10

Robert's Cross is a little tricky because it's not an odd size (2x2 rather than 3x3 or 5x5). I've done it using numpy+scipy using a padded 3x3 convolution mask. import sys import numpy as np from scipy import ndimage import Image roberts_cross_v = np.array( [[ 0, 0, 0 ], [ 0, 1, 0 ], [ 0, 0,-1 ]] ) ...


10

A random process is a collection of random variables, one random variable for each time instant. It is best to write the random process as $$\{X(t)\colon -\infty < t < \infty\} \tag{1}$$ where the $\{$ and $\}$ indicate that a set (or collection) of objects is being defined, and the interior says that a typical member of this set is denoted by $X(t)$ ...


8

For simplicity, let's talk about one-dimensional convolutions only. You need to understand that the result of a convolution is a function and the alleged convolution integral is actually an infinitude of integrals, one for each time instant. By this I mean that to compute the output $y(t)$ of a system with impulse response $h(t)$ to an input signal at ...


8

Here is the first one: By definition, $\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,-2y)e^{-j2\pi (ux+vy)}dxdy$ let $\tau = -2y$ and conversely $y=\frac{-\tau}{2}$ $\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,\tau)e^{-j2\pi (ux-\frac{v\tau}{2})}dxd(-\frac{\tau}{2})$ $\mathfrak{F}\{ f(x,-2y)\} = -\frac{1}{2}\iint_{- \infty}^{\...


8

Why is the fourier transform a special case of the laplace transform? The Laplace transform produces a 2D surface of complex values, while the Fourier transform produces a 1D line of complex values. The Fourier transform is what you get when you slice the Laplace transform along the jω axis. For instance, a simple lowpass filter $H(s)=\frac{1}{s+1}$ has a ...


8

Abhinav Jain, Welcome to DSP Community. I build for you a proper testing of the run time comparison. Few tips about timing in MATLAB: Never time in a script. Always call a function to do the heavy lifting. When you run something from script it runs in the global scope which means MATLAB can't optimize it as well as it could if it was in a function. When ...


7

A causal system does not need to know the future in order to compute its output. A memoryless system computes the output only from the current input. A memoryless system is always causal (as it doesn't depend on future input values), but a causal system doesn't need to be memoryless (because it may depend on past input or output values). The system $$y[n]=x[...


6

Here let me show you a simple procedure very similar to pole zero placement which will be helpful for your notch filter design. First, lets analyse the frequency response of a single zero and let $$ H(z) = 1 - b z^{-1} $$ be a first order system with a single zero at $z = b$ where $b$ is a complex constant with a radius $r$ and phase angle $\phi$ radians; ...


6

Note that option (b) is not correct, and that it is also not equal to what you came up with. Option (b) is just the multiplication of $x(t)$ and $y'(t)$, not the convolution. Your solution and option (c) are both correct, assuming that all derivatives exist and that the convolution integrals converge, because with that assumption the following holds: $$\...


5

You are mostly correct. The equation should be $w(t) = sin(2\pi\frac{1}{20}t)$. The variable in the sin should be $t$, not $f$. The frequency is determined by the $\frac{1}{20}$, not by the variable. Other than that you are correct. The period is 20s, which makes it a $\frac{1}{20}$ Hz sin wave.


5

While this is by your admission homework (and fairly basic), I'll bite. Recall the definition of the DTFT: $$ X(\omega) = \sum_{n=0}^{\infty}x[n] e^{-j\omega n} $$ And recall the definition of the frequency response $H(\omega)$: $$ H(\omega) = \frac{Y(\omega)}{X(\omega)} $$ where $x[n]$ is the input to the system and $y[n]$ is its output. Combine these ...


5

This to me looks exactly like the two-dimensional Haar wavelet transform. All four quadrants are a part of the transform, even though different operations have been performed on them. Upper left quadrant is a result of low-pass filtering with a 2x2 two-dimensional rect function (Haar wavelet) and decimation be a factor of 2, upper right is the output of a ...


5

Even though I realize that this is a very late response, I will nevertheless try to answer this question because I find it instructive and also because the number of upvotes suggests that this question is of general interest to the community. As already suggested in the question, let's define two signals $x(t)$ and $w(t)$ as $$x(t)=e^{-kt}u(t),\quad k>0\\...


5

However what is conceptually wrong with using say an infinite sum of time shifted rect functions to represent it? There is nothing conceptually wrong with it. Fourier transforms decompose a signal into a sum of complex sinusoids, but you can also decompose a signal into many other things, which may be more useful in certain applications. The Haar wavelet ...


5

You can achieve this result by using two combs filters : https://en.wikipedia.org/wiki/Comb_filter Put simply, the comb filter consists of adding a delayed version of the signal to itself, causing destructive or constructive interference. For instance, with $K = 20$ and a negative gain value after the delay line, you can significantly decrease or suppress ...


