New answers tagged homework
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 ...
1
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 ...
1
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.
2
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.
1
[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 ...
2
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
$$I(k-n,W)=\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\...
0
As you know, we don't solve homework problems here, but you can get some hints. I don't know if you're really required to solve the problem in the frequency domain, but I would suggest to compute the output of the integrator in the time domain:
$$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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