ECE 252 Introduction to Electrical Engineering

Lesson 15. Frequency Response, Filters, and Resonance


Frequency Response

Circuits with inductors and capacitors change their responses as the driver frequency changes. If the frequency is 0, we get DC.

We will investigate two important classes of simple circuits that respond to changes in frequency: RC filters and RLC resonant circuits.

RC Filters

Circuits with inductors, capacitors, or both change their response as the driver frequency changes. RC filters are especially useful. You will find them in all sorts of electronic equipment, including touch-tone telephone decoders, audio tone controls, and graphic equalizers.

Low-Pass Filter

Below is a circuit for an RC low-pass filter. It tends to pass low frequency signals and block high frequency signals.

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We can solve the circuit for the phasor current I, using Ohm's Law:

I = Vin/(R + 1/jωC)

The output voltage is also found from Ohm's Law:

Vout = IZC = [Vin/(R + 1/jωC)](1/jωC) = Vin/(1 +jωRC)

Now let's define the transfer function H or gain G of the circuit this way:

H = G = Vout/Vin

For this low-pass filter, the result is:

G = Vout/Vin

G = [Vin/(1 +jωRC)]/Vin = 1/(1 +jωRC)

We see that the gain is one when the frequency is zero. The gain falls as the frequency increase. That's why we call this a low-pass filter.

G is a complex quantity, so it contains both magnitude and phase information. It can be expressed in polar form as follows:

G = √1/(1 +(ωRC)2) /-arctan(ωRC)

Usually, the magnitude is the most important part of this function:

G = √1/(1 +( ωRC)2)

Now let's define ωc:

ωc = 1/RC

This is the so-called cutoff frequency, the frequency at which the output is 1/√2 of its maximum value. It is also called the 3 dB point or the half-power point. It is called the 3 dB point because at this point the output is about 3 decibels lower than its maximum value. It's called the half-power point because, at this frequency, the power consumed by the resistor is half of its maximum.

Using this definition in the formula for G:

G = √1/(1 +( ω/ωc)2)

Also note that ω = 2πf, so:

G = √1/(1 +( f/fc)2)

It is frequently more convenient to analyze the circuit in Hz rather than radians/s. Let us do this now. The graph below shows the behavior of this low-pass filter as a function of frequency.

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Notice that at f = fc (the cutoff frequency) the gain has dropped to1/√2 of its maximum value.

Using Decibels Correctly

Gain is often given in decibels, rather than as a ratio. A decibel is one tenth of a bel, and a bel is the common log (base 10 log) of a ratio of powers.
Power gain in bels = log10(Pout/Pin)

Decibels are more commonly used than bels. The equation then becomes:
Power gain in decibels = 10log10(Pout/Pin)

Voltage gain is more commonly used than power gain, and power is proportional to the square of the voltage.
Gain in decibels = 10log10(vout2/vin2)

Taking the squares out of the log gives us the most comon form of the decibel equation:
Gain in decibels = 20log10(Vout/Vin)

Examples:
Vout = 32. Vin = 4. Gain in decibels = 20log10(32/4) = 18.06 dB
Vout = 4. Vin = 32. Gain in decibels =20log10(4/32) = -18.06 dB
Vout = 1/√2. Vin = 1. Gain in decibels =20log10(1/√2) = -3.01 dB

Given gain in decibels, the gain can be found:
G = 10gain in dB/20

Examples:
Gain in dB = 12. G = 1012/20 = 3.98
Gain in dB = -5. G = 10-5/20 = .562

When devices, such as amplifiers or filters, are connected in series, the gains multiply:
GTot = G1 G2 G3 ...

If the gains are given in dB, the gains add:
GTotdb = G1dB + G2dB + G3dB + ...

Low-Pass Filter Practice Problem

Design and sketch a low-pass filter with a cutoff frequency of 1000 Hz. Use a 10 μF capacitor and an appropriate resistor.

fc = 1000 Hz, so ωc = 2π1000 = 6283 radians/s

ωc = 1/RC

R = 1/ωcC = 1/(62831010-6) = 15.9 Ω

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High-Pass Filter

A high-pass filter tends to block low frequency signals and pass high frequency signals. A high-pass filter is shown below.

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The cutoff frequency of the high-pass filter is calculated exactly the same as for the low-pass filter:

ωc = 1/RC

The gain G is also quite similar:

G = √1/(1 +( ωc/ω)2)

G = √1/(1 +( fc/f)2)

Loading of Filters

The filter analysis and equations above will only work if the filter is unloaded. This assumes that a pure voltage source is connected to the input and that nothing is connected to the output. This is never completely the case with real circuits, but it can be approximately true. Consider the low pass filter circuit below.

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This is a low pass filter fed by a device that is represented by its Thevenin equivalent circuit. The filter will have a cutoff frequency given by ωc = 1/RC. C is obviously CLO, but what is R? Notice that RT is in series with RLO. This series combination should be used for the R in calculating ωc = 1/RC. If and only if RLO >> RT, RLO can be used for the R, and RT can be ignored.

Now consider the high pass filter circuit below.

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This is a high pass filter loaded by some device. The device could be an amplifier with an input impedance of RLoad. The high pass filter will have a cutoff frequency found from this equation: ωc = 1/RC. It should be clear that the C in this equation will be CH, but what is R? Notice that RH is in parallel with RLoad. Therefore, the parallel combination of RH and RLoad should be used for the R. If RLoad >> RH, then RLoad can be ignored, and RH can be used as the value of R.

