ($N = 4$) gives $b_k = \frac{1}{4}(1 + e^{-j\pi k}) = \{0.5, 0, 0.5, 0\}$, which is $\frac{1}{2}\{a_0, a_1, a_0, a_1\}$.