SIGNAL ANALYSIS

These undesired sounds occur in the frequency spectrum above the 10kHz-frequency range. To eliminate these undesired sounds would therefore require a frequency filter allowing only the frequencies below 10kHz to be reproduced for the listener. This type of filter is described as a 'Low Pass Filter' as it is required to pass all frequencies below 10kHz but block all frequencies above this frequency from passing. The frequency of 10kHz is referred to as the 'Frequency Cut Off Point', as this is the frequency at which we want to cut all other frequencies. The Frequency Cut Off Point is usually denoted as 'fc'or '?c'.

Below is the response graph or Bode Plot of an 'Ideal' Low Pass filter. A Bode plot has output Gain(dB) against Angular Frequency(?).

Both of these quantities can be obtained from the following formulas.

Gain dB = 20 log (Voltage Out / Voltage In)

Angular Frequency (?) = 2?f or Where: f = Frequency (Hz)

T = Time Constant (s)

A Time Constant is the time it takes for the voltage across a capacitor, within an RC combination to reach 63% of an applied voltage.

The Ideal Filter below shows that the pass stage of the filter is at a fixed level of 0dB's indicating that there are no gains and no losses. Therefore the input signal is equal in amplitude to that of the output signal.

When the frequency exceeds the cut off frequency ?c, The output of the filter becomes infinitely small, Zero. Therefore nothing above ?c has passed through the filter.

Fig 1 Ideal Filter

In reality the response of a filter is as in the following diagram.

Fig 2 Typical Filter

It can be seen that the frequency does not cut off immediately after the cut off frequency ?c has been exceeded. Nor does the cut off frequency occur at a 0dB level. The cut off now occurs once the amplitude reaches -3dB. At -3dB the signal is at 50% of its power.

This -3dB point is also referred to as a 'Corner Point' and the 'Roll Off Point'. The reference as a 'corner point' is obvious from Fig1. This is the point where we would expect a change in direction. The term 'roll off' is apparent from Fig2 as the level clearly rolls off but also the gradient of the roll off increases after the cut off frequency. The roll off is given as Decibels per decade (on a logarithmic scale). For a single stage filter the roll off is 20dB per decade. By increasing the order of the filters i.e. two low pass filters in series we have a second order filter. This will improve the response and cause the curve to focus in around the cut off frequency as required.

Fig 3 Typical 2-stage Filter

It must also be noted that as the stages are put in series the power lost at each stage must be accounted for and therefore a buffer should be used to ensure the signal is strong enough for the next stage.

Transfer Functions and the Laplace Transform:

A circuit contains components with the properties of resistance, capacitance and inductance. A combination of these in a circuit or network can be represented mathematically with the inclusion of either voltage or current. The describing equations for resistance, capacitance and inductance are as follows:

Resistor =

Capacitor =

Inductor =

Using these equations in a circuit, enable the circuit to be represented as a mathematical model called a system model. This system model will be described as a differential equation. Differential equations can be simplified into algebraic equations by the use of a method called the Laplace Transform named after the mathematician who invented the technique. The technique involves the integration of a function with an exponential. The formula is given as:

(Laplace Transform)

The Laplace transform takes a function from the time (t) domain and represents it the (S) domain or Laplace domain.

Now with a circuit or system in the Laplace domain, if a signal is applied to the system model it too must be transformed from the time domain and represented in the (S) domain or Laplace domain. Note that when trying to analyse frequency and phase, 'S' can be substituted by 'j?' The signal is then applied to the system model and an output will be achieved. The gain i.e. the ratio between the output and the input is represented by the system model. The transformed representation of the system model is referred to as the Transfer Function.

E.g.

The Low Pass filter circuit on the previous page can be written as:

Which when rearranged gives the transfer function of:

If the Transfer Function for a filter is calculated, then it is possible to calculate the cut off frequency of the filter by the use of simple mathematics. When the denominator of a Transfer function is equal to zero, it is known as a 'pole'.

By calculating the poles of a filters Transfer function, the cut off frequency is obtained. As illustrated below:

Now substituting S for j?, we get:

Fourier Analysis:

Fourier, a French mathematician discovered that any periodic function could be represented by the summation of a series of Sine waves. This method is known as Fourier Series and can be obtained by the following formula.

