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1

Although this question is over 2 years old now, I think it's interesting to consider the solution, assuming that the original interpretation was incorrect, and that $u[n]$ represents the system input, not the unit step sequence. If so, then the first thing to recognize is that the system is just an accumulator. For a time-domain demonstration, simply make ...


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The question can only be answered if it is clarified what it is that the function $G(z)$ describes. If $G(z)$ is the system's transfer function then we're done immediately, because only linear time-invariant (LTI) systems can be characterized by a transfer function of that form. If $G(z)$ is the transfer function, the output sequence $y[n]$ is given by the ...


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Your solution looks correct. In any case, the given solution $y(t)=5e^{-3t}-4e^{-4t}$, $t>0$, must be wrong because it doesn't satisfy the initial condition on the derivative ($y'(0^+)=0$): $$y'(t)=-15e^{-3t}+16e^{-4t},\qquad t>0\tag{1}$$ So we have $y'(0^+)=1$. Your solution satisfies both initial conditions.


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HINT: The frequency response has the form $$H(e^{j\omega})=a+be^{-j\omega}+ce^{-2j\omega}+be^{-3j\omega}+ae^{-4j\omega}\tag{1}$$ which can be rewritten as $$H(e^{j\omega})=e^{-2j\omega}\big[ae^{2j\omega}+be^{j\omega}+c+be^{-j\omega}+ae^{-2j\omega}\big]\tag{2}$$ Now note that the term in brackets is purely real-valued. I trust that you can take it from here.


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[Apologizing: most of the following is already covered by Matt L.] Expression $$X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x[n]e^{-j\omega n}$$ is not valid in general, because we do not know whether it exists. One (classical) sufficient condition for the DTFT (discrete-time Fourier transform) is that the sequence $x[n]$ is summable. In other words, let us ...


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HINT: The error energy can be written as $$\begin{align}\sum_k\big|d[k]\big|^2&=\sum_k\left|x[k]-\frac{1}{2\pi}\int_{-W}^W\sum_nx[n]e^{-jn\omega}e^{jk\omega}d\omega\right|^2\\&=\sum_k\left|x[k]-\sum_nx[n]\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\omega}d\omega\right|^2\tag{1}\end{align}$$ Now compute the integral $$I(k-n,W)=\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\...


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As you know, we don't solve homework problems here, but you can get some hints. I don't know if you're really required to solve the problem in the frequency domain, but I would suggest to compute the output of the integrator in the time domain: $$y(t)=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau=\int_{-\infty}^{\infty}e^{-\tau/T}u(\tau)u(t-\tau)d\tau\tag{1}$...


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