# Continuous-time convolution of signals with negative amplitudes

While preparing for a mid-term exam, I encountered negative amplitudes for the first time while convolving two signals. I've already solved the problem, but my result and results from others conflict in the leading sign of the output signal.

Given the input signal $$x(t) = u(t-4) - u(t-2)$$ and the impulse response of the system $$h(t) = (t+1)[u(t)-u(t-1)]$$, where $$u$$ denotes Heaviside step function, compute the output $$y := x \star h$$.

Doing the convolution, I get the following results:

$$y (t) = \begin{cases} 0 & \text{if } t<2 \\ \frac12 t(2-t) & \text{if } 2 \le t < 3 \\ -\frac32 & \text{if } 3 \le t < 4\\ \frac12 (t-5) (t-1) & \text{if } 4 \le t < 5 \\ \end{cases}$$

Others have gotten the following results:

$$t<2; y(t) = 0$$ $$2 \le t < 3 ; y(t) = \frac{t^2-2t}{2}$$ $$3 \le t < 4 ; y(t) = \frac{3}{2}$$ $$4 \le t < 5 ; y(t) = -\frac{(t-5)\cdot(t-1)}{2}$$

In my calculations, I took the negative amplitude in account, and I put it in front of the integral. When they were doing the problem, they omitted the fact that $$x(t)$$ is a signal with negative amplitude. So, they are saying that we can omit the fact that the $$x(t)$$ is a negative amplitude signal, and that it doesn't affect the result. That makes no sense to me, and my way of thinking is that by their way of doing it the LTI system would have same output for two different input signals. I might be mistaken and not understand it properly, but if we omit the negative areas, in the case of, e.g., $$\sin(x)$$ on the input, wouldn't we get a rectified all positive sinus signal at the output?

What is the correct way of dealing with this kind of signals, does the negative amplitude effect the results and does it change the output of LTI system, am I right or mistaken?

$$(a\cdot x(t))*(b \cdot h(t)) = a\cdot b \cdot (x(t)*h(t))$$
That's complete nonsense. If you multiply the input with $$-1$$ the output will also be multiplied with $$-1$$.