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 ...


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 ...


4

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 ...


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

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 ...


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}...


2

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.


2

You can use differentiator filter that acts as a differentiator in the band of interest, and as an attenuator at all other frequencies, effectively removing high frequency noise. Take a look at the matlab example


2

If there was no noise in the input, then the causal filter: $$y[k] = \frac{\frac{3}{2}x[k] - 2x[k-1] + \frac{1}{2}x[k-2]}{T_s}$$ would give the asymptotically optimal 3-sample linear approximation of the first derivative in the limit of an infinite sampling frequency assuming zero input noise. The asymptotic error as sampling period $T_s\to0$ is a good ...


2

Well, Since basically the Derivative Operation is Linear Filter you can chose your own optimal trade off between Noise Sensitivity and Bandwidth. If you look at Finite Differences Coefficients page at Wikipedia you can see you can chose higher "Accuracy" filters for 1st Derivative. You can also chose Forward / Central / Backward method.


2

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 ...


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)...


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 ...


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

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

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 theory, M[0] doesn't exist. Your discrete derivative is only defined on [1,127]. As a practical solution, you can define M[0] as a null element of your signal. For example if you expect values between -1 and 1, take M[0] = 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[0] = 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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