Difference between revisions of "Lab 6 RS"
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! scope="col" width="50" | <math>\tau \approx 60.0\ ms </math> | ! scope="col" width="50" | <math>\tau \approx 60.0\ ms </math> | ||
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− | | [[File: | + | | [[File:Tek00012.png | 350 px]] || [[File:Tek00010.png | 350 px]] || [[File:Tek00011.png | 350 px]] |
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The times constant <math>\tau_{RC} = \frac{1}{(R/2)(2C)} = \frac{1}{RC}</math> becomes unchangeable so nothing is really happened | The times constant <math>\tau_{RC} = \frac{1}{(R/2)(2C)} = \frac{1}{RC}</math> becomes unchangeable so nothing is really happened | ||
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=Pulse Sharpener= | =Pulse Sharpener= |
Revision as of 23:53, 10 February 2011
Lab 6 Pulses and RC Filters
Differentiator
Construct the circuit below selecting an RC combination such that RC s
Adjust the pulse generator to output square pulses which at RC/10, RC and 10 RC
1)2) 3)
Measure and . Sketch a picture comparing and . ( 3*5 = 15 pnts.)
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From the plots above:
Questions
What happens if the amplitude of is doubled.( 5 pnts.)
The amplitude of
is doubled as well.
What happens if R is doubled and C is halved?( 5 pnts.)
The times constant
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.
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From the plots above:
Questions
What happens if the amplitude of is doubled.( 5 pnts.)
The amplitude of
is doubled as well.
What happens if R is doubled and C is halved?( 5 pnts.)
The times constant
becomes unchangeable so nothing is really happened
Pulse Sharpener
The goal of this section is to demonstrate how well the circuit below can sharpen an input pulse
1.) The first step is to create an input pulse which is rounded, similar to the output of the integrator circuit when RC = 10
. 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
where
- = impedance of the function generator at output which produces V_{in}
- = capacitor shorting the function generator output to ground (not shown in the above picture)
2.) The output should be given by
where
3.) Make measurements of the rise time
and . 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
to what you expected based on your measured values of , and .( 15 pnts.)Questions
1.) Qualitatify, why is
?( 10 pnts.)2.) How is
worse than ( 10 pnts.)