Lab 8
Time and Frequency Responses of
Low-Pass and High-Pass Filters
In this lab you will apply the concept of AC impedance to the frequency response
of a low-pass filter, and explore the relationship between frequency response and
time response.
8.1 LR Filter
A simple low-pass filter can be made from an inductor and a resistor as shown in
Fig. 8.1. In this configuration, the gain is defined to be G ≡ VR /Vin . The magnitude
of the gain works out to be (see example 7.1 with C → ∞)
R
|G| = p
R 2 + (ωL)2
(low-pass filter) (8.1)
which exhibits the frequency dependence shown in Fig. 8.2. Only low-frequency
signals are able to pass from the input to the output unimpeded; high frequencies
are strongly attenuated. Observe that the trend with frequency is reversed from
that of the high-pass filter studied in the previous lab.
p
The cutoff frequency is again defined as the frequency where |G| = 1/ 2 ≈
0.707. We solve for f 0 as follows:
R
1
q
¡
¢2 = p2
R 2 + 2π f 0 L
⇒
¡
¢2
2R 2 = R 2 + 2π f 0 L
⇒
f0 =
R
2πL
(low-pass filter) (8.2)
Example 8.1
Write the magnitude of the gain (8.1) in terms of f and f 0 .
Solution: From (8.2), we have 2πL = R/ f 0 . When this is inserted into (8.1), it
55
!
Figure 8.1 LR circuit.
8.2 Decibel Gain
56
simplifies as follows:
|G| = q
R
R
1
¡
¢2 = q
¡
¢2 = q
¡
¢2 (low-pass filter) (8.3)
R 2 + 2π f L
R 2 + R f / f0
1 + f / f0
8.2 Decibel Gain
It is not uncommon to express the gain of a circuit using decibel units. The decibel
gain is defined as
G db ≡ 10 log10 |G|2 = 20 log10 |G|
(8.4)
We use |G|2 in the definition because oft times it is power that is of primary
interest. The time-averaged power delivered to the resistor is |VR |2 / (2R) ∝ |G|2 ,
as will be discussed in the next lab.
An alternative ‘cutoff frequency’ is sometimes expressed as the frequency
where the decibel gain drops to G db = −3.0. This frequency, which we will call
f 3db , is very close to f 0 , although not precisely the same.1
Example 8.2
Show that for practical purposes, f 3db and f 0 are the same frequency.
Solution: When f = f 3db , we have
−3.0 = 10 log10 |G|2
⇒
|G|2 = 10−0.3 ≈ 0.5012
When f = f 0 , we have by definition |G|2 = 1/2, so clearly f 3db and f 0 are nearly
equal. Solving for the exact connection between f 3db and f 0 yields
f 3db = f 0
p
100.3 − 1 = 0.995 f 0 ∼
= f0
(low-pass filter) (8.5)
8.3 Time Response of Circuits
A time response study addresses the significant question, How does the output
vary with time when the input experiences an instantaneous change? We can be
fairly certain that a real circuit will not respond instantaneously. Even a circuit
that responds very quickly will take a finite amount of time to complete its response. This is because every electronic device has some capacitance and some
inductance with which to store electric and magnetic energy, respectively. The
flow of energy to or from a capacitor or inductor takes time.
1 It is fortuitous that 100.3 is surprisingly close to 2, which is why 3 db is a handy attenuation
factor.
©2011 Campbell & Peatross
!
Figure 8.2 Frequency-response
function of a low-pass filter with
!
f 3db = 1 kHz.
8.3 Time Response of Circuits
57
Fig. 8.3 illustrates what happens when a square-wave voltage is applied to our
low-pass LR circuit. Relative to the input, the output voltage measured across the
resistor is strongly modified. Because a square wave alternately increases and
decreases sharply, we can observe both the rise time and the fall time in a single
experiment. These turn out to be the same for our LR circuit. However, for more
complicated instruments, involving amplifiers and such, the rise and fall times
can be quite different.
Consider an LR circuit as depicted in Fig. 8.1. From (6.6), the voltage across
the circuit is
dI
V = RI +L
(8.6)
dt
Suppose a constant voltage V0 is applied to the circuit, which eventually causes
current I 0 = V0 /R to flow. If the voltage is suddenly switched off (i.e. V = 0) at
t = 0, the equation governing the current thereafter is
L
dI
= −R I
dt
R
I = I0e − L t
which has the solution
(8.7)
R
The voltage across the resistor is then VR = I R = V0 e − L t , which decays exponentially towards zero in time.
On the other hand, if the applied voltage and current are initially zero before
a constant voltage V0 is applied at t = 0, the equation governing the current
thereafter is
L
dI
= V0 − R I
dt
which has the solution
I=
´
R
V0 ³
1 − e− L t
R
(8.8)
³
´
R
The voltage across the resistor is then VR = V0 1 − e − L t , which rises exponentially
towards V0 in time.
We define the response time τ to be time required for the output to complete
¡
¢
1 − e −1 = 63% of the full response to a sudden change in the input.
Example 8.3
Find the response time and fall time for an LR circuit (call it τLR ).
