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

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?

• The TF looks like a typical high-Q band-pass filter at about 36 Hz - do you not have solutions for those? – Andy aka Oct 26 '16 at 10:13
• I care more about how the maths is performed than solutions. I probably do have solutions locked away in the thousands of pages in my text books. – Supernovah Oct 26 '16 at 10:32
• I hate to say this… but the engineering way of math is looking it up. Better ask this at mathematics@stackexchange. – Janka Oct 26 '16 at 14:03
• You can probably split the transfer function to a sum of two transfer functions of which the inverse Laplace transform will be in the form of sin() + cos() – Mike Oct 26 '16 at 14:09

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

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

$A=2cos(0.7)$

$B=-2sin(0.7)$

$a=0$

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

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

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

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

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.