ECE 213Lab 10 – An Introduction to Filters

University of IdahoSpring 2017

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Lab 10 – An introduction to Filters

Objective

The objective of this experiment is to determine the characteristics of two simple filters: Low Pass filter and High Pass Filter.

Theory

Historically, the first filters used extensively were in sound systems such as audio amplifiers, radios, and telephones. For this reason many of the conventions used in describing filters are based on the properties of human hearing. The first of these properties is that human hearing is logarithmic. This makes it possible for us to hear sounds with very low volumes while not being completely overwhelmed by loud sounds. This also results the average person not being able to recognize small changes in the power of a sound. The average person will not hear a change in sound until its power has decreased to half of its original value or doubled. Second, the commonly accepted frequency range for human hearing is from 20 Hertz to 20 kilo Hertz. This varies from person to person and very few people have this complete range. Third, people cannot hear a phase change in a sinusoid or in one sinusoid in relation to another. The effects of these hearing characteristics on the conventions used in filter theory will be indicated where they show up in the following discussion.

Either a low-pass or a high-pass filter can be constructed using two components: a resistor and a reactive element. Basically the circuit acts as a voltage divider. Since the impedance of one of the components is frequency dependent the ratio of the voltage division varies with frequency. The voltage is divided between the two components. As one voltage in the divider increases, the other decreases. Consider a series resistive-capacitive filter. If the output potential is taken across the capacitor, the output impedance and therefore the output voltage will be high at low frequencies and decrease as the frequency increases. On the other hand if the output is taken across the resistor the output voltage will be low at low frequencies and will increase as the reactance of the capacitor decreases. The size of the components will determine the frequency at which the filter becomes effective. In keeping with the characteristic of the human ear, that it does not recognize a change in sound volume until the power has been reduced to half of its original value, the term "half-power point" is used to describe the effective point of a filter. This half-power point has two other commonly used names: the "corner frequency" and the "3-dB-down point". Because the power can be calculated as:

The voltage at the half-power point is 0.707 times the band-pass voltage. The gain in decibels is given by:

Gain in Voltage =

Gain in dB = , Then at the half-power point.

Gain in dB = 20log (0.707) = -3dB. This is usually just called -3 dB or 3 dB down.

The Gain (which is less than or equal to in a passive circuit) is often plotted in the logarithmic units of decibels(dB). This is reasonable since the ear responds to the volume of sound in a logarithmic way.

If an audio amplifier has a half-power point at 20 Hertz and another at 20 kilo-Hertz, with a band pass filter in between, it is said to have a bandwidth from 20 Hertz to 20 kilo-Hertz. If the gain is plotted as a function of frequency on a linear frequency scale to a high enough frequency to show the high end drop-off ofthe amplifier, both the areas of gain drop-off will be very small and not show much detail. These are the regions of primary interest. For this reason the frequency-response curve is plotted on a semi-log plot, on which the frequency (on the x axis) is plotted on a logarithmic scale. The frequency is plotted in ether Hertz or radians per second and the gain on a linear scale in decibels. This type of plot is called a Bode plot after H. W. Bode who developed much of this theory while working for Bell Laboratories.

In addition to the changes in the amplitude caused by these filters there will be a phase shift between the input and output potentials. The human ear does not hear changes in phase, so the change in amplitude caused by the filter is of primary importance when dealing with sound. There are areas of study where both phase shift and gain are important. An example of this would be feedback control systems. In this case the phase of the feedback can be very important.

Procedure:

In this circuit,
R2= 330Ω
C1=480nF

Write down the measured value below:

R2______C1______

  1. Solve symbolically the Vout/Vin for Figure 1 below. Result should in terms of f.
  2. Setup the circuit of Figure 1 using a function generator as the sinusoidal source.
  3. Connect an oscilloscope to read the voltage at the terminals of the function generator and the voltage across the capacitor.The voltage at the terminals ofthe function generator isthe input to the filter.The potential across the capacitor isthe filter output.Set the input voltage to a 3Vpeak.
  4. Adjust the frequency in figure 1and record output voltage and phasein Table D-1on the data sheet. ***You can use excel to record the measurement if prefer.
    (The reason for using the oscilloscope for the amplitude measurements rather than the digital multimeter since none ofthe digital multimeters in the laboratory are specified to measure frequencies over 50,000 Hz.)
    i.e. The oscilloscope should be read carefully in ordered to get a good curve.
  5. Redraw the circuit below by switching the position of resistor and capacitor. This is the high pass filter. Specify the input and output.
  1. The circuit drawn above is the High pass filter. Calculate symbolically the |Vout/Vin| of the high pass filter. Remember if Z=a+jb, |Z|=√(a^2+b^2)
  1. Construct the High Pass filter and record the measurement results in D2.
  2. Simulate the voltage and phase verse frequency for the Low pass and high pass filter using PSpice/LTSpice/Cadence/TINA. Simulate in a range from 100Hz to 10000Hz.

Result Analysis

All the measured value for the following tables should be calculated from D1 and D2

Low Pass Filter Analysis

Frequency (Hz) / Low Pass Filter
Calculated Gain |Vout/Vin|(V/V) / Measured Gain |Vout/Vin| (V/V) / 20 log of measured Vo/Vi (dB) / Measured Phase Shift Degree / Percent Err. (Calc. Gain - Meas. Gain)
100
200
500
1000
2000
5000
10000

High Pass Filter Analysis

Frequency (Hz) / High Pass Filter
Calculated Gain |Vout/Vin|(V/V) / Measured Gain |Vout/Vin| (V/V) / 20 log of measured Vo/Vi (dB) / Measured Phase Shift Degree / Percent Err. (Calc. Gain - Meas. Gain)
100
200
500
1000
2000
5000
10000
  1. Use the data above, plot the voltage vs frequency in excel for both low pass and high pass filter.
  2. Use the data above, plot the phase vs frequency in the excel for both low pass and high pass filter.

Question:

  1. What are the use of low pass filter and the use of high pass filter? Give two examples.
  1. Why do engineer in recent year try to use resistor and capacitor and avoid using inductor?
  1. What is the maximum gain andlowest gain in voltage of the low pass filter?
  1. In Both low pass and high pass circuit, what is the 3dB frequency? 3dB frequency means 3dB down voltage from maximum gain. You can calculate the Vout using Gain in dB equation.

Datasheet

Table D1 – Output and phase potential for high pass filter

Approx. Input Frequency (Hz) / Actual Input Frequency (Hz) / Output Potential / Phase
Scale (1x/10x/100x) (so you don’t get the wrong voltage) / Peak to Peak Potential (V) / Time between Vin and Vout
100
200
500
1000
2000
5000
10000

Table D2 – Output and phase for the high pass filter

Approx. Input Frequency (Hz) / Actual Input Frequency (Hz) / Output Potential / Phase
Scale (1x/10x/100x) (so you don’t get the wrong voltage) / Vout Peak to Peak Potential (V) / Time between Vin and Vout
100
200
500
1000
2000
5000
10000

Write a conclusion.

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October 12, 2018