TF EIMLab6 Writeup: Difference between revisions
| Line 8: | Line 8: | ||
: <math>C_1 = 1 \times 10^{-6} F</math> | : <math>C_1 = 1 \times 10^{-6} F</math> | ||
:<math>C_2 = 10. | :<math>C_2 = 10.24 \times 10^{-9} F</math> | ||
Possible Resistors | Possible Resistors | ||
:<math>R_1 = 1 \times 10^3 \Omega</math> | :<math>R_1 = 1 \times 10^3 \Omega</math> | ||
:<math>R_2 = 14 \times 10^{3} \Omega</math> | :<math>R_2 = 14.35 \times 10^{3} \Omega</math> | ||
: <math>R_1 C_1 = \tau_1 = (1 \times 10^3 \Omega ) (1 \times 10^{-6} F )= 0.001</math> s | : <math>R_1 C_1 = \tau_1 = (1 \times 10^3 \Omega ) (1 \times 10^{-6} F )= 0.001</math> s | ||
: <math>R_1 C_1 = \tau_2 = (14 \times 10^{3} \Omega)(10. | : <math>R_1 C_1 = \tau_2 = (14.35 \times 10^{3} \Omega)(10.24 \times 10^{-9} F )= 1.46 \times 10^{-4}</math> | ||
:<math>\omega_1 = 100 rad/sec \Rightarrow \nu_1 = 100/2\pi = 16 Hz</math> | :<math>\omega_1 = 100 rad/sec \Rightarrow \nu_1 = 100/2\pi = 16 Hz</math> | ||
:<math>\omega_2 = | :<math>\omega_2 = 6805 rad/sec \Rightarrow \nu_2 = 1083 Hz</math> | ||
tek012 | tek012 | ||
Revision as of 23:08, 8 February 2011
Lab 6 Pulses and RC Filters
Differentiator
1.) Adjust the pulse generator to output square pulses which at [math]\displaystyle{ \tau }[/math] sec in time.
Possible capacitors
- [math]\displaystyle{ C_1 = 1 \times 10^{-6} F }[/math]
- [math]\displaystyle{ C_2 = 10.24 \times 10^{-9} F }[/math]
Possible Resistors
- [math]\displaystyle{ R_1 = 1 \times 10^3 \Omega }[/math]
- [math]\displaystyle{ R_2 = 14.35 \times 10^{3} \Omega }[/math]
- [math]\displaystyle{ R_1 C_1 = \tau_1 = (1 \times 10^3 \Omega ) (1 \times 10^{-6} F )= 0.001 }[/math] s
- [math]\displaystyle{ R_1 C_1 = \tau_2 = (14.35 \times 10^{3} \Omega)(10.24 \times 10^{-9} F )= 1.46 \times 10^{-4} }[/math]
- [math]\displaystyle{ \omega_1 = 100 rad/sec \Rightarrow \nu_1 = 100/2\pi = 16 Hz }[/math]
- [math]\displaystyle{ \omega_2 = 6805 rad/sec \Rightarrow \nu_2 = 1083 Hz }[/math]
tek012
2.)Construct the circuit below selecting an RC combination such that RC [math]\displaystyle{ \approx }[/math] 1/10
3.)Measure[math]\displaystyle{ V_{in} }[/math] and [math]\displaystyle{ V_{out} }[/math]. Sketch a picture comparing[math]\displaystyle{ V_{out} }[/math] and [math]\displaystyle{ V_{in} }[/math].
4.) Change the pulse width such that [math]\displaystyle{ RC = \tau }[/math]
5.)Measure[math]\displaystyle{ V_{in} }[/math] and [math]\displaystyle{ V_{out} }[/math].Sketch a picture comparing[math]\displaystyle{ V_{out} }[/math] and [math]\displaystyle{ V_{in} }[/math].
6.) Change the pulse width such that[math]\displaystyle{ RC = 10 \tau }[/math]
7.)Measure [math]\displaystyle{ V_{in} and V_{out} }[/math].Sketch a picture comparing[math]\displaystyle{ V_{out} }[/math] and [math]\displaystyle{ V_{in} }[/math].
Questions
1.) What happens if than amplitude of [math]\displaystyle{ V_{in} }[/math] is doubled.
2.) What happens if R is doubled and C is halved?
Integrator
To illustrate the integrator circuit we need to have an input pulse which looks like the output of the above differentiator circuit. In other words, input a pulse whose output is obviously the integral of the input pulse.