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5

In general there is no straightforward analytical solution. As you know, you need to solve $$\left|H(e^{j\omega_c})\right|=\frac{1}{\sqrt{2}}\tag{1}$$ for $\omega_c$, where it is assumed that the maximum filter gain equals $1$. For Butterworth filters, the specified cut-off frequency always equals the $3\textrm{ dB}$ frequency. This is not the case for other ...

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.

2

Look at the equation for the frequency magnitude response of a sinc filter. It will be of the form: $$\frac{\sin(\alpha)}{ \alpha}$$ Look carefully at the $\alpha$ term and determine for what frequencies the $\sin(\alpha)$ numerator will be equal to zero. The magnitude response notches occur at those frequencies.

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I've used this method in Octave: function [c] = Hz_at_3dB(b, a, fs, samples) [H,W] = freqz(b,a,samples); magresp = 20*log10(abs(H)); maxresp = max(magresp); [I,~] = find(magresp < maxresp-3.0103,3,'first'); c=(fs/2 * W(I(1)))/pi; EDIT: SECTION 8.4: STANDARD RESPONSES - https://www.analog.com/media/en/training-seminars/design-...

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.

1

Rick's answer is a good frequency-domain answer: the sinc function has zeros, so the filter's frequency-domain response has zeros. It may be easier to do this in the time domain, though. A $1^{st}$ order sinc filter in the frequency domain is a "boxcar" filter in the time domain, with an impulse response of $$h(\tau) = \begin{cases} \frac{1}{T} &... 1 The equation of the filter is given as$$H(z) = 0.2 + 0.8z^{-1}$$This can be equivalently described with positive powers of z as:$$H(z) = \frac{0.2z+0.8}{z}$$The two equations above are mathematically equivalent. The function freqz specifies that the numerator and denominator polynomials are to be entered in decreasing (negative) powers of z, while ... 1 The Fourier Transform of stationary white noise exists: in fact "white" here refers to the characteristics of the Fourier Transform, in that the power spectral density will be constant over the entire spectrum, and also implies that each sample in the time domain is independent of all other samples. The distribution of the magnitude of the samples ... 1 The Laplace transform of the original differential equation is$$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{...

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