Why does the CT Fourier series sum run from k = -infinity to infinity?

Each e^{jk w0 t} is a distinct signal, and a signal with sharp jumps may need infinitely many

Because CT signals are always infinite in duration

Because the period of a CT signal is infinite

Because the coefficients are periodic in k

Why does a DT Fourier series need only N terms?

Because e^{j(k+N)(2 pi/N)n} = e^{jk(2 pi/N)n}, so only N distinct exponentials exist

Because DT signals have no high frequencies

Because the DTFS ignores the phase

Which statement about convergence is right?

The DTFS is a finite sum so it always exists, while the CT series needs convergence checks

The DTFS needs the Dirichlet conditions

Neither series ever has convergence issues

2077 Chaitra · 2 marks

Why DTFS uses only one period

Short answer

In continuous time, all harmonics e^{j k ω0 t} are distinct, requiring an infinite sum to represent signals with sharp transitions. In discrete time, e^{j (k+N) (2π/N) n} = e^{j k (2π/N) n}, so only N distinct harmonics exist; the series is a finite sum over one period.

A continuous time Fourier Series is evaluated over infinite duration while a discrete time Fourier series is evaluated over one time period only. Why?

The difference comes from how many distinct harmonics each signal can have:

Continuous time: the exponentials $e^{jk\omega_0 t}$ are all different for every integer $k$ ; a higher $k$ is always a faster oscillation. A CT periodic signal (e.g. a square wave with sharp jumps) may need infinitely many harmonics, so the synthesis sum runs from $k = -\infty$ to $\infty$ , and convergence must be examined (Dirichlet conditions).
Discrete time: $e^{j(k+N)(2\pi/N)n} = e^{jk(2\pi/N)n}e^{j2\pi n} = e^{jk(2\pi/N)n}$ , because $n$ is an integer. Only $N$ distinct exponentials exist, so both the synthesis and analysis sums need only one period ( $N$ terms) and the coefficients repeat, $a_{k+N} = a_k$ . The DTFS is a finite sum, so it always exists and has no convergence problem.

In both cases the analysis is done over one period of the signal; the "infinite" range in CT refers to the infinite number of harmonic terms needed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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