Test 12.1 Impulse Function

Evaluate $\displaystyle y=\int_0^4(e^{-4t}+1)\;\delta(t-2)\;dt $

y = e^-8/4 + 2
y = e⁸ + 1
y = e⁸/4 - 1
y = e^-8 + 1

Test 12.2 Impulse Function

Evaluate $\displaystyle y=\int_0^1(e^{-4t}+1)\;\delta(t-2)\;dt $

y = e^-8/4 + 2
y = 0
y = e⁸/4 + 1
y = e^-8 + 1

Test 12.3 Laplace Transform

Given the Laplace transform pair:   $\displaystyle u(t-T) \;{\color{red}\Leftrightarrow}\; \frac{e^{-T{\bf s}}}{\bf s}\;, $ obtain the Laplace transform of f(t):

${\bf F}({\bf s})=\frac4{\bf s}-\frac6{\bf s}\;e^{-5\bf s}+2 e^{-10\bf s}$
${\bf F}({\bf s})=\frac1{\bf s}(4e^{-5\bf s}-6 e^{-10\bf s}+2)$
${\bf F}({\bf s})=4e^{-5\bf s}-6e^{-10\bf s}+2 e^{-15\bf s}$
${\bf F}({\bf s})=\frac1{\bf s}\;e^{-5\bf s}(4-6e^{-5\bf s}+2e^{-10\bf s})$

Test 12.4 Laplace Transform

Given the Laplace transform pair:   $\displaystyle t^{n-1}e^{-at}\;u(t) \;{\color{red}\Leftrightarrow}\; \frac{(n-1)!}{({\bf s}+a)^n}\;, $ obtain the Laplace transform of $f(t)=5t^2 e^{-3t}\;u(t)$:

${\bf F}({\bf s})=\frac5{({\bf s}+2)^2}$
${\bf F}({\bf s})=\frac{10}{({\bf s}+3)^2}$
${\bf F}({\bf s})=\frac{10}{({\bf s}+3)^3}$
${\bf F}({\bf s})=\frac5{({\bf s}+2)^3}$

Test 12.5 Laplace Transform

Given the Laplace transform pairs:

$\displaystyle e^{-at}\sin\omega t\;u(t) \;{\color{red}\Leftrightarrow}\; \frac\omega{({\bf s}+a)^2+\omega^2} $
$\displaystyle t\;f(t) \;{\color{red}\Leftrightarrow}\; -\frac d{d\bf s}\;{\bf F}({\bf s}), $
Obtain the Laplace transform of $g(t)=5te^{-3t}\sin2t\;u(t)$.

${\bf G}({\bf s})=\frac{20({\bf s}+3)}{[({\bf s}+3)^2+4]^2}$
${\bf G}({\bf s})=\frac{10({\bf s}+2)}{({\bf s}+2)^2+9}$
${\bf G}({\bf s})=\frac{20}{[({\bf s}+3)^2+4]^2}$
${\bf G}({\bf s})=\frac{20({\bf s}+3)}{({\bf s}+3)^2+4}$

Test 12.6 Inverse Laplace Transform

Obtain the inverse Laplace transform of $\displaystyle {\bf F}({\bf s})=\frac2{\bf s}-\frac{8{\bf s}}{{\bf s}^2+16}\;. $

$f(t)=2-4\cos2t$
$f(t)=[2-8\cos4t]\;u(t)$
$f(t)=[2+8\sin4t]\;u(t)$
$f(t)=[2-4\sin4t]\;u(t)$

Test 12.7 Inverse Laplace Transform

Obtain the inverse Laplace transform of $\displaystyle {\bf F}({\bf s})=3e^{-2\bf s}+\frac4{{\bf s}+6}\;. $

$f(t)=3u(t-2)+4e^{-6t}\;u(t)$
$f(t)=3\delta(t+2)+6e^{-4t}\;u(t)$
$f(t)=3\delta(t-2)+4e^{-6t}\;u(t)$
$f(t)=3\delta(t-2)+6e^{-4t}\;u(t)$

Test 12.8 RL Circuit

The circuit is excited by a 10 V, 2 s long rectangular pulse. Given that R = 1 Ω and L = 1/3 H, determine i(t).

$i(t)=[10(1-e^{-3t})\;u(t)-10(1-e^{-3(t-2)})\;u(t-2)]$ A
$i(t)=[10e^{-3t}\;u(t)-10e^{-3(t-2)})\;u(t-2)]$ A
$i(t)=[10(1-e^{-t/3})\;u(t)-10(1-e^{-(t-2)/3})\;u(t-2)]$ A
$i(t)=[10(1-e^{-t})-e^{-(t-2)}]$ A

Test 12.9 RC Circuit

The circuit is excited by a 10 V, 2 s long rectangular pulse. Given that R = 4 Ω and C = 0.5 F, determine i(t).