5

Two principles here: When dealing with a differential equation, you define intermediate state variables so everything is in terms of first derivatives. This system is nonlinear, so the state-space equations won't be in terms of matrices. Applying these principles, we define a state vector: $$ \mathbf x = [x_1, x_2]^T, $$ where: $$ x_1 = y \\ x_2 = \dot y $$...


4

If $g(x)= f(N-1-x)$ then $$\begin{aligned} (\operatorname{DCT}\:g)(u) =& c(u)\cdot\sum_{x=0}^{N-1}g(x)\cdot\cos\bigl((2x+1)u\cdot\tfrac\pi{2N}\bigr) \\=& c(u)\cdot\sum_{x=0}^{N-1}f(N-1-x) \cdot\cos\bigl((2N-2N+1+2x)u\cdot\tfrac\pi{2N}\bigr) \\=& c(u)\cdot\sum_{(N-1-x)=(N-1)}^{0}f(N-1-x) \cdot\cos\Bigl(\bigl(2N-1-2(...


4

The main reason for that is that series of cosines and sines form an orthogonal basis. Then, you can use it to represent it in other "space" (frequency "space", for example). Other things, just to understand other things related to Fourier series and Tranform: A sine or cosine is just 2 delta functions in frequency representation (Fourier Transform). A ...


4

I'll go through it in the $z$-domain. First, we find the transfer function $H_1(z) = \frac{V(z)}{X(z)}$. As you noted, in the time domain, $x[n]$ and $v[n]$ are related as follows: $$ v[n] = x[n] + g * v[n - M] $$ Take the $z$-transform of the above and you get: $$ V(z) = X(z) + z^{-M}G(z) V(z) $$ taking advantage of the convolution property, which ...


4

This doesn't require any complicated computations! Look up some tables of identities about the Fourier Series. Which operation would you have to apply to $f$ so that its Fourier series would become $\sum_{-\infty}^{\infty}(c_k)^2 e^{2\pi ikx}$? For example, we know that derivating in the time domain is equivalent to multiplying the Fourier series ...


4

You're probably best off asking your professor for clarification. As you reasoned, it sounds like you're being asked to multiply the image in the frequency domain by either a rectangular or circular mask that eliminates all frequencies outside the mask. This is not typically how you would apply a filter, however, as "ideal" frequency-domain filter masks have ...


4

Polyphonic pitch estimation is still a difficult problem. Search on the MIREX forum for current research papers. Simplistic spectrum based approaches usually require component pitches with extremely simple timbres (nearly pure unmodulated sine waves, etc.) rarely found in real-life polyphonic music. More typical polyphonic music can contain a dense set of ...


4

The issue is resolved: the butter() function automatically does bilinear transform, you just need to specify the frequencies normalised to the Nyqyist freq.


4

I don't feel like it's a very well-posed question, as multi-channel signals are sometimes represented as vector-valued, which would make them multidimensional. However, I assume the answer that is being fished for is: This is a one-dimensional, two-channel setup. The two channels are the child's height and weight. Each signal is one-dimensional, measured ...


4

There are a few things going on here. First, one of the fundamental properties of the Fourier transform is that it conserves energy, i.e. the energy of the time domain is exactly equal to the energy of its Fourier transform. What is the power of a sinusoid? Assuming we restrict the calculation to integer multiples of cycles it is as follows (I suggest ...


4

If you have an understanding of Fourier transforms then you probably already have a conceptual model of transforming signals into the frequency domain. The Laplace transform provides an alternative frequency domain representation of the signal - usually referred to as the "S domain" to differentiate it from other frequency domain transforms (such as the Z ...


4

Hint: define a bound for $|x(t)|$, i.e., $|x(t)|\le A$; now try to find a positive number $B$ such that $|y(t)|\le B$ for any $|x(t)|\le A$ (that's simply the definition of BIBO stability). For the system $y(t)=1/x(t)$ it should be easy to show that the above cannot be satisfied (because $|x(t)|$ can get arbitrarily small).


4

Your work is correct. First just think about how causal digital filters produce an output as a function of the current and previous inputs. Right? Now think about the case where a 'filter' only produces an output as a function of the current input (i.e. not influenced by the previous inputs). We don't typically classify these as filters, instead we ...


4

In easy words: A process is stationary if its stochastic properties are independent of the time you look at it. Think of it like this: A stochastic process is just a Random Variable (RV) that, instead of giving you e.g. a real value gives you a function every time you look at it. We call that realizations. If you now take a lot of these realizations, and ...


4

For BIBO stability in the case of discrete time, there is a necessary and sufficient condition given by $\sum |h[n]| < \infty$ that is if the impulse response is absolute summable then the system is BIBO stable. Clearly $$h[n] = \delta[n] - \delta[n-1]$$ and it has a finite support, the impulse response is absolutely summable and therefore BIBO stable.


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