The Fourier Coefficients ao, an and bn can be found with the following formulas:

Each sine or cosine in the Fourier series represents the harmonic content of the signal. The first in the series is the first harmonic or fundamental frequency. This will usually have the largest amplitude and the lowest frequency.

Components of a sine function:

The Process of the Fourier Transform allows a signal/function to be represented as a spectra. This is obtained by using the following formula:

RESULTS

Below are the results for part 1.

Low Pass Filter

R = 470 ? C = 0.33?F

The Laplace Transform for this filter is 1/(RCs + 1).

Taking the poles to zero gives a cut off frequency of 1026Hz.

Table of Results:

Frequency (Hz)

Output Voltage (v)

Output Decibels(dB)

250

7.7

-0.33

500

7

-1.16

750

6.3

-2.07

500

4.4

-5.19

2000

3.5

-7.18

3000

2.5

-10.1

4000

.9

-12.49

5000

.6

-13.98

Graph of Results:

Bode Plot for the Low Pass Filter:

High Pass Filter

R = 47 ? C = 0.33?F

The Laplace Transform for this filter is 1/(RCs + 1).

Taking the poles to zero gives a cut off frequency of 10.26KHz.

Table of Results:

Frequency

Output Voltage

Output Decibels

500

0.4

-26.02

000

0.7

-21.16

2000

.4

-15.14

5000

3.4

-7.43

7000

4.3

-5.39

8000

4.7

-4.62

9000

5

-4.08

0000

5.4

-3.41

2000

6

-2.5

Graph of Results:

Bode Plot for the High Pass Filter:

Band Pass Filter

R1 = 47 ? R2 = 470 ? C = 0.33?F

The Laplace Transform for this filter is 1/(RCs + 1).

Taking the poles to zero gives a cut off frequencies of 1026Hz and 10.26KHz

Table of Results:

Frequency

Output Voltage

Output Decibels

0

0

----------

00

0

-40

000

64

-23.88

500

76

-22.38

2000

80

-21.94

3000

80

-21.94

4000

80

-21.94

5000

80

-21.94

6000

80

-21.94

7000

78

-22.16

8000

76

-22.38

9000

72

-22.85

0000

68

-23.35

1000

64

-23.88

2000

62

-24.12

5000

56

-25.04

Graph of Results:

Bode Plot for the Band Pass Filter:

Stop Band Filter

R = 1k ? C = 0.33?F

The Laplace Transform for this filter is 1/(RCs + 1).

Taking the poles to zero gives a cut off frequency of 1.04KHz.

Table of Results:

Frequency

Output Voltage

Output Decibels

50

0

00

0

250

0.68

-3.35

500

0.35

-9.12

750

0.155

-16.19

800

0.125

-18.06

900

0.074

-22.62

000

0.02

-33.98

100

0.022

-33.15

250

0.08

-21.94

500

0.17

-15.39

2000

0.3

-10.46

3000

0.62

-4.15

4000

0.64

-3.88

5000

0.7

-3.1

0000

0.88

-1.11

20000

0.96

-0.35

50000

0

Graph of Results:

Bode Plot for the Band Stop Filter:

2-Stage Low Pass Filter

R = 470 ? C = 0.33?F

The Laplace Transform for this filter is 1/(RCs + 1).

Taking the poles to zero gives a cut off frequency of 1026Hz.

Table of Results:

Frequency

Output Voltage

Output Decibels

00

8

0

250

6.4

-1.94

500

4.8

-4.44

750

3.8

-6.47

500

.8

-12.96

2000

.3

-15.78

3000

0.8

-20

4000

0.5

-24.08

5000

0.3

-28.6

Graph of Results:

Bode Plot for the 2-StageLow Pass Filter:

RESULTS FOR PART 2

Square Wave:

Due to the time required to reproduce equations on the computer, the hand written Fourier Series for this function is included in the appendices of this report.

Ramp Function:

The function for the Ramp Waveform can be given as:

Note: T = 1ms

The formula for the Fourier Series representation of a function is given as:

As this function is even, we only need to obtain the Fourier coefficients ao and an.

a0 = 0

So ao comes to zero now we need to calculate an.

Now the first part of the equation goes to zero and the second part requires integration by parts.

This will therefore give:

an

Which:

The Fourier Series for the Ramp function is therefore:

For 'n' being ODD

e.g.

