Refreshing stuff - transfer function

  • Thread starter liquidFuzz
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I'm trying to determine the transfer function to a simple circuit, see pdf. Is the simplest way to describe H(jω) by finding Y(jω) and U(jω) in H = Y/U. Further using the impedances for each component. If so, is each function Y(jω) and U(jω) determine by 'walking' through each impedance path giving something like this.

[itex]Y = \frac{1}{\frac{1}{Z_L} + \frac{1}{Z_C} }[/itex] and [itex]U = R +\frac{1}{\frac{1}{Z_L} + \frac{1}{Z_C} }[/itex]

[itex]Y(jω) = \frac{1}{\frac{1}{jωL} + \frac{1}{1/jωC} }[/itex] and [itex]U(jω) = R +\frac{1}{\frac{1}{jωL} + \frac{1}{1/jωC} }[/itex]

And next step simplify H(jω) to get the transfer function?

I'd really appreciate some help or pointers on this. Is this how it's done or..?
 

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  • #2
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I'm trying to determine the transfer function to a simple circuit, see pdf. Is the simplest way to describe H(jω) by finding Y(jω) and U(jω) in H = Y/U. Further using the impedances for each component. If so, is each function Y(jω) and U(jω) determine by 'walking' through each impedance path giving something like this.

[itex]Y = \frac{1}{\frac{1}{Z_L} + \frac{1}{Z_C} }[/itex] and [itex]U = R +\frac{1}{\frac{1}{Z_L} + \frac{1}{Z_C} }[/itex]

[itex]Y(jω) = \frac{1}{\frac{1}{jωL} + \frac{1}{1/jωC} }[/itex] and [itex]U(jω) = R +\frac{1}{\frac{1}{jωL} + \frac{1}{1/jωC} }[/itex]

And next step simplify H(jω) to get the transfer function?

I'd really appreciate some help or pointers on this. Is this how it's done or..?
Hello,

Assuming your Y and U functions are voltages, then your Y function is incorrect for starters. You have equated your Y and U voltages to resistances, which is an incorrect approach (it doesnt make sense to have volts on the left side and ohms on the right side). You got steps 1 and 2 mixed up:

Step 1: Define a complex impedance Z, which is equal to the C and L impedance in parallel. Basically lump your C and L into an equivalent single component. You attempted to do this, but you mistakenly equated the impedance to your voltage function Y.

Step 2: Define your output voltage Y as a function of your input voltage U. The C and L in parallel, Z, acts as a voltage divider with R on U(jw): Z/(Z+R). this should be straight forward: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

Step 3: Solve the transfer function from the above equation by dividing out U from both sides.
 
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  • #3
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liquidFuzz,

What is u(t), current or voltage? What is y(t), output admittance? The transfer function is determined by the components, not by the input voltage/current.

Ratch
 
  • #4
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Sorry! For violating Ohms law. :shy:

I didn't post any result in my last post, I get: [itex]H(jω) = \frac{ {jωL} }{{(jω)^2 CLR + R+jωL}}[/itex]

If I like to perform a Laplace transformation to get into the time domain, H(s) to h(t) to examining a pulse. How do I go from H(jω) to H(s)?
 
  • #5
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You are already there when you define s as s = jw. There is actually a real part too (s = jω + σ), but the sigma is ignored.
 
  • #6
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Sorry! For violating Ohms law. :shy:

I didn't post any result in my last post, I get: [itex]H(jω) = \frac{ {jωL} }{{(jω)^2 CLR + R+jωL}}[/itex]

If I like to perform a Laplace transformation to get into the time domain, H(s) to h(t) to examining a pulse. How do I go from H(jω) to H(s)?
That transfer function result is the same that I got :)
 

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