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One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function.

The transfer function:

$$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100} $$

This could be re-written as:

$$(s^2+12s+100)Y(s)=(6s+100)r(s) $$

Back to the time domain:

$$y''+12y'+100y=6r'+100r $$

Here, \$r(t)\$ is the unit step function. The derivative of the step function is the dirac delta function, so the differential equation becomes:

$$y''+12y'+100y=6\delta(t)+100, \text{for } t>0 $$

Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at t=0\$t=0\$. Then for t>0\$t>0\$:

$$y''+12y'+100y=100, \text{for } t>0 $$

After solving the differential equation (homogeneous and particular solutions added together),

$$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) $$$$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) \tag1$$

NowAnd the constants are found by using the initial conditions.

Here is the trick, you have to somehow account for the impulse (dirac delta function), right at t=0 (we have disregarded it so far). Laplace transform are usually defined from \$t=0^-\$ in orderin order to include the impulses (such as the dirac delta function). So we need to include this impulse if we want to get the same result one would with the Laplace transform.

This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includesincludes a dirac delta, you need to find the solution to the case when t>0\$t>0\$ (just what we did before in (1)), with the initial conditions \$y(0)=0\$, \$y'(0)=a/m\$ (for this case, \$a=6\$ and \$m=1\$). This is what brings in the fact the you have a dirac delta as part of the forcing function.

\$a\$ is just the coefficient multiplying the dirac delta function, 6function—6 in this problem. \$m\$ is the coefficient for the highest order derivative (\$y''\$ is highest order and its coefficient is 1).

With those initial conditions (\$y(0)=0\$, \$y'(0)=6\$), you can find the values of the constants, \$K_1\$ and \$K_2\$. These turn out to be \$K_1=-1\$ and \$K_2=0\$. Therefore:

$$y(t)=1-e^{-6t}\cos(8t) $$

A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis. Also, the nature of the dirac delta function is what makes this problem a bit trickier.

One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function.

The transfer function:

$$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100} $$

This could be re-written as:

$$(s^2+12s+100)Y(s)=(6s+100)r(s) $$

Back to the time domain:

$$y''+12y'+100y=6r'+100r $$

The derivative of the step function is the dirac delta function, so:

$$y''+12y'+100y=6\delta(t)+100, \text{for } t>0 $$

Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at t=0. Then for t>0:

$$y''+12y'+100y=100, \text{for } t>0 $$

After solving the differential equation,

$$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) $$

Now, you have to somehow account for the impulse (dirac delta function), right at t=0. Laplace transform are usually defined from \$t=0^-\$ in order to include the impulses. So we need this to get the same result one would with the Laplace transform.

This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includes a dirac delta, you need to find the solution to the case when t>0, with the initial conditions \$y(0)=0\$, \$y'(0)=a/m\$ (for this case, \$a=6\$ and \$m=1\$).

\$a\$ is just the coefficient multiplying the dirac delta function, 6 in this problem. \$m\$ is the coefficient for the highest order derivative (\$y''\$ is highest order and coefficient is 1).

With those initial conditions (\$y(0)=0\$, \$y'(0)=6\$), you can find the values of the constants, \$K_1\$ and \$K_2\$. These turn out to be \$K_1=-1\$ and \$K_2=0\$. Therefore:

$$y(t)=1-e^{-6t}\cos(8t) $$

A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis.

One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function.

The transfer function:

$$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100} $$

This could be re-written as:

$$(s^2+12s+100)Y(s)=(6s+100)r(s) $$

Back to the time domain:

$$y''+12y'+100y=6r'+100r $$

Here, \$r(t)\$ is the unit step function. The derivative of the step function is the dirac delta function, so the differential equation becomes:

$$y''+12y'+100y=6\delta(t)+100, \text{for } t>0 $$

Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at \$t=0\$. Then for \$t>0\$:

$$y''+12y'+100y=100, \text{for } t>0 $$

After solving the differential equation (homogeneous and particular solutions added together),

$$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) \tag1$$

And the constants are found by using the initial conditions.

Here is the trick, you have to somehow account for the impulse (dirac delta function), right at t=0 (we have disregarded it so far). Laplace transform are usually defined from \$t=0^-\$ in order to include the impulses (such as the dirac delta function). So we need to include this impulse if we want to get the same result one would with the Laplace transform.

This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includes a dirac delta, you need to find the solution to the case when \$t>0\$ (just what we did before in (1)), with the initial conditions \$y(0)=0\$, \$y'(0)=a/m\$ (for this case, \$a=6\$ and \$m=1\$). This is what brings in the fact the you have a dirac delta as part of the forcing function.

\$a\$ is just the coefficient multiplying the dirac delta function—6 in this problem. \$m\$ is the coefficient for the highest order derivative (\$y''\$ is highest order and its coefficient is 1).

With those initial conditions (\$y(0)=0\$, \$y'(0)=6\$), you can find the values of the constants, \$K_1\$ and \$K_2\$. These turn out to be \$K_1=-1\$ and \$K_2=0\$. Therefore:

$$y(t)=1-e^{-6t}\cos(8t) $$

A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis. Also, the nature of the dirac delta function is what makes this problem a bit trickier.

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One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function.

The transfer function:

$$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100} $$

This could be re-written as:

$$(s^2+12s+100)Y(s)=(6s+100)r(s) $$

Back to the time domain:

$$y''+12y'+100y=6r'+100r $$

The derivative of the step function is the dirac delta function, so:

$$y''+12y'+100y=6\delta(t)+100, \text{for } t>0 $$

Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at t=0. Then for t>0:

$$y''+12y'+100y=100, \text{for } t>0 $$

After solving the differential equation,

$$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) $$

Now, you have to somehow account for the impulse (dirac delta function), right at t=0. Laplace transform are usually defined from \$t=0^-\$ in order to include the impulses. So we need this to get the same result one would with the Laplace transform.

This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includes a dirac delta, you need to find the solution to the case when t>0, with the initial conditions \$y(0)=0\$, \$y'(0)=a/m\$ (for this case, \$a=6\$ and \$m=1\$).

\$a\$ is just the coefficient multiplying the dirac delta function, 6 in this problem. \$m\$ is the coefficient for the highest order derivative (\$y''\$ is highest order and coefficient is 1).

With those initial conditions (\$y(0)=0\$, \$y'(0)=6\$), you can find the values of the constants, \$K_1\$ and \$K_2\$. These turn out to be \$K_1=-1\$ and \$K_2=0\$. Therefore:

$$y(t)=1-e^{-6t}\cos(8t) $$

A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis.