PHYSICS LAB #2 Passive Low-pass and High-pass Filter Circuits and Integrator and Differentiator Circuits

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PHYSICS 360 - LAB #2

Passive Low-pass and High-pass Filter Circuits and Integrator and Differentiator Circuits

Objective:

 Study the behavior of low-pass and high-pass filters.  Study the differentiator and integrator circuits.

Introduction:

Resistors respond to ac sources as they do to dc. Capacitors and inductors, however, respond to ac sources quite differently. In this lab you will study the behavior of RC circuits to sinusoidal input signals and to input signals in the form of square pulses. Because of the frequency-dependent behavior of capacitor, the output voltage depends on the frequency of the voltage sources. RC circuits are used widely in electronics to perform a variety of functions, such as frequency filtering, differentiation and integration.

2-1. Low-pass Filter Circuit.

Let’s say you want to design a filter that will block all frequencies above 1 kHz and pass all frequencies below 1 kHz. This would be called a low pass filter. If a perfect filter could be made to achieve this task what would its response curve (gain versus frequency) look like? In other words, what would a graph of vout/vin versus frequency look like? Sketch the perfect

response below. Add a linear scale to the y-axis.

10 100 1000 10000 100000 Frequency (Hz) Ga in ( vout /vin )

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The RC circuit shown below is a low pass filter, but its response curve is not perfect.

Fig. 2-1 Low-pass filter

Think of it as a voltage divider where the resistance of one of the resistors depends on frequency. (In this case, we say the impedance of the capacitor depends on frequency.) At high frequency the capacitor behaves like a short, so the voltage across it will be a small fraction of the input voltage. The circuit is not passing, it is blocking, high frequencies. At low frequencies the capacitor behaves like an open, so the output will be a large fraction of the input voltage. The circuit is passing low frequencies. Thus the low pass filter has vo vin at low frequencies and v_o 0 at high frequencies. These are the extreme cases. The gain (and the phase shift) as a function of frequency over the entire range of frequencies for this low-pass filter circuit are given by:

Phase shift = _{tan-1(RC) (That is, vc lags vin by } (2)

In the following exercise you will measure the gain and phase shift as a function of frequency for a low pass filter. Not only is it a study of the properties of a low pass filter, it will develop your skill at measuring gain and phase shift from an oscilloscope trace. These are very common measurements. Part of the skill in making measurements is in achieving the highest precision possible when necessary. Answer these questions for a reminder on how to maximize precision with an oscilloscope.

Estimate the uncertainty in measuring the peak-to-peak amplitude of a sinusoidal signal on the oscilloscope. Give your answer as plus-or-minus so many divisions. ___________________ Calculate the percent uncertainty in the measured amplitude of a 5 Vp-p signal when the

oscilloscope is set to display 10 V/division. _______________________________________ Calculate the percent uncertainty in the measured amplitude of a 5 Vp-p signal when the

oscilloscope is set to display 1 V/division. _______________________________________   Gain = A_v = _vvo in = 1 1+ 2R 2C





v

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What lesson is made apparent by your answers? The same principle applies when you are measuring time (phase shifts) using the horizontal divisions and the times/division display setting.

Wire the low-pass filter circuit shown in Fig. 2-1 on the breadboard. Use R = 15 kand C = 0.01 µF. Measure the input and output voltages as a function of the sine wave frequency as well as the phase shift of the output sine wave as compared to the input sine wave on the oscilloscope. If necessary, adjust the amplitude of function generator to maintain a constant vin. Use an input voltage of 10 Vp-p.

 Transfer your data to Excel/Origin. Create additional columns for calculated gain and phase using formulas (1) and (2). Plot a graph that overlays the measured and calculated gain versus frequency, using a log scale for the frequency axis. Do this twice, once using the simple out/in ratio and again plotting the gain in dB.

 Determine the cut-off frequency fo (also called -3dB frequency or break point frequency). This is the frequency at which the gain is 0.707. Mark it on printouts of your graphs.

f_{o (measured) = _________ Hz} f_{o (calculated) = 1/(2RC) = _________ Hz}  Show that the output attenuates at 20 dB/decade and 6 dB/octave at f >> fo.

