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If x(t) is the curve shown in the image

enter image description here

how can I find (x(2t)+x(2t+2))?

I know how to find the individual terms but don't know how to add them!

And how can I express my answer in terms of the unit step function?

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    $\begingroup$ Add them point-wise; i.e. if the solution is $y(t)$ and $t_0$ is a time instant, then $y(t_0) = x(2t_0)+x(2t_0+2)$. Note that $x(2t_0)$ and $x(2t_0+2)$ are scalars and can be added like any pair of numbers. $\endgroup$ – MBaz Aug 10 '17 at 17:39
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First express $x(t)$ in piece-wise form:

$$ x(t) = \begin{cases} 0 ~&, \text{ for} & t < -2 \\ 0.5(t+2) &, \text{ for} &-2 < t < 0 \\ -t+1 &, \text{ for} & 0 < t < 1 \\ 0 &, \text{ for } &1 < t \\ \end{cases} $$

Now, express $x_1(t)=x(2t)$ in a similar form: $$ x_1(t) = x(2t) = \begin{cases} 0 ~&, \text{ for} & 2t < -2 \\ 0.5(2t+2) &, \text{ for} &-2 < 2t < 0 \\ -2t+1 &, \text{ for} & 0 < 2t < 1 \\ 0 &, \text{ for } &1 < 2t \\ \end{cases} $$

Recognize how the formula and intervals are adjusted for $x_1(t)$ $$ x_1(t) = x(2t) = \begin{cases} 0 ~&, \text{ for} & t < -1 \\ t+1 &, \text{ for} &-1 < t < 0 \\ -2t+1 &, \text{ for} & 0 < t < 0.5 \\ 0 &, \text{ for } &0.5 < t \\ \end{cases} $$

Now, express $x_2(t)=x(2t+2)$ in a similar form: $$ x_2(t) = x(2t+2) = \begin{cases} 0 ~&, \text{ for} & 2t+2 < -2 \\ 0.5(2t+2+2) &, \text{ for} &-2 < 2t+2 < 0 \\ -(2t+2)+1 &, \text{ for} & 0 < 2t+2 < 1 \\ 0 &, \text{ for } &1 < 2t+2 \\ \end{cases} $$

Recognize how the formula and intervals are adjusted for $x_2(t)$ as well $$ x_2(t) = x(2t+2) = \begin{cases} 0 ~&, \text{ for} & t < -2 \\ t+2 &, \text{ for} &-2 < t < -1 \\ -2t-1 &, \text{ for} & -1 < t < -.50 \\ 0 &, \text{ for } &-0.5 < t \\ \end{cases} $$

Now add those two piecewise defined signals $x_1(t)$ and $x_2(t)$, considering the intervals carefully:

$$ y(t) = x_1(t) + x_2(t) = x(2t) + x(2t + 2) = \begin{cases} 0 ~&, \text{ for} & t < -2 \\ t+2 &, \text{ for} &-2 < t < -1 \\ -t&, \text{ for} & -1 < t < -0.5 \\ t+1 &, \text{ for } &-0.5 < t < 0 \\ -2t+1 &, \text{ for} & 0 < t < 0.5 \\ 0 &, \text{ for } &0.5 < t \\ \end{cases} $$

Expressing this using unit steps (alone) is not possible, you need the ramp functions. Then you can provide a representation using scaled and shifted ramps.

A MATLAB plot of the resulting signal is provided below ; enter image description here

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