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Feed forward refers to the direction of the signal flow. For feed forward, the direction is, well, forward :-) I think it is easier to show an example. I know that many "sigma-delta" ADCs (analog to digital converters) use a combination of feedback and feed forward. I found an example of a block diagram of such an ADC here in this article, about Higher-...

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Raising m(t) to the power 2 is of course squaring the signal and if the signal consisted of two sine waves spaced at f1 and f2 , f1 would transform to 2f1 and f2 would transform to 2f2. So if f1 were 100 Hz and f2 were 1000 Hz your original bandwidth would be 900 Hz and your bandwidth (after squaring) would be: - 2000 Hz - 200 Hz = 1800 Hz i.e. your ...

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Assuming that with "associate" you mean "put in series" that is you hook S1 output to S2 input then you are not doing anything wrong. Problem is that sometimes (plenty of times) solving a convolution integral is not a piece of cake thus you prefere the Laplace transform. Let's start by computing $G_1(s)$ and $G_2(s)$: $g_2(t)=e^{-\alpha t}u(t)$, but \...

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Convolution basically tells you how similar two signals are as one of them is shifted in the x-axis and reflected. So consider taking your signal $g$ and shifting it by various amounts. The convolution will have a peak when the two signals are mirror images across the y-axis. In this case, they already are, so the peak convolution occurs with a shift of ...

3

The problem of processing a 2-d kernel of data over a large dataset (not just convolution) comes up so regularly in HD video processing that I came up with a generic way of handling it that I use all the time. I developed a generic "kernel generator" that uses line buffers and registers to present all of the input data for a given output pixel in parallel. ...

3

The functions to be convolved are $h(t)$ and $u(t)$, as shown in the top two diagrams in the figure below: Figure source: I drew it. The third diagram shows the folded $u(t)$, i.e., $u(-t)$ and the fourth diagram shows the folded and shifted $u(t)$, i.e., $u(-t-\tau)$. Note that $\tau$ is the shift variable. There will be two non-zero ...

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Here is your problem: "We know that squaring a signal doubles its bandwidth." That is not true in general. It is true for sinusoidal signals. Probably you assume it must be true also for linear combinations of sinusoidal signal (i.e. Fourier sums/integerals which can be used to approximate a rectangular signal), but this is not the case. This reasoning ...

3

I think you'd better not split the functions. If you want to calculate the fourier transform of $0.5^t u(t)$, then you might put all that into the integral. Therefore, your integral limits will become $0$ to $+\infty$ because of $u(t)$. And your integral result should not be infinity because your function $0.5^t$ goes to zero. Solving the ...

3

Depending on the system equation behavior of the curve of input versus output can be of any shape. System is said to be linear if it satisfies these two conditions: Superposition - if input applied is (x1+x2), then the output obtained will be y1+y2 .(equivalently we say that if x1 and x2 are applied simultaneously then out put will be the sum of the ...

3

Hint: You have to combine the resulting complex exponentials into sine and cosine terms: $$\sin x=\frac{e^{jx}-e^{-jx}}{2j}\\ \cos x=\frac{e^{jx}+e^{-jx}}{2}$$ If you use $h(t)=e^{-at}u(t)$, you should end up with the expression $$y(t)=\frac{1}{a^2+25}\left[a\sin(5t)-5\cos(5t)+5e^{-at}\right]u(t)$$

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The picture below represents a velocity controller that you can usually find in AC servo motor. It's just a fraction of the entire cascade loop. The low frequency signals (slow response) go through lower path, this is: velocity setpoint ($\Omega{set}$) , actual velocity ($\Omega{m}$) from encoder feedback, lowpass filter ($T_{f}$), PI controller ($... 3 Let me step back first...A control system needs an input and output, and some method to measure feedback. Open loop examples include magnetic stepper motors, while closed loop tend to be linearized for one or more variables. In between, there are signal conditioners or processors to feed an improved response in both time and frequency domain. This can be ... 3 Cross-correlation and convolution both have an integral of a product of 2 signals. But they have totally different base ideas. Convolution makes a new signal, a function of time. Cross-correlation compares two signals over their whole lengths. The result is not a function of time, but a function of the delay parameter. Cross-correlation is a measure for "do ... 3 GIVEN: x[n] = [3,2,1,0] y[n] = [1,2,3,0] v[n] = x[n] * y[n] is... v= xy = 3 * 1 = 3 v= xy + xy = 3 * 2+2 * 1 = 8 v= xy + xy + xy = 3 * 3+2 * 2+1 * 1 = 14 v= xy + xy + xy + xy = 3 * 0+2 * 3+1 * 2+0 * 1 = 8 v= xy + xy + xy + xy + xy = 3 * 0+2 * 0+1 * 3+0 *... 2 Convolution simply gives you the output of a system given an input. In the case of a periodic input, you'll end up with a periodic output. This implies that the steady state response to this input will be a periodic "steady state" response. I put steady state in quotes because it's not really steady in terms of DC analysis. It is steady though in terms of ... 2 Your bandwidth will double. You can derive this using trig identities: DC and 2*theta components will form unless you filter them out. I ran a quick simulation in MATLAB to confirm. F(m(t)) on the top-left, F(m(t)*m(t)) on top-right. Bottom-left is the the top-left zoomed in, and vice verse. conclusions: m(t) has a passband from 4kHz to 6kHz m(t)*m(t) has ... 2$\delta (t)$is the unit impulse and is zero everywhere except at$t=0$, and$sin\:(\omega _0 t)=0$at$t=0$. The product of these two is zero everywhere.$u(t)$is the unit step function and its derivative is infinite at$t=0$and zero everywhere else, and this is the definition of the unit impulse. 2 Writing down convolution product results in  (h * u)(t) = \int_{-\infty}^{+\infty}h(t - v)\cdot u(v) dv  The parts of the domain we're interested in are: t-v\ge 0 \Rightarrow v \leq t for$h(t-v)$0 \leq v \leq 1 for$u(v)$We can then find that for$0 \leq t \leq 1: \begin{align} (h*u)(t) &= \int_0^th(t - v)\cdot u(v) dv \\ ... 2 It should not be difficult to verify thatR(\alpha) $is periodic with period$T$. Here, the value of$ R_1(\alpha) = T- 4\alpha $looks correct, for$ R_2(\alpha) $I calculate it to be$4\alpha - 3T$. Both$R_1,\ R_2$define$R(\alpha)$over one period, as$R_1$is defined for$0<\alpha<T/2$whereas$R_2$is for$T/2<\alpha&...