 Plot the phase angle difference () between vo and vi versus log frequency.

 Do your results agree with calculations based on Eqns. (1) and (2) for a low-pass filter? Frequency vo(p-p) Gain Measured (vo/vin) Gain Measured (in dB) Phase Measured (φ) in deg. 40 Hz 100 Hz 200 Hz 400 Hz 500 Hz 800 Hz 1 kHz 4 kHz 8 kHz 10 kHz 40 kHz 100 kHz 400 kHz

4 2-2. High-pass Filter Circuit.

Behavior of a high-pass filter is opposite to that of a low-pass filter. The circuit below is a simple RC high-pass filter. Here the output is taken across the resistor.

Fig. 2-2 High-pass filter

With a sinusoidal applied ac input the output voltage across the resistor is very small (vo 0 ) at low frequencies (cap. behaves like an open) and at vo vin at high frequencies (cap. behave like a short).

The gain and the phase as a function of frequency for the high-pass filter circuit are given by:

Phase shift =  tan-1 (1/RC) (Note that v_R leads v_in by  (4)

Let’s say you want to design a filter that will pass all frequencies above 5 kHz and block all frequencies below 5 kHz. This would be called a high-pass filter. If a perfect filter could be made to achieve this task what would its response curve (gain versus frequency) look like? In other words, what would a graph of vout/vin versus frequency look like? Sketch the perfect

response below. Add a linear scale to the y-axis.

v_i vo C R   Gain = A_v = v_vo i = RC 1+ 2R 2C 3 10 100 1000 10000 100000 Ga in ( vout /vin )

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The frequency response of the real RC high-pass filter is not perfect. In fact, it is just like that of the low pass filter, but reversed. With the measured low-pass frequency response from the previous activity as a model, sketch the frequency response of an RC high pass filter that has a cut-off frequency of 1000 Hz.

Let’s say there is a signal that is the sum of a 10 Vp-p 1 kHz sinusoid and a 10 Vp-p 12 kHz sinusoid. Design an RC high pass filter that will pass the 12 kHz signal and block the 1 kHz signal. There are two choices to make.

First, choose the cut-off frequency. For high-pass filters, any frequency above the cut-off frequency is in the pass-band. Any frequency below the cut-of frequency is in the stop-band. Choose it so as to minimize attenuation of the frequency you want to pass, and maximize attenuation of the frequency you want to block. In other words, the signals are equal (

1 1 , 12 , kHz in kHz  in v

v ) at the input, but you want the 12 kHz to be much larger than the 1 kHz ( possible as big as 1 , 12 , kHz out kHz  out v

v ) at the output. Shift the frequency response transparency

horizontally over the graph below to help make your selection.

0 0.2 0.4 0.6 0.8 1 10 100 1000 10000 100000 -60 -40 -20 0 10 100 1000 10000 100000 Frequency (Hz) Frequency (Hz) Ga in ( vout /vin ) Ga in ( dB ) -60 -40 -20 0 Ga in ( dB ) 1000 Hz 12000 Hz

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What cut-off frequency did you choose? ______________________________

Your second design choice is the resistor value. Its value should be much greater than the output resistance of the stage that precedes the filter (the source), and much less than the input

resistance of the stage that follows the filter (the load). This ensures that the source and load, when connected to the filter, do not significantly alter the cut-off frequency from the theoretical value of f₀ 1 2RC.

What is the output resistance of the source (in this case, the function generator)? ____________ What is the input resistance of the load (in this case, the oscilloscope)? ________________ What resistance did you choose? _____________________

Now calculate the required capacitance. Show your work. You’ll probably have to adjust your choices to the nearest stock values. Use only one capacitor. (Exact values are not necessary. A circuit that absolutely has to have a 1245.57 Ω resistor and a 0.385 µF capacitor is poorly designed. Resistors are typically ±5% and capacitors are even worse, often ±20%. And everything drifts over time and with temperature. A well designed circuit tolerates these variations.)

