What Is the Transfer Function H(ω) for This Circuit?

In summary, the conversation was about finding the transfer function and determining the cutoff frequency and slope for a circuit with two energy storing elements. The participants discussed the correct form for the expression, the units, and how to approach finding the cutoff frequency. They also debated whether the circuit was a bandpass filter or not.
  • #1
gfd43tg
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Homework Statement


For the circuit shown below:
(a) Obtain an expression for H(ω) = Vo/Vi in standard form.
(b) Generate spectral plots for the magnitude and phase of H(ω), given that R1 = 1 Ω, R2 = 2 Ω, C1 = 1 μF, and C2 = 2 μF.
(c) Determine the cutoff frequency ωc and the slope of the magnitude (in dB) when ω/ωc ≪1 and when ω/ωc ≫1.

Homework Equations


The Attempt at a Solution


Hello, I am working on this problem part (a). I believe what they mean by ''standard form'' is that I need it to include (ω/ω_c), where ω_c is the corner frequency. This expression is so ugly that I don't know how I will ever find an expression for the corner frequency. Is there a way in general to approach finding the corner frequency?
 

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  • #2
Here is my attempt

ImageUploadedByPhysics Forums1397084399.482648.jpg
 
  • #3
EDIT: my table was wrong.

This is a bandpass transfer function with two corners so I don't know what they mean by "the cutoff frequency ". There is no one cutoff frequency ωc.

But you can do the Bode plots OK. I have the two corner frequencies & mid-band gain if you want to compare results.
 
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  • #4
I haven't done the bode plots because I don't know if my part (a) is correct
 
  • #5
Your units are not correct, therefore I know your answer is not correct.
The transfer function should be unit less. Yours is not.

your first two equations look correct.
Go through your work and check the units. You should be able to find your mistake.

Hint: get everything over one common denominator. your units are 1/(Ω^2). you want your units to be Ω/Ω
 
  • #6
I see now where I made that mistake, I will have to fix that. Also, once I get an expression, how will I be able to determine what my expression should be for the cut off frequency?
 
  • #7
edit:Due to the fact that there are two energy storing elements, there will be two break frequencies.

Also when presenting your equation reduce it so everything is over one common denominator. Then you can see the poles and zeros.
 
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  • #8
note: this is a band pass system. you will have both a low and high cutoff frequency.
 
  • #9
Maylis said:
I haven't done the bode plots because I don't know if my part (a) is correct

I would use s rather than jω in your transfer function. Bode plots are easier that way.
You should have used s to begin with in your analysis rather than jω anyway. I personally get a headache with all those complex numbers floating around!
 
  • #10
ImageUploadedByPhysics Forums1397232839.445913.jpg


ImageUploadedByPhysics Forums1397232860.228226.jpg


I have no idea how I would find 2 cutoff frequencies, I just tried to put my pole into the form of a quadratic and my Zero as a simple zero
 
  • #11
By the way guys, as far as I can tell and what the TA has pointed me to, this isn't a bandpass filter.
 
  • #12
Maylis said:
View attachment 68550

View attachment 68551

I have no idea how I would find 2 cutoff frequencies, I just tried to put my pole into the form of a quadratic and my Zero as a simple zero

A quadratic has how many solutions?
Your zero does not contribute to either corner frequency.
 
  • #13
Maylis said:
By the way guys, as far as I can tell and what the TA has pointed me to, this isn't a bandpass filter.

Just look at the diagram.
Put dc into the input. What is the dc output? (Hint: think C2).

Put a very high frequency into the input. What is the output? (Hint: think C1).

Either you have the wrong problem statement or the wrong TA!
 
  • #14
Maylis said:
By the way guys, as far as I can tell and what the TA has pointed me to, this isn't a bandpass filter.
If it is not a bandpass filter, the bode plot will show what it is.
 
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FAQ: What Is the Transfer Function H(ω) for This Circuit?

What is a transfer function?

A transfer function is a mathematical representation that describes the relationship between the input and output of a system or circuit. It is typically represented as a ratio of the output to the input, and it allows for analysis and prediction of the system's behavior.

How is the transfer function of a circuit calculated?

The transfer function of a circuit is calculated by taking the Laplace transform of the circuit's differential equations. This results in a function that represents the input-output relationship of the circuit in the frequency domain. The transfer function can also be found experimentally by applying different input signals to the circuit and measuring the resulting output.

What does the transfer function tell us about a circuit?

The transfer function provides important information about a circuit's behavior, such as its frequency response, stability, and damping. It can also be used to analyze the circuit's performance and design filters to achieve specific frequency responses.

Can the transfer function of a circuit change?

Yes, the transfer function of a circuit can change depending on the components and parameters of the circuit. For example, changing the resistance, capacitance, or inductance values in a circuit will result in a different transfer function. Additionally, non-linear components in a circuit can also affect the transfer function.

How is the transfer function used in circuit analysis?

The transfer function is used in circuit analysis to determine the response of a circuit to different input signals. By taking the inverse Laplace transform of the transfer function, the time-domain output of the circuit can be obtained. This allows for the prediction and optimization of a circuit's performance.

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