The floor indicator for an elevator is what type of signal?
Solution: The floor number is an integer, but it varies with continuous time.
Test 1.2
If $x(t)$ is 2 seconds long, how long is $y(t)=x(t/2)$?
Solution: See Section 1-2.2. Note that $y(4)=x(2)$ and $y(0)=x(0)$.
Test 1.3
$x(t)=5\cos(t/2+3)$ is a pure cosine $5\cos(t/2)$ delayed by
Solution: $5\cos((t-(-6))/2)=5\cos(t/2+3)=x(t)$. Note that the delay is $-6$, not 6.
Test 1.4
If $x(t)$ is a sinusoid with a nonzero frequency $f_0$ Hz,
$y(t)=x(3t)$ is a sinusoid with frequency (in Hz):
Solution: If $x(t)=A\cos(2\pi f_0t+\theta)$, $y(t)=A\cos(2\pi f_0(3t)+\theta)=A\cos(2\pi(3f_0)t+\theta)$.
Test 1.5
Given an expression for $x(t)$, to obtain an expression for
$y(t)=x(at-b)$, where $a>1$ and $b>0$, you should:
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.
Test 1.6
The even part of $\sqrt{2}\cos(t-45^\circ )$ is (Appendix C may
be useful):
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.
Test 1.7
The period of $4\cos(2t-1)$ is:
Solution: From Section 1-3.4, the period of
$A\cos(\omega_0t+\theta)$ is $2\pi/\omega_0$. Here, $2\pi/2=\pi$.
Test 1.8
The period of $3\cos(2\pi t+1)+5\cos(6\pi t+2)$ is:
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.
Test 1.9
Which of the following most completely relates a ramp $r(t)$ to
a step $u(t)$?
Solution:
See Eq. (1.18):
$\displaystyle
r(t)=\int_{-\infty}^t u(\tau)\;d\tau=t\;u(t).
$
Test 1.10
A triangle with height 1 centered at $t=0$, extending from
$t=-1$ to $t=1$, can be represented using ramps $r(t)$ as:
Solution: $-2r(t)$ levels off the ramp and
starts it going down and $r(t-1)$ levels it off.
Test 1.11
Which of the following most completely relates an impulse
$\delta(t)$ to a step $u(t)$?
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*}
Test 1.12
$\int_{-\infty}^{\infty}t\delta(2t-4)\;dt=$?
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.
$
Test 1.13
$x(t)=e^t\;u(-t)$ does which of these:
Solution:
$e^t$ blows up as $t\to\infty$,
but decays to 0 as $t\to-\infty$. See Fig. 1-22(a):
Test 1.14
The energy of $x(t)=r(t)-r(t-1)-u(t)$ is:
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}.
$
Test 1.15
The average power of $4\cos(3t-1)$ is:
Solution: From Eq. (1.38) the average power
of a sinusoid with a nonzero frequency is half the square of its
amplitude. Here $(4)^2/2=8$.
Test 1.16
If $x(t)$ is:
then which plot represents $y(t)=x(-t/2+2)$?
Solution: $x(-t/2+2)=x(\frac{-1}{2}(t$--4)).
Expand by 2 and reverse, then delay by 4.
Test 1.17
If $x(t)$ is:
then which plot represents $y(t)=x(-2t+1)$?
Solution: Shift then scale: Advance by 1,
then compress by 2, then reverse.
Test 1.18
The energy of the signal $x(t)=5t^2$, 0<t<2 with $x(t)=0$ otherwise, is:
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.
$
Test 1.19
The average power of the periodic signal (maximum value=6) shown
below is:
Solution: Energy of one period is
$\displaystyle
\int_0^1|6t|^2\;dt=36\int_0^1t^2\;dt=12.
$
Average Power $=\frac{\rm Energy}{\rm Period}={{12}\over 1}=12$.
Test 1.20
An LTI system has an impulse response $h(t)=e^{-|t|}$. The
system is:
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:
then which plot represents $y(t)=x(-t/2+2)$?
then which plot represents $y(t)=x(-2t+1)$?
