Lesson 19: Power Gain & Introduction to Filters

[Pages:4]Lesson 19: Power Gain & Introduction to Filters

Objectives: (a) Define gain and attenuation and describe their application to wireless communication. (b) Convert between unitless gain and gain in decibels. (c) Draw the frequency response plots for ideal low-pass and high-pass filters and determine an output signal for a given input signal. (d) Given an RC circuit, determine the cutoff frequency and whether it is a low-pass or high-pass filter.

1. Gain/Attenuation

When we talk about transmitting signals, we're very interested in the power of the signal that is transmitted (and the power that is received). As you know, power is measured in units of Watt (W). To get a sense of relative amounts of power, consider the following

A typical household incandescent light bulb consumes 60W An exercise bike average output is approx. 200W A microwave uses 1100W AM radio stations transmit 10,000-50,000W

Now, consider a typical situation where a radio receiver receives signal from a radio station as shown in the illustration to the right.

Note that: The signal power decreases significantly at the car receiver signal loss (attenuation). The signal power received is too low signal must be amplified first (gain).

The term gain usually refers to amplification. For an amplifier, the gain (A) is the ratio of the output to the input. So, to calculate power gain (AP) where Pin is the power input and Pout is the power output:

= / (unitless)

Reduction or signal loss is termed attenuation. This occurs when we have a gain that is less than 1. Although gain/attenuation is usually associated with the signal power, it is sometimes used to described the gain/attenuation of the signal voltages as well.

2. Decibels

For many reasons of history and practicality, engineers usually use a logarithmic unit called the decibel (dB) to express gain and attenuation figures. To express power gain or attenuation with decibels (dB):

= ()

If we're dealing with current or voltage gain such as = /, then the gain in decibels is given by

1

 = ()

Notes:

Gains greater than 1 (amplification) will result in a positive (+) dB value.

Gains less than 1 (attenuation) will result in a negative (-) dB value.

A few decibel values that you should know

o 0 dB =

1

o 3 dB =

2 (approximate)

o -3 dB =

? (approximate)

o 10 dB =

10

If you want to convert dB to power gain ratio:

=

()

=

If you want to convert dB to voltage/current gain ratio:

=

()

=

Practice Problem 1: Convert the two gains to decibels (dB). = 1000 = 10 log(1000) = 30 = 0.0001 = 10 log(0.0001) = -40

Practice Problem 2: Convert the following from decibels to ratios.

25

25 = 1010 = 316.2

-6

-6 = 1010 = 0.251

3. Exercises

a. What is the value of the voltage at point X and Y in the circuit below? What is the gain in dB for the 3rd amplifier?

= =

3

595 = 119

5

=

2

=

119 17 = 7

Vin = 1V

y

Vout = 595V X

A1 = 7

A2 = 17

A3 = 5

b. A power amplifier has an output of 10 W and a gain of 36 dB. What is the power into the amplifier, Pin?

=

=

10 36 = 2.5

1010

c. What is the overall power gain (both in dB and ratio) of the following multi-stage amplifier? What is the power at the output of the last stage, Pout?

24

,1 = 1010 = 251.19 1 = 1 ? 251.19 = 251.19

6

,2 = 1010 = 3.98 = 251.19 ? 3.98 = 1000 = 1

Pin = 1nW G = 24dB

Pout = G = 6dB

4. Introduction to Filters 2

Consider the signal shown below, in the time-domain and in the frequency domain.

15 10

5 0 -5 -10 -15

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

10

9

8

7

6

5

4

3

2

1

0

0

5

10

15

20

25

Freq(Hz)

The time-domain signal looks complicated! But what does the frequency-domain representation tell us about this signal?

There are three sin components at frequencies 2 Hz, 15 Hz, and 22Hz.

How can we separate out different frequency components of this signal? We can use filters.

Filters are designed to pass some frequencies and reject others. A filter is a frequency-selective circuit. Today we'll introduce two types of filters: the low-pass filter and the high-pass filter.

Low-pass (LP) filter. Passes frequencies below a critical frequency, called the cutoff frequency (fco), and rejects those above. An ideal frequency response curve for LP filter is shown below.

High-pass (HP) filter. Passes frequencies above the cutoff frequency but rejects those below. An ideal frequency response curve for LP filter is shown below.

Example: Suppose the signal at the top of the page was passed through an ideal low-pass filter with a cut-off frequency of 7 Hz. Write out the time-domain equation for the resulting output signal.

() = 10 sin(22)

Example: Suppose the signal at the top of the page was passed through an ideal high-pass filter with a cut-off frequency of 7 Hz. Write out the time-domain equation for the resulting output signal.

() = 2 sin(215) + 1sin( 222)

Real filters. In actuality, we can't build filters that match the ideal filter response curve, but we can build some effective filters out of basic components such as resistors and capacitors.

Building a filter (we'll find out which kind later)

Filters are intended to be used on circuits that have time-varying inputs. In our next lab, we'll be working with the circuit shown below, and the input will be a single sinusoid, i.e. we're working with an AC circuit.

3

Using the voltage divider equation in terms of impedance (i.e. R and capacitor impedance ), we can determine the relationship / as follows:

Recall that:

ZC

1 jC

1 j2fC

=

-

1 2

=

-

1

=

+

=

2

+

1 2

1 = 2 + 1

For

very

high

frequencies,

__0______

For very low frequencies, __1______

This is a low-pass filter

Cut-off frequency. An ideal low-pass filter will attenuate all frequencies above the cut-off frequency . In practical filters, the transition is more gradual (see figure below).

For an RC filter (as shown above), the cut-off

frequency is the frequency at which the impedance of the capacitor is equal to the impedance of the resistor, i.e. || = , i.e.:

|| = or

1 =

2

The cut-off frequency fco corresponds to the

half-power point.

= 0.5 or -3dB

So, the cut-off frequency for an RC filter (both low-pass and high-pass) is given by:

1 = 2

Magnitude in dB

5

0

-5

-10

-15

-20

-25

-30

-35

-40 2 10

3

10

4

5

10 Frequency in Hz 10

Graph 1-PlotforCircuit1

4

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