TF EIMLab6 Writeup: Difference between revisions

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: <math>C_1 = 1 \times 10^{-6} F</math>
: <math>C_1 = 1 \times 10^{-6} F</math>
:<math>C_2 = 10.27 \times 10^{-12} F</math>
:<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)(10.27 \times 10^{-12} F )= 1.46 \times 10^{-5}</math>
: <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 = 68,291 rad/sec \Rightarrow \nu_2 = 10,868 Hz</math>
:<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.


Forest_Electronic_Instrumentation_and_Measurement