Convert back Integral with 2 functions: x(t) = [2/(pi*t)^2] int [y(f) sin^4(pi*f*t)] df
- Thread starter okto
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Zermelo
Junior Member
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- Jan 7, 2021
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Hello, where exactly are you stuck? We can’t solve your problems for you, can only help ?
Try to solve for y(f) like you would in a regular equation. Tell us at what step are you stuck
Sorry, I got stucked in the last line, since x(t) has no f variable, meaning it's constant & the diffrentiation of x(t) will be zero then?
[math]\begin{aligned} & \int y(f) \operatorname{Sin}^4(\pi f t) d f=\frac{x(t)(\pi t)^2}{2} \\ & y(f) \operatorname{Sin}^4(\pi f t)=\frac{d}{d f}\left[\frac{x(t)(\pi t)}{2}\right] \\ & y(f)=\frac{1}{\operatorname{Sin}^4(\pi f t)} \frac{d}{d f}\left[\frac{x(t)(\pi t)^2}{2}\right] \end{aligned}[/math]
[math]\begin{aligned} & \int y(f) \operatorname{Sin}^4(\pi f t) d f=\frac{x(t)(\pi t)^2}{2} \\ & y(f) \operatorname{Sin}^4(\pi f t)=\frac{d}{d f}\left[\frac{x(t)(\pi t)}{2}\right] \\ & y(f)=\frac{1}{\operatorname{Sin}^4(\pi f t)} \frac{d}{d f}\left[\frac{x(t)(\pi t)^2}{2}\right] \end{aligned}[/math]
Steven G
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It could be possible that that t is a function of f (maybe t = f+4), so x(t) might be a function of f (even if you see no f's!)