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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 $(2)$?

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  • $\begingroup$ Are you sure your equation for an LFM chirp is correct? shouldnt $j$ apply to both terms in the exponential? $$Ae^{j \left(\omega_o(t-\tau)+\alpha(t-\tau)^2\right)}$$ $\endgroup$
    – Jdip
    Commented Aug 17, 2023 at 9:35
  • $\begingroup$ Sorry for the typo, you are right $\endgroup$
    – user68882
    Commented Aug 17, 2023 at 10:37
  • $\begingroup$ I think it does indeed shift the spectrum, but until you include the effects of the $\Pi$ function, aren't you just shifting the spectrum of a rectangle function with infinite support? That is, is there really a difference between $\mathrm{rect}(f / \infty)$ and $\mathrm{rect}((f-2\alpha\tau) / \infty)$? $\endgroup$ Commented Aug 21, 2023 at 23:55
  • $\begingroup$ Hay you might be interested in this answer or this one. $\endgroup$ Commented Aug 23, 2023 at 5:37

2 Answers 2

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Let me get rid of $\Pi_{\frac{T}{2}}(t)$ in the derivation:

$$s(t) = Ae^{j\left(\omega_o t + \alpha t^2\right)}$$

Let's show that $$\mathcal{F}\left\{s(t-\tau)\right\} = e^{-j\omega \tau}\mathcal{F}\left\{s(t)\right\}$$

  1. Start with the Fourier Transform of the chirp (I'm omitting the integral's limits, they are $-\infty$ and $\infty$): \begin{align} \mathcal{F}\left\{s(t)\right\} &= \int s(t)e^{-j\omega t}dt\\ &= \int Ae^{j\left(\omega_o t + \alpha t^2\right)}e^{-j\omega t}dt \end{align}
  2. Now let's compute the Fourier Transform of the delayed chirp (notice the substitution $\tilde{t} = t-\tau$): \begin{align} \mathcal{F}\left\{s(t-\tau)\right\} &=\int s(t-\tau)e^{-j\omega t}dt\\ &=\int Ae^{j\left(\omega_o(t-\tau) + \alpha(t-\tau)^2\right)}e^{-j\omega t}dt\\ &= \int Ae^{j\left(\omega_o\tilde{t} + \alpha\tilde{t}^2\right)}e^{-j\omega (\tilde{t}+\tau)}d\tilde{t}\\ &=e^{-j\omega \tau} \int Ae^{j\left(\omega_o\tilde{t} + \alpha\tilde{t}^2\right)}e^{-j\omega \tilde{t}}d\tilde{t}\\ &= e^{-j\omega \tau}\mathcal{F}\left\{s(t)\right\} \end{align}
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  • $\begingroup$ Here the Shift theorem is used, but I was mentioning a direct derivation from the delayed chirp, without the use of that theorem.. $\endgroup$
    – user68882
    Commented Aug 18, 2023 at 9:19
  • $\begingroup$ Where do you see the shift theorem being used? I use the delayed chirp, compute its Fourier transform (step 2) and show that it indeed equals the Fourier transform of the chirp (step 1) multiplied by $e^{j\omega \tau}$, effectively deriving the shift theorem from these two steps. $\endgroup$
    – Jdip
    Commented Aug 18, 2023 at 9:35
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    $\begingroup$ @user68882 You should accept this answer to close it out. Jdip addressed your question in it's entirety. He proved the time-shift property, he didn't use it directly to show that it works. $\endgroup$
    – Envidia
    Commented Aug 19, 2023 at 22:19
  • $\begingroup$ I think the OPs confusion arises from an infinite duration assumption on the LFM. Once the finite nature of the signal is accounted for by the inclusion of the $\Pi$ term, I think the shift will take care of itself. At least I think so. $\endgroup$ Commented Aug 21, 2023 at 23:58
  • $\begingroup$ I don’t think that’s where the confusion comes from, but I agree that the inclusion of $\Pi$ doesn’t change the derivation. $\endgroup$
    – Jdip
    Commented Aug 22, 2023 at 5:40
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Ignoring constant terms, the FT of an infinite duration LFM is given by

$$ e^{j\pi\alpha t^2} \leftrightarrow e^{-j\pi f^2/ \alpha}. \tag{1} $$

Using the delay property of FTs, we know that

$$ e^{j\pi\alpha (t-\tau)^2} = e^{j\pi\alpha t^2} \ast \delta (t-\tau)\leftrightarrow e^{-j\pi f^2/ \alpha} e^{-j2\pi f \tau}. \tag{2} $$

If we expand the phase term of the delayed signal prior to the FT, we get

$$ e^{j\pi\alpha (t-\tau)^2} = e^{j\pi\alpha \tau^2} e^{j\pi\alpha t^2} e^{-j2\pi\alpha t \tau}, \tag{3} $$

which is the original LFM multiplied by a constant and modulated by the linear phase signal $\exp(-j2\pi \alpha t \tau)$, which, as you correctly point out, must shift the spectrum of original LFM by $-\alpha \tau$. This seems to be the cause of some confusion because a delay in time shouldn't lead to a shift in frequency, but for an LFM it does. This is due to the one-to-one relationship between time and frequency for LFMs.

Let us look at what happens when we shift the spectrum of the undelayed LFM as given in $(1)$ by $-\alpha \tau$:

$$ e^{-j\pi f^2/ \alpha} \rightarrow e^{-j\pi (f + \alpha \tau )^2 / \alpha} = e^{-j\pi / \alpha (f^2 + 2 f \alpha \tau + \alpha^2 \tau^2 ) } = e^{-j\pi f^2 / \alpha } e^{-j2\pi f \tau} e^{-j\pi \alpha \tau^2}, \tag{4} $$

which, aside from the constant, is the same as $(2)$. If you went through a careful derivation that accounted for the constant phase terms, you would see that the two are indeed equal.

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