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7

The task is to filter x(t) when given y(t), where y(t) = x(t) + n(t). Great but first we need to build an appropriate filter. At this point: No. The task is to filter $y[n]$ to achieve $x[n]$. Your filter acts on the available signal $y[n]$ to get a best estimate of $x[n]$ from a noisy observation of it. BTW. discrete-time uses $y[n]$ or at least $y(n)$ ...


5

It's actually quite straightforward: positive powers of $s$ (or, in discrete-time, $z$), correspond to poles at infinity. Negative powers give you zeros at infinity. Let's look at some examples. In continuous time, an ideal differentiator has the transfer function $$H(s)=s\tag{1}$$ Clearly, $\lim_{s\to\infty}H(s)=\infty$, hence you have a pole at infinity ...


5

$$2s+1=2\left(s+\frac12\right)$$ That's all I can say.


4

We know that $\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ has value $0$ for $n < -1$ since $u[n+1]=0$ for $n < -1$. At $n=-1$, $u[n+1]$ jumps to value $1$, but $\sin(\theta (n+1))\bigr|_{n=-1} = \sin(\theta (-1+1))$ has value $0$, and so $\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ just happens to equal $\...


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 ...


3

This happens frequently if your poles are reasonably close to the unit circle. Consider the following example %% TF2ZP is problematic fs = 44100; % 6th order lowpass, fc = 50Hz, sampled at 44.1kHz [z,p,k] = cheby2(6,80,50*2/fs); % to transfer function [b,a] = zp2tf(z,p,k); % back to zpk [z1,p1,k1] = tf2zp(b,a); display([p p1]); Displaying the poles side ...


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 ...


3

It's often easier to design FIR filters for compensating group delay. At the same time they could also compensate the magnitude if necessary. The easiest method is to use a complex least squares method, which boils down to solving a system of linear equations for the filter coefficients. The difficult part is to choose an appropriate bulk delay in the ...


3

In some applications, such as channel equalization, one often uses a training sequence known to the receiver to compute the optimum filter coefficients. During normal operation, the training sequence is replaced by the decisions of the receiver ("decision-directed mode"). In other applications, such as speech denoising, one attempts to estimate the ...


3

Your two transfer functions are in parallel, i.e they simply add up. So your feedback transfer function is simply $G(z) = H_1(z)+H_2(z)$. You want to makes sure that the magnitude of $G(z)$ is smaller than one. and the overall closed loop transfer function is $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1+H_1(z)+H_2(z)}$$


3

$$H(s) = \frac{R C s}{R C s + 1}$$ If $R C s$ is much lower than 1 (i.e., $R C s \ll 1$, in Math), then you can make an approximation: $$H(s) \simeq \frac{R C s}{1}$$ Basically, the effect of the $R C s$ in the denominator becomes insignificant. So, just pick a radian frequency well below $1 / R C$ and compute the magnitude of $H(s)$ at that point -- mark ...


2

Your final answer looks good, but the derivation and notation are a bit mixed. Remember the capitalized $H()$ refers to the frequency domain transfer function, so writing $H(n)$ with the time-domain index $n$ doesn't make sense. Also, it is the time-domain impulse response that is a delta function, not the frequency domain transfer function, so $H(f)=-\...


2

You cannot solve this problem using the Laplace transform. The reason is that the Laplace transform of the input signal doesn't exist. You could use the Fourier transform, but in this case there's an even simpler way to determine the output signal. You need to know one important property of linear time-invariant (LTI) systems: their response to a sinusoidal ...


2

We do not define it in that manner. Note the difference between the time and Laplace domains. Generally, we characterize a system by the response produced when the system is excited by an impulse response $\delta(t)$. Given the output $y(t)$ for such signal we ask which function $h(t)$ did we convolve with the $\delta(t)$ to produce it. Formulation: $$\...


2

Obviously, you might need to use the given frequency somehow to solve the problem. Guessing alone will usually not suffice. Since this is a homework type problem, and since you still need to learn a lot of basics, I'll give you a few hints to get you started. First, the frequency domain behavior can be obtained from a (stable) transfer function by ...


2

You give a transfer function in $z^{-1}$, but for the purposes of making a state-space model, you need a transfer function in $z$ (more on that later). So your $H$ becomes $$H(z) = \frac{b_0 z + b_1}{z^2 + a_1 z + a_2}$$ When you invert that it's improper, which shows that you need to know information one time step into the future. That's not possible (or, ...


2

Another way to see how the forward Euler method approximates a continuous-time system is by considering the "ideal" mapping of the $s$-plane to the $z$-plane (why?): $$z=e^{sT}\tag{1}$$ For frequencies that are much smaller than the sampling frequency (i.e., $|s|T\ll 1$) we can approximate $e^{sT}$ by its first order Taylor series: $$z\approx 1+sT\tag{2}$$...


