I know that for a given signal $s(t)$ and a given delay $ \tau $, by the shift theorem:
$$ \mathcal{F}\{s(t-\tau)(f) \} = e^{-j 2 \pi f \tau} \mathcal{F}\{s(t)(f) \} \tag{1} $$
However, when I try to derive the Fourier Transform of a delayed LFM chirp, written as:
$$ s_{LFM}(t-\tau) = A e^{j (\omega_o (t-\tau) + \alpha (t-\tau)^2)} \Pi_{\frac{T}{2}}(t-\tau) $$ it gives:
$$ s_{LFM}(t-\tau) = A e^{j( \omega_o (t-\tau) + \alpha (t-\tau)^2)} \Pi_{\frac{T}{2}}(t-\tau) \\ = A e^{j \omega_o t} \ e^{-j \omega_o \tau} e^{j \alpha t^2} e^{j \alpha \tau^2} e^{j \alpha 2t\tau} \Pi_{\frac{T}{2}}(t-\tau)$$ so: $$ \mathcal{F}\{s_{LFM}(t-\tau)(f) \} = A e^{-j \omega_o \tau} e^{j \alpha {\tau}^2} \ \mathcal{F}\{ e^{j \omega_o t} e^{j \alpha t^2} e^{j \alpha 2t{\tau}} \Pi_{\frac{T}{2}}(t-\tau)(f) \} \tag{2} $$ The term $ e^{j \alpha 2t{\tau} } $ shifts the spectrum of $s_{LFM}$.
Why do not we get the same result between $(1)$ and $(2)$?