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## Hot answers tagged impulse-response

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We know that $\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ has value $0$ for $n < -1$ since $u[n+1]=0$ for $n < -1$. At $n=-1$, $u[n+1]$ jumps to value $1$, but $\sin(\theta (n+1))\bigr|_{n=-1} = \sin(\theta (-1+1))$ has value $0$, and so $\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ just happens to equal $\... 3 The sequences$e^{j\omega_0n}are eigensequences of discrete-time LTI systems, i.e., the response to such a sequence is the same sequence scaled by a complex constant (the eigenvalue). This can be shown as follows: \begin{align}y[n]&=(h\star x)[n]\\&=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\\&=\sum_{k=-\infty}^{\infty}h[k]e^{j\omega_0(n-k)}\\&=... 3 In a discrete-time system such as the one that you have, the number y[n_0] (here n_0 is a fixed integer) is a sum of the form\sum_{k=-\infty}^\infty h[k]x[n_0-k]$$which can be re-arranged via a change of variables (replace k by n_0-\ell) to$$\sum_{\ell=-\infty}^\infty h[n_0-\ell]x[\ell].$$So, the commutativity of the convolution is trivial. ... 3 So dealing with generalized functions like the Dirac delta requires some care, and when dealing with N-dimensional versions you need to be very explicit with your notation to keep things straight. I'll denote the 2 dimensional delta function in polar coordinates at the origin as {}^2\delta(r, \theta) = {}^2\delta(r), since for the special case of the ... 3 (Adapted from this answer on dsp.SE) The reason that the impulse response (also called the unit pulse response for discrete-time systems) determines the output for arbitrary input x to an LTI system is that The output of a linear time-invariant system in response to input x is the sum of scaled and time-delayed versions of the impulse response. ... 2 A digital system with complex input and complex output has a block diagram in which the various arrows carry complex signals, each register holds a complex number, each multiplier computes the product of two complex numbers, each adder computes the sum of two complex signals, etc. Consider, for illustration, a simple FIR filter with complex input x and ... 2 You are probably confused by the memoryless systems whose output due to an impulse is just another impulse; a single value just as you expect. However, there are systems with memory too, and some of them have infinite memories; their current output is affected by an impulse applied infinite samples ago. Incidentally they have infinite length impulse ... 2 Imagine a system that accepts a single number x as its input, and it multiplies that number with another number h. Would it surprise you that another system which multiplies its input with the number x gives the same output as the first system when fed with the number h as input? If not, then it also shouldn't come as a surprise that the output of an ... 1 It's not possible to recover the original signal X based on h perfectly !! Filter or conv are usually performed in time domain, so it's similar "But not exactly" to y = conv(X,h) , So in that case after you know h or in other words after you estimate h , you need to use any equalizer like ZF or MMSE to get back X. Of course, ... 1 In general, what you want is impossible! Not every operation is reversible. In frequency domain, that becomes obvious: a system's frequency response H(f) might have actual zeros at some points, and thus, no inverse filter exists. Multiplication with zero can't be undone, since no matter what was at that frequency before filtering, it's 0 afterwards and we ... 1 The scaling difference between your two output is exactly 32767= 2^{15}-1 which is exactly the maximum amplitude of a 16-bit signed integer. I'm guessing, the difference is how you import the audio into the program: Matlab's audioread() typically normalize the data to [-1,1], i.e. divide by 32767. It would seem that whatever Python method you are using ... 1 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\\&... 1 Arbitrary shapes is A LOT more difficult than rectangular rooms. You either need to do image method with visibility checking or some sort of ray, particle or cone tracing algorithm. There are few commercial programs that do this (Ease, Odeon, Bose Modeller, etc. ) and I doubt you'll find a free one. In addition the much more complicated algorithms, these ... 1 A simple way to derive the differential equation from the impulse response is to transform the latter to the frequency domain, rewrite the input/output relation, and then transform the resulting equation back to the time domain. The Laplace transform of the given impulse responseh(t)is \begin{align} H(s) &=\frac{A}{\tau}\frac{1}{\left(s+\frac{1}{\... 