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# Math Problem Assignment: Circuit and Network Theory III (Math Problem Sample)

Instructions:

a). Design a two stage low pass filter with a d.c gain of 8, a bandwidth of 2000 Hz and a suitable roll- off rate (dB/dec).
b). Draw the circuit and analyse how the specs will be achieved.
c). Calculate the practical values of the components using the standard values of resistors and capacitors.
d). Calculate the voltages at each node as output for an input sinusoidal signal of 10V.
e). Draw the Bode plots and discuss whether the requirements have been achieved.
NB: Ensure that the cut-off gain is not below -3dB point.

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Content:

Name:
Instructor:
Course Title: Electrical Engineering
Unit Title: Circuit and Network Theory III
Assignment: Filter Design
Date:
QUESTION
a). Design a two stage low pass filter with a d.c gain of 8, a bandwidth of 2000 Hz and a suitable roll- off rate (dB/dec).
b). Draw the circuit and analyse how the specs will be achieved.
c). Calculate the practical values of the components using the standard values of resistors and capacitors.
d). Calculate the voltages at each node as output for an input sinusoidal signal of 10V.
e). Draw the Bode plots and discuss whether the requirements have been achieved.
NB: Ensure that the cut-off gain is not below -3dB point.
SOLUTIONS
a). Designing a two stage low pass filter with a d.c gain of 8, a bandwidth of 2000 Hz and a suitable roll- off rate (dB/dec);
* Choose two low pass filters with a roll-off rate of -20 dB/dec. These two low pass filters will be cascaded to give a roll-off rate of -40dB/dec, each contributing 20dB/dec.
* The Op-Amp, an active component, is used to ensure that a cut-off of -3dB is achieved.
* The passive components of the circuit will consist of resistors and capacitors. To obtain a d.c gain of 8, the d.c gain of filter A is set to 2 and that of filter B to 4. The two d.c gains are multiplied to give the desired d.c gain of 8.
b). Drawing the circuit;
Analysing how the specifications will be achieved;
A). Analysis of the first filter (Filter A);
The transfer function T(s) will be given by:
T
T
Where K = 1R2C1 and P = 1R1C1
Since K = 1CR=2,000, the two poles will be both at 2,000Hz
B). Analysis of the second filter (Filter B);
The transfer function T(s) will be given by:
T(s) = R1'//C1'R2'= VoVi
T(s) = 1R2'C1'R1'C1'+1= 1R2'C1'S+1R1'C1' = KS+P
Where P = 1R1'C1' and K = 1R2'C1'
When we cascade the two filters A and B, the overall transfer function will be given by;
T(s) overall = (T(s of Filter A) Ã— (T(s) of Filter B) which is expressed as
c). Calculating the practical values of the components using the standard values of resistors and capacitors;
To obtain the parameters of the first filter, we let K = 4, C1 = 1, Kf = 104, and Km = 105 .
Therefore, the transfer function will be; T(s) = KS+P= KS+1.26
The resistances will be; R1' = 1PC1'= 11.26Ã—1=0.794Î©
Applying the scaling factor Km = 105, we obtain the real values as
New R1' = KmR1'Old = 105Ã—0.794=79.4 kÎ©
New R2' = KmR2'Old= 105Ã—0.25=25 kÎ©
New C1' = 1KmKfC1'Old = 1105Ã—104Ã—1=10-9F=1 nano Farads
From the expression of the transfer function of filter A, we evaluate its transfer function to be
T(s) = 1R2'C1'S+1R1'C1' = 1(25 Ã— 103)(10-9)S+1(79.4 Ã— 103)(10-9) = 4 Ã—103S+(1.259 Ã— 104)
The frequency of operation is given by Ï‰c=2Ï€fc=2Ï€ Ã—2000=1.26 Ã— 104 rad/s
Next we evaluate the parameters of the second filter (Filter B).
Let K = 2, C1 = 1, Kf = 104, and Km = 105
The transfer function will reduce toT(s) = KS+P= KS+1.26
The values of the resistances will be; R1' = 1PC1'= 11.26Ã—1=0.794Î©
Applying the scaling factor Km = 105, we obtain the real values as
New C1 = 1KmKfC1Old= 1104Ã—105Ã—1=10-9F
From the expression of the transfer function of filter B, we evaluate the transfer function of filter B to be T(s) = 1R2C1S+1R1C1 = 1(50 Ã— 103)(10-9)S+1(79.4 Ã— 103)(10-9) = 2 Ã—104S+(1.259 Ã— 104)
Lastly, from the overall transfer function expression T(s) = T(s) of Filter A Ã— T(s) of Filter B, we obtain the overall transfer function as T(s) 2 Ã—104S+(1.259 Ã— 104) Ã— 4 Ã—104S+(1.259 Ã— 104) =8 Ã—108S2+25200s+1.59 Ã—108)
d). Calculating the voltages at each node as output for an input sinusoidal signal of 10V;
Voltage at Node A;
V(a) = R1 R1 +R2 Ã— Vin= 79.4 79.4 +25 Ã—10=6.136 Volts
Voltage at Node B;
V(b) = R1' R1' +R2' Ã— Vin= 79.4 79.4 +25 Ã—6.136 = 4.667 Volts
e). Drawing the Bode plots and discussing whether the requirements have been achieved;
We use the expression for magnitude T(s) = 8 Ã—108Ï‰2 + 25200Ï‰ + 1.59Ã—108) to obtain values for plotting the magnitude Bode plot by substituting with different values of frequency.
For example, frequency, w = (0, 100, 400, 1000, 5000) rads/s and so on.
We get the plotting points to be; Point 1 = 14....
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