# Tag Info

5

In general there is no straightforward analytical solution. As you know, you need to solve $$\left|H(e^{j\omega_c})\right|=\frac{1}{\sqrt{2}}\tag{1}$$ for $\omega_c$, where it is assumed that the maximum filter gain equals $1$. For Butterworth filters, the specified cut-off frequency always equals the $3\textrm{ dB}$ frequency. This is not the case for other ...

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 ...

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

$$1 - e^{-4j\omega} = e^{-2j\omega}(e^{2j\omega} - e^{-2j\omega}) \tag{1}$$ Now, $$\sin(2\omega) = \frac{e^{2j\omega} - e^{-2j\omega}}{2j} \tag{2}$$ Equation 2 is a consequence of Euler's formula. Multiply and divide by $2j$ in (1) and use the identity (2) in equation 1 we have: $$1 - e^{-4j\omega} = 2je^{-2j\omega}\sin(2\omega) \tag{3}$$ Now $j = e^{... 3 You are right that the system$h[n] = 2^n.u[n]$is unstable in absolute summability sense. And absolute summability of impulse response of an LTI system is the required condition for BIBO stability. Meaning Bounded Input bounded output stability. Here the input is bounded and it is only corresponding to 4 frequencies in digital frequency domain$\omega = -\...

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}$$Now you can solve Eqs (1) and (2) to get the frequency response H(f)=Y(f)/X(f). 2 Your math looks correct, thanks for including what you have done. Also for such a block diagram of a linear system, you can rearrange each of the three blocks in any order at the points where the nodes come together (can't break loops) - For example, you can move the derivative to the end without changing the overall result. This may make it even more ... 2 You are essentially seeing a Cascade-Integrator-Comb (CIC) response which is identical to a moving average filter (Aliased Sinc function magnitude response) as seen with CIC filter structures. Consider what is happening in units of phase: You start with a white noise signal which is translated from magnitude directly to units of frequency in the FM ... 2 For an LTI system with impulse response h(t) and transfer function H(s)=\mathcal{L}\{h(t)\}, an input signal x(t)=e^{st} (note: no multiplication with the unit step function u(t)), results in an output$$\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\\&...

2

Frequency Response of Unknown System from Freq Chirp and FFT's My understanding from further discussion with the OP that he wants to specifically use an approach of providing a swept sine wave stimulus and use the FFT of this input and system output response to derive the transfer function. This may be for a system identification problem where the swept ...

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The region of Convergence of the system is $\vert z \vert >2$. Sinusoids are eigen functions of LTI systems over infinite time, since the ROC does not include the unit circle hence system is not stable and hence an infintely running periodic signal will not converge. More info after comments The BIBO property of the system, if the input is unbounded ...

2

No, at least not the way it's written. $U_1(\omega)$ contains two frequency peaks (10Hz and 0Hz) and $U_2(\omega)$ only one at $\omega _2$. $G(\omega)$ can change the relative height of the peaks but the overall but not create new ones. $A_2$ and $U_2(\omega)$ can vary the height and position of the peak but not the number of peaks.

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I've used this method in Octave: function [c] = Hz_at_3dB(b, a, fs, samples) [H,W] = freqz(b,a,samples); magresp = 20*log10(abs(H)); maxresp = max(magresp); [I,~] = find(magresp < maxresp-3.0103,3,'first'); c=(fs/2 * W(I(1)))/pi; EDIT: SECTION 8.4: STANDARD RESPONSES - https://www.analog.com/media/en/training-seminars/design-...

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Look at the equation for the frequency magnitude response of a sinc filter. It will be of the form: $$\frac{\sin(\alpha)}{ \alpha}$$ Look carefully at the $\alpha$ term and determine for what frequencies the $\sin(\alpha)$ numerator will be equal to zero. The magnitude response notches occur at those frequencies.

