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I'm having trouble understanding how this system is supposed to work. My sketch (all my sketches are for fourier transforms) for \$X_b(jw)\$ are two pyramids centered at 7pi and -7pi with height 1/2. Once it goes through the sampler, I have an infinite number of pyramids centered at odd pi frequencies and height 1. Since K=1, nothing happens with the gain.

What's supposed to happen once it goes through the sample-to-impulse? I thought that a signal going through the sample to impulse will cause the signal to repeat itself. So if I had one pyramid at 0 then the resulting signal through the sample-to-impulse will repeat itself every even pi frequency with height 1 (assuming T = 1).

BUT my \$X_d(jw)\$ sketch already has infinite pyramids so wouldn't this make their heights infinite?

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  • \$\begingroup\$ Isn't a triangular frequency impulse like \$X_a\left(j\omega\right)\$ the same as convolving two rectangular frequency impulses? I recall that the Fourier of a rectangle (ignoring shift) is a \$\operatorname{sinc}\$ function. Given duality, doesn't this imply that \$x_a\left(t\right)\$ must be the product of two \$\operatorname{sinc}\$ functions? So I imagine, in the time domain, two \$\operatorname{sinc}\$ functions then also multiplied by a cosine to get \$x_b\left(t\right)\$. \$\endgroup\$ Aug 9, 2023 at 2:52

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Once it goes through the sampler, I have an infinite number of pyramids centered at odd pi frequencies and height 1. Since K=1, nothing happens with the gain.

This isn't quite correct. The way your book seems to be presenting it, after going through the sampler, you have a mathematical sequence of measurements. These numbers aren't actual voltages on an actual wire occurring at actual points in time.

Therefore you can't use the Fourier transform to obtain a spectrum from them. Depending on how your text presents it, you may be able to use the Discrete Time Fourier Transform (DTFT) or Discrete Fourier Transform (DFT) on the \$x_d[n]\$ discrete-time signal, but it's also possible you haven't learned the DTFT and DFT yet.

However, what's supposed to happen once it goes through the sample-to-impulse?

The sample-to-impulse block is what takes the mathematical list of samples and turns it into an actual time domain voltage waveform. After doing that you can take the waveform's Fourier transform and get your spectrum consisting of an infinitely repeating triangular function.

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  • \$\begingroup\$ Sorry, I meant DFT of \$x_d\$ which is equal 1/2pi times the convolution of \$X_b(jw)\$ and the fourier transform of an impulse train \$P(jw)\$ both evaluated at \$ w = \Omega T\$. The convolution with the impulse train is what gave me an infinite repeating triangular function for the DFT of \$x_d\$ \$\endgroup\$
    – Ray
    Aug 9, 2023 at 13:20
  • \$\begingroup\$ Ok I see what my mistake was. When going through the sample-to-impulse, I ended up convolving the DFT \$x_d\$ with an impulse train when it really should have been a convolution with the CFT of \$x_b\$ with an impulse train. \$\endgroup\$
    – Ray
    Aug 9, 2023 at 13:35
  • \$\begingroup\$ The DFT doesn't produce a function \$X(\omega)\$, it produces another mathematical sequence of complex numbers \$X_n\$ (which happen to be samples of the DTFT at certain frequency values). \$\endgroup\$
    – The Photon
    Aug 9, 2023 at 14:46

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