Question

For a discrete-time periodic signal with period $N$, how many distinct Fourier coefficients exist?

• A. Infinite
• B. $N/2$
• C. $N$
• D. $2N$

Hint:

Unlike continuous-time periodic signals (which have an infinite number of harmonic coefficients $c_k$), discrete-time periodic signals with period $N$ have a periodic spectrum, leaving exactly $N$ distinct coefficients.

Answer 1

The Correct Option is C

Understanding the Concept: The Discrete-Time Fourier Series (DTFS) represents a discrete-time periodic sequence $x[n]$ with a fundamental period $N$. The standard analysis formulation to calculate the spectral coefficients $a_k$ is: \[ a_k = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-j k \left(\frac{2\pi}{N}\right) n} \]

Step 1: Check the periodicity of the spectral coefficients.
Let us calculate the value of the coefficient at a shifted frequency index position $(k + N)$: \[ a_{k+N} = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-j(k+N)\left(\frac{2\pi}{N}\right)n} \] Expanding the exponential term: \[ a_{k+N} = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-j k \left(\frac{2\pi}{N}\right) n} \cdot e^{-j \left(\frac{2\pi}{N}\right) N n} \] Simplifying the second exponential term: \[ e^{-j \left(\frac{2\pi}{N}\right) N n} = e^{-j 2\pi n} = 1 \quad (\text{since } n \text{ is an integer}) \] Substituting this back into the equation yields: \[ a_{k+N} = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-j k \left(\frac{2\pi}{N}\right) n} = a_k \]

Step 2: Determine the total number of independent values.
Since $a_{k+N} = a_k$, the sequence of Fourier coefficients $a_k$ is periodic with the exact same fundamental period $N$. Consequently, values outside any consecutive block of length $N$ (e.g., $k = 0, 1, \dots, N-1$) are simply repeating replicas. Thus, there are exactly $N$ unique, distinct Fourier coefficients.