Skip to main content

Frequency shifting property of DTFS

Short answer

Multiplying by e^{j M (2π/N) n} shifts coefficients: e^{j M ω0 n} x[n] ↔ a_{k-M}. Proof: factor e^{-j(k-M)ω0 n} in the DTFS sum.

State and prove frequency shifting property of discrete time Fourier series.

Statement

If x[n]x[n] is periodic with period NN and x[n]⟷FSakx[n] \overset{FS}{\longleftrightarrow} a_k, then for an integer MM:

ejM(2π/N)n x[n]⟷FSak−Me^{jM(2\pi/N)n}\,x[n] \overset{FS}{\longleftrightarrow} a_{k-M}

Multiplying by a complex exponential at the MMth harmonic shifts the coefficient sequence by MM.

Proof

y[n]=ejM(2π/N)nx[n]y[n] = e^{jM(2\pi/N)n}x[n] is also periodic with period NN (both factors are). Its coefficients are

bk=1N∑n=⟨N⟩ejM(2π/N)nx[n] e−jk(2π/N)n=1N∑n=⟨N⟩x[n] e−j(k−M)(2π/N)n=ak−M\begin{aligned} b_k &= \frac{1}{N}\sum_{n=\langle N\rangle}e^{jM(2\pi/N)n}x[n]\,e^{-jk(2\pi/N)n} \\ &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-j(k-M)(2\pi/N)n} \\ &= a_{k-M} \end{aligned}

Example: for N=4N = 4, multiplying by ejπn/2e^{j\pi n/2} (M=1M = 1) moves a0a_0 to position k=1k = 1, a1a_1 to k=2k = 2, and so on (cyclically, since aka_k has period 4).

✓ Answer checked by Claude Opus 5.5 (high effort)

All Signals and Systems chapter 2 questions →