Difference between revisions of "Lab 6 TF EIM"

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=Differentiator=
 
=Differentiator=
  
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1.) Adjust the pulse generator to output square pulses which at <math>\tau</math> sec in time.
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1.)Construct the circuit below selecting an RC combination such that RC <math>< 10^{-4} </math> s
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2.)Construct the circuit below selecting an RC combination such that RC <math>\approx</math> 1/10
 
  
 +
[[File:TF_EIM_Lab6b.png| 200 px]]
  
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3.)Measure<math> V_{in}</math> and <math>V_{out}</math>.  Sketch a picture comparing<math> V_{out}</math> and <math>V_{in}</math>.
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2.) Adjust the pulse generator to output square pulses which at <math>\tau \approx</math> RC/10.
  
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4.) Change the pulse width such that <math>RC = \tau</math>
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3.)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.)
  
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5.)Measure<math> V_{in}</math> and <math>V_{out}</math>.Sketch a picture comparing<math> V_{out}</math> and <math>V_{in}</math>.
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4.) Change the pulse width such that <math>\tau=</math>RC
  
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6.) Change the pulse width such that<math> RC = 10 \tau</math>
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5.)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.)
  
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7.)Measure <math>V_{in} and V_{out}</math>.Sketch a picture comparing<math> V_{out}</math> and <math>V_{in}</math>.
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6.) Change the pulse width such that <math>  \tau</math>=10 RC
 +
 
 +
7.)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==
 
==Questions==
  
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1.) What happens if than amplitude of <math>V_{in}</math> is doubled.
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1.) What happens if the amplitude of <math>V_{in}</math> is doubled.( 5 pnts.)
  
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2.) What happens if R is doubled and C is halved?
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2.) What happens if R is doubled and C is halved?( 5 pnts.)
  
 +
=Integrator=
  
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Now repeat the above experiment with the resistor and capacitor swapped to form the low pass circuit below.
  
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=Integrator=
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[[File:TF_EIM_PulsedRCLowpass.png | 400 px]]
 +
 
 +
=Pulse Sharpener=
 +
 
 +
The goal of this section is to demonstrate how well the circuit below can sharpen an input pulse
 +
 
 +
[[File:TF_EIM_Lab6a.png| 200 px]]
 +
 
 +
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 takes the pulse to go from 0 to <math>(1-e^{-1})</math> = 0.63 of its max value.( 5 pnts.)
  
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To illustrate the integrator circuit we need to have an input pulse which will be the output of the above differentiator circuit. In other words, input a pulse whose output is obviously the integral of the input pulse.
+
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} < \tau</math>?( 10 pnts.)
  
 +
2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)
  
 
[[Forest_Electronic_Instrumentation_and_Measurement]]
 
[[Forest_Electronic_Instrumentation_and_Measurement]]

Latest revision as of 04:19, 4 March 2011

Lab 6 Pulses and RC Filters

Differentiator

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

TF EIM Lab6b.png

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

3.)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.)

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

5.)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.)

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

7.)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

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

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

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 takes the pulse to go from 0 to [math](1-e^{-1})[/math] = 0.63 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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