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You are seeing the difference between "wide band FM" and "narrow band FM" in your modulated waveform. Observe your modulation index for each case and then review the sideband levels versus mod index using Bessel functions. (The modulation index is the ratio of the frequency deviation over the frequency modulation). The first case is wide-band FM as your ...

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Depending on the context, the use of the complex form could be for mathematical convenience or for a no-kidding need for both real and imaginary parts. When you factor the expression, you get $$u(t) = e^{{\eta}t^2}e^{j{\beta}t^2}$$ Where the first exponential is a generic magnitude envelope, in this case Gaussian. The second exponential is the chirp itself ...

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$i$ is the symbol for $\sqrt{-1}$ There is a very important formula called Euler's Equation. $$e^{i\theta}=\cos(\theta) + i \sin(\theta) = (e^i)^\theta$$ "$e^i$" is a point on the unit circle one radian along the circumference. Any point on the unit circle raised to a power will stay on the unit circle and its distance along the circumference ...

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If you have a signal $$g(t)=\cos(2\pi \hat{f}(t)t)\tag{1}$$ then the function $\hat{f}(t)$ is not the instantaneous frequency of $g(t)$ (unless $\hat{f}(t)$ is constant). If you want an instantaneous frequency $f(t)$, then the equation $$\frac{\phi'(t)}{2\pi}=f(t)\tag{2}$$ must be satisfied, where $\phi(t)$ is the phase of the signal $g(t)$. So in order ...

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Instead of phase delay $td(f)$, use group delay $\tau(f)$: $$\tau(f) = -\frac{d}{d\omega}\phi(f),\tag{1}$$ calculated as the negative of the derivative of the phase $\phi(f)$ with respect to the angular frequency $\omega$, which is defined by: $$\omega = 2\pi f\quad\Leftrightarrow\quad f = \frac{\omega}{2\pi}.\tag{2}$$ Using your phase shift $\phi(f)$: $$\... 2 The book is not wrong, but it does present the concepts on LFM in a clunky manner and can be misleading. The book presents the analytical expression for the LFM spectrum, which is an approximation. It also plays with the plot views and most likely unwraps the phase angles, which is usually required to see the phases you expect. Usually when you're ... 2 You still only get one range and one velocity line per radar observation. You can, however, "scan" a region (that's why airport towers and ships have these rotating antennas) and get another axis of information. The same can be achieved using digital beamforming, which, however, requires multiple signal chains, and specific antenna arrays. For a two-... 1 It is easy to calculate IDFT using DFT, by applying the property:$$\operatorname{IDFT}(x) = \frac{1}{N}\big(\operatorname{DFT}(x^*)\big)^*,\tag{1} where $^*$ denotes elementwise complex conjugation. You can find this equation in a more verbose form in Eq. 13-55 of Richard G. Lyons "Understanding Digital Signal Processing" 3rd edition. Probably ...

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With pulse compression, the SNR improves by a factor equal to the time-bandwidth product (synonymous with pulse compression gain). In the equation you listed, $\tau$ can be interpreted as the uncompressed pulse width, in which case that you'd need to add a factor for the pulse compression gain.

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No you don't need to zero-pad each chirp. MATLAB's cspd function divide the signal into segments of length nfft. Would you want is to zero-pad each segment. Luckily for you, the cspd function can do it for you. Take a look here: If $\tt nfft$ is greater than the signal length, the data is zero-padded. If $\tt nfft$ is less than the signal length, the ...

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The Python implementation looks almost exactly like the definitions in Wikipedia. One will need to define functions in order to make it work, like this: # necessary imports: import numpy as np # start by defining a function that returns a sine wave with time-dependent frequency # f_func is a function f(t, f0, k) def chirp(t, f_func, f0, k): return np....

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You need to resample the signal to simulate the Doppler induced dilation. The resampling factor is I=Td/Ts where Td = Duration of the signal after dilation Ts = Actual duration of the transmitted signal In Matlab you can use the function resample(), but you need to find the resampling parameters P and Q from I. Also, I=1+v/c for signal expansion and I=1-v/...

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I assume by "shift" you mean "shift in time". No, the first correlation peak should not be shifted in time, you should just get multiple peaks, assuming the time difference of arrival is not much shorter than the chirp length. If it is shorter then you will get "blurring" of the cross-correlation peak. Having said that, there are a couple of other ...

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Assuming they all have the same $\omega$ and $\alpha$, meaning they are all following the same frequency ramp, then you can multiply by a ramp in the opposite direction of a starting guess while measuring the derivative of the phase of each carrier (a simple approach to see this would be to observe the result with an Short-Time Fourier Transform). Point is ...

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As an alternative to @hotpaw2 ‘s answer. The frequency domain resolution $df$ equals $1/T$, with $T = NFFT / f_s$ the length of your FFT block in seconds. So, in order to increase the resolution between $f_0$ and $f_1$, you could also increase the length of your signal.

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Zero-padding before the FFT (which is equivalent to Sinc interpolation after the FFT) will give you a higher resolution plot. Symmetric zero-padding (centered or both edges) will give you similar FFT phase results to the original FFT.

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