Difference between revisions of "Lab 6 RS"

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[[File:Lab6 formula1_m1.png | 600 px]]
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[[File:Lab6 1_m1.png | 600 px]]
 
 
  
 
==Adjust the pulse generator to output square pulses which at <math>\tau \approx</math> RC/10.==
 
==Adjust the pulse generator to output square pulses which at <math>\tau \approx</math> RC/10.==

Revision as of 16:19, 10 February 2011

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Lab 6 Pulses and RC Filters

Differentiator

Construct the circuit below selecting an RC combination such that RC [math]\lt 10^{-4} [/math] s

TF EIM Lab6b.png


600 px

Adjust the pulse generator to output square pulses which at [math]\tau \approx[/math] RC/10.

[math]\tau \approx \frac{6.99 \times 10^{-5}}{10} \approx 7.0\ \mu s [/math]

Measure[math] V_{in}[/math] and [math]V_{out}[/math]. Sketch a picture comparing[math] V_{out}[/math] and [math]V_{in}[/math]. ( 5 pnts.)

Change the pulse width such that [math]\tau=[/math]RC

[math]\tau \approx 6.99 \times 10^{-5} \approx 70.0\ \mu s [/math]

Measure[math] V_{in}[/math] and [math]V_{out}[/math].Sketch a picture comparing[math] V_{out}[/math] and [math]V_{in}[/math].( 5 pnts.)

Change the pulse width such that [math] \tau[/math]=10 RC

[math]\tau \approx 10 \times \left(6.99 \times 10^{-5}\right) \approx 700.0\ \mu s [/math]

Measure [math]V_{in}[/math] and [math]V_{out}[/math]. Sketch a picture comparing[math] V_{out}[/math] and [math]V_{in}[/math].( 5 pnts.)

Questions

What happens if the amplitude of [math]V_{in}[/math] is doubled.( 5 pnts.)

What happens if R is doubled and C is halved?( 5 pnts.)

The times constant [math]\tau_{RC} = \frac{1}{(R/2)(2C)} = \frac{1}{RC}[/math] becomes unchangeable so nothing is really happened

Integrator

Now repeat the above experiment with the resistor and capacitor swapped to form the low pass circuit below.

TF EIM PulsedRCLowpass.png

Pulse Sharpener

The goal of this section is to demonstrate how well the circuit below can sharpen an input pulse

TF EIM Lab6a.png

1.) The first step is to create an input pulse which is rounded, similar to the output of the integrator circuit when RC = 10 [math]\tau[/math]. You can do this using a capacitor shorted across the output of the pulse generator. This will essential be coupled to the input impedance of the pulse generator and form a low pass circuit.

As a result the input voltage is given as

[math]V_{in} = V_0 \left ( 1 - e^{-t/\tau}\right )[/math]

where

[math]\tau=R_{out} C_{out}[/math]
[math]R_{out}[/math] = impedance of the function generator at output which produces V_{in}
[math]C_{out}[/math] = capacitor shorting the function generator output to ground (not shown in the above picture)

2.) The output should be given by

[math]V_{out} = V_0^{\prime} \left ( 1 - e^{-t/\tau^{\prime}}\right )[/math]

where

[math]\tau^{\prime} =\left ( \frac{R_1 R}{R_1+ R}\right )C_1[/math]

3.) Make measurements of the rise time [math]\tau[/math] and [math]\tau^{\prime}[/math]. The rise time is defined as the time it take the pulse to go from 10% of its max value to 90% of its max value.( 5 pnts.)

4.) Compare the measurement of [math]\tau^{\prime}[/math] to what you expected based on your measured values of [math]C_1[/math], [math]R_1[/math] and[math] R[/math].( 15 pnts.)

Questions

1.) Qualitatify, why is [math]\tau^{\prime} \lt \tau[/math]?( 10 pnts.)

2.) How is [math]V_{out}[/math] worse than [math]V_{in}[/math]( 10 pnts.)



Forest_Electronic_Instrumentation_and_Measurement

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