# Tag Info

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Is there any trade-off in numerical precision or speed? Yes. For delays that are integer multiples of the sampling period method 1 is far superior: it's computationally efficient, it's bit-exact, it's easy to implement and it's almost fool proof. Method 2 is computationally expensive, you need to pick an FFT length (which is not trivial) it's subject to ...

6

Your problem comes about because you're pondering the convolution of a signal by itself: $$g(x) = f(x) * f(x)$$ So, yes, if you had a block diagram where you collect f(x) from the æther and then somehow magically convolve it by itself (which can't be done physically because it would require time-reversing $f(x)$, which requires looking into the future), then ...

4

I did it for the DCT-I and DCT-II. At first I thought it is about circular convolution, but it is not. After desperate attempts to do it by myself I found this article: https://dr.ntu.edu.sg/bitstream/10356/94137/1/Convolution%20Using%20Discrete%20Sine%20and%20Cosine%20Transforms.pdf. In this paper there is a derivation of circular convolution using DTT's, ...

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Some general hints for this type of problem If the function is defined piecewise than chances are your calculation and solution will also have to be done piecewise in sections. Determine first where the result is zero, i.e. where $x(\tau)$ and $h(t-\tau)$ don't overlap. This determines the start and end of the sections that you need to consider. Hint: if ...

4

At the risk of blowing my own trumpet and that of my co-author, Bob Williamson, there is also this paper which shows the equivalence of three techniques referred to in the FIR link in Ben's answer and also referred to in the paper linked to from Hilmar's answer. The two results that are of particular interest are Proposition's 2 and 3 of the paper, ...

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As Hilmar pointed out, for delays that are integer multiples of the sampling period, method 1 is far superior. Also, Method 1 is more suitable for real-time operations as you don't need to buffer the data to perform the FFT. For delays that are non-integer multiples of the sampling period you can adapt method 1 by using an FIR or IIR filter using Lagrange ...

2

Considering the Laplace transform of a function $x(t)$ as : $$X(s) = \mathcal{L}\{x(t)\} = \int_{-\infty}^{\infty} x(t) e^{-st} dt \tag{0}$$ then to find the following Laplace transform $$\mathcal{L} \{ \int_{-\infty}^{t} e(\tau) d\tau \} \tag{1}$$ you can write the integral as a convolution of $e(t)$ with the unit-step function $u(t)$ as $$\mathcal{L}\{ \... 1 This exercise is meant to help the student appreciate the fact that if the response y_1(t) of an LTI system to an input x_1(t) is known, then the response to an input$$x_2(t)=\sum_{k=1}^Ka_kx_1(t-t_k)\tag{1}$$is given by$$y_2(t)=\sum_{k=1}^Ka_ky_1(t-t_k)\tag{2}$$which is a direct consequence of linearity and time-invariance. Consequently, if an input ... 1 It should be clear that a property of a system, such as causality, cannot be determined by looking at its input signals. For a linear time-invariant system, it is its impulse response h(t) from which properties such as causality or stability can be determined. Only the second definition in the question is correct: a causal LTI system has an impulse ... 1 The key is being even, notice that f_{N+k} = f_{k}, and when you do a circular wrapping of an infinite function you should have f_{k} = \sum_{i=-\infty}^\infty f((i \cdot N + k) T_s), then naturally f_{N-k} = f_{k}, in practice you want to choose f and N so that f vanishes before N/2, then you are safe to use only one term of the summation that ... 1 Well, you know that u[n]=0 for all n<0. So from the two terms in your sum, all terms will be zero where either k+2 < 0 (i.e., k < -2) or n-2-k<0 (i.e., k>n-2). From this your sum will need to run from -2 to n-2. Can you take it from here? 1 Standard multiplication by a scalar s is a linear operation, since:$$ s\times(a\times x+b\times y)= a\times (s\times x)+ b\times(s\times y) In other words, multiplying a linear combination by a scalar is the same as doing a linear combination of multiplies. Here, the multiplication is a very simple example of a system(a amplifier). Your question amounts ...

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Convolution can be understood as the effect one signal has on another signal i.e., you pass a signal through a system (defined by its impulse response) then what is the output? This is answered through convolution. Likewise, Correlation answers the similarities between two signals. The output of correlation can be positive, zero, or negative. zero: the two ...

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