PURPOSE
Design an active second order Low-pass filter and characterize its frequency response.EQUIPMENT AND COMPONENTS
Oscilloscope
Power Supply
Function Generator
UA741 or LM741 OP AMP
4.7 nF Capacitors
Resistors
PRELAB
1. The RC circuit is shown in the following figure. Node Voltage Analysis is done in order to generate a transfer function Tvs=V2sV1s. In the circuit, voltage at node A is also the input voltage (V1=VA), and voltage at node D is also the output voltage (V2=VD).
KCL at node B:
V1-VBR1=VB-V21/sC1+VB-VCR2 (1)
KCL at node C:
VB-VCR2=VC1/sC2 (2)
V2=uVC→VC=V2u (3)
Simplifying (2):
VB=VC1+sR2C2 (4)
Then, substituting (3) to (4):
VB=V2u1+sR2C2 (5)
Substituting (3) and (5) to (1):
V1-V2u1+sR2C2R1=V2u1+sR2C2-V21sC1+V2u1+sR2C2-V2uR2
V1-V2u1+sR2C2R1=V2u1-u+sR2C21sC1+V2usR2C2R2
V1-V2u1+sR2C2R1=sC1V2u1-u+sR2C2+sC2V2uV1-V2u1+sR2C2R1=sV2uC1+C2-uC1+sR2C1C2
V1=sR1V2uC1+C2-uC1+sR2C1C2+V2u1+sR2C2
V1=V2u1+sR2C2+sR1C1+sR1C2-usR1C1+s2R1R2C1C2
V2V1=u1+sR2C2+sR1C1+sR1C2-usR1C1+s2R1R2C1C2
V2V1=us2R1R2C1C2+sR2C2+R1C1+R1C2-uR1C1+1
2. The equivalent op-amp based circuit derived from the previous model is shown in the next figure.
Let RA=R4, RB=R3. The transfer function for this circuit is:
Tvs=RA+RBRBs2R1R2C1C2+sR1C1+R2C1+R1C2-RARB+1
3. Simplifying the transfer function:
V2V1=u/R1R2C1C2s2R1R2C1C2+sR2C2+R1C1+R1C2-uR1C1+1/R1R2C1C2 =uR1R2C1C2s2+sR2C2+R1C1+R1C2-uR1C1R1R2C1C2+1R1R2C1C2
Comparing this to the generic form:
K=uR1R2C1C2
2ςω0=R2C2+R1C1+R1C2-uR1C1R1R2C1C2
ω02 = 1R1R2C1C2
4. The expressions ς and ω0 can be derived.
ω0= 1R1R2C1C2
2ςω0=R2C2+R1C1+R1C2-uR1C1R1R2C1C2
2ςω02ω0=R2C2+R1C1+R1C2-uR1C1R1R2C1C2 ×12R1R2C1C2
ς=R2C2+R1C1+R1C2-uR1C12R1R2C1C2
5. Unity gain K = 1.
K=1=uR1R2C1C2→u=R1R2C1C2
a. ω0= 1R1R2C1C2
b. ς=R2C2+R1C1+R1C2-R1R2C1C2R1C12R1R2C1C2=R2C2+R1C1+R1C2-R12R2C12C22R1R2C1C2
6. Equal R, C method: R1=R2, C1=C2, u=R+RR=2RR=2
a. ω0= 1R2C2
ω0=1RC
b. ς=RC+RC+RC-2RC2R2C2=RC2RC
ς=12
PROCEDURE, RESULTS AND ANALYSIS
Part A
1. Design an active Low Pass Filter with an OP-amp, resistors and capacitors.
The op-amp design will use the following circuit diagram.
2. All capacitor values should be limited to 4.7 nF.
3. The design should meet the following specifications:
i. ω0=7880 rad/s
ii. ς=0.5
Since ς=0.5, the equal R, C method will be used. To achieve the frequency,
ω0=7880=1RC=1R4.7nF→R=178804.7n=27kΩ
The design parameters are:
R1=R2=27kΩ, C1=C2=4.7nF, R3=R4=27kΩ
4. Find the Transfer Function and then design the circuit with components that are available in the lab.
The transfer function is:
Tvs=2s227k×4.7n2+s27k×4.7n+27k×4.7n+27k×4.7n-1+1
Tvs=21.61×10-8s2+1.27×10-4s+1
5. What is the value of the linear DC gain ‘k’?
The DC gain is k = 2.
6. For a second order filter design, what is more important, “k” or “ς”? Explain your answer.
For a second order filter design, ς is more important because it is directly related to the stability of the system. If ς is chosen carefully, the other parameters can work accordingly.
7. What is the difference in a narrowband and a broadband filter? Explain in terms of the damping ratio.
A narrowband filter has a small damping ratio, which gives way to a sharper response. A broadband filter has a large damping ratio, which gives way to a flatter and wider response.
8. Simulate the circuit first on MULTISIM. (Run the AC analysis from 1Hz to 100kHz. On your graph, make sure that both your axes are on the dB scale.)
9. Build the circuit and make sure to bias your Op-Amp with a ±12Vdc or lower.
10 What happens if you lower the DC input below ±12V?
The output waveform becomes distorted.
11. Keeping the input amplitude constant at 1V, measure the output and calculate the gain for frequencies given in Table 8.1.
12. Fill out Table 8.1.
Part B
1. Use MULTISIM analysis to show that the given circuit implements a band-stop gain response.
Use a Vac Source with a 1V input.
2. Run an AC Sweep from 1Hz to 50kHz.
3. Show the four graphs on four separate plots.
4. Find ωC1 and ωC2 from your graph.
In Hz,
fC1=920 Hz, fC2=1.57 kHz
In rad/s,
ωC1=2π920=5.78krads, ωC2=2π1.57k=9.86krads
5. Find the Stop-Band from your graph.
The stop-band is from fC1=920 Hz to fC2=1.57 kHz.
6. Explain in detail each of the three Op-Amp circuits’ function. {Analyze the individual outputs of the Op-Amp}
The first Op-amp network functions as a low-pass filter. The second Op-amp network functions as a high-pass filter. The two filters have the same cut-off frequencies, thus the resulting response is a stop band with a very small bandwidth. The third Op-amp network functions as a summing amplifier with gain of 1, in which the outputs of the first two filters are summed.
7. Determine the Transfer function of the gain response.The transfer function is simply the sum of the low pass filter transfer function and the high pass filter transfer function.
Tvs=2s2RC2+sRC+1+21+2RC+RC1-2R2C2RC∙1s +11RC2RC2∙1s2
Tvs=2s2RC2+sRC+1+21+1s+1s2=2s2RC2+sRC+1+2s2s2+s+1=
Tvs=21.61×10-8s2+1.27×10-4s+1+2s2s2+s+1
8. Run a MATLAB simulation using your Transfer function to draw a Bode Plot.The MATLAB code used is:
s = tf('s');
Tv = 2/(1.61e-8*s^2 + 1.27e-4*s+1)+ 2*s^2/(s^2+s+1);
bode(Tv)
The resulting bode plot is:
A band-stop filter response is observed.CONCLUSION
The methods of 2nd order active filter design has been implemented and analyzed in this laboratory. The computation of the component values for the function prototype is straightforward especially when using the equal R, C method. The band-stop filter is easily achieved by combining the outputs of a low-pass filter and a high-pass filter with the same cut-off frequency, and then summing them through a summing amplifier.
