Test 13.1 Fourier Series

In the Fourier series representation of the waveform, $\displaystyle f(t)=a_0+\sum_{n=1}^\infty(a_n\cos n\omega_0 t+b_n\sin n\omega_0 t), $

symmetry considerations dictate that:

a₀ = 0 and a_n = 0 for all n ≥ 1.
a₀ = 0 and b_n = 0 for all n.
a₀ ≠ 0 and b_n = 0 for all n.
a₀ ≠ 0 and a_n = 0 for all n ≥ 1.

Test 13.2 Fourier Series

In the Fourier series representation of the waveform, $\displaystyle f(t)=a_0+\sum_{n=1}^\infty(a_n\cos n\omega_0 t+b_n\sin n\omega_0 t), $

symmetry considerations dictate that:

a₀ = 0 and a_n = 0 for all n ≥ 1.
a₀ = 0 and b_n = 0 for all n.
a₀ ≠ 0 and b_n = 0 for all n.
a₀ ≠ 0 and a_n = 0 for all n ≥ 1.

Test 13.3 Fourier Series

In the Fourier series representation of the waveform, $\displaystyle f(t)=a_0+\sum_{n=1}^\infty(a_n\cos n\omega_0 t+b_n\sin n\omega_0 t), $

symmetry considerations dictate that:

a₀ = 0 and a_n = 0 for all n ≥ 1
a₀ = 0 and b_n = 0 for all n.
a₀ ≠ 0 and b_n = 0 for all n.
a₀ ≠ 0 and a_n = 0 for all n ≥ 1.

Test 13.4 Gibbs Phenomenon

The Gibbs phenomenon refers to:

circuits that exhibit an oscillatory behavior
symmetry considerations in waveforms of input signals
oscillations associated with the Fourier transform of waveforms with discontinuities
energy considerations of electric circuits

Test 13.5 Power Consumption

The current flowing through a 12 Ω resistor is $\displaystyle i(t)=[2+4\cos(377t-30^\circ)] A. $ What is the average power consumed by the resistor?

$P_{\rm av}=144$ W
$P_{\rm av}=72$ W
$P_{\rm av}=36$ W
$P_{\rm av}=18$ W

Test 13.6 Power Consumption

The current flowing through a 12 Ω resistor is $\displaystyle i(t)=[2+4\cos(377t-30^\circ)] A. $ What is the ac power fraction?

ac power fraction = 33.33%
ac power fraction = 50%
ac power fraction = 66.67%
ac power fraction = 100%

Test 13.7 Fourier Transform

Obtain the Fourier transform for the waveform shown, with A = 10 and T = 2 s.

${\bf F}(\omega)=\frac{10}{\omega^2}[e^{-j2\omega}(1+j2\omega)-1]$
${\bf F}(\omega)=\frac{5}{\omega^2}[e^{-j2\omega}+j2\omega e^{-j2\omega}-1]$
${\bf F}(\omega)=5[e^{-j2\omega}+j2\omega e^{-j2\omega}-1]$
${\bf F}(\omega)=10[e^{-j2\omega}+j2\omega e^{-j2\omega}+1]$

Test 13.8 Fourier Transform

Obtain the Fourier transform of the waveform shown, with A = 5 and T = 4 s.

${\bf F}(\omega)=5[1-\cos(2\omega)]$
${\bf F}(\omega)=5[1-\cos(4\omega)]$
${\bf F}(\omega)=\frac5{\omega^2}[1-\cos(4\omega)]$
${\bf F}(\omega)=\frac{10}{\omega^2}[1-\cos(2\omega)]$

Test 13.9 Fourier Transform Synthesis

Given the following Fourier transform pairs:

$\displaystyle {\color{red}\Leftrightarrow}\;{\bf F}(\omega) =\frac{4A}{T\omega^2}\left[ 1-\cos\left(\frac{\omega T}2\right)\right], $
synthesize the Fourier transform ${\bf G}(\omega)$ for the following waveform with $B=6$ and T = 4 s.

${\bf G}(\omega)=6\;{\rm sinc}(2\omega)+\frac6{\omega^2}\;[1-\cos(2\omega)]$
${\bf G}(\omega)=6\;{\rm sinc}(3\omega)+\frac2{\omega^2}\;[1-\cos(3\omega)]$
${\bf G}(\omega)=12\;{\rm sinc}(3\omega)+\frac3{\omega^2}\;[1-\cos(3\omega)]$
${\bf G}(\omega)=12\;{\rm sinc}(2\omega)+\frac3{\omega^2}\;[1-\cos(2\omega)]$

Test 13.10 Fourier Transform Synthesis

Given the following Fourier transform pairs:

$\displaystyle {\color{red}\Leftrightarrow}\;{\bf F}(\omega) =A\,\cos\left(\frac{\omega T}4\right) \left[\frac{4\pi/T}{(2\pi/T)^2-\omega^2}\right], $
synthesize the Fourier transform ${\bf G}(\omega)$ for the following waveform with $B=6$ and T = 8 s.

