The question asks to identify signals from a given list whose Continuous-Time Fourier Transform (CTFT) is purely imaginary. We are given a real-valued signal $x(t)$ with finite energy and $x(t)=0$ for $t<0$.

Understanding CTFT Properties

Let $X(\omega)$ be the CTFT of a real signal $x(t)$. We know the following properties:

• The CTFT of an even real signal is real.
• The CTFT of an odd real signal is purely imaginary.

Let's analyze the symmetry of each option and determine the nature of its CTFT.

Analysis of Options

Let $y(t)$ be the signal in each option, and $Y(\omega)$ its CTFT.

Option 1: $y_1(t) = x(t) + x(-t)$

• Symmetry: Since $x(t)$ is causal ($x(t)=0$ for $t<0$), $x(-t)=0$ for $t>0$.
  • For $t>0$, $y_1(t) = x(t) + 0 = x(t)$.
  • For $t<0$, let $t = -a$ where $a>0$. Then $y_1(-a) = x(-a) + x(-(-a)) = 0 + x(a) = x(a)$. Also, $y_1(a) = x(a) + x(-a) = x(a) + 0 = x(a)$. Thus, $y_1(-a) = y_1(a)$.
  • For $t=0$, $y_1(0) = x(0) + x(0) = 2x(0)$.
  Therefore, $y_1(t)$ is an even function.
• CTFT: The CTFT of a real even function is real. $Y_1(\omega) = \text{Re}\{X(\omega)\}$. This is not purely imaginary.

Option 2: $y_2(t) = x(t) - x(-t)$

• Symmetry:
  • For $t>0$, $y_2(t) = x(t) - 0 = x(t)$.
  • For $t<0$, let $t = -a$ where $a>0$. Then $y_2(-a) = x(-a) - x(-(-a)) = 0 - x(a) = -x(a)$. Also, $y_2(a) = x(a) - x(-a) = x(a) - 0 = x(a)$. Thus, $y_2(-a) = -y_2(a)$.
  • For $t=0$, $y_2(0) = x(0) - x(0) = 0$.
  Therefore, $y_2(t)$ is an odd function.
• CTFT: The CTFT of a real odd function is purely imaginary. $Y_2(\omega) = j \text{Im}\{X(\omega)\}$. This is purely imaginary.

Option 3: $y_3(t) = j(x(t) + x(-t))$

• Symmetry: $y_3(t) = j \cdot y_1(t)$, where $y_1(t)$ is the even function from Option 1.
• CTFT: $Y_3(\omega) = j \cdot Y_1(\omega)$. Since $Y_1(\omega)$ is real (from Option 1 analysis), $Y_3(\omega)$ is $j$ times a real number, which is purely imaginary.

Option 4: $y_4(t) = j(x(t) - x(-t))$

• Symmetry: $y_4(t) = j \cdot y_2(t)$, where $y_2(t)$ is the odd function from Option 2.
• CTFT: $Y_4(\omega) = j \cdot Y_2(\omega)$. Since $Y_2(\omega)$ is purely imaginary, let $Y_2(\omega) = j K(\omega)$ where $K(\omega)$ is real. Then $Y_4(\omega) = j (j K(\omega)) = -K(\omega)$. Since $K(\omega)$ is real, $-K(\omega)$ is real. This is not purely imaginary.

Conclusion

Based on the analysis, the signals whose CTFT is purely imaginary are:

• Option 2: $x(t) - x(-t)$ (an odd function)
• Option 3: $j(x(t) + x(-t))$ (j times the CTFT of an even function)

Therefore, the correct options are 2 and 3.