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HINT: The frequency response has the form $$H(e^{j\omega})=a+be^{-j\omega}+ce^{-2j\omega}+be^{-3j\omega}+ae^{-4j\omega}\tag{1}$$ which can be rewritten as $$H(e^{j\omega})=e^{-2j\omega}\big[ae^{2j\omega}+be^{j\omega}+c+be^{-j\omega}+ae^{-2j\omega}\big]\tag{2}$$ Now note that the term in brackets is purely real-valued. I trust that you can take it from here.

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\begin{align} 1-e^{-j4\omega} &=1-(\cos(4\omega)-j\sin(4\omega))\\ &=1-\cos(4\omega)+j\sin(4\omega)\\ &=1-(1-2\sin^2(2\omega))+j2\sin(2\omega)\cos(2\omega)\\ &=2\sin(2\omega)(\sin(2\omega)+j\cos(2\omega))\\ &=2\sin(2\omega)(j)(\cos(2\omega)-j\sin(2\omega))\\ &=2\sin(2\omega)e^{j\pi /2}e^{-j2\omega} \end{align}

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You need to zero pad the FFT so that it reveals more of the frequency response (increasing the resolution of the plot). I corrected your code (while taking fft, I took 1024 point FFT for windows as well as h_lpf) %% Low Pass Filter Fs=10*10^3; cutoff = 1000; %100 Hz fc = cutoff / Fs ; b = 800/Fs ;%Transition bandwidth of 800 Hz M = 51 ; %Length of Filter n =...

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Since this is not an LTI system, we cannot have $H(\omega) = K\Theta_1(\omega)$, where $K$ is a scalar complex number. But as OP mentioned in the comment he has knowledge of $\Theta_1(\omega)$, we can compute $\Theta_2(\omega)$ and $\Theta_3(\omega)$. As $x^2(t)=x(t)\times x(t)$,  \Theta_2(\omega)= \Theta_1(\omega)*\Theta_1(\omega) =\int_{-\infty}^{+\infty}...

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These are all frequency selective filters. Note the zeros on the unit circle, which correspond to zeros of the frequency response. The filter's passband is determined by range of the angles of the poles, and the stopband is determined by the range of the angles of the zeros. As an example, let's consider PZ-diagram $\#1$: since the stopband is around DC (i....

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These terms refer to the general duality properties of these two domains. It should be obvious that higher frequency components mean faster changes in time and lower rise/fall times. In most natural systems, some of these fundamental properties oppose each other: Shorter rise time means more overshoot and higher settling times and vice versa; lower ...

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The key is to apply 'shift and scale'. So for $x(-2t + 4)$, you would first do $x_1(t) = x(t+4)$, then $x_2(t) = x_1(-2t) = x(-2t + 4)$. So $X_1(f) = e^{-j2\pi f(-4)}X(f)$ = $e^{j8\pi f}X(f)$. Then $X_2(f) = \frac{1}{|-2|}X_1(f/-2)$ = $\frac{1}{2}e^{j-4\pi f}X(f/-2)$ In your answer how did you assume it is symmetric? Also, in your third step, for $x(-t)$...

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When we talk about Doppler, we typically refer to time-varying systems. A popular model in system theory is to assume systems to be LTI (linear time-invariant). This means that beside being linear, their input-output transformation does not change in time, it does not matter when we apply a certain input signal $x(t)$, we'll always get the same $y(t)$. ...

1

Normalizing the filterbanks by their widths is optional and totally up to you (similarly to the warping scale Mel/Bark). Depending on your application, you can start without normalization and see what results you are getting. Personally I prefer to keep it fixed and have one knob less for turning. There are more important parameters to tune, such as warping ...

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 ...

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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 ...

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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)...

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I agree with @user28715 answer. The best method is to apply a filter to your timeseries to get calibrated timeseries. Filter You did not specify which language you are using, but in Matlab I use the designfilt function. https://www.mathworks.com/help/signal/ref/designfilt.html d = designfilt('arbmagfir',...); a = 1; b = d.Coefficients; or you ...

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