Help Finding the Transfer Function H(s)=Vi(s)/Ii(s)

[Captain1024]

Homework Statement

Find the transfer function $H(s)=\frac{Vi(s)}{Ii(s)}$

The circuit consists of a voltage source and a 5H Inductor in series with a 10Ohm Resistor which are in parallel with a 10uF Capacitor in series with a 500Ohm Resistor. $\rightarrow$ Diagram here.

Homework Equations

Parallel Impedances: $Z_{eq}=\frac{Z1Z2}{Z1+Z2}$

$s=j\omega$

The Attempt at a Solution

[/B]
$5H\ \rightarrow\ j\omega5$
$10\mu F\ \rightarrow\ \frac{1}{j\omega10^{-5}}$
$Z_1=j\omega5+10$
$Z_2=\frac{1}{j\omega10^{-5}}+500$
$V_i=\frac{Z_1Z_2}{Z_1+Z_2}I$

$V_i=\frac{\frac{j\omega5}{j\omega10^{-5}}+j\omega2500+\frac{10}{j\omega10^{-5}}+5000}{j\omega5+\frac{1}{j\omega10^{-5}}+510}I$

$H(s)=\frac{V_i}{I}=\frac{5.05e5+(1.0025e6)s}{5s+\frac{10e5}{s}+510}$

 

[NascentOxygen]

Before you replace jw by s, multiply both numerator and denominator by jw to get a tidier form.

 

[Captain1024]

Before you replace jw by s, multiply both numerator and denominator by jw to get a tidier form.

I have to go to sleep now. I will be back at it tomorrow afternoon.

 

[NascentOxygen]

You could have used s right from the start.
5H → s5
10uF → 1/(s10^-5)

Then V(s) = Z(s).I(s)

 

[Captain1024]

Problem reworked:

$j\omega5\ \rightarrow\ 5s$
$\frac{1}{j\omega10^{-5}}\ \rightarrow\ \frac{1}{10^{-5}} \frac{1}{s}\ \rightarrow\ \frac{10^5}{s}$

$Z_1=5s+10$
$Z_2=\frac{10^5}{s}+500$
$V_i=Z_{eq} I_i$
$\frac{V_i}{I_i}=\frac{(5s+10)(\frac{10^5}{s}+500)}{5s+10+\frac{10^5}{s}+500}$

$=\frac{5e5+2500s+\frac{10^6}{s}+5000}{5s+\frac{10^5}{s}+510}$

$=\frac{5.05e5+2500s+\frac{10^6}{s}}{5s+\frac{10^5}{s}+510}(\frac{s}{s})$

$=\frac{2500s^2+(5.05e5)s+10^6}{5s^2+510s+10^5}$

Correct answer: $H(s)=\frac{0.25s^2+5.05s+10}{(5e-5)s^2+(5.1e-3)s+1}$

These two answers do not seem equivalent to me? Where am I going wrong?

 

[Captain1024]

I just verified by finding the zeros and poles. My answer matches. Thanks for the guidance.

 

[NascentOxygen]

When you don't have the correct answer to compare yours with, a simple check you can do is ask: what is H(s) when s=0. This is the DC value, which inspection of your circuit indicates is the inductor branch with its 10 ohms. Also, when s → infinity you have the high frequency impedance, here being the 500 ohms in the capacitive branch. If these two extremes of H(s) check okay, there's good reason to be optimistic the rest of your work is right.

 

[rude man]

Rather than introduce numbers early in the solution, you should retain the component symbols until the end.