7

You can apply a so-called all-pass transformation to a discrete-time low-pass prototype filter in order to convert it to other standard filters (such as high-pass, band-pass, and band-stop). This is accomplished by transforming the complex variable $z$ in the transfer function of the prototype filter by a function $G(z)$ which satisfies $|G(e^{j\omega})|=1$, ...


5

Computing your DFT with a 0.5 Hz spacing is equivalent to appending your signal with 1000 zeros and then computing a 2000 point DFT. It should be clear that this adds no real information to the signal. Yet it can sometimes be useful. What really happens by this zero padding is that you're seeing an interpolated version of your spectrum. This means that ...


5

This appears functionally identical to the traditional interpolation approach of zero-insert and filtering (and in the form of a polyphase interpolator would be identical in processing as detailed even further below), in this case the OP's filter is a 5 tap filter with one coefficient zeroed: $$coeff = [factor2, 1, factor1, 0, -factor2]$$ Below is the most ...


4

If your signal is really as simple as $$x(t)=A\sin(\omega_0t)\tag{1}$$ with known $\omega_0$, and you have observations $y(t_i)$, which are noisy samples of $x(t)$ at known time instances $t_i$, then a simple solution would be the least squares estimate $$\hat{A}=\frac{\displaystyle\sum_iy(t_i)}{\displaystyle\sum_i\sin(\omega_0t_i)}\tag{2}$$


4

Assuming you meant to produce something similar to the green line: What about $$\text{output}[n] = \max\{\text{input}[n-k], \text{input}[n-k+1], \ldots ,\text{input}[n]\}$$ i.e. you just find the maximum along a sliding window over the last $k$ input values?


3

Cool facts about the Gaussian surface: It is a rotation: $$ G(x,y) = \frac{1}{2\pi \alpha}e^{-\frac{x^{2}+y^{2}}{2\alpha}} = \frac{1}{2\pi \alpha}e^{-\frac{r^{2}}{2\alpha}} = G(r) $$ where $ r = \sqrt{x^2 + y^2} $ It is separable: $$ G(x,y) = \frac{1}{2\pi \alpha}e^{-\frac{x^{2}+y^{2}}{2\alpha}} = \frac{1}{2\pi \alpha} e^{-\frac{x^2}{2\alpha} } e^{-\frac{y^2}...


3

The way I understand the problem is each sample of the output is a linear combination of the samples of the input. Hence it is modeled by: $$ \boldsymbol{y} = H \boldsymbol{x} $$ Where the $ i $ -th row of $ H $ is basically the instantaneous kernel of the $ i $ -th sample of $ \boldsymbol{y} $. The problem above is highly ill poised. In the classic ...


3

There is a very important point that is being glossed over in this question (which follows how this topic is conventinally taught) which is: The DFT does not care what your sampling rate is. Ultimately, every DFT calculation boils down to these parameters using conventional naming: $$ \begin{aligned} N &= SamplesPerFrame = \frac{Samples}{Frame}\\ k &...


2

@Henrique Luna. Please forgive me. In Part (b) of the problem the words "sequence time" should "time sequence". Sorry for the confusion! Years ago when I created that Part (b) question I was thinking about the answer to the question; "What is the time duration of an N-sample time-domain sequence?" Back then I believed the time ...


2

Yes and no. In principle, you can use the peak of your correlation function. However, it is not at 10000. The correlation function is symmetric around 0, so your peak is actually much closer to zero than you think. This is one of the reasons why xcorr returns two paramters, one for the lags at which the function is calculated. The correct way of plotting the ...


2

For a single pure tone across the frame, the frequencies below 1 cycle per frame (the DFT doesn't care about the sampling rate) represent fractions of a single cycle across the frame. Theoretically, they can be measured just as accurately as a tone with with more cycles across the frame. As long as there is just a single pure tone, the frequency (then the ...


