1. MM3FC Mathematical Modeling 3 LECTURE 2 Times Weeks 7,8 & 9. Lectures : Mon,Tues,Wed 10-11am, Rm.1439 Tutorials : Thurs, 10am, Rm. ULT. Clinics : Fri, 8am, Rm.4.503 Dr. Charles Unsworth, Department of Engineering Science, Rm. 4.611 Tel : 373-7599 ext. 2461 Email : c.unsworth@auckland.ac.nz

2. This Lecture What are we going to cover & Why ? Discrete representation of continuous signals. (because this is how we digitize a signal) The Running Average Filter. (the simplest form of FIR filter) The General FIR filter. Impulse response of a filter. (a way to classify a filter’s characteristics)

3. Real world (analogue) signals are “ continuous ” in time. A ‘continuous sinusoid’ x(t) is represented : x(t) = Acos(ωt + φ) Any recorded signal is said to be “discrete ”. A ‘discrete (digital) signal’ is a ‘snap-shot’ x[n] of a continuous signal taken every (T s ) secs. x[n] = x(t) with t= nT s x[n] = Acos(ω(nT s ) + φ) Where, (n) is an integer indicating the position of the values in the sequence. Continuous & Discrete Signals … (2.1)

4. Discrete-Time (Digital) Filters A ‘digital filter’ - takes in a digital ‘input signal’ x[n]. - alters it in some way, by an operation T{ }. - And creates a digital output signal y[n] = T{x[n]} x[n] Input y[n] = T{x[n]} Ouput Discrete Time System T{ }

5. A discrete-time (digital) signal is just a sequence of numbers. Thus, one can compute the values of the output sequence y[n] from it’s input sequence x[n]. Example 1. y[n] = (x[n]) 2 The y[n] value is just the square of the x[n] value. (y[1] = x[1] 2, y[2] = x[2] 2, y[3] = x[3] 2, …, y[n] = x[n] 2 ). Example 2. y[n] = max{x[n], x[n-1], x[n-2]} The y[n] value is the largest of either x[n], x[n-1] or x[n-2]. Obviously, there are an infinite number of systems that can be created.

6. Example 3 : The digital signal x[n] = {1,2,3,4,5, …, N} is passed through the filter y[n] =(x[n]) 2. Determine the new filtered sequence y[n].

7. The ‘Running Average’ filter A ‘Finite Impulse Response’ (FIR) Filter takes a finite length input sequence x[n] and produces a finite length output sequence y[n]. The simplest FIR filter is the ‘running average’ filter. It computes the ‘moving average’ of two or more consecutive numbers in a sequence. The ‘FIR’ filter is a generalisation of the idea of the running average filter. Example 4 : A ‘3-point running average’ filter creates 1 output value from 3 consecutive input values divided by 3. y[0] = 1/3( x[0] + x[1] + x[2] ) y[1] = 1/3( x[1] + x[2] + x[3] ) Which generalises to : y[n] = 1/3( x[n] + x[n+1] + x[n+2] ) This is known as a ‘difference equation’. It completely describes the entire ‘output signal’ for all indexed values –inf to +inf. … (2.2)

8. Let’s consider a finite input signal x[n] which is triangular in shape. Which is the discrete-time (digital) sequence ; n n 5 x[n] 0 0 0246420 0 Using the difference equation : y[n] = 1/3( x[n] + x[n+1] + x[n+2] ) We can create a ‘difference table’ to calculate the filters output. 2 4 6 4 2

9. N n 5 x[n] 0 0 0246420 0 y[n] 02/3 24 14/342 2/30 0 2 4 6 4 2 2 4 14/3 4 2 2/3 Note : How the output sequence is longer and smoother than the input sequence. Note : That the output sequence starts before the input sequence starts. This is because we are using inputs from the present and future values.