Band Pass Filter

Band pass filters are intended to block low and high frequencies, while allowing signals in a midrange of frequencies to pass through without attenuation. Loading considerations are important for such filters. Below is a typical band pass filter.

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In order to keep the math easy, we would like to be able to consider the hi-pass section as independent from the lo-pass section. This can be approximately true if the lo-pass section looks like an infinite load to the hi-pass section. Notice that the output of the hi-pass section consists of the RHi resistor, and the input to the lo-pass section starts with the RLo resistor. We can therefore make the lo-pass section look like a very high impedance load by making RLo >> RHi. The values chosen for RHi and RLo will depend on the precision of the results required, but a good rule of thumb is to make RLo > 50 RHi.

The example below shows how to use these ideas to analyze a band pass filter.

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Notice that RLo >> RHi. Therefore the two filters can be treated as though they are independent, and the usual equations will apply.

For the hi-pass section:
ωc = 1/RC = 1/(2.2 kΩ 2 F) = 227 rad/s.
fc = ωc/2π = 36.2 Hz
GHi = √1/(1 +( fc/f)2) = √1/(1 +(36.2/f)2)

For the lo-pass section:
ωc = 1/RC = 1/(150 kΩ .01 F) =667 rad/s.
fc = ωc/2π = 106.1 Hz
GLo = √1/(1 +( f/fc)2) = √1/(1 +(f/106.1)2)

The total gain is the product of the hi-pass and lo-pass gains:
GTot = GHi GLo = √ 1/(1 +(fc/f)2) 1/(1+(f/fc) 2) = √1/(1 +( 36.2/f)2)1/(1+( f/106.1)2)

Here are a few sample values for the total gain:
f = 10 Hz. GTot = GHi GLo = √ 1/(1 +(36.2/10)2) 1/(1+(10/106.1) 2) = .266 .996 = .265
f = 35 Hz. GTot = GHi GLo = √ 1/(1 +(36.2/35)2) 1/(1+(35/106.1) 2) = .692 .949 = .657
f = 100 Hz. GTot = GHi GLo = √ 1/(1 +(36.2/100)2) 1/(1+(100/106.1) 2) = .940 .728 = .684
f = 500 Hz. GTot = GHi GLo = √ 1/(1 +(36.2/500)2) 1/(1+(500/106.1) 2) = .997 .208 = .207

It should be clear that the gain is (relatively) high between the low and high cutoff frequencies, but falls off rapidly for frequencies higher and lower than these limits.


Before going on, you should complete Tutorial 15 on RC filters.


Resonance

A series resonant circuit has a resistor, an inductor, and a capacitor in series, as shown below.

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A parallel resonant circuit has a resistor, an inductor, and a capacitor in parallel, as shown below.

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In either case, the resonant frequency is given by:

ω0 = 1/√LC

At this resonant frequency, the effect of the inductor exactly cancels out the effect of the capacitor, and the impedance is just the impedance of the resistor alone. At resonance the inductive reactance XL is jω0L, and capacitive reactance XC is 1/jω0C. Math tells us that XL = -XC at ω0. That's why the inductor and capacitor cancel each other out at resonance.

Series Resonance

For the series resonant circuit:

Z = jωL + R + 1/jωC = R + j(ωL - 1/ωC)

The magnitude of this impedance is:

Z = √R2 + (ωL - 1/ωC)2

Although the phase information may occasionally be significant, the magnitude is usually of much greater importance.

At resonance, Z = R because the impedance of the inductor exactly cancels the impedance of the capacitor. At any other frequency, the impedance is greater.

If a sinusoidal voltage source of maximum magnitude Vm drives the series resonant circuit, it will result in a current of:

I = Vm/Z

The real power consumed by the series resonant circuit will all be consumed in the resistor. At resonance, Z = R, as explained above. This will result in the maximum current possible:

I0 = Vm/R

Series Resonant Circuit Practice Problem

Analyze the series resonant circuit below to find the resonant frequency, the impedance at the resonant frequency, the impedance at 2ω0, and the impedance at 0.5ω0.

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ω0 = 1/√LC = 1/√.025(4x10-6) = 3162 radians/s

f0 = ω0/2π = 503 Hz

At this frequency the impedance is purely resistive: 300Ω.

At ω = 2ω0 = 2 x 3162 = 6324 radians/s (1006 Hz):

Z = √R2 + (ωL - ((6324)(.025) - 1/((6324)(4x10-6))) 2 = 322.6 Ω

At ω = 0.5ω0 = 0.5 x 3162 = 1581 radians/s (252 Hz):

Z = √R2 + (ωL - 1/ωC)2 = √3002 + ((1581)(.025) - 1/((1581)(4x10-6))) 2 = 322.6 Ω

Notice that the impedances are the same at these two frequencies.

Parallel Resonance

A parallel resonant circuit is shown below.

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For this circuit:

1/Z = 1/R + jωC + 1/jωL = 1/R + j(ωC - 1/ωL)

The magnitude of this impedance is:

Z = 1/√1/R2 + (ωC - 1/ωL)2

At resonance, Z = R because the impedance of the inductor exactly cancels the impedance of the capacitor. At any other frequency, the impedance is less.

Parallel resonant circuits are frequently used for frequency selection (tuning) in circuits such as radios and TVs.


At this time you should complete Tutorial 15A on resonance.