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FS equations and existence conditions

Short answer

Synthesis: x(t)=∑ a_k e^{jkω0t}. Multiply by e^{-jnω0t} and integrate over one period; orthogonality yields a_k = (1/T)∫_T x(t)e^{-jkω0t} dt. Dirichlet conditions: absolute integrability, finite extrema, and finite discontinuities per period.

Derive the formula for synthesis equation and analysis equation of Fourier series. Mention the three conditions that guarantee the existence of Fourier series.

A periodic signal x(t)x(t) with period TT and fundamental frequency ω0=2π/T\omega_0 = 2\pi/T can be written as a sum of harmonically related complex exponentials (the exponential Fourier series).

Synthesis equation

The functions ejkω0te^{jk\omega_0 t}, k=0,±1,±2,…k = 0, \pm1, \pm2, \dots, all repeat every TT, so

x(t)=∑k=−∞∞ak ejkω0tx(t) = \sum_{k=-\infty}^{\infty}a_k\,e^{jk\omega_0 t}

is periodic with period TT. This is the synthesis equation; aka_k are the Fourier series coefficients.

Analysis equation

Multiply both sides by e−jnω0te^{-jn\omega_0 t} and integrate over one period:

∫0Tx(t)e−jnω0tdt=∑k=−∞∞ak∫0Tej(k−n)ω0tdt\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt

Orthogonality of the exponentials:

∫0Tej(k−n)ω0tdt={T,k=n0,k≠n\int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases}

(for k≠nk \ne n, the cosine and sine parts complete a whole number of cycles and integrate to zero). Hence

∫0Tx(t)e−jnω0tdt=anT\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T ak=1T∫Tx(t) e−jkω0t dta_k = \frac{1}{T}\int_T x(t)\,e^{-jk\omega_0 t}\,dt

This is the analysis equation; a0=1T∫Tx(t)dta_0 = \frac{1}{T}\int_T x(t)dt is the dc value.

EquationFormula
Synthesisx(t)=∑kakejkω0tx(t) = \sum_k a_k e^{jk\omega_0 t}
Analysisak=1T∫Tx(t)e−jkω0tdta_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt

Conditions for existence (Dirichlet conditions)

Over any one period:

  1. Absolutely integrable: ∫T∣x(t)∣ dt<∞\int_T |x(t)|\,dt < \infty, which guarantees every ∣ak∣|a_k| is finite.
  2. Finite number of maxima and minima (bounded variation) in one period.
  3. Finite number of discontinuities in one period, and each discontinuity is finite.

When these hold, the series converges to x(t)x(t) at every point where x(t)x(t) is continuous, and to the average 12[x(t−)+x(t+)]\frac{1}{2}[x(t^-) + x(t^+)] at a discontinuity. (A weaker condition, finite energy over one period, ∫T∣x(t)∣2dt<∞\int_T|x(t)|^2dt < \infty, guarantees convergence in the mean-square sense.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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