Sign In
Not register? Register Now!
Essay Available:
You are here: HomeLab ReportEngineering
5 pages/≈1375 words
3 Sources
Lab Report
English (U.S.)
MS Word
Total cost:
$ 27

Simulation and Filter Design (Lab Report Sample)

the task is ABOUT a lab report. source..
Name: Instructor: Course Title: Electrical Systems. Unit Title: Analog Filters. Lab: Low Pass Filter Network. Date: QUESTIONS 1 Obtain the transfer function V1V2 for the filter network shown below, plot its magnitude response and hence determine the kind of filter it implements. 2 Simulate the circuit in any software of your choice with V1 = 1 V peak to peak and frequencies 10 Hz to 20 kHz. Plot the frequency response versus input signal frequency for the output 3 31432547879000Compare your results for 1 and 2 above. right867410V200V2-1301751009015V100V1C1=C2=0.1 μF, R1=10 kΩ, R2=R4=1.6 kΩ, R3=5.66 kΩ OBJECTIVES The main aim of the 2 experiments is: 1 To obtain the transfer function V2/V1 for the filter network. 2 To plot the magnitude response of the filter network. 3 To determine the kind of filter that the network implements. SOLUTIONS Obtain the transfer function V2/V1 for the filter network shown below, plot its magnitude response and hence determine the kind of filter it implements. Circuit Diagram. OBTAINING THE TRANSFER FUNCTION OF THE FILTER Considering Node1, we have: Vin-VaR2 + Vo-Va1sC2 = Va-VbR4 Va-VbR4= Vin-VaR2 + sC2Vo-Va This becomes VinR2 = 1R2 + 1R4 + sC2Va - VbR4- sC2Vo…………………Equation 1 Considering Node2 we have Vb at the same potential. Therefore: Va-VbR4 = Vb1sC1 = sC1Vb Va-Vb = sC1R4Vb Va = Vb + sC1R4Vb = Vb (1+ sC1R4)……………………………Equation 2 Considering Node3 we have: VbR1=Vo-VbR3 VoR3= VbR1 + VbR3= Vb1R1 + 1R3= VbR1 + R3R1R3 Therefore, Vb=VoR3R1R3R1 + R3=VoR1R1 + R3………………………….…….Equation 3 Replacing Equation 2 in Equation 1 we get: VinR2 = 1R2 + 1R4 + sC2Va - VbR4- sC2Vo VinR2 = 1R2 + 1R4 + sC2Vb(1+ sC1R4) - VbR4- sC2Vo…………..Equation 4 Replacing Equation 3 in Equation 4 we get: VinR2 = 1R2 + 1R4 + sC2Vb (1+ sC1R4) - VbR4- sC2Vo VinR2 = 1R2 + 1R4 + sC2VoR1R1 + R3 (1+ sC1R4) - VoR1R1 + R3R4- sC2Vo VinR2 =Vo 1R2 + 1R4+ sC2R1R1 + R3 (1+ sC1R4)- R1R4 ((VoR1 + R3)-sC2Vo Collecting like terms and factoring out Vin and Vout from the above equation gives: Vout Vin=1 + R3R1R2R1+R3 . { S2(C1C2R2R4R4) +S C1R4 R2+R4 +C2R2R4-C2 R1 + R2+R4 -1R4} The values of the circuit components in the filter network are: R1=10k, R2=1.6k, R3=5.66k, R4=1.6k, C1=0.1µF, C2=0.1µF Replacing the above values in our transfer function equation we obtain the transfer function as: H{s} = 1.56625.056 x 10^6( 0.00004096 S2 + 0.768 S + 3,200.000625) H{s} = 1.5661,026.3 S2 + 19,243k S + 8.02X10^10) Looking at the above transfer function, it is similar to that of a second order Sallen-key low pass filter. The gain of the filter network is 1.566. PLOT OF MAGNITUDE RESPONSE. The transfer function was simulated to plot the magnitude response using the following commands: >>w=100:100:20000; >>s=j*w; >> h = (1.566)./(1026*s.^2 + 19243000*s + (8.02*10.^10)) >>plot (w,abs(h)) >> grid >> title ('A graph of Magnitude, H[s] against Frequeny, [Hz] ') >>ylabel ('Magnitude, H[s] ') >>xlabel ('Frequency, [Hz] ') The magnitude response curve above depicts the graph of a low pass filter. In this case, the network was Sallen-key topology. Thus, the filter network implements an active second order Sallen-key low-pass filter. SIMULATION 1 PROCEDURE: 1 The simulation circuit was connected as shown in the diagram below. 2 An input signal voltage V1 of 1 Volt (peak to peak) was applied to the circuit. 3 The output was measured for different input frequencies (100Hz to 10,000Hz). 4 For appropriate operation, increment steps of 100HZ for the range (100Hz -1,000Hz) and increment steps of 1,000Hz for (1,000Hz – 10,000Hz) are used. 5 The measured output voltages for the different frequencies are shown below: SIMULATION RESULTS. The following table shows the simulation results for the output voltage and the calculated frequency response in decibels (dB). Frequency (Hz) Output Voltage, V peak-peak Frequency Response (20log(Vo/Vi)) [dB] 100 0.5535 -5.138 200 0.5526 -5.152 300 0.5501 -5.191 400 0.5443 -5.283 500 0.5334 -5.459 600 0.5159 -5.749 700 0.4909 -6.180 800 0.4592 -6.760 900 0.4227 -7.479 1,000 0.3840 -8.313 2,000 0.1318 -17.602 3,000 0.0602 -24.408 4,000 0.0340 -29.370 5,000 0.0218 -33.231 6,000 0.0152 -36.363 7,000 0.0111 -39.094 8,000 0.0085 -41.412 9,000 0.0067 -43.479 10,000 0.0055 -45.193 THE PLOT OF MAGNITUDE RESPONSE Using AC Analysis in Multisim simulation, the graph of magnitude was obtained. The graph looks as shown below: COMPARISON OF RESULTS FOR 1 AND 2 ABOVE 1 COMPARISON OF THE TRANSFER FUNCTION From Multisim transfer function analysis, the input impedance of the circuit is 890.30253 M and the output impedance is 590.79288 u. These two impedance parameters on the input voltage side and the output voltage side give the transfer function as T.F= [Vout ÷ Vin] = 1.56594. The computed transfer function by derivation is: G = [1+ R3R1 ] = 1 + 5,660/10,000 = 1 + 0.566 = 1.566 Therefore, the transfer function from all the results is the same. 2 COMPARISON OF THE MAGNITUDE RESPONSE The graph from the simulation results depicts a realistic roll-off of a low pass high filter as compared to the graph from the experimental results. This is because there are minimal errors involved in simulation as compared to the actual experiment. In the actual experiment, operation of the circuit at higher frequencies in the range of (10^3) caused the output voltage to decrease steadily. This is because of the effect of the high impedance on the circuit loading which affects the outp...
Get the Whole Paper!
Not exactly what you need?
Do you need a custom essay? Order right now:

Other Topics:

Need a Custom Essay Written?
First time 15% Discount!