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1.1 | 0 / 20 |

1.2 | 0 / 20 |

1.3 | 0 / 20 |

1.4 | 0 / 20 |

1.5 | 0 / 20 |

1.6 | 0 / 20 |

1.7 | 0 / 20 |

1.8 | 0 / 20 |

1.9 | 0 / 20 |

1.10 | 0 / 20 |

1.11 | 0 / 20 |

1.12 | 0 / 20 |

1.13 | 0 / 20 |

1.14 | 0 / 20 |

1.15 | 0 / 20 |

1.16 | 0 / 20 |

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1.18 | 0 / 20 |

1.19 | 0 / 20 |

1.20 | 0 / 20 |

Total | 0 / 100 |

Solution:

The floor number is an integer, but it varies with continuous time.

The floor number is an integer, but it varies with continuous time.

Solution:

See Section 1-2.2. Note that $y(4)=x(2)$ and $y(0)=x(0)$.

See Section 1-2.2. Note that $y(4)=x(2)$ and $y(0)=x(0)$.

Solution:

$5\cos((t-(-6))/2)=5\cos(t/2+3)=x(t)$.

Note that the delay is $-6$, not 6.

$5\cos((t-(-6))/2)=5\cos(t/2+3)=x(t)$.

Note that the delay is $-6$, not 6.

Solution:

If $x(t)=A\cos(2\pi f_0t+\theta)$, $y(t)=A\cos(2\pi f_0(3t)+\theta)=A\cos(2\pi(3f_0)t+\theta)$.

If $x(t)=A\cos(2\pi f_0t+\theta)$, $y(t)=A\cos(2\pi f_0(3t)+\theta)=A\cos(2\pi(3f_0)t+\theta)$.

Solution:

Shift, then scale. Shift by $b$ means shift right, and multiplying time $t$ by $a$ compresses by $a$. See Section 1-2.4 for details.

Shift, then scale. Shift by $b$ means shift right, and multiplying time $t$ by $a$ compresses by $a$. See Section 1-2.4 for details.

Solution:

$\sqrt{2}\cos(t-45^\circ)=\sqrt{2}\cos(t)\cos(45^\circ )+\sqrt{2}\sin(t)\sin(45^\circ )$.

The even part is $\sqrt{2}\cos(t)\cos(45^\circ )=\cos(t)$.

Note that (c) and (d) are odd, so they couldn't be right.

$\sqrt{2}\cos(t-45^\circ)=\sqrt{2}\cos(t)\cos(45^\circ )+\sqrt{2}\sin(t)\sin(45^\circ )$.

The even part is $\sqrt{2}\cos(t)\cos(45^\circ )=\cos(t)$.

Note that (c) and (d) are odd, so they couldn't be right.

Solution:

From Section 1-3.4, the period of $A\cos(\omega_0t+\theta)$ is $2\pi/\omega_0$.

Here, $2\pi/2=\pi$.

From Section 1-3.4, the period of $A\cos(\omega_0t+\theta)$ is $2\pi/\omega_0$.

Here, $2\pi/2=\pi$.

Solution:

The period of $3\cos(2\pi t+1)$ is $\frac{2\pi}{2\pi}=1$.

The period of $5\cos(6\pi t+2)$ is $\frac{2\pi}{6\pi}=\frac{1}{3}$.

The 2nd term repeats 3 times while the first term repeats once, so the period of the combined expression is 1.

The period of $3\cos(2\pi t+1)$ is $\frac{2\pi}{2\pi}=1$.

The period of $5\cos(6\pi t+2)$ is $\frac{2\pi}{6\pi}=\frac{1}{3}$.

The 2nd term repeats 3 times while the first term repeats once, so the period of the combined expression is 1.

Solution:

See Eq. (1.18): $\displaystyle r(t)=\int_{-\infty}^t u(\tau)\;d\tau=t\;u(t). $

See Eq. (1.18): $\displaystyle r(t)=\int_{-\infty}^t u(\tau)\;d\tau=t\;u(t). $

Solution:

$-2r(t)$ levels off the ramp and starts it going down and $r(t-1)$ levels it off.

$-2r(t)$ levels off the ramp and starts it going down and $r(t-1)$ levels it off.

