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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 ...


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

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 $\phi(f)=-2\pi ft_0$?


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)}\\&=...


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 ...


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 can describe a frequency response in terms of its real-valued amplitude and its phase: $$H(f)=A(f)e^{j\phi(f)}\tag{1}$$ Note that $A(f)$ is not the magnitude, but a bipolar amplitude function. Equivalently, you can express $H(f)$ in terms of its magnitude and its phase: $$H(f)=M(f)e^{j\tilde{\phi}(f)}\tag{2}$$ Now we have $M(f)=|A(f)|\ge 0$, and, ...


2

With feedback systems such as the one given it's often easy to define an additional signal at the output of the adder. This gives the following equations: $$U(f)= X(f)-H_2(f)Y(f)\tag{1}$$ and $$Y(f)=U(f)H_1(f)\tag{2}$$ Now you can solve Eqs $(1)$ and $(2)$ to get the frequency response $H(f)=Y(f)/X(f)$.


2

Your math looks correct, thanks for including what you have done. Also for such a block diagram of a linear system, you can rearrange each of the three blocks in any order at the points where the nodes come together (can't break loops) - For example, you can move the derivative to the end without changing the overall result. This may make it even more ...


2

You are essentially seeing a Cascade-Integrator-Comb (CIC) response which is identical to a moving average filter (Aliased Sinc function magnitude response) as seen with CIC filter structures. Consider what is happening in units of phase: You start with a white noise signal which is translated from magnitude directly to units of frequency in the FM ...


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

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

LTI means Linear Time-Invariant systems. The system has to satisfy two conditions. (1) Linear and (2) Time-Invariant. (1) Linear means, if the response of the system due to load Px and Py is Rx and Ry respectively, then for the load (Px+Py), the response of the system will be (Rx+Ry). (2) Time-Invariant means, the parameters of the system does not vary with ...


1

First of all, I wouldn't worry too much about the speaker response since it is relatively flat and the microphone has a much bigger roll-off. Since you've captured the frequency response using sweep, why not to skip the whole part of designing the filter that mimics the frequency response and use the original impulse response? I don't know what kind of ...


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

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

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

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

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

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

True that a chirp signal helps to get the FRF, but every time we change the frequency we can't reach the steady state, so this will cause bias in the estimation. As an advice try to use the multisine excitation, they are more suitable for such cases.


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