ELE_292_F23_PreLab5

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Dec 6, 2023

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ELE 292 – Linear Systems Lab Prelab 5 – Filter design and Bode plots Name: Goals: 1) Find the values of R and C, given a cutoff frequency and some constraints. 2) Find the output of an RC system at some frequencies, using the phasor method. 3) Produce the Bode plot (for both the magnitude and the phase) of the system. Designing a low-pass filter A low-pass filter is a system that responds strongly to low frequencies but responds weakly to high frequencies. A filter is typically a circuit, an electrical system, but it could be any type of system. The question that should pop in your mind is “what defines a low frequency versus a high frequency?” The definition between low and high frequencies is called the cut-off frequency. It turns out we are familiar with the cut-off frequency, even though we don’t know it yet! For first order systems, the cut-off frequency is the magnitude of the natural frequency of the system. Often, you will want to design a filter, which means you need to choose the components. Let’s work with the most common low-pass filter students may encounter. The transfer function is shown below. Assume we want to build a filter that has a cut-off frequency of 1500 Hz. This filter will allow frequencies below 1500 Hz to pass through, but frequencies above 1500 Hz will be attenuated. Given the cut-off frequency, the design becomes easy. The cut-off frequency is the absolute value of the natural frequency. That means we need a natural frequency of 1500 Hz, or ω = 2 πf = 2 π ∙ 1500 rad/s. Now we know the numerical values of the transfer function.

H ( s ) = 9425 s + 9425 The first design criteria we need to consider is the current. Resistors determine the current, they are the components that introduce loss into the system. Assume we are going to apply voltages between 1V and 5V, and we want to limit the current to values between 0.1mA and 1 mA. Simply use Ohm’s law, V = IR, to choose one single resistor value that will fit the constraints. Type the Resistor value you chose that fits the specs 10000 ohms The next step in the design is to calculate the capacitance. Use the value of R above to calculate a capacitance that makes the absolute value of the natural frequency match the desired value. Type the Capacitor value you calculated that gives the proper natural frequency 100uf Because this is going to be a lab experiment, you need to choose a capacitor that we can actually get off the shelf. From the list below, choose the closest capacitor that you can get in the lab. 10p 15p 22p 33p 47p 68p 100p 150p 220p 330p 470p 680p 1.0n 1.5n 2.2n 3.3n 4.7n 6.8n 10n 15n 22n 33n 47n 68n 100 n 150n 220n 330n 470n 680n Type the Capacitor value from the list above that is closest to your calculated value 100nf Finally, we need to re-calculate the required resistance, because of the limited choices of capacitance. Once you re-calculate the resistance, choose the closest one in the list below. 1.0 1.5 2.2 3.3 4.7 6.8 10 15 22 33 47 68 100 150 220 330 470 680 1.0k 1.5k 2.2k 3.3k 4.7k 6.8k 10k 15k 22k 33k 47k 68k 100k 150k 220k 330k 470k 680k 1M 1.5 M 2.2 M 3.3 M 4.7 M 6.8 M 10M Type the Resistor value from the list above that gives you the best available circuit design 10kohms

Now that you have a design, calculate the transfer function based on the R and C available in the lab. Type the transfer function of your design below H_s = 1/(10000*(100*10^-6))/(s + (1/(10000)*(100*10^-6)) verifying the low-pass filter design Let’s compute the value of the transfer function at various frequencies. Here are the steps.  Choose a frequency, f, in Hz  Find the radial frequency, ω = 2πf, in rad/s  Find s = jω, in 1/s  Plug the value of s into the transfer function Complete the table below Frequency Hz H(s) Rectangular form |H(s)| magnitude ∠H(s) Phase in degrees |H(s)| Magnitude in decibels 300 2.814e-7 - .00053j 0.000530515 1617417991 - 89.96960371 883867 - 65.50604399 52262 1000 2.59e-10-1.61e-5j 5.305152363 975894e-12 - 89.999999999 69604 - 75.96334130 433335 1500 1.125102e-08- 0.008j 0.000106103 27183987824 - 89.99392073 031964 - 79.48542447 65262 10,000 2.534e- 10-.5913e-05j 1.591549508 5268835e-05 89.99908810 930272 - 95.96359694 361661 We could perform these calculations for many frequencies and make a very long table. However, Scipy has a method that does just that. It is called Bode for Bode plot. Bode is the name of a famous engineer. Hendrik Wade Bode - Wikipedia As described by Bode, the Bode plot is really an approximation that allows one to quickly plot the transfer function, magnitude or phase, versus frequency. Today we use

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