Finding H(s) for Circuit in s-Domain

  • Thread starter VinnyCee
  • Start date
  • Tags
    Circuit
In summary, the homework statement is to find the H() function for a circuit with an inductor. The attempt at a solution is to find the j\omega(Inductor Value) using V_1 and V_2 and the KCL equation. The equation for V_1 is V_1\left(2\,+\,j\omega\right)\,+\,\frac{V_1\left(1\,+\,j\omega\right)}{j\omega}, and the equation for V_2 is i_2\,=\,i_3\,+\,I_0(\omega).
  • #1
VinnyCee
489
0

Homework Statement



For the circuit below, find H(s) = [tex]\frac{I_0(s)}{I_S(s)}[/tex]

http://img108.imageshack.us/img108/7136/problem146td5.jpg

Homework Equations



Inductor in the s-domain: [itex]j\omega[/itex](Inductor Value)

The Attempt at a Solution



I made a new circuit diagram for the s-domain:

http://img293.imageshack.us/img293/7284/problem146part2gg0.jpg

[tex]i_1\,=\,\frac{V_1}{1\Omega}[/tex] [tex]i_2\,=\,\frac{V_1\,-\,V_2}{j\omega}[/tex] [tex]i_3\,=\,\frac{V_2}{j\omega}[/tex] [tex]I_0(\omega)\,=\,\frac{V_2}{1\Omega}[/tex]KCL @ [itex]V_1[/itex]:

[tex]I_S(\omega)\,=\,i_1\,+\,i_2[/tex]

[tex]I_S(\omega)\,=\,V_1\,+\,\frac{V_1\,-\,V_2}{j\omega}[/tex]KCL @ [itex]V_2[/itex]:

[tex]i_2\,=\,i_3\,+\,I_0(\omega)[/tex]

[tex]\frac{V_1\,-\,V_2}{j\omega}\,=\,\frac{V_2}{j\omega}\,+\,V_2[/tex]

After a couple of algebra moves(is it safe to multiply both sides by jw?), I get for [itex]V_1[/itex]:

[tex]V_1\,=\,V_2\left(2\,+\,j\omega\right)[/tex]

Substituting into the [itex]V_1[/itex] KCL equation:

[tex]I_S(\omega)\,=\,V_2\left(2\,+\,j\omega\right)\,+\,\frac{V_2\left(1\,+\,j\omega\right)}{j\omega}[/tex]

I also have this:

[tex]I_S(\omega)\,=\,i_1\,+\,i_3\,+\,I_0(\omega)[/tex]

[tex]\frac{I_0(\omega)}{I_S(\omega)}\,=\,1\,-\,\frac{i_1}{I_S(\omega)}\,-\,\frac{i_3}{I_S(\omega)}[/tex]

But how do I get rid of the i1 and i3? That would require ridding the equations of the V1 and V2 right? How to do that? I bet there is a much easier way of going about this. Does anyone know of an easier way?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
note [tex]I_0=V_2[/tex]
alternatively, you can use current divider formula a few times to solve this problem.
 
  • #3
Ok, let's try again...

I am going to use s in place of jw.

[tex]I_S\,=\,V_1\,+\,\frac{V_1\,-\,V_2}{s}[/tex]

[tex]I_o\,=\,\frac{V_1\,-\,V_2}{s}\,-\,\frac{V_2}{s}[/tex]

What do I do now? This problem really sucks, and it's the first one in the assignment! Please help.
 
  • #4
previously you arrived at
[tex]I_S(\omega)\,=\,V_2\left(2\,+\,j\omega\right)\,+\, \frac{V_2\left(1\,+\,j\omega\right)}{j\omega}[/tex]

while I haven't check your final answer, but methods are all good. from here as I said before [tex]I_0=V_2[/tex]... so sub in and solve for [tex]I_0/I_s[/tex]
will get same answer as your latest trial once you set [tex]I_0=V_2[/tex]
 
  • #5
[tex]I_S(\omega)\,=\,I_0\left(2\,+\,j\omega\right)\,+\, \frac{I_0\left(1\,+\,j\omega\right)}{j\omega}[/tex]

[tex]I_S(s)\,=\,I_0\left(2\,+\,s\right)\,+\,\frac{I_0\left(1\,+\,s\right)}{s}[/tex]

[tex]I_S(s)\,=\,I_0\left(3\,+\,s\,+\,\frac{1}{s}\right)[/tex]

Finally, I get:

[tex]\frac{I_0(s)}{I_S(s)}\,=\,\frac{1}{3}\,+\,\frac{1}{s}\,+\,s[/tex]

Is that the correct transfer function?
 
Last edited:
  • #6
one should be confident with one's work...again I don't see a problem with the methods... to avoid silly mistakes like 2+3=6, just check your workings...
 

1. What is H(s) in the s-domain?

H(s) is the transfer function of a circuit in the s-domain, which is a mathematical representation of the relationship between the input and output signals in the frequency domain.

2. How do I find H(s) for a circuit in the s-domain?

To find H(s), you will need to use circuit analysis techniques such as Kirchhoff's laws, mesh analysis, or nodal analysis to determine the circuit's impedance. Then, you can use this impedance to calculate the transfer function H(s) = Vout/Vin, where Vout is the output voltage and Vin is the input voltage.

3. Why is it important to find H(s) for a circuit in the s-domain?

Knowing the transfer function H(s) allows us to understand the behavior of a circuit in the frequency domain. This is crucial for designing and analyzing circuits in applications such as signal processing, communication systems, and control systems.

4. Can H(s) be found for any circuit in the s-domain?

Yes, H(s) can be found for any linear time-invariant (LTI) circuit in the s-domain. LTI circuits have the property that their response to a given input signal is the same regardless of when the input is applied, as long as the circuit is in a steady state.

5. What can we learn from the poles and zeros of H(s)?

The poles and zeros of H(s) provide important information about the stability and frequency response of a circuit. The location of poles and zeros in the s-plane can indicate the presence of resonances, which can affect the circuit's performance. Additionally, the number of poles and zeros can also determine the order of the transfer function and the complexity of the circuit.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
429
  • Introductory Physics Homework Help
Replies
2
Views
846
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
982
  • Engineering and Comp Sci Homework Help
Replies
6
Views
730
  • Introductory Physics Homework Help
Replies
2
Views
293
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
853
Back
Top