Simple transfer function of circuit

In summary, the op amp provides a gain of 3 and the transfer function is 3*(8/(8+1/sC1)) or [24sC1/(8sC1 + 1)]. C2 and R5 have no effect on the transfer function, but the professor is asking for two corner frequencies at -3dB points at 80Hz and 28kHz. The output across the speaker is simply a high-pass filter, but adding a output resistor before the C2 leg will allow for two corner frequencies.
  • #1
Rob119
3
0
http://www.flickr.com/photos/61432156@N02/5593767202/

Ok so I'm trying to find the transfer function relating voltage across the 8 ohm resistor (speaker) with the input voltage for the circuit shown above...

The two op amps simply provide a gain of 3... and by simple voltage division I am getting the transfer function (Vspeaker / Vinput) to be 3*(8/(8+1/sC1)) or [24sC1/(8sC1 + 1)].

It looks to me as if C2 and R5 have no effect on the transfer function...

But my professor is asking to design C1 and C2 such that there are TWO corner frequencies (-3dB points) at 80Hz and 28kHz... how can this be possible if the output across the speaker is simply a high pass filter? Or did I calculate the transfer function wrong??
 
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  • #2
Rob119 said:
http://www.flickr.com/photos/61432156@N02/5593767202/

Ok so I'm trying to find the transfer function relating voltage across the 8 ohm resistor (speaker) with the input voltage for the circuit shown above...

The two op amps simply provide a gain of 3... and by simple voltage division I am getting the transfer function (Vspeaker / Vinput) to be 3*(8/(8+1/sC1)) or [24sC1/(8sC1 + 1)].

It looks to me as if C2 and R5 have no effect on the transfer function...

But my professor is asking to design C1 and C2 such that there are TWO corner frequencies (-3dB points) at 80Hz and 28kHz... how can this be possible if the output across the speaker is simply a high pass filter? Or did I calculate the transfer function wrong??

There does appear to be something missing. For the C2 leg to have an effect and roll off gain at high frequencies, the preamp has to have a finite and known output impedance. Are you given datasheet info for the preamp part? If it were ideal with zero output impedance, you'd need to add your own output resistor before the C2 leg.
 
  • #3
interesting didnt think of that... an lm741 has ~75 ohms output impedence (typically) until after like 30 kHz when it becomes much larger.

ill see if I can redo some calculations taking that output impedance into consideration.
 
  • #4
They want you to drive an 8 Ohm speaker with an LM741 opamp? Not practical. Maybe you can earn bonus points on the assignment pointing that out...
 
  • #5
well I am just modeling the speaker with an 8 ohm resistor and taking measurements, but I will point that out. After redoing the calculations with the 75 ohm output resistance my transfer function is a mess... i don't know what to make of it.

the lab is for a class but you do it at home.. and the professor doesn't answer emails. very frustrating.
 
  • #6
Rob119 said:
well I am just modeling the speaker with an 8 ohm resistor and taking measurements, but I will point that out. After redoing the calculations with the 75 ohm output resistance my transfer function is a mess... i don't know what to make of it.

the lab is for a class but you do it at home.. and the professor doesn't answer emails. very frustrating.

Here's a hint that may help with the transfer function. The low-frequency and high-frequency corners are far enough apart that you may be able to treat them seperately in fairly accurate approximations. Can you see any simplifications in your messy transfer function for low-frequency and high-frequency approximations?
 

FAQ: Simple transfer function of circuit

1. What is a simple transfer function of a circuit?

A simple transfer function of a circuit is a mathematical representation of the relationship between the input and output signals of a circuit. It describes how the output signal changes in response to changes in the input signal.

2. How is a simple transfer function calculated?

A simple transfer function is calculated by taking the ratio of the output signal to the input signal. This can be done using the Laplace transform, which converts the input and output signals from the time domain to the frequency domain.

3. What does a simple transfer function tell us about a circuit?

A simple transfer function provides information about the behavior of a circuit. It shows how the circuit amplifies or attenuates certain frequencies, and can also reveal the presence of resonance or instability in the circuit.

4. Can a simple transfer function be used to design a circuit?

Yes, a simple transfer function can be used to design a circuit. By manipulating the transfer function, designers can determine the necessary values for components such as resistors, capacitors, and inductors to achieve a desired output.

5. How does the complexity of a circuit affect its simple transfer function?

The complexity of a circuit can affect its simple transfer function in several ways. A more complex circuit may have a more complicated transfer function, making it more difficult to analyze and design. Additionally, the presence of nonlinear components in a circuit can make the transfer function nonlinear, which can complicate the analysis process.

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