2

Here is one example I use to help me understand some of it. By no means is it complete. In x-ray imaging, the x-ray source is a distributed source. You can consider it as many point sources next to each other. If there was only one point source, the shadow that is cast from the object (patient) would be very sharp. If you placed your hand on the xray ...

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A brute-force implementation of a 500,000-tap FIR filter @ 44.1 ksps requires about 22 G operations per second. Large FPGAs have hundreds of DSP units capable of doing a multiply-add in one clock, so the resulting clock frequency of a few hundred MHz is quite reasonable. A slightly more interesting problem is managing the data. Let's be conservative and ...

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In addition there should be somewhere mentioned the integration period, it's from time=0 to current time t. $$\int_0^t{x(\tau)\cdot h(t-\tau)d\tau }$$ Tau is the formal variable for integration process, t is the running physical time. The integral should be calculated completely from blank board for every t. That's quite a job for a signal processor. ...

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HINT The transfer function in the s-domain (using Laplace transform) is given by: $$\mathcal{H}\left(\text{s}\right):=\frac{\text{v}_\text{o}\left(\text{s}\right)}{\text{v}_\text{i}\left(\text{s}\right)}=\frac{\text{R}}{\text{R}+\text{sL}}\tag1$$ Using the definition of the Laplace transform we know: $$\text{v}_\text{i}\left(\text{s}\right)=\mathcal{L}_t\... 2 I was hoping someone could have a look through what I've done so far, and also help me with the final integration. You don't need to perform that integration. In the s-domain the unit impulse is unity hence, you have calculated the response already and did the inverse transform correctly. The answer is $t\cdot e^{-t}$. The great thing about transferring ... 1 You can say that the impulse response of an ideal buffer is a Dirac pulse. Which is actually quite obvious, because the impulse response has to return the impulse response again in its entirety. Other variations on the matter:$$\begin{align} y(t) &= u(t)\\ &\Updownarrow \mathcal{L}\{\}\\ Y(s) &= U(s)\\ &\Downarrow\\ G(s) &= \frac{Y(s)}{...

1

What you're describing is not really convolution per se, although the results can be used with convolution to get solutions for more complicated input signals. Maybe you're slightly misunderstanding what the book is trying to do? The usual method for solving this for arbitrary $x(t)$ is to first solve it for $x(t) = \delta(t)$, which is what the ...

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The main thing I see wrong with your approach is that as drawn, your $x(t)$ is actually $x(t)= u(t)e^{-t}$, not $e^{-t}$. Making this change should remove one the infinite limits on your integration (when the exponential goes to infinity).

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Thanks to @ThePhoton and @EugeneSh., it turns out there are different definitions of Fourier Transforms. https://www.physicsforums.com/threads/why-the-1-sqrt-2-pi-in-the-definition-of-the-fourier-transform.487312/

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How can the multiplication with the dirac function give x(t-t0) of every point in the convolution integral? It can't, that is what the integral does. -- I try to imagine the convolution graphically. This is very helpful, but inside the integral, it is unnecessary. Inside the integral it is just multiplication. The integral operation handles the infinite ...

1

Doing this graphically, or conceptually, is quite easy. This problem seems to have been designed to make it easy. Imagine the top signal sliding along, with the bottom signal staying fixed. For each position of the top signal, produce the product of the two signals, then add up the resulting area under the curve. For example, at T = 0, the top signal is ...

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