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Given the following,

$$f(l)=\sum_{k}f_{k}(l)e^{-i\phi_{k}l}$$

$$W_{f}(\tau,v)=\int_{\mathbb{R}}f(l+\tau/2)f^{*}(l-\tau/2)e^{-ivl}\partial l$$ $$W_{g}(\tau-t,v-\omega)=\int_{\mathbb{R}}g(l+(\tau-t)/2)g^{*}(l-(\tau-t)/2)e^{-i(v-\omega)l}\partial l$$

$$S_{f}^{g}(t,\omega)=\frac{1}{2\pi}\iint_{\mathbb{R}}W_{g}(\tau-t,v-\omega)W_{f}(\tau,v)\partial\tau\partial v$$

$$V_{f}^{g}(t,\omega)=\int_{\mathbb{R}}f(l)g\text{*}(l-t)e^{-i\omega l}dl$$

$$\widehat{\omega}_{f}(t,\omega)=\omega-Im\left\{ \dfrac{V_{f}^{g'}(t,\omega)}{V_{f}^{g}(t,\omega)}\right\} $$

$$ \widehat{t}_{f}(t,\omega)=t+Re\left\{ \dfrac{V_{f}^{lg}(t,\omega)}{V_{f}^{g}(t,\omega)}\right\} $$

$$ g' = \partial g(l)/\partial l$$ $$ lg = l*g(l)$$

Can somebody help me simplify the following equation?

$$\widetilde{S}_{f}^{g}(t,\omega)=\frac{1}{2\pi}\iint_{\mathbb{R}}S_{f}^{g}(\tau,v)δ(ω\text{−}\widehat{ω}_{f}(τ,ν))δ(t\text{−}\widehat{t}_{f}(τ,ν))\partial\tau\partial v $$

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  • $\begingroup$ I'm voting to close this question as off-topic because it seems a better fit for math.stackexchange.com $\endgroup$ – MBaz Sep 11 at 14:48

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