${\bf G}(\omega)=6\cos(2\omega) \left[\frac{\pi/2}{(\pi/4)^2-\omega^2}\right]-12\;{\rm sinc}(2\omega)$
${\bf G}(\omega)=3\cos(2\omega) \left[\frac{\pi/2}{(\pi/4)^2-\omega^2}\right]+12\;{\rm sinc}(2\omega)$
${\bf G}(\omega)=6\cos(3\omega) \left[\frac{\pi/2}{(\pi/4)^2-\omega^2}\right]-12\;{\rm sinc}(3\omega)$
${\bf G}(\omega)=3\cos(3\omega) \left[\frac{\pi/2}{(\pi/4)^2-\omega^2}\right]+12\;{\rm sinc}(3\omega)$

Test 13.11 Fourier Transform

Given that
$\displaystyle {\color{red}\Leftrightarrow} \frac{A}{T\omega^2}[e^{-j\omega T}(1+j\omega T)-1] $
and
$\displaystyle f(t-t_0) \;{\color{red}\Leftrightarrow}\; e^{-j\omega t_0}\;{\bf F}(\omega), $
what is the Fourier transform ${\bf G}(\omega)$ of the waveform

${\bf G}(\omega)= \displaystyle\frac A{2T\omega^2}\;[e^{-j\omega T}(1+j\omega T)-1]$
${\bf G}(\omega)= \displaystyle\frac A{T\omega^2}\;[e^{-j\omega T/2}(1+j\omega T)-1]$
${\bf G}(\omega)= \displaystyle\frac A{T\omega^2}\;[e^{-j3T/2}(1+j\omega T)-1]$
${\bf G}(\omega)= \displaystyle\frac A{2T\omega^2}\;[e^{-j3T/2}(1+j\omega T)-1]$

Test 13.12 Fourier Transform

Given that
$\displaystyle {\color{red}\Leftrightarrow} \frac A\omega\sin\left(\frac{\omega T}2\right) +\frac{2A}{T\omega^2} \;\left[1-\cos\left(\frac{\omega T}2\right)\right] $
and
$\displaystyle f(t-t_0) \;{\color{red}\Leftrightarrow}\; e^{-j\omega t_0}\;{\bf F}(\omega), $
what is the Fourier transform ${\bf G}(\omega)$ of the waveform

${\bf G}(\omega)=\displaystyle e^{-j\omega T/2}\left[\frac A\omega\sin\left(\frac{\omega T}2\right)+\frac{2A}{T\omega^2}\;\left[1-\cos\left(\frac{\omega T}2\right)\right]\right]$
${\bf G}(\omega)=\displaystyle e^{j\omega T/2}\left[\frac A\omega\sin\left(\frac{\omega T}2\right)+\frac{2A}{T\omega^2}\;\left[1-\cos\left(\frac{\omega T}2\right)\right]\right]$
${\bf G}(\omega)=\displaystyle e^{j\omega T/2}\left[\frac A\omega\sin\left(\frac{\omega T}2\right)-\frac{2A}{T\omega^2}\;\left[1-\cos\left(\frac{\omega T}2\right)\right]\right]$
${\bf G}(\omega)=\displaystyle e^{-j\omega T/2}\left[\frac A\omega\sin\left(\frac{\omega T}2\right)-\frac{2A}{T\omega^2}\;\left[1-\cos\left(\frac{\omega T}2\right)\right]\right]$

Test 13.13 Fourier Transform

What is the Fourier transform of $\displaystyle f(t)=[e^{-2(t-3)}\cos(4t)]\;u(t) $?

${\bf F}(\omega)=[2+j(\omega+3)]/\{[2+j(\omega+3)]^2+16\}$
${\bf F}(\omega)=e^6\{(2+j\omega)/[(2+j\omega)^2+16]\}$
${\bf F}(\omega)=e^{-6}\{(2+j\omega)/[(2+j\omega)^2+16]\}$
${\bf F}(\omega)=(2+j\omega/3)/[(2+j\omega/3)^2+16]$