# Tag Info

6

Here I expected $y(n)$ is to be computed by convolving $x(n)$ with $h(n)$, but in the equation given by Wikipedia it is shown as a matrix multiplication $y(n) = h^H(n).x(n)$. Are these two operations(convolution and matrix multiplication) same here?. The system is an FIR system, so the vector multiplication here is equivalent to convolution --- for ...

6

The most straightforward way to see this is to note that for $k=mN$ $$W_N^{kn}=e^{j2\pi mnN/N}=e^{j2\pi mn}=1$$ So the sum for the case $k=mN$ is simply $$\sum_{n=0}^{N-1}1=1+1+\ldots+1=N$$ Note that the solution using L'Hopital's rule is a bit dubious because for $k=mN$ the formula for the geometric series is not valid because the terms are all equal to ...

5

The estimation of derivative is straightforward: $$x'(n)~=\frac{x(n+1)-x(n-1)}{2}$$ $$x''(n)~={x(n+1)-2*x({n})+x({n-1})}$$ or if you have a signal sampled at $t_i=i\Delta t$, it is $$x'(t_{i})~=\frac{x(t_{i+1})-x(t_{i-1})}{2*\Delta t}$$ $$x''(t_{i})~=\frac{x(t_{i+1})-2*x(t_{i})+x(t_{i-1})}{(\Delta t)^2}$$ What you are interested in may be how to ...

4

The key is in the last step of your work: $$\frac{1 -e^{j2\pi k}}{1-e{\frac{j2\pi k}{N}}}$$ If $k$ is some integer multiple of $N$, then the exponents in the numerator and denominator are both some integer multiple of $j2\pi$. In this case, both exponential functions are equal to 1, meaning that the expression above is equal to $\frac{0}{0}$ for $k$ an ...

3

so with an LMS filter, we have a time-variant $N$-tap FIR filter: $$y[n] = \sum\limits_{k=0}^{N-1} h_n[k] \, x[n-k]$$ $x[n]$ is the input signal, $y[n]$ is the FIR output, and $h_n[k]$ are the FIR tap coefficients at the time of sample $n$. with an LMS filter, we also have another input called the desired signal: $d[n]$. we want our LMS filter to adapt ...

3

Your interpretation is correct: directional derivation operators highlight variation in a given direction. Here, you use the $2$-point discrete derivative in the $x$-direction (along image rows). It may emphasize vertical features. First, such operators indeed extend the initial image range. However, one often uses the absolute value of the derivative to ...

3

As shown in the answers to this question, for continuous-time signals, the bound predicted by Bernstein's inequality is achieved with equality by a sinusoidal signal with a frequency equal to the upper frequency limit. In analogy to this, in this answer I'll show a bound on $\big|x[n]-x[n-1]\big|$ for a discrete-time sinusoidal signal $x[n]$ at angular ...

3

The problem is in the way you apply the bilinear transform. You have to use the appropriate (pre-)warping of the frequencies. Since the bilinear transform warps the frequency axis, you have to make sure that the corner frequency of the discrete-time filter is correct. One way to do that is as follows. The bilinear transform is defined as $$s=k\frac{z-1}{z+1}... 3 There is a good article in Wikipedia: Numerical Differentiation Since you'd expect differentiation to be (Anti?)symmetric, using IIR filters might not be the wisest of ideas. 3 So, first of all: A differentiator is really just a high-pass filter. Digitally (and this is dsp.SE, so I presume this is the case), that typically means a differentiator is really just FIR with taps [1;-1]. A butterworth filter being a low-pass IIR, the combination of both does sound like you might want to build a bandpass, or are doing something that ... 2 The problem in your derivation is the way you discard the high frequency components. Note that \hat{s}(t), the Hilbert transform of the bandpass signal s(t), has frequency components around \omega_c as well as around -\omega_c. So you can't just discard one of the two terms \hat{s}(t)e^{j\omega_ct} and \hat{s}(t)e^{-j\omega_ct}. The low pass ... 2 Non-zero pixels after derivation provide you with two informations: a potential location (where the pixel is), and a potential strength (in magnitude) of an edge-pixel. Linear edge detectors are linear filters. A 3\times3 kernel adds a smoothing effect in the orthogonal direction. It will favor edges with a certain spatial extend, over short-lengthed ... 2 So if you continue the substitution you get: R_t = \sqrt{((-R_BK_X/\sqrt{(K_R^2 - K_X^2)} + X_t - X_t)^2 + R_B^2)}  R_t = \sqrt{((-R_BK_X/\sqrt{(K_R^2 - K_X^2)} )^2 + R_B^2)}  R_t = \sqrt{(R_B^2K_X^2/(K_R^2 - K_X^2) + R_B^2)}  R_t = \sqrt{(R_B^2K_X^2/(K_R^2 - K_X^2) + R_B^2 (K_R^2 - K_X^2)/(K_R^2 -K_X^2))} R_t = \sqrt{(R_B^2K_R^2/(K_R^2 - K_X^2)... 1 The impulse response doesn't change because it is the output of the filter when the input is an impulse. When the complex input is applied to a complex filter the output is still determined by the convolution of the input and the impulse response of the filter. The addition/integration and multiplication operations that take place inside a convolution ... 1 The expression$$s_d[m] = \frac{1}{D}\sum_{k=0}^{D-1}\exp(j2\pi \frac{km}{D})$$is the Fourier series expansion of the periodic function s_d[m] in the discrete case (i.e. only up to a certain amount of frequencies). s_d[m] is a also called comb function, and I have written more about it (including the relation you ask for) in one of my articles. ... 1 For spatial filtering there are numerous beamforming techniques and it will be difficult to compare all here, there are numerous resource available on internet. Just to give you insight I am giving brief description of well known beamforming techniques Phased shift beamformer also known as conventional beamformer: This beamforming techniques tries to add ... 1 In order to derive equation (20) you can use the following steps: From substitution of equation (16) into equation (18)$$ P_n = P_{n|n-1} - K_nP^t_{x_ny_n}$$Now plugging right side equation of (19) into the last equation$$ P_n = P_{n|n-1} - K_n(P_{y_ny_n} - R_n)= =P_{n|n-1} - K_nP_{y_ny_n} + K_nR_n $$Now Pluging (16) again we know that$$ ...

1

Note that the Butterworth filter is only one of many possible approximations of an ideal low pass filter. It is a Taylor series approximation (at $\omega=0$) of the desired response, and, consequently, the magnitude response of the Butterworth filter is maximally flat at $\omega=0$. Approximating the squared magnitude response (instead of the complex ...

1

In theory, M doesn't exist. Your discrete derivative is only defined on [1,127]. As a practical solution, you can define M as a null element of your signal. For example if you expect values between -1 and 1, take M = 0. In the case of real-time, there is two options: Either you save the last value of your buffer so you can take M = last for ...

1

Just wanted to supplement an excellent answer by Matt L., since it was not very clear to me how he calculated the numerical value of $k$ in equation $(3)$. After reading the book "Introduction to signal processing" by Orfandis I have found the formula $$k = \frac{1}{\tan\left(\frac{\omega_c}{2}\right)}$$ where $\omega_c$ is the so called digital cutoff ...

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