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It's probably more straightforward to first compute the total transfer function in terms of $H(z)$, then plug in the actual expression for $H(z)$, and from that write down the difference equation: $$Y(z)=X(z)+\alpha z^{-M}H(z)Y(z)\tag{1}$$ From $(1)$ we can derive the total transfer function: $$\frac{Y(z)}{X(z)}=\frac{1}{1-\alpha z^{-M}H(z)}\tag{2}$$ ...

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This question could be rephrased as is $y(t) = T{x(t)} = x(t) ^ 2$ LTI system. This is a textbook case of a non-linear system. Here is a quick proof: For a system to be linear the following must be true: $T{a \cdot x(t)} = a \cdot T{x(t)}$ . i.e. if you input scaled signal into linear system you should get the same output as if you inputted that same signal ...

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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 ...

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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 ...

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It must be added to the problem that $R(\omega)$ is a real-valued, possibly bipolar function. In that case, its inverse discrete-time Fourier transform must be even: $$r[n]=r[-n]\tag{1}$$ From the given relation between $H(e^{j\omega})$ and $R(\omega)$ it is clear that $$h[n]=r[n-25]\tag{2}$$ must hold. I'm sure that you'll manage to combine $(1)$ and $(2)$ ...

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Given the system I/O definition: $$y[n] = \mathcal{H}\{x[n]\} = x[n^2] \tag{1}$$ you can easily show that it's a linear (but time-varying) system. Following the standard procedure let $$y_1[n] = \mathcal{H}\{x_1[n]\} = x_1[n^2] \tag{2.1}$$ and $$y_2[n] = \mathcal{H}\{x_2[n]\} = x_2[n^2] \tag{2.2}$$ then define $$x_3[n] = a x_1[n] + b x_2[n] \tag{3}$$ and $$... 1 Summary You can have only one (or no) primary sensor which is used in the prediction process. All other sensors are secondary and only used for corrections. Each correction takes the current state and a measurement from one of the secondary sensors and outputs the corrected state. You can run corrections as often as you want, typically each time one of the ... 1 The solution to the differential equation is given by the sum of a particular solution and the solution of the homogeneous differential equation. The particular solution is a solution to the non-homogeneous equation$$T\dot{x}(t)+x(t)=k_vu(t)\tag{1}$$which is easily found as$$x_p(t)=k_v,\qquad t>0\tag{2}$$The homogeneous solution is the solution of the ... 1 Given a transfer function$$G_v(s) = \frac{k_v}{1 + sT} \tag{1}$$the corresponding LCCDE, with y(t) being the solution, and x(t) being the input, will be$$ T ~\dot{y}(t) + y(t) = k_v ~x(t) \tag{2} $$Your formulation replaces x(t) with a unit-step u(t), and y(t) with x(t), yielding$$ T ~\dot{x}(t) + x(t) = k_v ~u(t) \tag{3} $$or equivalently ... 1 I'm actually starting this out not knowing the answer, although I have my suspicions. I think it'll be a linear system, but I'm not sure. Here's your starting point: you define the system y = h(x) as:$$y = h(x) \mid y(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} x(k_1,k_2)$$So let$$\begin{aligned} y_1(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\...

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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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