Question

A signal \(x(t)\) is shown below.

The plot of \(y(t)=1-x(4-t)\) is shown in ______.

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

To sketch \(y(t)=1-x(4-t)\), first time-reverse \(x(t)\) to get \(x(-t)\), shift it right by \(4\) to get \(x(4-t)\), then flip the amplitude and add \(1\).

Answer 1

The Correct Option is B

Step 1: Break the transformation \(y(t)=1-x(4-t)\) into simple operations.
Write the argument as \(4-t=-(t-4)\), so \(x(4-t)=x(-(t-4))\). This tells us to first time-reverse \(x(t)\) to get \(x(-t)\) (flip the graph about the vertical axis), then shift that flipped graph to the right by \(4\) units. After we have \(x(4-t)\), the full expression \(y(t)=1-x(4-t)\) turns it upside down (multiply by \(-1\)) and raises the whole curve by \(1\) unit.

Step 2: Read the given plot of \(x(t)\) as a piecewise function.
From the graph:
\[ x(t)= \begin{cases} 0, & t\le -3 \\ -(t+3), & -3\le t\le -1 \\ 1, & -1\le t\le 3 \\ 1-1.5(t-3), & 3\le t\le 5 \\ 0, & t\ge 5 \end{cases} \]
The curve ramps from \(0\) at \(t=-3\) down to \(-2\) at \(t=-1\), jumps up to \(1\), stays flat at \(1\) till \(t=3\), then ramps down from \(1\) to \(-2\) by \(t=5\).

Step 3: Substitute \(s=4-t\) into each piece to build \(x(4-t)\).
We check, for every value of \(t\), which piece of \(x(s)\) applies when \(s=4-t\).
For the left ramp, \(-3\le s\le -1\) means \(5\le t\le 7\), and there \(x(4-t)=t-7\).
For the flat middle, \(-1\le s\le 3\) means \(1\le t\le 5\), and there \(x(4-t)=1\).
For the right ramp, \(3\le s\le 5\) means \(-1\le t\le 1\), and there \(x(4-t)=1.5t-0.5\).
Outside \(-3\le s\le 5\), that is \(t<-1\) or \(t>7\), \(x(4-t)=0\).

Step 4: Form \(y(t)=1-x(4-t)\) on each piece.
For \(t<-1\): \(y(t)=1-0=1\) (flat baseline).
For \(-1\le t\le 1\): \(y(t)=1-(1.5t-0.5)=1.5-1.5t\), a line falling from \(y(-1)=3\) to \(y(1)=0\).
For \(1\le t\le 5\): \(y(t)=1-1=0\) (flat at zero).
For \(5\le t\le 7\): \(y(t)=1-(t-7)=8-t\), a line falling from \(y(5)=3\) to \(y(7)=1\).
For \(t>7\): \(y(t)=1-0=1\) (flat baseline).

Step 5: Note the jumps that come from the discontinuities in \(x(t)\).
At \(t=-1\), \(y\) jumps up from the baseline \(1\) to \(3\), because this is where \(s=4-t=5\), the point where \(x(t)\) itself drops from \(-2\) to \(0\).
At \(t=5\), \(y\) jumps up from \(0\) to \(3\), because this is where \(s=4-t=-1\), the point where \(x(t)\) jumps from \(-2\) to \(1\).

Step 6: Match with the given options.
Putting the pieces together, \(y(t)\) is: flat at \(1\) for \(t<-1\), a jump to \(3\) at \(t=-1\) followed by a straight fall to \(0\) at \(t=1\), flat at \(0\) from \(t=1\) to \(t=5\), a jump to \(3\) at \(t=5\) followed by a straight fall to \(1\) at \(t=7\), and flat at \(1\) for \(t>7\). The breakpoints \(t=-1,1,5,7\) and peak value \(3\) match exactly the shape drawn in option (B), whose axes are marked at these same values.
Option (A) keeps a peak of only \(1\) and dips near \(-2\), so it is really a plot of a reflected and shifted \(x(t)\) without the required amplitude flip and offset. Options (C) and (D) place their breakpoints near \(t=-4,-2,2,4\) instead of \(-1,1,5,7\), which is what you get from a sign slip in the shift, so they do not match our derivation either.

Final Answer:
The plot of \(y(t)=1-x(4-t)\) is shown in option (B).
\[ \boxed{\text{Option (B)}} \]