Design Op Amp Circuit: Low-Pass Filter to G(s)=10/(s+10)

In summary, the conversation is about designing an op amp circuit that acts as a low-pass filter with a transfer function of G(s) = 10/(s+10). The conversation mentions using either the inverting or non-inverting equation, and the possibility of using a passive R-C low-pass filter with a voltage follower. It is suggested that the circuit would require 4 resistors and 2 capacitors with specific values, and the input would be through the positive input pin of the op amp.
  • #1
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Homework Statement


Design an op amp circuit that acts as low-pass filter with the following transfer function

G(s) = 10/(s +10)

Homework Equations



Inverting operational amplifier Vo(s)/Vi(s) = Z2(s)/Z1(s)

Non-inverting operational amplifier Vo(s)/Vi(s) = Z1(s) + Z2(s)/ Z1(s)

The Attempt at a Solution



The fact that there is no negative sign in the transfer function leads me to believe I need to use the non-inverting equation, but I can't seem to find any combination of components that cancel out and gives me the given transfer function.
 
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  • #2


I think you can get that single pole with a passive R-C low-pass filter. Can you buffer its output with a voltage follower?
 
  • #3


Takes 4 R's and 2 C's. The two C's have the same value, and 2 of the R's are 10 times the value of the other two. Yes, the input is via the + op amp input pin.
 

Related to Design Op Amp Circuit: Low-Pass Filter to G(s)=10/(s+10)

1. What is a design op amp circuit?

A design op amp circuit is a circuit that uses an operational amplifier (op amp) to perform a specific function, such as amplification, filtering, or signal conditioning. Design op amp circuits are commonly used in various electronic devices, including audio equipment, sensors, and control systems.

2. What is a low-pass filter?

A low-pass filter is a type of electronic filter that allows low-frequency signals to pass through while attenuating higher frequency signals. This is achieved by using a combination of resistors, capacitors, and/or inductors in the circuit. Low-pass filters are commonly used to remove noise from signals and to smooth out signal waveforms.

3. What does the transfer function G(s)=10/(s+10) represent in this circuit?

The transfer function G(s)=10/(s+10) represents the frequency response of the low-pass filter circuit. It describes how the output voltage of the filter changes in relation to the input voltage at different frequencies. In this case, the filter has a gain of 10 and a cutoff frequency of 10 Hz, meaning that it attenuates signals above 10 Hz and amplifies signals below 10 Hz.

4. How is the cutoff frequency of a low-pass filter determined?

The cutoff frequency of a low-pass filter is determined by the values of the resistors and capacitors used in the circuit. Generally, the cutoff frequency is equal to 1/(2πRC), where R is the resistance and C is the capacitance. In the case of the given transfer function, the cutoff frequency is 10 Hz, which can be achieved by using a resistor and capacitor with appropriate values.

5. What are the advantages of using a design op amp circuit for a low-pass filter?

Design op amp circuits offer several advantages for low-pass filters, including high accuracy, low distortion, and high stability. They also allow for flexibility in choosing the cutoff frequency and gain of the filter by adjusting the values of the components. Additionally, op amps have high input impedance and low output impedance, making them ideal for use in signal processing applications.

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