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I assume this would be as follows: s*exp(j*2*pi*f*n) Where: n: sample index s: baseband signal, magnitude and phase for each sample f: carrier frequency in cycles/sample Note how the carrier as a discrete signal is given in cycles per sample. This is the normalized frequency and is related to the desired frequency $F$ in Hz by dividing by the ...


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The filter function assumes this form (for your case): $$ \tilde{H}(z) = \frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}} $$ where the $1$ on the bottom is from the assumption that $a_0=1$, and call it $\tilde{H}(z)$ so not to be confused with your $H(z)$. Since you have written the denominator with the signs of $a_1$ and $a_2$ inverted, then you just ...


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I believe this will do: x=a.^((0:40)+(0:40)’)


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Could you do something like: (1/2)*unwrap(2*theta)? edit: More concretely, I think about this. I would think that the numerical example and asserts are sufficient to see if it can be adapted to your situation. It is standard usage of the unwrap() function to replace large jumps with the "phase alias" that cause the smallest jump in phase, applied to a 2-d ...


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It looks like your source us using a pretty unusual convention. Most books or courses use the "+" sign in the denominator and that's the convention Matlab follows. See for example https://ccrma.stanford.edu/~jos/filters/Z_Transform_Difference_Equations.html The reason for that is simple: This way zeros are roots of the numerator polynomial and the poles are ...


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You don't. An ideal low pass FIR filter has infinite length so the requirements of "ideal low pass" and "64 taps" are mutually exclusive. You can approximate an ideal filter, but the best way to do this depends on the specific requirement and trade-offs of your application. This being said, Richard's answer is a really good starting point :-)


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b = fir1(63, 0.25) figure(1) freqz(b,1,256)


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Answer: Your derivation is absolutely correct and You will see 0 for both $\omega = -\pi$ and $\omega = \pi$, if you use fvtool() function of MATLAB instead of fft(). The reason is simple : FFT does not calculate values of $H(e^{j\omega})$ at continuous $\omega \in [-\pi, \pi]$, but it calculates DFT and for that the digital frequency resolution is $\omega = ...


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You can iterate for each channel: data = double(imread('lena.jpg')); data = data/255; % Potentially optional dataFilt = zeros(size(data)); nChan = size(data,3); kernelFilter = ones(11,11)/121; for iChan = 1:nChan dataFilt(:,:,iChan) = filter2(kernelFilter,data(:,:,iChan)); end subplot(1,2,1) imagesc(data) xlabel('Picture') subplot(1,2,2) imagesc(...


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Looks like a simple sign error in the denominator. If you use a = [1 -1.866685 .91058], you get something that looks notch-like at around 380 Hz. The Form is Direct Form II This is numerically the worst. Shouldn't matter here, but it's good practice to either stick with Direct Form I or Transposed Form II.


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Phase modulation is demodulated using phase discriminators otherwise called phase detectors- anything that can translate a change in phase to a change in magnitude. The most common for real functions is a simple multiplier followed by a low pass filter as given by the product of two sinusoids in quadrature: From the relationship $$\sin(\alpha)\cos(\beta) =...


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Yes, it changes the sampling rate. Just like with audio, 1 kHz tone is 1 kHz tone after resampling from 48kHz to 96kHz. So the image has same content, but with more pixels.


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There is no best rule, it depends on what we know about the signal. Energy Detector: The energy Detector makes a decision on presence or absence of a signal based on sum of squared samples. This rule comes from the fact that the signal to be detected is inherently assumed to be random following a wide sense stationary process with a known PDF, the ...


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Yes you are correct that you are indeed seeing the equivalent of what would be the 15th harmonic of the fundamental frequency of the equivalent Fourier Series Expansion of the resulting equivalent continuous time waveform that was truncated in time to 40 samples (2 ms). According to the Fourier Series Expansion, $T$ would be 2 ms and the fundamental ...


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Natural images can seem complicated. However, the set of humanly interpretable images is relatively limited, with respect to all possible images ($24^{x\textrm{millions of pixels}}$). There is a belief that their structure can be better summarized. Indeed, lossless compaction can achieve 2-3 factor ratios, and lossy compression about 8-16 with little ...


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Relative to the RMS level of a full-scale sine wave, so the calculation is: value_dBFS = 20*log10(rms(signal) * sqrt(2)) = 20*log10(rms(signal)) + 3.0103 Need some clarity about - > value_dBFS = 20*log10(rms(signal) * sqrt(2)) why in sin wave multiplying by sqrt(2). Please share your valuable information.


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This question is old, but for the benefit of the next person stumbling across it: You have to multiply transfer functions in order to "add" them in the bode plot.


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