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The Laplace transform of the original differential equation is $$Cs V_o(s) = V_i(s) \left( \frac{1}{R_1} \right) - V_o(s) \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$ $$R_1 R_2 C s V_o(s) = R_2 V_i(s) - (R_1 + R_2) V_o(s).$$ In the Laplace domain, the ratio $V_o(s)/V_i(s)$ is the transfer function for the circuit, $$H(s) = \frac{ V_o(s) }{ V_i(s) } = \frac{... 2 HINT: The frequency response has the form$$H(e^{j\omega})=a+be^{-j\omega}+ce^{-2j\omega}+be^{-3j\omega}+ae^{-4j\omega}\tag{1}$$which can be rewritten as$$H(e^{j\omega})=e^{-2j\omega}\big[ae^{2j\omega}+be^{j\omega}+c+be^{-j\omega}+ae^{-2j\omega}\big]\tag{2}$$Now note that the term in brackets is purely real-valued. I trust that you can take it from here. 0 Here you go: b = [-0.0625, 0.25, 0.625, 0.25, 0, 0, 0, 0, -0.0625]; a = 1; % **** Plot impulse response ***** [ImpResp,T] = impz(b,a,31); figure(1), clf plot(T,real(ImpResp),':bs','MarkerFaceColor','b','markersize',4) title('Filter Impulse Response'), grid on, zoom on % **** Plot poles & zeros ***** [Z,P,Q] = tf2zpl(b,a); % Calc poles and ... 0 Two hints: The standard way of writing a transfer function in the \mathcal Z domain is a rational function in z^{-1}, not in z. Try this Use tf2zp() instead of residuez() 1 It looks like a Homework problem so I will only provide small hints: Check from the block diagram implementation of the system whether the current output sample y[n] depends on only current input sample x[n] and past input & output samples x[n-n_0] and y[n-m_0]. If so, the system is causal, else it will be non-causal. You can use the ... 1 I suspect that the reason for the difference is the number of samples used in the plot, since the extreme points are likely going to infinity (or very large numbers) such that if less points are used, those large values are simply missed in the computation. If the OP uses the same number of points in each, then the same result should be achieved- and for all ... 0 In general the filter you want to implement is an infinite impulse response (IIR) filter, unless all poles are at the origin of the complex plane (assuming causality), in which case it is a finite impulse response (FIR) filter. In the general (IIR) case, your suggested method will not result in an exact implementation of the filter. There are two reasons for ... 1 Use fvtool and freqz in Matlab. b=[0.5b 0 1]; a=[2 0 b]; freqz(b,a); 3 The given transfer function has a specific form. With the denominator polynomial A(z)=2+bz^{-2}, you can write the transfer function as$$H(z)=\frac{z^{-2}}{2}\frac{A\left(\frac{1}{z}\right)}{A(z)}\tag{1}$$Since for z=e^{j\omega} (i.e., on the unit circle, where we evaluate the frequency response), we have |A(z)|=|A(z^{-1})|, the magnitude of H(e^{... 1 I am not sure how one would avoid complex math or why one would want to given the simplification complex math provides (an oxymoron due to poor naming or perhaps to scare those that don't know yet how easy it is away) but perhaps the graphical explanation can help provide further intuitive insight. First some necessary background: z is the domain of all ... 0 You can use the lsim command, it works for discrete and continuous systems. https://www.mathworks.com/help/control/ref/lsim.html In your case, the 'sys' variable should replaced by tf(num,den) 0 One approach could be to get a finite number of values of the impulse response of your system using the 'impulse' function, and then convolve this with your signal (taking care of the actual sampling frequency you are using). % sampling time of 1ms, 20s of time signal T = 0.001; t = 0:T:20; % low pass filter with cutoff frequency at sine frequency wo = 1; ... 2 No, at least not the way it's written. U_1(\omega) contains two frequency peaks (10Hz and 0Hz) and U_2(\omega) only one at \omega _2. G(\omega) can change the relative height of the peaks but the overall but not create new ones. A_2 and U_2(\omega) can vary the height and position of the peak but not the number of peaks. 1 With the help of the two answers above, I think I finally understood what the paper was all about. Let Z_n = \sum_{i=1}^n X_i where X_i \sim \text{Bernoulli}(p). The sum of n Bernoulli random variables can be found by convolution in the time domain. Then Z_n is a binomial variable. In the document, the authors mistakenly wrote B_2 = \frac{1}{4}\... 2 Explicitly making the imaginary part of your FFT coefficients 0 will change your magnitude response. It is very clear if you consider this:$$ X_k = a_k + jb_k\\ |X_k| = \sqrt{a_k^2 + b_k^2}  Now, you're explicitly setting new coefficients to be $X'_k = a_k$, and hence $|X'_k| = |a_k|$. Taking the IFFT of your new frequency response will obviously not be ...