A real-valued signal x(t) limited to the frequency band $$\left| f \right| \le {W \over 2}$$ is passed through a linear time invariant system whose frequency response is $$H\left( f \right) = \left\{ {\matrix{ {{e^{ – j4\pi f,}}} & {\left| f \right| \le {W \over 2}} \cr {0,} & {\left| f \right| > {W \over 2}} \cr } } \right.$$ The output of the system is

[amp_mcq option1=”x(t + 4)” option2=”x(t – 4)” option3=”x(t + 2)” option4=”x(t – 2)” correct=”option1″]

The correct answer is A. $x(t+4)$.

The frequency response of the system is a delayed version of the unit impulse response, with a delay of $4$ samples. This means that the output of the system is the input signal delayed by $4$ samples.

To see this, consider the following example. Let the input signal be $x(t) = u(t)$, where $u(t)$ is the unit step function. The frequency response of the system is then

$$H(f) = e^{-j4\pi f}$$

The output of the system is then

$$y(t) = \int_{-\infty}^{\infty} x(t-\tau) H(\tau) d\tau = \int_{-\infty}^{\infty} u(t-\tau) e^{-j4\pi \tau} d\tau = e^{-j4\pi t} u(t) = x(t+4)$$

Therefore, the output of the system is the input signal delayed by $4$ samples.

The other options are incorrect because they do not take into account the delay of the system.