C = _______________

Obtain your components. Now let’s calculate how well we expect the filter to perform. Based on the values of the actual components you will use in your filter, recalculate the cut-off frequency.

f0 = ________________

Accurately sketch the frequency response of the filter on the graph above.

From the frequency response plot, determine the output voltage of the 12 kHz and 1 kHz signals. (You’ll have to convert the value read from the graph in dB to a value in volts. Show your work.)

7 Graphical vout,1kHz = _________________

Next, calculate a theoretical value for the output amplitudes using equation (3). Show your work.

Theory vout,12kHz = ________________

Theory vout,1kHz = _________________

Finally, build your filter. (If your capacitor is electrolytic be careful with the polarization.) Measure the output amplitude for a 10 Vp-p input at 12 kHz and at 1 kHz.

Measured vout,12kHz = ________________

Measured vout,1kHz = _________________

Compile all these results and calculations in the tables below.

FRACTIONS kHz in kHz out v v 12 , 12 , kHz in kHz out v v 1 , 1 , kHz out kHz out v v 1 , 12 , Graphical Theory Measured dB _       kHz in kHz out v v 12 , 12 , log 20 _       kHz in kHz out v v 1 , 1 , log 20 _       kHz out kHz out v v 1 , 12 , log 20 Graphical Theory Measured

8 Integrating and Differentiating Circuits

The shapes of electrical signals must often be modified to be in a suitable form for operation of circuits. The simple RC circuit often plays a part in the formation of suitable waveforms. Here you will study the effects of low-pass and high-pass filter circuits on the input square voltage pulses of differing pulse widths.

2.3. Integrator Circuit:

From theory we know that the voltage across the capacitor in an RC circuit is given by:

(Vin = Vp in this case and  is the pulse width)

This is why the RC circuit used in this configuration is called an integrator circuit.

Wire the circuit shown in Fig. 2-3 with R = 10 k and C = 0.1 F. Time constant of the below circuit _{c = RC = _____________}

Charging: v_c = V_p (1 - e - _RC t ) (5) Discharging: v_c = V_p e - _RCt (6) Integrator: For RC >> , v_c   RC_

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Fig. 2-3 RC Integrator

Using a 5 V square wave input signal (use TTL output from the function generator on the design station), sketch the wave forms for the input and output signals for the following cases. Connect your input to CH 1 and output to CH 2 of the oscilloscope.

i) RC >>  f = __________Hz, Period (T) = _________sec

ii) f = 100 Hz, Period (T) = _________sec

C R V_o 0  V_p CH 1 _________ V/div CH 2 _________ V/div Time Base _________ s/div

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Calculate the time constant of the circuit from the figure in (ii). Recall that the time constant of the RC circuit is the time it takes for the capacitor to charge to 0.632Vp during the charging process.

2.4. Differentiator Circuit:

From theory we know that the voltage across the resistor in an RC circuit is given by:

(Vin = Vp in this case and  is the pulse width)

This is why the RC circuit used in this configuration is called a differentiator circuit. Wire the circuit shown in Fig. 2-4 with R = 10 k and C = 0.1 F.

Time constant of the below circuit _{c = RC = _____________}

CH 1 _________ V/div CH 2 _________ V/div Time Base _________ s/div Charging: v R = Vp e - _RCt (8) Discharging: v R = - Vp e - _RCt (9) Differentiator: For RC <<, v R  RC dV in dt (10)

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Fig. 2-4 RC Differentiator Circuit   C R V_o 0  V_p

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Using a 5 V square wave input signal (use TTL output from the function generator on the design station), sketch the wave forms for the input and output signals for the following cases. Connect your input to CH 1 and output to CH 2 of the oscilloscope.

i) RC <<  f = __________Hz, Period (T) = _________sec

ii). f = 100 Hz, Period (T) = _________sec

CH 1 _________ V/div CH 2 _________ V/div Time Base _________ s/div CH 1 _________ V/div CH 2 _________ V/div Time Base _________ s/div

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Calculate the time constant of the circuit from the figure in (ii). Recall that the time constant of the RC circuit is the time it takes for the resistor to charge to 0.368Vp during the charging process.