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1

Real-valued and causal FIR filters have a (generalized) linear phase response if and only if their coefficients satisfy either $$h[n]=h[N-1-n]\tag{1}$$ or $$h[n]=-h[N-1-n]\tag{2}$$ where $N$ is the filter length (number of taps). There are four types of linear phase FIR filters, depending on the type of symmetry (even as in $(1)$ or odd as in $(2)$), ...


4

It's true that the algebraic expression for the $\mathcal{Z}$-transform is generally not sufficient for computing the corresponding time-domain sequence. The additional information we need is the region of convergence (ROC). The ROC together with the expression for the $\mathcal{Z}$-transform uniquely determines the corresponding time-domain sequence. For ...


1

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: $$y_{ZS}[n]=(h\star u)[n]$$


1

You may know that one important property of linear time-invariant (LTI) systems is that the complex exponential $e^{j\omega_0}$ is an eigenfunction, and the corresponding eigenvalue is given by the system's frequency response evaluated at $\omega_0$. So the response to an input signal $$x[n]=e^{jn\omega_0}$$ is given by $$y[n]=H(e^{j\omega_0})e^{jn\omega_0}$...


1

$$ H(s) = {K*\frac{s-z}{s-p}}\\ \\ $$ when |z| < |p|, you have a lead compensator. It adds phase between a certain band, and can help you improve your phase margin. It can also increase your bandwidth. You typically use it to improve your transient as you mentionned. You can think of it as a PD controller cascaded with a low-pass filter. The gain K is ...


0

The cut-off frequency is whatever you define it to be. One standard definition would be the frequency $\omega_c$ at which the attenuation is $3$dB, i.e., $\big|H(\omega_c)\big|=\frac{1}{\sqrt{2}}$. As far as we know there is no analytical formula for the exact computation of the $3$-dB cut-off frequency of a moving average filter. More than you ever might ...


3

What you have are not the poles and zeros, but simply the filter coefficients, i.e., the coefficients of the numerator and denominator polynomials. The poles are the roots of the denominator polynomial, and the zeros are the roots of the numerator polynomial. In Matlab they can be found by using the roots command: p = roots(a); z = roots(b); Note that in ...


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I don't think that using a Dirac impulse really helps to see what's going on in this problem. Using the notation of the given solution we have $$\mathcal{F}\left\{g^2(t)\right\}=A(\omega)\;\Longrightarrow\;A(0)=E_g$$ and $$Y(\omega)=A(\omega)H(\omega)$$ where $H(\omega)$ is the frequency response of the ideal low pass filter. The output signal is given ...


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One can see that the given expression can be decomposed as $$ \begin{align} H(j\omega) &= \frac{1+0.5 e^{-j\omega}}{1-1.8 \cos(\frac{\pi}{16}) e^{-j\omega}+0.81 e^{-j2\omega}} \\ &= \frac{A }{1-0.9 e^{j\frac{\pi}{16}} e^{-j\omega}} + \frac{B}{1-0.9 e^{-j\frac{\pi}{16}} e^{-j\omega}} \end{align}$$ where $A = B^* = 0.5 - j 3.9375 = 8/16 - j 63/16 $. ...


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