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A rectangular window of $N = 4$ samples, at a sampling rate of $32$ Hz would provide about $$ \Delta f = \frac{4\pi}{4} \frac{32}{2\pi} = 16 ~Hz$$ of frequency resolution... On the other hand, the apparent DFT bin frequencies, will be : $$ f_k = [0,8,16,24] $$ Hz. Interpret the last frequency as $24-32 = -8$ Hz due to the sampling theorem.


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Carry the units and it becomes understandable. T = 0.03125 seconds per sample Thus your sampling rate is 1/T = 32 samples per second Each bin corresponds to the cycles per frame. Your frame has four samples. Suppose the bin index is k. k (cycles per frame) * 32 (samples per second) / 4 (samples per frame) = 8k cycles per second = 8k Hz So the ...


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There is no real shortcut for computing the impulse response. Partial fractions is the standard way to do it. However, in the case of a second-order transfer function as in your example, the result can be found in tables of $\mathcal{Z}$-transform pairs. E.g., if you use the last two correspondence of this table, you can pretty quickly write down the result. ...


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I suspect you are aiming at what is known as eigenimage or eigenface analysis. A key step resides in computing the eigenvectors, often performing with preprocessing (e.g. mean subtraction) followed by vectorizing 2D images (aligning pixels in a 1D vector), before concatenating vectors and computing the eignevectors, with fast alternatives.


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HINT: The (zero-state) step response is just the cumulative sum of the impulse response $h[n]$: $$y_{ZS}[n]=u[n]\sum_{k=0}^nh[k]$$ This follows in a straightforward manner from the convolution of a unit step $u[n]$ with the impulse response: $$y_{ZS}[n]=(h\star u)[n]$$


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You're getting yourself into unnecessary trouble by choosing the integration limits exactly at those values of $t$ where you have Dirac impulses. Note that you can choose any integration limit as long as you integrate over one period of the given function. E.g., if you choose some positive $\epsilon$ satisfying $0<\epsilon\le\frac16$ and you integrate ...


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It seems that you've correctly found the two poles: $z_{\infty,0}=2$ and $z_{\infty,1}=-\frac13$. The only other thing you need to know is that the possible ROCs are limited by the pole radii, and that there can be no poles inside an ROC. Note that $z<2$ doesn't make much sense because $z$ is a complex number. If you mean $|z|<2$ then the pole at $z=-\...


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You're overcomplicating things here. There's no need for sines and cosines and squares. Note that $\omega=0$ corresponds to $z=1$, and $\omega=\pi$ corresponds to $z=-1$. From the definition of the $\mathcal{Z}$-transform you should be able to figure out that the DC term of the transfer function $H(z)$ is given by $$H(1)=h_1+h_2+h_3\tag{1}$$ and the value ...


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If you check graph of h(t) which is underdamped oscillations presrnt in 1st and 4th quadrant only. And integration of this h(t) from limits (-infinity to + infinity ) Is finite value. therefore, system is stable


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A real coefficient, minimum phase, FIR filter will have the following property : $$ H(z) = H^*(z^*) = H(\frac{1}{z}) = H^*(\frac{1}{z^*}) $$ which implies that for every zero $z_0$ of $H(z)$ there will be three more zeros at $z_0^*$, $1/z_0$, and $1/z_0^*$; at conjugate, reciprocal, and conjugate-reciprocal locations respectively. Note that the reciprocal ...


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In order for H(z) to be a linear phase filter, it must zeros both on the inside of the unit circle and at the complementary locations (1/z) which are outside the unit circle. Therefore a linear phase circle has no stable causal inverse (since this would necessitate poles outside of the unit circle.) Note that a linear phase filter can be decomposed into ...


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The best answer I could give is just a hint: think about what is the inverse discrete time Fourier transform of $X(e^{j\omega})$. Constants in the frequency domain are what in the time domain? Answering that will lead you to the answer of this question.


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