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Let $x(t) = \cos(2\pi\times15t)+\cos(2\pi\times22t)+\cos(2\pi\times35t)+\cos(2\pi\times42t)$ and $\forall t\in\mathbb{R} :w(t) = 1$. We sample $x(t)$ and $w(t)$ with $F_s = 92 \ \text{Hz}$. So we obtain the discrete-time signals $x[n]$ and $w[n]$. The product $y[n] = x[n]w[n]$ is the windowed version of $x[n]$. Let $N$ be the length of $w[n]$ which means: $$w[n] = \left\{ \begin{array}{ll} 1 & \quad |n| \leq N \\ 0 & \quad |n| > N \end{array} \right.$$ According to the multiplication property of DTFT, we have: $$Y(e^{j\omega}) = \frac{1}{2\pi}\int_{-\pi}^{+\pi}X(e^{j\theta})W(e^{j(\omega - \theta)})d\theta$$Because $x[n]$ is sum of cosine functions, its DTFT $X(e^{j\omega})$ consists of deltas at $\omega_1,\omega_2=\pm2\pi \frac{15}{92}\approx \pm0.326\pi, \ \omega_3,\omega_4\approx\pm0.478\pi, \ \omega_5,\omega_6\approx\pm0.761\pi$ and $\omega_7,\omega_8\approx\pm0.913\pi$. So $Y(e^{j\omega})$ can be calculated: $$Y(e^{j\omega}) =\frac{1}{2}\sum_{k=1}^{8}W(e^{j(\omega-\omega_k)})$$ It's clear that $w[n]$ is the rectangle window and its DTFT is known: $$W(e^{j\omega}) = \frac{\sin\left(\omega(N+\frac{1}{2})\right)}{\sin\left(\frac{\omega}{2}\right)}$$ We can see that the main lobe width for this window is: $$BW=\frac{2\pi}{N + \frac{1}{2}} = \frac{4\pi}{2N+1}=\frac{4\pi}{L}$$ The goal is to find minimum $N$ such that frequencies $\omega_k$ can be resolved. We select this criterion for the resolving two adjacent frequencies $\omega_i$ and $\omega_j$: The main lobe of the $W(e^{j(\omega-\omega_i)})$ and $W(e^{j(\omega-\omega_j)})$ shouldn't interfere with each other (see this for further details).

Obviously, the smallest frequency separation determines the window length. So we should look for $\Delta \omega_{min}$. I think the main constraint here is because of $W(e^{j(\omega-\omega_7)})=W(e^{j(\omega-0.913\pi)})$, since DTFT is $2\pi$-periodic and it repeats itself after $\omega = \pi$. So we have: $$\Delta = \pi - 0.913\pi= 0.087\pi \\ \Delta\ge \frac{BW}{2} \implies L\ge\frac{2\pi}{\Delta}\approx 23$$ I'm not sure whether $\Delta$ is the right minimum frequency separation because $Y(e^{j\omega})$ is sum of DTFTs which are $2\pi$-periodic. This periodicity makes difficult to plot $Y(e^{j\omega})$ and find $\Delta \omega_{min}$.

S.H.W
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  • What's your end goal here? You do the analysis using the DTFT but if you want to eventually implement something on a computer you will have to use the DFT and in this regard the two behave somewhat differently. – Hilmar Mar 03 '23 at 00:02
  • @Hilmar I want to use DTFT not DFT. It's a theoretical question only. – S.H.W Mar 03 '23 at 00:30
  • Well if it's theoretical: $N = 2310 \cdot k, k \in \mathbb{I}$ will give you perfect resolution. :-). – Hilmar Mar 03 '23 at 05:18
  • @Hilmar But that's not the minimum value for $N$ :-). – S.H.W Mar 03 '23 at 09:06
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    Does this answer your question? What happens when N increases in N-point DFT. Note that zero padding the DFT gives you more samples on the DTFT. The resolution BW of a rectangular windowed DTFT is 1 bin – Dan Boschen Mar 03 '23 at 12:56
  • @DanBoschen Thanks. I think my question is something else. I just want to know the minimum value for $N$ such that the main lobes don't interfere with each other. The confusing part for me is the periodicity of DTFT. I mean what's $\Delta \omega_{min}$ here? It's $\pi - 0.913\pi$ or $0.761\pi - 0.478\pi$? – S.H.W Mar 03 '23 at 14:22
  • @S.H.W I am showing in the link how close two tones can be spaced and still be distinguished from each other: the Dirichlet Kernel from the rectangular window convolves with each of the tones, isn't that the main lobe you refer to? In the link I show two tones spaced .02 apart where 1 is the sampling rate, and show how with 100 samples in time we can distinguish the two tones in frequency (the bin spacing of the DFT is 0.01 in this case, and the mainlobe of the DTFT has a null to null BW of 0.02). – Dan Boschen Mar 03 '23 at 18:36

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