Using the Fourier series from the other page:-

The amplitude of the fundamental is 3.24 volts and the second harmonic is 0.36 volts.

Half Wave Rectifier:

Due to the time required to reproduce equations on the computer, the hand written Fourier Series for this function is included in the appendices of this report.

Full Wave Rectifier:

Due to the time required to reproduce equations on the computer, the hand written Fourier Series for this function is included in the appendices of this report.

Modulated Sine Wave:

Modulated Square Wave

Due to the time required to reproduce equations on the computer, a general hand written explanation for modulation is included in the appendices of this report.

Discussion

The majority of the objectives have been achieved within this report as has the theory section of this report.

The graph of the low pass filter results clearly shows that the cut off frequency does coincide with the -3dB point. The result is very close to 1kHz (the selected frequency ). The Bode plot for the low pass filter also supports this claim. Note that the phase also changes around this point. This is due to the capacitive element of the circuit, as the capacitors reactance to ac signals is dependant on the frequency. This will therefore affect the phase of the output signal. It is also apparent from the Bode plot that the roll off does follow the 20dB per decade fall.

The same applies for the results of the high pass filter.

The Band Pass filter results do appear to have a problem. It is difficult to identify the cut off point as the signal is very weak. This is because as mentioned in the theory, a buffer should be used to ensure the output level is maintained. Because of this I feel it is not fair to make a defined statement regarding these results although It should be noted that it does appear that the main part of the Pass Band is between 1kHz and 10kHz as selected.

The results for the notch filter are very precise with very little error. The graph clearly shows the output attenuates at 1kHz. This filter is very effective as the drop is -35dB in only one decade. This however is not as good as an ideal model.

The graph of the two-stage low pass filter, demonstrates the problem of using this method. Although the roll off is greater, it starts to roll off sooner i.e. before the -3dB point. This means that the roll off coincides with the selected cut off frequency at -9dBs. Note that from the Bode Plot, the roll off is now just greater than 40dB per decade.

The Square wave function had amplitude of eight volts peak to peak The time period is the time taken to complete one of the functions periodic cycles. In this case the time period is one millisecond. The Fourier series was calculated giving the Series to be:

Now by referring to the sine description from the theory section, we have:

Now by taking the first term of the series (the fundamental) the comparison can be made with that of the spectrum graph. The amplitude is defined as:

16/? which is 5.09 to 2dp.

If this value is then converted to decibels we get:

20log (5.09) = 14.14 Volts which corresponds to the graph.

Repeating this for the next term (the second harmonic) we get:

20log (1.70) = 4.60 Volts which also corresponds to the graph.

Now if we look at the Frequency of both of these terms we see that the frequencies are given as angular frequencies. These must be converted to frequency in Hertz by dividing by 2?. For the first harmonic this gives a frequency value of 1000Hz and for the second harmonic a value of 3000Hz. Again this corresponds to the graph.

This method was then repeated for the Ramp Function and again the results of the Fourier series correspond to the graph.

The results of this clearly show that indeed a function can be represented by a summation of sine waves but also that the Fourier series can provide the harmonic content of any periodic function.

The modulated signal demonstrates that a signal can be multiplied by another signal to produce a new signal combining both sets of harmonics. The Modulated ~Sine wave demonstrated that if we have a frequency ? and modulate this with a carrier frequency ?c, we actually get a symmetry of the carrier frequency ?c about the frequency ?. In the case of the Modulated Sine wave the frequency ?c was 500Hz and the carrier frequency ? was 30kHz. If we look at the graph we can see that about the 30kHz spike, are two other spikes that are 500Hz away from the 30kHz one. The smaller spikes on the graph are noise. This symmetry is also supported in the maths of the Fourier Transform (in the appendices). The text in green demonstrates the reason why this symmetry exists.

Conclusion

All of the results from this experiment have met the requirements of the objectives of this report and also proves the theory section of the report. The lab has proved to be a valuable learning experience that has joined both the mathematical representation of filter circuits and signals in a visualised manor, which has allowed an understanding and interest to be gained.

Errors occurring in this lab are due to component values and tolerances, mistakes in the mathematics and human error in obtaining readings and measurements.

Appendices

ASSIGNMENT 1&2 LAB REPORT BEng Electronic Engineering Year '2'

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