# Tag Info

4

Your first suggestion is correct, the derivatives of the local polynomials are being sampled. From the horse's mouth (Savitzky & Golay 1964): Again, if we restrict ourselves to evaluating the function only at the center point of a set of equally spaced observations, then there esist sets of convoluting integers for the first derivative as well. (...

4

Wouldn't it be much easier to just compare the left-hand side and the right-hand side of $$e^{-j2\pi ft_0}=|H(f)|e^{j\phi(f)}\tag{1}$$ to see that $|H(f)|=1$ and $\phi(f)=-2\pi ft_0$?

4

Yes, that's definitely possible. The key would be to find recordings where it's reasonably easy to find an acoustic reference. Great candidates for this would be classical or big band recordings. Let's say you have wax cylinder of Beethoven's 5th, you can try to create a reference by analyzing a dozen or so recent recordings of the same piece. The standard ...

4

The amount of distortion by e.g. saturation within the mechanical recording process can probably not be inferred just from recordings? Actually, amplitude and rate saturation would be fairly easy -- just play them back, and look for the waveform flat-topping (which tells you about amplitude saturation), then look for the slopes never going beyond a certain ...

3

The system is not just a wire, it is in fact time-varying, so it has no frequency response in the conventional sense. The sequence $w[n]$ is of course just a low-pass filtered version of the input signal. The filtered signal is band-limited to half the Nyquist frequency. Now just write the output $y[n]$ in the time domain as the sum of $w[n]$ and the ...

3

The frequency response can exist, even if there are poles in the right half-plane, as long as there are no poles on the $j\omega$-axis. You cannot use the unilateral Laplace transform, instead you must use the bilateral Laplace transform (with the lower integration limit at $-\infty$), or, simply, the Fourier transform. The corresponding time domain ...

3

The sequences $e^{j\omega_0n}$ are eigensequences of discrete-time LTI systems, i.e., the response to such a sequence is the same sequence scaled by a complex constant (the eigenvalue). This can be shown as follows: \begin{align}y[n]&=(h\star x)[n]\\&=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\\&=\sum_{k=-\infty}^{\infty}h[k]e^{j\omega_0(n-k)}\\&=... 2 Polynomials are not integrable, hence their Fourier interpretation is complicated. However, locally, there are a few works. Under least-squares, polynomial interpolation's linearity works well with frequency analysis. The class of Savitsky-Golay filters performs least-squares polynomial interpolation, and yields different derivative orders. As they seemed ... 2 Define a frequency response\tilde{H}(e^{j\omega})=\frac{e^{-j\omega}-|\alpha|}{1-|\alpha|e^{-j\omega}}\tag{1}$$and note that the group delay \tilde{\tau}(\omega) corresponding to (1) is related to the group delay \tau(\omega) of the original frequency response H(e^{j\omega}) by$$\tau(\omega)=\tilde{\tau}(\omega-\theta)\tag{2}$$because (1) ... 2 This is easier to see graphically, on an s-plane, knowing that the phase is from the frequency response of the system and the frequency response is determined when we restrict s to be the j\omega axis. Consider the simple ratio of a single pole and a single zero expressed as$$H(s) = \frac{s-z_1}{s-p_1}$$Given s as any point on the s-plane, and z_1 ... 2 ok let me give you a practical (discrete-time) example. Consider a system with input-output relation of the form$$ y[n] = x[n] + 0.5 x[n-1] + 0.25 x[n-2] $$This input-output relation satisfies linearity and time-invariance properties and hence defines an LTI (linear time invariant) system and therefore it has an impulse response h[n] found to be :$$...

2

If $T$ is a positive real number such that $$x(t) = x(t+T) ~\text{for all}~t \in \mathbb R\tag{1}$$ for continuous-time signals, or if $T$ is a positive integer and $$x[n] = x[n+T]~\text{for all}~n \in \mathbb Z \tag{2}$$ for discrete-time signals, then $x$ is said to be periodic with period $T$ or to have period $T$. A signal $x$ that has period $T$ also ...

2

The easiest way to solve the problem is using the knowledge of eigenfunctions of LTI system and the consequence that an LTI system's response to a sinusoidal input $x(t)=A\cos(\omega_0t+\phi)$ is given by $$y(t)=A\big|H(\omega_0)\big|\cos\big(\omega_0t+\phi+\angle H(\omega_0)\big)\tag{1}$$ where $H(\omega)$ is the system's frequency response. This is ...

