If square waves are square integrable, then why does expanding on a fourier basis not recover the equation?
Fourier Series – Why Square Waves Are Not Square Integrable
fourier series
fourier series
If square waves are square integrable, then why does expanding on a fourier basis not recover the equation?
Best Answer
The Fourier Series converges in $L^2$ and pointwise except at the discontinuities, where it converges to the average of the limits from either side. Since the function to which the partial sums converge is not continuous, the convergence cannot be uniform. In fact, the Gibbs phenomenon guarantees an over shoot of about $8.95\%$ of the discontinuity on either side of the discontinuity
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Since the Dirichlet kernel for the $n^\text{th}$ partial sum is $$ \sum_{k=-n}^ne^{ikt}=\frac{\sin\left(\frac{2n+1}{2}t\right)}{\sin\left(\frac12t\right)} $$ The jump from minimum to maximum for an intended jump of $2\pi$ is $$ \int_{-2\pi/(2n+1)}^{2\pi/(2n+1)}\frac{\sin\left(\frac{2n+1}{2}t\right)}{\sin\left(\frac12t\right)}\,\mathrm{d}t $$ As $n\to\infty$, the limit of the overshot divided by the jump is therefore $$ \begin{align} &\lim_{n\to\infty}\frac1{4\pi}\left(\int_{-2\pi/(2n+1)}^{2\pi/(2n+1)}\frac{\sin\left(\frac{2n+1}{2}t\right)}{\sin\left(\frac12t\right)}\,\mathrm{d}t-2\pi\right)\\ &=\lim_{n\to\infty}\frac1{2\pi}\int_0^{2\pi/(2n+1)}\frac{\sin\left(\frac{2n+1}{2}t\right)}{\sin\left(\frac12t\right)}\,\mathrm{d}t-\frac12\\ &=\lim_{n\to\infty}\frac1{(2n+1)\pi}\int_0^\pi\frac{\sin(t)}{\sin\left(\frac1{2n+1}t\right)}\,\mathrm{d}t-\frac12\\ &=\frac1\pi\int_0^\pi\frac{\sin(t)}{t}\mathrm{d}t-\frac12\\[18pt] &\doteq0.089489872236 \end{align} $$
Subsequent peak-to-peak amplitude differences in terms of the jump discontinuity are $$ \begin{array}{l} \text{first drop: }&-0.138078205446&=\frac1\pi\int_\pi^{2\pi}\frac{\sin(t)}{t}\mathrm{d}t\\ \text{second rise: }&\hphantom{-}0.081681570828&=\frac1\pi\int_{2\pi}^{3\pi}\frac{\sin(t)}{t}\mathrm{d}t\\ \text{second drop: }&-0.058123567735&=\frac1\pi\int_{3\pi}^{4\pi}\frac{\sin(t)}{t}\mathrm{d}t\\ \text{third rise: }&\hphantom{-}0.045137494308&=\frac1\pi\int_{4\pi}^{5\pi}\frac{\sin(t)}{t}\mathrm{d}t\\ \text{third drop: }&-0.036901946694&=\frac1\pi\int_{5\pi}^{6\pi}\frac{\sin(t)}{t}\mathrm{d}t \end{array} $$