Inverse Laplace Transform of an LTI with complex roots and s + real term in the numerator

Question

I have this transfer function

$$H(s)=\frac{10(s+300)}{s^2+20s+50000}$$

and want to find the steady state response to an input of

$$x(t)=2\cos(2\pi30t+0.7)u(t)$$

But I am struggling to do anything with the transfer function.

I am familiar when I can get it to the form

$$\mathcal{L}\left[Ae^{-at}\cos{{\omega}t}+Be^{-at}\sin{{\omega}t}\right] = \frac{A(s+a) + B\omega}{(s+a)^2+\omega^2}$$

But clearly the \$a\$ terms will not match in the numerator and denominator. What do I do to find this response?

I have succesfully found \$h(t)\$ and I know it to be correct, but I'd like to know the steady state response. How is that determined?

Answer 1

\$x(t) = 2 cos( 2 \pi 30 t + 0.7) u( t )\$

\$x(t) = 2 cos( 0.7 )cos( 2 \pi 30 t + 0.7) u( t ) – 2 sin( 0.7 )sin( 2 \pi 30 t) u( t )\$

\$A=2cos(0.7)\$

\$B=-2sin(0.7)\$

\$a=0\$

\$X(s) = \frac{2 \cdot s \cdot cos(0.7) -2sin(0.7) \omega}{s^2+\omega^2} \cdot \frac{1}{s}\$

\$X(s) = 2 \frac{ cos(0.7)s -sin(0.7) \omega}{s^3+\omega^2 \cdot s}\$

Answer 2

I thought I'd post an answer since I figured this one out an it had a few views:

$$H(s)=\frac{10(s+300)}{s^2+20s+50000}$$

It should be immediately obvious the poles are complex, but we only need to know them to determine \$h(t)\$ later.

The input has a frequency \$\omega=2\pi30\$ in \$x(t)=2\cos(2{\pi}30t + 0.7)u(t)\$

Substituting for \$s=j\omega\$ we get

$$H(j\omega)=\frac{10j\omega+3000}{20j\omega+50000-\omega^2}$$

The magnitude and phase of this response at a chosen frequency affects the forced response's magnitude and phase. ie for a sinusoidal input:

$$u(t)=M_i\cos({\omega}t+{\phi}_i)\tag{1}$$ $$y(t)=M_o(\omega)\cos({\omega}t+{\phi}_o(\omega))\tag{2}\label{2}$$

Where the subscripts \$i, o\$ denote the input and output magnitudes and phases.

Substituting \$\omega=2{\pi}30\$ gives

$$\frac{3000+j600\pi}{50000-3600\pi+j1200\pi}\tag{3}$$ $$=0.09114{\angle}0.4639=M_o(\omega)\angle{\phi}_o(\omega)\tag{4}\label{4}$$

With \$\ref{4}\$ substituted into \$\ref{2}\$ The sinusoidal steady-state response is

$$x(t)=0.1823\cos(2{\pi}30t + 1.1622)u(t)$$

Which is the desired result.

The more interesting part is the time domain for the transfer funtion \$h(t)\$.

I will employ Laplace transforms

$$\mathcal{L}\left[Ae^{-at}\cos{{\omega}t}\right]=\frac{A(s+a)}{(s+a)^2+\omega^2}\tag{5}\label{5}$$

and

$$\mathcal{L}\left[Be^{-at}\sin{{\omega}t}\right]=\frac{B\omega}{(s+a)^2+\omega^2}\tag{6}\label{6}$$

or

$$\mathcal{L}\left[Ae^{-at}\cos{{\omega}t}+Be^{-at}\sin{{\omega}t}\right] = \frac{A(s+a) + B\omega}{(s+a)^2+\omega^2}\tag{7}\label{7}$$

Completing the square for \$D(s)\$ in \$H(s)=\frac{N(s)}{D(s)}\$ to reach the form of \$\ref{7}\$

$$H(s)=10\frac{(s+10) + 290}{(s+10)^2+49900}\tag{8}$$

And figuring out \$B\$ with \$A=1\$ gives

$$B=\frac{290}{\sqrt{49900}}\tag{9}$$

Then the time domain of the transfer by substituting values into the left side of \$\ref{7}\$ becomes

$$h(t)=10e^{-10t}\left(\cos(\sqrt{49900}t)+B\sin(\sqrt{49900}t)\right)\tag{10}$$

Letting \$C=\sqrt{A^2+B^2}=\sqrt{1^2+B^2}\$

$$h(t)=10e^{-10t}C\left(\frac{1}{C}\cos(\sqrt{49900}t)+\frac{B}{C}\sin(\sqrt{49900}t)\right)\tag{11}$$

Then \$\cos{\phi}=\frac{1}{C}\$ and \$\sin{\phi}=\frac{B}{C}\$

$$h(t)=10e^{-10t}C\left(\cos{\phi}\cos(\sqrt{49900}t)+\sin{\phi}\sin(\sqrt{49900}t)\right)\tag{12}$$

Using a trigonometric identity becomes

$$h(t)=16.3871e^{-10t}\cos(223.3831t - 0.9144)\tag{13}$$

Answer 3

Get the Bode plot.

From the plot we can see that at the frequency \$2 \pi 30 \ (=188.5)\$ of the input signal, the magnitude is about -12.5 db and the phase is about 17.5 degrees.

Thus we can say that the steady-state output value is

$$ 10^{-\frac{12.5}{20}} 2 \cos \left(2\pi30 t+0.7+\frac{17.5 \pi }{180}\right)$$

$$ = 0.474275 \cos (60 \pi t+1.00543)$$

This can be confirmed from actual simulations.

A zoomed-in view.