$i(t)=2.5[e^{-0.5t}\;u(t)-e^{-0.5(t-2)}\;u(t-2)]$ A
$i(t)=[2.5e^{-0.5t}-e^{-0.5t}\;u(t-2)]$ A
$i(t)=5[e^{-t}\;u(t)-e^{-(t-2)}\;u(t-2)]$ A
$i(t)=5[e^{-2t}\;u(t)-e^{-2(t-2)}\;u(t-2)]$ A

Test 12.10 RC Circuit

The circuit is excited by current waveform $\displaystyle i_{\rm s}(t)=4[1-e^{-2t}]\text{ A}. $ Given that RC = 0.2, and C = 0.5 F, determine i(t).

$i(t)=\frac83(e^{-2t}-e^{-5t})\;u(t)]$ A
$i(t)=4(1-e^{-5t})\;u(t)]$ A
$i(t)=\frac13(e^{-2t}-1)\;u(t)]$ A
$i(t)=\frac23(e^{-2t}-1)\;u(t)]$ A

Test 12.11 Impulse Response

Determine $\upsilon_{\rm C}(t)$, given that R = 200 Ω and C = 1 mF.

$\upsilon_{\rm C}(t)=50e^{-2t}$ V
$\upsilon_{\rm C}(t)=250e^{-5t}$ V
$\upsilon_{\rm C}(t)=50e^{-5t}$ V
$\upsilon_{\rm C}(t)=250e^{-2t}$ V

Test 12.12 Impulse Response

Determine $\upsilon_{\rm R}(t)$, given that R = 200 Ω and C = 1 mF.

$\upsilon_{\rm R}(t)=25e^{-5t}$ V
$\upsilon_{\rm R}(t)=50e^{-5t}$ V
$\upsilon_{\rm R}(t)=-250e^{-5t}$ V
$\upsilon_{\rm R}(t)=-50e^{-5t}$ V

Test 12.13 RC Circuit

The circuit is excited by current waveform $\displaystyle i_{\rm s}(t)=4[1-e^{-2t}]\text{ A}. $ Given that RC = 0.2, and C = 0.5 F, determine i(t).

$i(t)=\frac83(e^{-2t}-e^{-5t})\;u(t)]$ A
$i(t)=4(1-e^{-5t})\;u(t)]$ A
$i(t)=\frac13(e^{-2t}-1)\;u(t)]$ A
$i(t)=\left[4-\frac{20}3e^{-2t}+\frac83e^{-5t}\right]\;u(t)]$ A

Test 12.14 Delayed Response

When excited by an input voltage source given by $\displaystyle \upsilon_{\rm in}(t)=5u(t), $ a linear circuit's output voltage is $\displaystyle \upsilon_{\rm out}(t)=[1.5-1.56e^{-4t}+0.072e^{-12t}]\;u(t)\text{ V}. $
If the input is delayed by 2 s such that $\displaystyle \upsilon'_{\rm in}(t)=5u(t-2), $ what would be the corresponding output voltage $\upsilon'_{\rm out}(t)$?

$\upsilon'_{\rm out}(t)=[1.5-1.56e^{-4t}+0.072e^{-12t}]\;u(t-2)$ V
$\upsilon'_{\rm out}(t)=[1.5-1.56e^{-4(t-2)}+0.072e^{-12(t-2)}]\;u(t)$ V
$\upsilon'_{\rm out}(t)= [1.5-1.56e^{-4(t-2)}+0.072e^{-12(t-2)}]\;u(t-2)$ V
$\upsilon'_{\rm out}(t)=[1.5e^{-(t-2)}-1.56e^{-4(t-2)}+0.072e^{-12(t-2)}]\;u(t)$ V

Test 12.15 Delayed Response

When excited by a sinusoidal current source that started at t = 0, the output response of a linear circuit is given by $\displaystyle i_{\rm out}(t)=[0.3+0.2e^{-15t}+0.4\cos(4t-15^\circ)]\;u(t)\text{ A}. $
If the input excitation were to be delayed by 2 s, what would be the corresponding output current $i'_{\rm out}(t)$?

$i'_{\rm out}(t)=[0.3+0.2e^{-15t}+0.4\cos(4t-15^\circ)]\;u(t-2)$ A
$i'_{\rm out}(t)=[0.3+0.2e^{-15(t-2)}+0.4\cos(4(t-2)-15^\circ)]\;u(t-2)$ A
$i'_{\rm out}(t)=[0.3+0.2e^{-15(t-2)}+0.4\cos(4(t-2)-15^\circ)]\;u(t)$ A
$i'_{\rm out}(t)=[0.3+0.2e^{-15(t-2)}+0.4\cos(4t-2-15^\circ)]\;u(t-2)$ A