# How to find analytic description of filtered signal

I am looking for an exact analytic description of a filtered signal. I have an electronic circuit whose input is a monoexponential decay. First (1) the signal gets filtered by a simple RC-Lowpass. Then (2) a second order Bessel Low Pass Filter is applied.

The respective transfer functions are: \begin{align} TF_{\rm RC}(s) &= \frac{R}{s+\frac{1}{R C}}\\ TF_{\rm Bessel}(s) &= \frac{a_1}{a_1+b_1s+s^2} \end{align}

The input signal is: $$I_{\rm in}(t) = \frac{V}{R_1 + R_2} + R_1\cdot V_{\rm cmd}\cdot e^{-t\frac{R_1 + R_2}{CR_1R_2}}\frac{1}{R_2(R_1 + R_2)}$$

My idea was to calculate the output in the $s$-domain: $$V_{\rm out}(s) = I_{\rm in}(s)\cdot TF_{RC}(s)\cdot TF_{Bessel}(s)$$ and then apply the inverse laplace transform to obtain $V_{\rm out}$(t). The inverse transform is what troubles me. I tried to calculate it with MATLAB, but it doesn't return an analytic function. I guess this comes from the complex poles of the bessel filter. Could you give me a hint on how to calculate the output?

Edit: $$TF(s) = \frac{6.092e12}{s^3 + 5.367e04*s^2 + 1.038e09*s + 6.074e12}$$

$$I_{in}(t) = \frac{10e-3}{500e6 + 10e6} + 500e6*10e-3*e^{-t*\frac{500e6 + 10e6}{10e-12*500e6*10e6}}\frac{1}{10e6*(500e6 + 10e6)}$$

$$TF(s)*I_{in}(s) = \frac{4.9e-21*(1.2e24*s + 2.5e26)}{s*(s + 1.0e4)*(s^3 + 5.4e4*s^2 + 1.0e9*s + 6.1e12)}$$

PFE yields:

$$TF(s)*I_{in}(s) = \frac{1.1e-9 - 4.8e-11i}{s + 2.2e4 - 1.1e4i} + \frac{1.1e-9 + 4.8e-11i}{s + 2.2e4 + 1.1e4i} + \frac{6.6223e-7}{s + 10200.0} + \frac{-6.6441e-7}{s + 10235.0} + \frac{1.9666e-11}{s}$$

Inverse Laplace transform of the upper term will yield a complex result. I know that I can simplify the first two terms by bringing them to a common denominator and thereby get rid of the complex numbers. The inverse transform is however still complex.

Edit: calculation with symbolic parameters yields: $$H(s) = \frac{a*s+b}{c*s^2+d*s+f}$$ $$H(t) = a*dirac(1, t) + (2*b*sin((t*(- d^2 + 4*c*f)^{1/2})/(2*c))*exp(-\frac{d*t}{2*c}))\frac{1}{(4*c*f - d^2)^{1/2}}$$

I don't quite understand where the dirac comes from, but this is basically the same result as doing the calculation with complex numbers and the taking the real part.

• Use partial fraction expansion and apply inverse Laplace transform to each term separately. – Matt L. Mar 4 '17 at 10:53
• Thanks for the hint. I did as you said, however the symbolic function I obtain is complex, whereas the actual output signal is of course real. Should I just take the real part? – luis Mar 4 '17 at 15:32
• No, maybe you forgot to combine the two complex conjugate poles, which must result in a real-valued signal. If you do things right, the result must be real-valued because your input signal and both impulse responses are real-valued. – Matt L. Mar 4 '17 at 16:42
• What do you mean with 'combine' ? Obviously there are roots which are complex, so the signal in the time domain also contains complex numbers. The real part of this result however yields the correct signal, can't be that wrong. – luis Mar 4 '17 at 18:14
• Allright so instead of having to terms with complex conjugate residues and poles I multiplied them to obtain one term with a second order polynomial in the denominator. Inverse Laplace transform still yields a complex number. – luis Mar 5 '17 at 9:29

Assuming your algebra is correct, there is nothing wrong with complex numbers in the Laplace domain. Complex poles reflect an oscillatory signal - there is no oscillation at all if all poles are strictly real, just exponential growth or decay. The variable $s$ is itself a complex number, and the transfer function is defined over all $s$, i.e. the entire complex plane.
• Which terms, exactly, yield the complex results? I am seeing that you have two terms whose denominators are complex conjugates. Combine these two terms into a single one to remove the $i$ from any denominator. Now, this term will correspond to the oscillatory mode. I am actually having trouble finding a table of Laplace transforms that contains complex numbers in the real domain. – Radu S. Visina Mar 5 '17 at 16:10