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## Hot answers tagged transfer-function

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}$-transform uniquely determines the corresponding time-domain sequence. For ... 3 Your two transfer functions are in parallel, i.e they simply add up. So your feedback transfer function is simply$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 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 In short:$(0.937)^2=0.878A bit more detail: \begin{align}(z-re^{j\theta})(z-re^{-j\theta})&=z^2-r\left(e^{j\theta}+e^{-j\theta}\right)z+r^2\\&=z^2-2r\cos(\theta)z+r^2\end{align} 3 This happens frequently if your poles are reasonably close to the unit circle. Consider the following example %% TF2ZP is problematic fs = 44100; % 6th order lowpass, fc = 50Hz, sampled at 44.1kHz [z,p,k] = cheby2(6,80,50*2/fs); % to transfer function [b,a] = zp2tf(z,p,k); % back to zpk [z1,p1,k1] = tf2zp(b,a); display([p p1]); Displaying the poles side ... 3 What you have are not the poles and zeros, but simply the filter coefficients, i.e., the coefficients of the numerator and denominator polynomials. The poles are the roots of the denominator polynomial, and the zeros are the roots of the numerator polynomial. In Matlab they can be found by using the roots command: p = roots(a); z = roots(b); Note that in ... 2 I wasn't familiar with that term in the context of signal processing. (Instead, I've seen the term being used in the context of the Riemann zeta function.) But I've found a document and this book where the term is used in a DSP context. The (obvious) definition is that trivial poles and zeros are the ones at the originz=0$and at infinity$|z|=\infty$. ... 2 The last difference equation is not a linear system due to the addition of the constant$\gamma$, therefore it does not have a transfer function. 2 Considering a discrete time framework, the interpolation operation has two stages; first expand the signal by zero stuffing between the existing samples, and then lowpass filter the expanded signal to get the interpolated samples. The second operation is LTI, hence it's the first stage which is not LTI. The expansion block is shown like $$x[n]\... 2 Your final answer looks good, but the derivation and notation are a bit mixed. Remember the capitalized H() refers to the frequency domain transfer function, so writing H(n) with the time-domain index n doesn't make sense. Also, it is the time-domain impulse response that is a delta function, not the frequency domain transfer function, so H(f)=-\... 2 If the problem description is correct, i.e., if h[n] is zero for n<0 and n>7, and if we assume that h[7]\neq 0 (and why should we assume otherwise?), then the filter has 8 taps and it is a 7^{th} order FIR filter with 7 zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number ... 2 You cannot solve this problem using the Laplace transform. The reason is that the Laplace transform of the input signal doesn't exist. You could use the Fourier transform, but in this case there's an even simpler way to determine the output signal. You need to know one important property of linear time-invariant (LTI) systems: their response to a sinusoidal ... 2 We do not define it in that manner. Note the difference between the time and Laplace domains. Generally, we characterize a system by the response produced when the system is excited by an impulse response \delta(t). Given the output y(t) for such signal we ask which function h(t) did we convolve with the \delta(t) to produce it. Formulation:$$\... 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 $\delta(t)$ ...

1

Assuming that you denote the output by $y(t)$ and the input by $x(t)$, the initial conditions are given by the values of $y(t)$ and its derivatives at a certain time $t_0$ (choose $t_0=0^-$ if you like). A value $x(t_0)$ has nothing to do with an initial condition. For the zero-input response (ZIR) you just set $x(t)=0$ and compute the output caused by ...

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