Frequency Dependent Circuits
Why is Frequency Response Important?
An Example Circuit
Using Frequency Response Concepts
A Frequency Response Lab Problem

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:

vout(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
vout(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) = Cdvout/dt = CBw cos(wt)
• And then we can compute the voltage across the resistor, R, as:
vR(t) = Ri(t) = RCB w cos(wt)

Now, we can apply KVL to get the input voltage.

• The input voltage is given by,
vin(t) = vR(t) +vout(t)
• Or:
vin(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:
vin(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 -45o.  That is halfway between the low frequency phase (which tends toward 0o as the frequency tends to zero - i.e. DC) and the high frequency phase (which tends toward -90o 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).

NOTE:  This problem has a long time constant and a low frequency so that you can use a real time simulator to check your answer before you submit it. Click here to get the simulator in a separate window.

P5.   In an RC filter circuit , the parameters are:
• R = 1 kW
• C = 0.047 mf
Determine the frequency (in Hertz) for which the attenutation is 0.707.  In other words, the output is reduced by 29.3%.