Solution: From (8.8), we solve for the the rise time by setting
R
1 − e − L τLR = 1 − e −1
which manifestly requires
τRL = L/R
³
(8.9)
´
R
Similarly, from (8.7) the change in current during the fall time is I 0 −I = I 0 1 − e − L t ,
which gives the same characteristic time (8.9).
©2011 Campbell & Peatross
!
Figure 8.3 Typical time response
of an LR circuit to instantaneous
input changes.
8.4 Time Response vs Frequency-Response
58
8.4 Time Response vs Frequency-Response
If a circuit responds well to high frequencies (i.e. a high-pass filter), it will be able
to ‘keep up’ with sudden changes in time. However, it will not be able to cope well
with long sustained trends. On the other hand, if a circuit responds well to low
frequencies (i.e. low-pass filter), it will be capable tracking long sustained trends
(epitomized by DC output), but it will smooth over and ‘miss’ abrupt changes.
One can use the frequency response function (FRF) of a circuit to predict
its time response function (TRF). The TRF describes the manner in which, for
example, a circuit responds to a square step voltage. Ultimately, the TRF and the
FRF are just two different ways at looking at the same system. More intuitively,
they have a type of inverse or reciprocal relationship.
Time Frequency Product
The product of the cutoff frequency f 0 (for an LR circuit) and the response time
τRL is independent of the values of R and L. Multiplying (8.2) and (8.9), we get
τRL f 0 =
L R
1
=
R 2πL 2π
(8.10)
8.5 Equipment
LRC component board, LC meter, signal generator/oscilloscope stack, frequency
meter, amplifier, cables.
©2011 Campbell & Peatross
Quiz
59
Quiz
Q8.1
When a square wave (as opposed to a sinusoidal wave) is applied to
the low-pass LR filter, the current is (choose one)
(a) also a square wave.
(b) a sinusoidal wave.
(c) a wave composed of spliced exponential growth and decay curves.
(d) none of the above.
Q8.2
Consider an LR circuit with R = 40Ω and L = 50 mH that is driven by
an AC voltage source at frequency f = 160 Hz. Compute the following
quantities.
(a) ZL
(b) |Z |.
(c) φ (in degrees).
(d) |G| (output measured across R).
Q8.3
(a) For the LR low-pass filter in the previous question, compute the
cutoff frequency f 0 (in Hz).
(b) At what frequency does the decibel gain drop to G db = −6 db? And
to G db = −10 db?
Q8.4
For a high-pass RC filter, show that (7.11) can be written |G| =
where f 0 is given by (7.12).
©2011 Campbell & Peatross
1
q
,
2
1+( f 0 / f )
Exercises
60
Exercises
A. Measure and analyze the frequency response of a low-pass RL filter.
We anticipate that the exercises in this section will proceed smoothly and quickly
based on your previous experience with high-pass filters.
L8.1
Repeat exercise 7.3 for the RL circuit shown in Fig. 8.4.
L8.2
Repeat exercises 7.4-7.5 with the RL circuit.
L8.3
Repeat the measurements of 7.6-7.7 for this low-pass RL filter. When
you get to the part of the exercise that involves curve fitting, employ a
A
model of the form |G| = p
, with A and L as variables (fix R
2
to its measured value).
L8.4
1+(2π f L/R)
Using the fitted value of L, calculate the cutoff frequency f 0 for the
low-pass LR filter.
! Figure 8.4 LR circuit.
B. Measure the time response of an RL low-pass filter.
L8.5
Send the TTL output of your signal generator to your amplifier input,
and use the amplifier output to apply a sharp 5V square-wave signal to
your low-pass LR filter circuit and observe the resulting time response
with an oscilloscope. Set the gain of the the amplifier knob to 1 and
monitor the temperature at the back of the amplifier to avoid accidental
overheating. Adjust the period of the square wave to be long enough
that the circuit voltage appears to decay completely, but not much
longer. Sketch two periods of the output signal in your lab notbook.
Adjust the vertical scale and the horizontal sweep time so that a single
voltage-decay curve approximately fills the display region. Visually
estimate τRL , combine it with your value for f 0 , and verify (8.10). Your
conclusions should include an intuitive explanation regarding the features of the frequency-response and time-response functions.
C. Measure the time response of an RC high-pass filter.
L8.6
Reconstruct the high-pass RC filter from the previous lab (Fig. 8.5)
using the same values for R and C , and repeat exercise L8.5 to obtain
and estimate for τRC . Use the value of f 0 obtained in the previous lab
to demonstrate that 2π f 0 τRC = 1. In your conclusions, please compare
and contrast the high-pass and low-pass time response curves in terms
of their high-frequncy and low-frequency features.2
2 For your information, if initially the applied voltage and current in the RC circuit are zero,
and then a constant voltage V0 is switched on at t = 0, the voltage equation analogous to (8.8) is
Rt ¡ ¢
t
V
V0 = R I (t ) + C1 I t 0 d t 0 . The solution is I = R0 e − RC , which gives a response time τRC = RC .
0
©2011 Campbell & Peatross
!
Figure 8.5 RC circuit.