# Fourier transform of a delayed LFM chirp

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 $$(2)$$?

• Are you sure your equation for an LFM chirp is correct? shouldnt $j$ apply to both terms in the exponential? $$Ae^{j \left(\omega_o(t-\tau)+\alpha(t-\tau)^2\right)}$$
– Jdip
Aug 17 at 9:35
• Sorry for the typo, you are right Aug 17 at 10:37
• 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 $\mathrm{rect}((f-2\alpha\tau) / \infty)$? Aug 21 at 23:55
• Hay you might be interested in this answer or this one. Aug 23 at 5:37

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}
• Here the Shift theorem is used, but I was mentioning a direct derivation from the delayed chirp, without the use of that theorem.. Aug 18 at 9:19
• 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.
– Jdip
Aug 18 at 9:35
• @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. Aug 19 at 22:19
• 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. Aug 21 at 23:58
• I don’t think that’s where the confusion comes from, but I agree that the inclusion of $\Pi$ doesn’t change the derivation.
– Jdip
Aug 22 at 5:40

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.

• Hay you might be interested in this answer or this one. Aug 23 at 5:36