Did you ever buy audio equipment and look carefully at how the manufacturer
specified how well the equipment would work? (And audio equipment is one
of the few consumer items where people actually try to sell things on the
basis of how well they work!) If you looked at the specifications for audio
equipment you would probably find the following.
A frequency response for
the unit.
If the unit is a speaker
set, you'll find separate frequency responses for the different speakers
like the mid-range, or the woofer and the tweeter.
If the unit is a microphone,
you'll find a frequency response that tells you how the unit responds to
different frequencies.
Frequency response is an important concept in
many areas - within electrical engineering and outside of electrical engineering.
Having a good grasp of frequency response is important in many areas, so
our objectives in this lesson include the following.
Given a linear system
or circuit described mathematically,
Be able to compute a frequency
response for the system.
Be able to predict an
output signal from a given input sinusoidal signal.
An
Example Circuit
We are going to examine a simple circuit that has frequency dependent behavior,
a resistor-capacitor (RC) circuit. It is shown below. To illustrate how
this circuit responds to a sinusoidal signal input we can do any of the
following.
We can write the differential
equation relating the input and output voltages and solve for the output
assuming a sinusoidal signal input.
We can assume a sinusoidal
input and use LaPlace transform methods to compute the output voltage.
Since the input is a sinusoid,
we know that the output contains a sinusoid and terms that decay to zero.
We can work from there.
We
will use the third approach - and we will assume a steady state output
and work backwards from the output to compute the input.
Since the first thing we want to do is just to look at how a circuit can
affect sinsusoidal signals, we're going to assume a sinusoidal output and
work backwards to calculate the input voltage that produces that output.
That's not a very general approach, but it will get us what we want now,
and prepare us for other things to come. We will be able to do that
without too much algebraic pain, and we can learn some things from the
result.
So, we will assume that the output voltage is given by:
v_{out}(t)
= B sin(wt)
Be sure that you understand that B is the magnitudeof
the output signal
Now, what does that form for the output voltage imply?
If the output voltage
is given by
v_{out}(t)
= B sin(wt)
Then, since the output
voltage is across a capacitor, we can compute the current flowing through
R and C as:
i(t) = Cdv_{out}/dt
= CBw
cos(wt)
And then we can compute
the voltage across the resistor, R, as:
v_{R}(t)
= Ri(t) = RCB w
cos(wt)
Now, we can apply KVL to get the input voltage.
The input voltage is given
by,
v_{in}(t)
= v_{R}(t) +v_{out}(t)
Or:
v_{in}(t)
= B(RCw
cos(wt)
+ sin(wt))
At this point step back from this. It may not be obvious, but we can take
advantage of a trigonometric identity,
sin(x+y) = sin(x)cos(y)
+ cos(x)sin(y)
if only we can make the things that multiply
the sines and cosines in the second bullet above look like other sines
and cosines.
We know:
sin(x+y) = sin(x)cos(y)
+ cos(x)sin(y)
And, we know:
v_{in}(t)
= B(RCw
cos(wt)
+ sin(wt))
And the second expression
can be put into the form of the first
We need to refer to a little geometrical
construction - at the right. "Clearly" we have the relationships
indicated below for cos(f)
and sin(f)
So, now we can write:
Which reduces to:
There are two conclusions to draw from the
resulting expression for the input voltage.
The ratio of the amplitude
of the output to the input voltage is given by:
The output voltage lags
the input voltage by a phase angle, f.
You can use the expressions for the gain, B/A,
and the phase shift, f,
to predict behvarior of circuits like this. You should note the following
in these expressions.
In the expression for
B/A, there is a factor (wRC)
which determines the attenuation. There is a "critical frequency"
where that factor is 1. That frequency is:
w
= 1/RC
or f = 1/2pRC
When f = 1/2pRC
the attenuation is 0.707 (the reciprocal of the square root of 2).
You can consider that as the mid-point in freqency where the frequency
is between the high frequency range (where the circuit does not pass a
sinusoidal signal well) and the low frequency range (where a sinusoidal
signal tends to pass through the filter unchanged).
When f = 1/2pRC
the phase is -45^{o}. That is halfway between the
low frequency phase (which tends toward 0^{o} as the frequency
tends to zero - i.e. DC) and the high frequency phase (which tends toward
-90^{o} as the frequency gets very high).
So, there are two reasons
to think of that frequency (f = 1/2pRC)
as a critical frequency.
Note that that frequency
is sometimes referred to as the bandwidth of the circuit. It's one
way to measure the band of frequencies that get through the circuit relatively
unscathed.
With those thoughts you can think a little more
deeply using the simulator we have just below.
A Simulation
of the Circuit
Note:
- This simulator is real time. However, to let you see how the circuit
behaves, we have made the signals very slow - on the order of a few Hertz,
or even a fraction of a Hertz. The time constant (the R-C product)
should be correspondingly long - on the order of a second (from a fraction
of a second to a few seconds). You won't see much if you stray far
from these limits - even though these are long time constants and the bandwidths
are quite low. That's just for purposes of illustration. (However,
note that you could get a one second time constant using R
= 1.0 MW,
and C = 1.0mf.)
Here is the simulator.
Using this simulator,
you can do the following.
You can change the frequency.
Notice that higher
frequencies are attenuated more. (The
output is smaller.)
Lower frequencies are
attenuated less. (The output is larger.)
You can change the time
constant (The R-C product).
Notice that higher
time constants shift the bandwidth lower, and high frequencies are attenuated
more.
Lower time constants shift
the bandwidth higher, and high frequencies are attenuated less.
(The output is larger.) Actually, with
lower time constants, the bandwidth is higher and more frequencies get
through the circuit.
Problems
& Questions
P1.
Here is an RC filter circuit - the same one discussed above.
In this circuit, the parameters
are:
R = 10 kW
C = 0.1 mf
Determine the bandwidth
of the circuit.
P2.
In the RC filter circuit determine the resistance value that gives a bandwidth
ten times as large as in problem 1.
P3.
In the RC filter circuit , the parameters are:
R = 10 kW
C = 0.1 mf
Determine the phase shift
between output and input at 200Hz. Remember the sign and give your
answer in degrees (not radians).
P4.
Assume that you have an RC filter circuit with a 0.5 second time constant.
Determine the phase shift between output and input at 1.0Hz. Remember
the sign and give your answer in degrees (not radians).
Determine the frequency
(in Hertz) for which the attenutation is 0.707. In other words, the
output is reduced by 29.3%.
Reflections
In this lesson you have been introduced to a simple frequency dependent
circuit. We have used a very brute-force method - assuming a voltage
at the output and chasing that back to the input. That worked here,
but it won't work everywhere. Moreover, this is not the easiest way
to make such predictions, and before you attempt more complex circuits
you should learn a better way to analyze frequency dependent circuits.
You need to learn about impedance
and phasors. There are links below that
will take you to many other topics.