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1

The standard notation for the unit step function is $u(t)$. I've seen the notation $u_k(t)$, with integer $k$ in Oppenheim's Signals and Systems, where $k$ is the order of the derivative of the Dirac delta impulse. I.e., $u_1(t)=\delta'(t)$, etc. Using the same logic, negative values of $k$ indicate integrals of the Dirac delta impulse. So $u_{-2}(t)$ is the ...


7

The integral doesn't converge in the conventional sense, so you can't solve it with standard methods. Assuming that you know (or can look up) the Fourier transform of the unit step function $u(t)$, it is straightforward to compute the Fourier transform of $\cos(\omega_0t)\,u(t)$ using the modulation property: $$\mathcal{F}\big\{u(t)\big\}=U(\omega)=\pi\delta(...


1

Your first approach doesn't work because the original signal is simply not the integral of the signal you sketched below it. Just graphically integrate that signal to see this. Your second approach doesn't work for the very same reason. If you integrate the rectangular signal you don't get a signal that is zero for $t>1$. The value of the integral would ...


0

Since this is homework, I will only give hints: The antiderivative of a cosine is the sine of the same frequency (but different amplitude). $\sin\left(\frac{2\pi kt}{3}\right)\bigg|_{t=0.5} = \sin\left(\frac{\pi k}{3} \right)$ $\sin\left(\frac{2\pi kt}{3}\right)\bigg|_{t=1} = \sin\left(\frac{2\pi k}{3} \right)$ $\sin\left(\frac{2\pi kt}{3}\right)\bigg|_{t=2}...


2

As said by @Juancho, the expression let us suspect that the system could be non-linear. So we could look for a counter-example, yet it is not evident how to find a good one. Anyway, lt us try to better understand the system. Since there is something around sign change. The insight is there to have a first signal that has a sign change, and a second that ...


1

Suppose that you could write $x_.$ as $x_.(t) = a_1x_1(t-n_1)+a_2x_1(t-n_2)$, then by LTI hypothesis, you can derive $y_.(t) = a_1y_1(t-n_1)+a_2y_1(t-n_2)$ with very little computation. You can get a first insight to that method by looking at the shape of $x_2$ which is self-evident, and a little more imagination gives you the structure for $x_3$. I find ...


1

If you did want to grind through this, you could define $x_3(t) = A x_1(t) + B x_2(t)$, then substitute it into the left side of either of your top two system definitions above. like this: $$x_3(t) = \begin{cases}0 & x_3(t) < 0 \\ x_3(t)+x_3(t-2) & x_3(t) \ge 0\end{cases} = \\ \begin{cases}0 & A x_1(t) + B x_2(t) < 0 \\ A x_1(t) + B x_2(t)+...


3

First, you are not trying to find the linearity of a signal, but the linearity of a system defined by the transform above between input and output signals. If you suspect your system is not linear (the different expressions depending on the sign of $x(t)$ is a big red flag), then you only need a counterexample rather than a general proof of linearity. In ...


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