Signals Fundamentals: 15 MCQs with answers

0

The argument $-t+4$ is negative for $t > 4$. The unit step $u(\text{negative}) = 0$, so $x(t)=0$ for $t > 4$.

4 seconds

The standard rectangular pulse $\Pi(t)$ has duration 1. Scaling the argument by $\frac{1}{4}$ stretches the duration by a factor of 4, so the duration is 4 seconds.

6

Amplitude scaling by 3 multiplies the peak amplitude by 3. The original peak is 2, so the new peak is $3 \times 2 = 6$.

$\frac{1}{3}$ seconds

For $\cos(\omega_0 t)$, the fundamental period is $T_0 = \frac{2\pi}{\omega_0}$. Here $\omega_0 = 6\pi$, so $T_0 = \frac{2\pi}{6\pi} = \frac{1}{3}$ seconds.

$r(t-4)$

The unit ramp $r(t)$ is zero for $t

True

The constant 2 does not affect periodicity. The sine term has frequency $4\pi$ rad/s, so $T_0 = \frac{2\pi}{4\pi} = \frac{1}{2}$ seconds.

$x(t) = -t + 4$

For $2 \le t < 4$, $r(t) = t$ and $r(t-2) = t-2$, while $r(t-4) = 0$. So $x(t) = t - 2(t-2) = t - 2t + 4 = -t + 4$.

2

At $t = -1$, the argument is $2(-1) + 2 = 0$. Since $x(0) = 2$ (peak), $x(2t+2)$ at $t=-1$ equals 2.

$x(t) = 2u(t) + 3\cos(2\pi t)$

The unit step $u(t)$ is not periodic (it is constant for $t>0$ and zero for $t

$\frac{2}{25}$ seconds

The period of $\cos(50\pi t)$ is $\frac{2\pi}{50\pi} = \frac{1}{25}$ s. The period of $\sin(25\pi t)$ is $\frac{2\pi}{25\pi} = \frac{2}{25}$ s. The fundamental period is the LCM of $\frac{1}{25}$ and $\frac{2}{25}$, which is $\frac{2}{25}$ seconds.

1

At $n=0$: $u[-0+2] = u[2] = 1$ and $u[-0-2] = u[-2] = 0$. So $x[0] = 1 - 0 = 1$.

Zero for $t > 4$

The argument $-t+4$ is negative when $t > 4$. Since $r(\text{negative}) = 0$, $x(t) = 0$ for $t > 4$. For $t < 4$, the argument is positive, so $x(t) = -t+4$.

$x(t) = 0$

For $t > 3$, the argument of $u(-t+3)$ is negative, so $u(-t+3) = 0$. Multiplying by $r(t)$ gives $x(t) = 0$.

False

The unit step $u(t)$ is not periodic because it is not repetitive—it is zero for $t0$. Adding a periodic cosine does not make the sum periodic.

Periodic with $T_0 = \pi$ seconds