2

Apply the eigenfunction property of the LTI system after decomposing the sinusoidal input by Euler identity. The eigenfunction property of the LTI system states that $$x(t) = e^{j \omega_0 t} \implies y(t) = H(\omega_0) e^{j \omega_0 t}$$ where $H(w)$ is the frequency response of the LTI system. Euler identity states that : $$x(t) = \cos(t) = 0.5 \{... 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 ... 1 I guess it's for a homework. 1 - You want to attenuate the 50-Hz component. Zeroes placed at 50 Hz will cancel the 50-Hz component, thus you need to place your zeros at 50 Hz. 2 - You want your 300-Hz signal to be unaffected by the zeros, i.e. you want it to keep the same amplitude and the same phase. If you only added the complex pair of zeroes at 50 Hz, ... 1 Is it just element-wise multiplication of the FFT array with the FRF array, followed by an IFFT No. Filtering in the frequency domain is not trivial since it's equivalent to circular convolution (not linear) and will result in time domain aliasing. Google "overlap-add" or "overlap save" for a description of adequate algorithms To take it one step ... 1 The "Error" would be the actual phase and amplitude relative to the asymptotic plot shown. You computed the gain and phase, so now compare this to the asymptotic lines given by the plot to determine the error. 1 This may be a better question for https://physics.stackexchange.com/ but I'll give it a shot. I assume you show the FFT of the time waveform at the sensor. In order to get the actual frequency response you would need to divide by the FFT of the hammer response. Since the hammer is reasonably smooth, we can eyeball this. It also looks like your measurement ... 1 Some ideas: You could train the model using audio with randomly chosen speaker--mic system impulse responses applied. To get you started, there are microphone impulse response (IR) packs available, intended for music production purposes. Or you could perhaps generate synthetic impulse responses that cover the variability in real-world impulse responses. Or ... 1 Note that in general the Fourier transform of a function is a complex-valued function, so in general it is not only positive or negative. Roughly speaking, the magnitude of the Fourier transform says something about the presence of certain frequencies components in a signal, regardless of the phase (or sign, in the real-valued case). The phase determines ... 1 For an LTI system with impulse response h(t) and transfer function H(s)=\mathcal{L}\{h(t)\}, an input signal x(t)=e^{st} (note: no multiplication with the unit step function u(t)), results in an output$$\begin{align}y(t)&=(x\star h)(t)\\&=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\\&=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}d\tau\\&...

1

As far as I know there is no simple characterization of filters with a monotonic magnitude response. I could only come up with a weak time domain characterization of LTI systems with monotonically decreasing amplitude for positive frequencies (cf. Eq. $(6)$ below). Let $h(t)$ be the real-valued impulse response of an LTI system, and let $H(\omega)=A(\omega)... 1 We don't ignore poles on the contour. As mentioned in a comment, poles are avoided by modifying the contour as shown in the figure below, where a contour appropriate for a pole at$s=0$is shown. Fig. 1: Nyquist contour for a pole at$s=0$(from "Modern Control Engineering" by K. Ogata). The contour moves around the pole along a semi-circle centered at the ... 1 Your understanding is correct. There is no difference between a phase of$\pi$and a phase of$-\pi$. You can always add or subtract integer multiples of$2\pi$to the phase without changing anything because$e^{j2\pi n}=1$for$n\in\mathbb{Z}$. Clearly you have$e^{j\pi}=e^{-j\pi}=-1$. The probable reason why they used opposite signs in the phase plot for ... 1 Your suspicion is correct. You can show this as follows: $$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$ $$X^*(\omega)=\int_{-\infty}^{\infty}x^*(t)e^{j\omega t}dt\tag{2}$$ Now if$x(t)$is purely imaginary we have$x^*(t)=-x(t), and, consequently, $$X^*(\omega)=-\int_{-\infty}^{\infty}x(t)e^{j\omega t}dt=-X(-\omega)\tag{3}$$ 1 This is the classical problem of filter design: You've got a frequency response that you want, how to implement it using a FIR? I'm not going to lay out all the theory here, because it can easily be found by looking for Fourier approximation methods for FIR design or windowing methods, but the idea is easy enough: Take your frequency response. Sample ... 1 Just apply the definitions. \begin{align} H(e^{j\omega}) &= \Re\Big\{H(e^{j\omega})\Big\} + j \Im\Big\{H(e^{j\omega})\Big\} \\ \\ &= \Big| H(e^{j\omega}) \Big| e^{j \arg\{H(e^{j\omega})\}} \\ \\ &= \Big| H(e^{j\omega}) \Big| e^{j \phi(\omega)} \\ \end{align} Where $$\phi(\omega)\triangleq \arg\Big\{ H(e^{j\omega}) \Big\}$$ and\$\...

1

I would suggest the frequency sampling design method, which results in a linear phase FIR filter. You sample the desired magnitude on an equidistant grid, add a linear phase term, and then you compute the impulse response by taking an inverse DFT. The resulting impulse response is usually windowed. I've explained the details in this answer, which also ...

1

So, when you remember how the step response, impulse response and frequency response are related, you'll notice that if you, instead of an actual impulse integral (i.e. a step) use something that is wider, then the frequency response simply gets windowed to lower frequencies. In other words, and to put it as an information-gathering problem: if the ...

Only top voted, non community-wiki answers of a minimum length are eligible