1 HINTS: \cos(2)-\cos(2)=0 What remains is y(t)=\cos(t-1)-\cos(t). Can this really be the response of a causal system to an input signal that starts at t=0? Maybe you forgot some important detail in your answer. The correct answer must consist of 3 different expressions for the time intervals t<2, 2<t<3, and t>3. Your answer ... 1 It's pretty simple if you know a few basic things: a single pole on the real axis corresponds to an exponential sequence. The elements of that sequence either have the same sign, or their signs alternate, depending on the angle of the pole (0 or \pi). This should allow you to pair figures (c) and (e) with figures (1) and (6). a complex conjugate ... 1 Note that the DFT transforms a periodic sequence into a periodic sequence. Hence, the result of your IFFT is intrinsically periodic and you should view it as such. The parts you are seeing at the end of it are thus just as well behind the start as they are before it. Hence, this part may as well be some form of pre-ringing, stemming from a nonzero group ... 1 There are fundamental differences in concept between signals and systems. I will explain this through the idea of unit consistency (see for instance). However, for LTI systems, signals and systems become dual through convolution, since the latter is commutative. Two digressions first, due to the mention in @Dilip Sarwate answer. Digression 1: LTI systems ... 1 If you know for sure that the filter type is one of the four given ones, then the exercise is quite straightforward. Look at the zeros of the corresponding transfer functionH(z)=\sum_{n=0}^{N-1}h[n]z^{-n}\tag{1}$$where N is the filter length. The zeros at DC (z=1) and at Nyquist (z=-1) are given by$$H(1)=\sum_{n=0}^{N-1}h[n]\tag{2}$$and$$H(-... 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 following might help. Given the data of5$samples, $$x[n] = [1, 2, 3, 4, 5]$$ You will have possible choices for the orders$(p,q)$of$a[k]$and$b[k]$, such as$\{(4,0),(3,1),(2,2),(1,3),(0,4)\}$. You can solve or each of the possibilities and check whether they yield stable systems or not, by looking at the roots of$a[k]$... afaik they should ... 1 I would suggest the frequency sampling design method, which results in a linear phase FIR filter. You sample the desired magnitude on an equidistant grid, add a linear phase term, and then you compute the impulse response by taking an inverse DFT. The resulting impulse response is usually windowed. I've explained the details in this answer, which also ... 1 HINT: Note that $$\int_{-\infty}^{\infty}x(\tau)x(t+\tau)d\tau=\int_{-\infty}^{\infty}x(\tau-t)x(\tau)d\tau$$ 1 For this specific example, using a partial fraction expansion (PFE) would be an overkill, so yo should better consider using the formula : $$1 - z^{-N} = (1- z^{-1}) (1 + z^{-1} + z^{-2} + ... + z^{-N+1}$$ to simplify the Z-transform$Y(z)of your output as: \begin{align} Y(z) &= H(z) X(z) = \frac{1 - z^{-N} } {1-z^{-1} } \frac{ 1 - z^{-M} }{ 1-z^... 1 The required length of the impulse response is primarily determined by the frequency resolution of the filter. If you sample at 44.1 kHz and want to do something meaningful at 40 Hz, you will need 1000s of samples. If your required frequency resolution is reasonably large, than the windowing approach can work fine. I'd suggest a rectangular window with a ... 1 Any filter with a DC gain of zero and, for most other frequencies, a gain of (or close to) 1, is a DC blocking filter. So just make a high-pass filter (HPF) and put the corner frequency as close as you can to zero. Of course, when the system is switched on, there will be a transient from blocking DC of 0 to blocking DC of whatever value it is. Without ... 1 As a complement to the previous answers, barely looking at the output doesn't allow you to identify what you had as input/system-function. Take for example the following (interconnection of) systems producing all the same output. For the system above we have: y[n] = (x \star h) [n]  Y(z) = X(z). H(z) = H(z) . X(z) $$For the system above we have:$$... 1 I very strongly recommend1 defining convolution in a slightly different way, namely as: $$(x\star h)[n]=\sum_{i+j=n}x[i]h[j]$$ You can see that this is just the standard definition with a change of variables, i.e. choosej=n-i$. (You do need to be a bit clearer about the range$i$and$j$are running over, i.e. it makes a difference if$i,j\in\mathbb N$... 1 Think of the convolution$x\star h$as taking some delayed copies of$x$summed together, each with an amplitude read out from the entry of$h\$ at that delay. Let's picture it with sparse-impulse signals: (Ignore the vertical shift, that's just to de-clutter the plot) Now, if you turn around the convolution, all that changes is the notion of which time-...

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