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

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1. Nov 28, 2014

### Captain1024

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

$s=j\omega$

3. The attempt at a solution

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

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

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

2. Nov 28, 2014

### Staff: Mentor

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

3. Nov 28, 2014

### Captain1024

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

4. Nov 29, 2014

### Staff: Mentor

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

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

5. Nov 29, 2014

### 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?

6. Nov 29, 2014

### Captain1024

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

7. Nov 29, 2014

### Staff: Mentor

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.

8. Nov 30, 2014

### rude man

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