2

Systematically, you could just solve the problem by finding a solution $A$ that satisfies the following two equations: $$(A-1)^2=A-1\\A^2=A$$


2

HINT: If you rewrite $y[n]$ as $$y[n]=(0.8)^3(0.8)^{n-1}u[n-1]\tag{1}$$ does it become easier to find the $\mathcal{Z}$-transform?


2

Note that the given Gaussian achieves its maximum at $x=y=0$. So that value corresponds to the center of the matrix. The corner values are given by $G(2,2)$. Furthermore, the values are quantized. You can try to estimate the chosen value of $\alpha$ from the given values. EDIT: If you assume $\alpha=1$ and you evaluate the 2D-Gaussian, multiply it by $273$ ...


2

$u[n]$ is the unit step function with the definition $$ u[n] = \begin{cases}{ 1 ~~~, ~~~n \geq 0 \\ 0 ~~~,~~~ n < 0 } \end{cases} $$ Then you will have for $u[2k-3]$ as $$ u[2k-3] = \begin{cases}{ 1 ~~~, ~~~ 2k-3 \geq 0 \\ 0 ~~~,~~~ 2k-3 < 0 } \end{cases} $$ $$ u[2k-3] = \begin{cases}{ 1 ~~~, ~~~ k \geq 2 \\ 0 ~~~,~~~ k< 2 } \end{cases} $$ I think ...


1

Build a basis set with your frequency and match your signal. It is straightforward linear algebra: $C$ is portion of cosine $S$ is portion of the sine $U$ is a vector of ones (DC) $$ X = a C + b S + c U $$ $$ X \cdot C = a (C \cdot C) + b (S \cdot C) + c (U \cdot C) $$ $$ X \cdot S = a (C \cdot S) + b (S \cdot S) + c (U \cdot S) $$ $$ X \cdot U = a (C \cdot ...


1

Dealing with indices, shifts ($-3$) and scaling ($2k$) can be confusing. Most functions used in such exercices are made of combinations of basic signal functions like discrete $\delta$ Diracs, unit steps or discrete Heaviside, etc. Hence, in such exercices, they are often piece-wise constant or piece-wise linear. So you can determine the resulting signal ...


1

In EE, waveform mixing is used for wave modulation. It is implemented via mixer circuits whose operation principle can be loosely described as multiplication of currents. It does not apply to your scenario of SPL data acquisition and processing: you have no waves that modulate one another. Your sound sources being independent and also removed at distances ...


1

If your routine works correctly for converting positive numbers, you could just use it to convert a negative number by first adding the value of the sign bit (MSB) to the negative number (which must make it positive if it is not out of range), then convert the resulting positive number, and finally add the sign bit. Example: convert the number $-0.7$ to Q3.9 ...


1

Error analysis: Suppose you get the frequency wrong by a little bit. What is the impact? Every other point is the original signal, so no error introduced there. Therefore, the errors will occur at the newly inserted points. Define the signal being upsampled as: $$ y[n] = A \cos( \theta n + \rho ) $$ This relation still holds: $$ y[n+1] - y[n-1] = - 2 A \...


1

The problem is asking you to find an analytical expression given $x_1[n]$ and $x_2[n]$. There is no need to compute anything, let alone code it up! The problem wants you to identify that given the discrete-time domain functions, you use properties of the DTFT to write $X_2(e^{j\omega})$ in terms of $X_1(e^{j\omega})$. You've already identified that $x_2[n]$ ...


1

As I mentioned in my comment, I think there is a typographical error in the equation. I think it is supposed to be \begin{equation} y[k] = \sum_{i=1}^{M}a_iy[k-i] + \sum_{j=1}^{N}b_jx[k-j]. \end{equation} This is an example of a linear constant-coefficient difference equation. These are introduced very early in most textbooks about digital signal processing. ...


1

As an answer probably require to have more details on the look-up table (smoothed and regularity of the kernels), here is a couple of recent papers, including a review: Satellite image restoration in the context of a spatially varying point spread function, 2010 Efficient shift-variant image restoration using deformable filtering, 2012 Fast Approximations ...


1

I was going to tell a joke about the appendages of an absolutely normal cat, but it has a really fat tail. Source


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