Lecture Three

Question 1

What is the function of a filter”?

Answer 1

To remove unwanted frequency components from a signal and hence improve the signal to noise ratio

Question 2

What are the types of filters?

Answer 2

Hardware (analogue) filters (analogue signal)
Digital Filters (digital signal)

Question 3

What is the best domain representation of filtering?

Answer 3

Frequency Domain

Signal and noise can be well discriminated in this domain

Question 4

What is the amplitude response of the ideal filter?

Answer 4

The amplitude response of the ideal filter is a rectangular function (bandpass) to seperate signal from noise.

An ideal filter will produce no phase shift.

Question 5

How can the filter be mathematically represented?

Answer 5

The frequency domain representation of the filtered signal is the product of the fourier transform of the input signal X(f) and the filter response H(f)

i.e multiple desired frequency by one and noise by zero

Question 6

How is a filter on the frequency domain described?

Answer 6

Pass band (signal desired)
Stopband (noise removal)

Question 7

What are the other types of filters and their characteristics?

Answer 7

Low pass filter (stopband high Hz)

High pass filter (passband high Hz)

Band Pass filter (keep a band of Hz)

Band Stop (notch) Filter (remove a notch of noise)

Question 8

What are the characteristics of the ideal filter?

Answer 8

• Gain is frequency independent in the pass band and 0 in the stop band (i.e multiple by one and zero)
• Produce no phase shift, this may not always be possibly but a linear phase shift in the passband represents a simple delay. i.e the same delay at all frequencies. (radians in shift)
• Doesnt alter quality of signal

Question 9

What happens in reality with digital filters?

Answer 9

The filter is unlikely to remove all frequencies outside of the passband. Instead the response is likely to roll off at the cut off frequency.

Question 10

How do we generate the time domain properties of a filter?

Answer 10

Use fourier transform properties

Question 11

Describe the relation of the time and frequency domain using notation;

Answer 11

We stated that
Hz domain;
Y(f) = X(f) x H(f) Hz response

i.e multiple the Hz domain to remove noise

Time Domain;
y(t) = x (t) . h(t) Impulse response

i.e convolution with the impulse response to remove noise

y(t) = filtered signal
x(t) = unfiltered signal
h(t) = filter
. = process of convolution

The impulse response h(t) for a filter H(f) specified in the Hz domain is simply the inverse fourier transformation

Question 12

What is convolution?

Answer 12

The filter data set is formed by convolving the the raw data with the filter weights.

These filter weights are sampled representations of h(t) the impulse response of the filter.

i.e three points moving average

y(i) = 0.25x(i-1) + 0.5x(i) + 0.25x(i+1)

i.e it moves along the line of data multiplying it by x amount of numbers either side to average it. Removes noise between samples

DONT NEED TO REPRODUCE EQUATION

Question 13

Redescribe convolution in terms fo y values

Answer 13

Each value of Y is evaluated as the weighted average of the corresponding input data according to filter weights h(k)

h(k) = 1 for low pass and 0 for high pass

Question 14

Whats another name for a filter that involves convolution?

Answer 14

It is because NON-RECURSIVE filters are implemented using finite set of filter weights that they are also called finite impulse response (FIR) filters

Question 15

How do we analyse the performance of filters in the time domain?

Answer 15

We need to obtain the Hz response we fourier transform its impulse response

Question 16

What is a three point moving average filter useful for?

Answer 16

Mainly low frequencies because it attenuates the signal therefore high Hz noise will increase their y values.

Stretching the impulse response (covolution) will contract its Hz domain representation. May be some leakage at high Hz with this filter.

Question 17

When is the attenuation of convolution sharped up?

Answer 17

If there are a large number of input data points

Question 18

Because of the principles of convolution what does this mean for frequency representation following filtering?

Answer 18

Inverse fourier transformation can determine the impulse response

Question 19

Describe the characteristics of a rectangular or brick wall filter (low pass) on the time domain;

Answer 19

The impulse response of this filter has an infinite extent in the time domain . The filter weights do not drop evenly to zero, but exhibit widespread ripple.

Hence to approximate this perfect filter response , it is necessary to convolve the impulse response with a considerable number of input data points.

Question 20

What is a compromise to the brick wall filter?

Answer 20

The blackman filter

Question 21

Describe the blackman filter

Answer 21

Compromise in which the attenuation characteristics are acceptable but time domain implementation of the filter is considerably more efficient than for the brick wall filter.

• Sinusoid taper applied to the cut off Hz in Hz domain
• The corresponding impulse response resembles a gussian function which drops smoothly to zero in the time domain NO RIPPLE

Question 22

Benefits of the gussian impulse response in the blackman filter?

Answer 22

Can be implemented with far fewer convolutions operations than the black wall filter

As for the triangle moving average filter, Fc (cut off) is adjusted by time scaling the impulse response

Fewer terms in the impulse response will give a higher cut off Hz and vice versa.

Look at notes on impulse and frequency response characteristics. Should be able to draw and explain

Question 23

What does a 3 point moving average require?

Answer 23

Large seperation between signal and noise

Question 24

What does an n point moving average have over 3 point?

Answer 24

Sharpens signal if noise and signal are closer together

Question 25

Describe filtering in the Hz domain;

Answer 25

Extend principles of recursive filters into non-recursive.

• Multiplication is performed on both real and imaginary spectral components produced by Fourier transformation. Preserves the symmetry of the discrete fourier transformation
• B/C Hz components are reflected around the central Hz the Hz response of the indrect filter must have the same form (i.e reflected)