Solution:

See Eqs. (1.23) and (1.24) and Fig. 1-19(b), which are: \begin{eqnarray*} \frac{du(t)}{dt} &=& \delta(t),\\ u(t) &=& \int_{-\infty}^tu(\tau)\;d\tau. \end{eqnarray*}

See Eqs. (1.23) and (1.24) and Fig. 1-19(b), which are: \begin{eqnarray*} \frac{du(t)}{dt} &=& \delta(t),\\ u(t) &=& \int_{-\infty}^tu(\tau)\;d\tau. \end{eqnarray*}

Solution:

From Eq. (1.30), $\delta(2t-4)=\delta(2(t-2))=\frac{1}{2}\delta(t-2)$.

From Eq. (1.29), $\displaystyle \int_{-\infty}^{\infty}t\delta(2t-4)\;dt =\int_{-\infty}^{\infty}\frac{t}{2}\delta(t-2)\;dt= \int_{-\infty}^{\infty}\frac{2}{2}\delta(t-2)\;dt=1. $

From Eq. (1.30), $\delta(2t-4)=\delta(2(t-2))=\frac{1}{2}\delta(t-2)$.

From Eq. (1.29), $\displaystyle \int_{-\infty}^{\infty}t\delta(2t-4)\;dt =\int_{-\infty}^{\infty}\frac{t}{2}\delta(t-2)\;dt= \int_{-\infty}^{\infty}\frac{2}{2}\delta(t-2)\;dt=1. $

Solution:

$e^t$ blows up as $t\to\infty$, but decays to 0 as $t\to-\infty$. See Fig. 1-22(a):

$e^t$ blows up as $t\to\infty$, but decays to 0 as $t\to-\infty$. See Fig. 1-22(a):

Solution:

$x(t)=t$ for 0<t<1 and 0 elsewhere, since $r(t-1)$ levels off the initial ramp and $u(t-1)$ drops the result to 0. So the energy of $x(t)$ is $\displaystyle \int_0^1t^2\;dt=\frac{1}{3}. $

$x(t)=t$ for 0<t<1 and 0 elsewhere, since $r(t-1)$ levels off the initial ramp and $u(t-1)$ drops the result to 0. So the energy of $x(t)$ is $\displaystyle \int_0^1t^2\;dt=\frac{1}{3}. $

Solution:

From Eq. (1.38) the average power of a sinusoid with a nonzero frequency is half the square of its amplitude. Here $(4)^2/2=8$.

From Eq. (1.38) the average power of a sinusoid with a nonzero frequency is half the square of its amplitude. Here $(4)^2/2=8$.

Solution:

$x(-t/2+2)=x(\frac{-1}{2}(t$--4)). Expand by 2 and reverse, then delay by 4.

$x(-t/2+2)=x(\frac{-1}{2}(t$--4)). Expand by 2 and reverse, then delay by 4.

Solution:

Shift then scale: Advance by 1, then compress by 2, then reverse.

Shift then scale: Advance by 1, then compress by 2, then reverse.

Solution:

Energy of $x(t)$ is $\displaystyle \int_{-\infty}^{\infty}|x(t)|^2\;dt=\int_0^2(5t^2)^2\;dt =\int_0^2 25t^4\;dt=5t^5\big|_0^2=160. $

Energy of $x(t)$ is $\displaystyle \int_{-\infty}^{\infty}|x(t)|^2\;dt=\int_0^2(5t^2)^2\;dt =\int_0^2 25t^4\;dt=5t^5\big|_0^2=160. $

Solution:

Energy of one period is $\displaystyle \int_0^1|6t|^2\;dt=36\int_0^1t^2\;dt=12. $

Average Power $=\frac{\rm Energy}{\rm Period}={{12}\over 1}=12$.

Energy of one period is $\displaystyle \int_0^1|6t|^2\;dt=36\int_0^1t^2\;dt=12. $

Average Power $=\frac{\rm Energy}{\rm Period}={{12}\over 1}=12$.

Solution:

The system is stable because $\displaystyle \int_{-\infty}^{\infty}|h(t)|\;dt=2\int_0^{\infty}e^{-t}\;dt=2<\infty, $ but not causal because $h(t)\ne0$ for $t<0$. See Fig. 1-21:

The system is stable because $\displaystyle \int_{-\infty}^{\infty}|h(t)|\;dt=2\int_0^{\infty}e^{-t}\;dt=2<\infty, $ but not causal because $h(t)\ne0$ for $t<0$. See Fig. 1-21:

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