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Improved readability of fractions wit display style equations
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edit approved Mar 26, 2013 at 8:49
lxop
edit approved Mar 26, 2013 at 8:49
lxop
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$\frac{d^2y(t)}{dt^2}$+$\frac{3dy(t)}{dt}$ + $2y$ = $x(t)$

For the differential equation above, $$\frac{d^2y(t)}{dt^2}+\frac{3dy(t)}{dt} + 2y = x(t),$$

I was able to find the frequency response as $H(j\omega)=$ $\frac{1}{-\omega^2+3j\omega+2}$.$$H(j\omega)= \frac{1}{-\omega^2+3j\omega+2}.$$ However, I am not sure how to find the impulse response, which is the next part. Please help, and thank you in advance.

$\frac{d^2y(t)}{dt^2}$+$\frac{3dy(t)}{dt}$ + $2y$ = $x(t)$

For the equation above, I was able to find the frequency response as $H(j\omega)=$ $\frac{1}{-\omega^2+3j\omega+2}$. However, I am not sure how to find the impulse response, which is the next part. Please help, and thank you in advance.

For the differential equation $$\frac{d^2y(t)}{dt^2}+\frac{3dy(t)}{dt} + 2y = x(t),$$

I was able to find the frequency response as $$H(j\omega)= \frac{1}{-\omega^2+3j\omega+2}.$$ However, I am not sure how to find the impulse response, which is the next part. Please help, and thank you in advance.

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Shankar Kumar
Shankar Kumar
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Shankar Kumar
Shankar Kumar
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Impulse Response Question

$\frac{d^2y(t)}{dt^2}$+$\frac{3dy(t)}{dt}$ + $2y$ = $x(t)$

For the equation above, I was able to find the frequency response as $H(j\omega)=$ $\frac{1}{-\omega^2+3j\omega+2}$. However, I am not sure how to find the impulse response, which is the next part. Please help, and thank you in advance.