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 ...


1

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 ...


1

I suppose that you obtain a transfer function of the desired discrete-time system in the form $$H(z)=\frac{b_0+b_1z^{-1}+\ldots +b_Nz^{-N}}{1+a_1z^{-1}+\ldots +a_Nz^{-N}}\tag{1}$$ From that transfer function you compute the coefficients of the second-order sections. Before doing that, you can make sure that the DC gain of $(1)$ equals $1$ (which of course ...


1

Q1: You should ask yourself if $$\frac{z}{3z^2 - 4z + 1}\stackrel{?}{=}\frac{\frac 3 2}{z-1} - \frac{\frac 1 2 }{z-\frac 1 3}$$ really holds. You'll find out that you forgot to scale correctly. Q2: There is no reason to consider $X(z)/z$ instead of $X(z)$ in this case. And concerning your question about the anti-causal sequence, a multiplication by $z^{-...


1

I think your question has nothing to do with "software", so I'll ignore this part, and only discuss the calibration part of it. Then again, your question is very broad and cannot be answered with yes or no: It really depends. In some applications you might want to try to invert as much of the transfer function of your sensor as possible. In other ...


1

It does matter that $a=b=2$, because this gives a relatively nice looking solution. As mentioned in a comment, you should split up the given transfer function $H(s)$. But first, let's introduce a normalized variable $p$: $$p=\frac{s}{\omega_n}\tag{1}$$ Now we can write the given transfer function as $$\hat{H}(p)=\frac{2p^2+2p+1}{p^3+2p^2+2p+1}\tag{2}$$ ...


1

The polar representation of a nonzero complex number in the form $$ z = R e^{j \theta} $$ requires that $R > 0$ ; a positive real number. Setting $R = j\sqrt{\frac{1}{2}} $ violates this. Your transfer function is $$ H(z) = \frac{1}{2} - \frac{1}{2} z^{-2} $$ and you want to find its zeros. Let's find the zeros of the following transfer function ...


1

HINT: If you're looking for the zeros of $H(z)$, try $z_0=1$ or $z_0=-1$. If that works, try to figure out why, and derive it yourself. Then, think about what the magnitude and the phase of those zeros might be. EDIT (in reaction to your comment): If you have a second-order polynomial with real-valued coefficients, then two complex conjugated zeros are not ...


1

As MattL already mentioned, they are the same block diagrams as Direct Form II (canonical) implementation of a third order (on the FIR side) LCCDE. However, the adders in the first block diagram are two input adders, while the the ones in the second block diagram are multiple input adders. Theoretically there is no difference. In a popular DSP text, ...


1

The comments to the question already hint at the solution, but I'll write it out. $\tilde{H}(\omega)$ can converted from a product into a sum using Partial Fraction Expansion: $$\tilde{H}(\omega) = \frac{1}{1 + i \omega \tau_1} \frac{1}{1 + i \omega \tau_2} = \frac{A}{1 + i \omega \tau_1} +\frac{B}{1 + i \omega \tau_2}$$ $$ 1 \omega ^0 + 0 \omega ^1 = A(1 ...


1

The chosen cost function is the mean squared error, i.e., the integral over a squared magnitude of the difference between frequency responses. The function $$E(e^{j\omega})=H(e^{j\omega})-\frac{B(e^{j\omega})}{A(e^{j \omega})}\tag{1}$$ depends on frequency, so you can't minimize it directly, unless you want to minimize it for exactly one frequency $\omega$,...


1

If $x[n]$ is the input of your discrete-time system and $y[n]$ is the output, then the transfer fucntion H(z) is written as: $$H(z)= \frac{Y(z)}{X(z)}$$ where $$X(z)=Z(x[n])\; \;\; \;,\; \;\; \;\; \;Y(z)=Z(y[n])$$ So we get: $$\frac{Y(z)}{X(z)} = \frac{2(z-0.5)(z-0.6)}{z-1}\Rightarrow (z-1)Y(z)=2(z-0.5)(z-0.6)X(z)$$ $$\Rightarrow zY(z)-Y(z)=2z^2X(z)-2,2zX(...


1

$$ x(0^-) \ \triangleq \ \lim_{0<\epsilon \to 0} x(-\epsilon) $$ using the traditional undergraduate one-sided definition of the Laplace Transform, $$ \mathscr{L}\Big\{ x(t) \Big\} \triangleq X(s) = \int_{0^-}^{\infty} x(t) \, e^{-st} \, \mathrm{d}t$$ the purpose of the notation "$0^-$" is to make sure you include all of any dirac delta $\delta(t)$ ...


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