10. A ‘filter’ that uses ‘future’ values of the input is called ‘non-causal’. (Non-causal systems cannot be used in real-time applics. as the data is not available.) A ‘filter’ that uses only ‘present’ and ‘past’ values is called a ‘causal’ filter Example 5 : Write the difference equation for a ‘3-point running average’ causal filter. And draw out it’s difference table. Difference equation : y[n] = 1/3( x[n] + x[n-1] + x[n-2] ) N n 7 x[n] 0 0 024 6 42 0 0 0 0 y[n] 0 0 0 2/3 2 4 14/3 4 2 2/3 0 0 The ‘causal’ filter is also known as a ‘backward averager’ since all values are past and present. When ‘present’ and ‘past’ values are used, in a causal filter, the output does not start before the input starts.

11. Causal : Uses present & past and output doesn’t start before input Non-causal : Uses contains future and output starts before input.

12. The ‘support’ of a system is the set of values where the sequence becomes non-zero. Thus, the ‘support’ of the causal system’s input is : Finite in the range : Thus, the ‘support’ of the causal system’s output is : Finite in the range : What’s the support for the ‘non-causal’ systems input and output ?

13. The General FIR filter The ‘running averager’ was a special case of the ‘FIR general difference equation’ : Namely, when M = 2. This is the ‘order’ of the filter. b k = {1/3, 1/3, 1/3}. Is the value of the ‘filter coefficient’ for each (k). (If (b k ) is not the same then we say that the filter is a ‘weighted running averager’.) The ‘filter length’ is defined as L = M+1 = no.filter coefficients. There will be an interval of ‘M’ samples at the beginning where the sliding window engages with the data using less than M+1 points. There will be an interval of ‘M’ points at the end where the sliding window dis-engages with the data. The ouput sequence can be as much as ‘M’ samples longer that the input. Let k be –ve for a non-causal. … (2.3)

14. Example 6 : The FIR filter is completely defined once the set of filter co-efficients {b k } are known. 0 1 2 3 = k {b k } = { 3, -1, 2, 1 } The length of the FIR filter is : L = 4. The order of the filter is : M = L-1 = 3.

15. Example 7: Compute the output y[n] for this FIR filter in the form of a difference table for our previous input sequence in Example 5.

16. In a real-time system, we don’t have any data at time n< 0 ? But our filter requires x[n-1], x[n-2] & x[n-3] to be known. Intially, these values do not exist. The 1st ‘Initial Rest Assumption’ helps us alleviate this : Initial Rest Assumption 1) The input x[n] is assumed to be zero prior to some starting point (n 0 ) i.e x[n] = 0 for n< n 0 We say that the inputs are ‘suddenly’ applied.

17. The Unit Impulse Sequence The ‘unit impulse’ has the simplest sequence which consists of one nonzero value at n=0. The mathematical notation is the ‘kronecker delta’,  [n]. n n 7  [n] 0 0 010 0 00 0 0 0 0 { … (2.4)

18. The ‘Shifted’ Impulse response For example  [n-3], is nonzero when its argument is zero. i.e. n-3 = 0 n n 7  [n-3] 0 0 000 0 10 0 0 0 0 Why learn such a trivial signal ? … What is all this for … ?

19. Example 8 : The digital signal x[n] is respresented as the following series of unit impulses. Determine the original sequence of x[n]. x[n] = 2  [n] + 4  [n-1] + 6  [n-2] + 4  [n-3] + 2  [n-4] Rule : By multiplying a unit impulse we multiple its magnitude. Compute the signal x[n] from its series of unit impulses sequence. n n 7 2  [n] 4  [n-1] 6  [n-2] 4  [n-3] 2  [n-4] x[n] = Thus, we can express ‘any’ digital signal as a series of ‘shifted’ unit impulses which have be scaled by a multiplication factor. … (2.5)

20. The Unit Impulse Response Sequence When the input to an FIR filter is a unit impulse sequence, i.e. x[n] =  [n], the output is by definition a ‘unit impulse response‘ sequence, i.e y[n] = h[n]. Then the ‘general difference equation’ becomes. x[n]  [n] y[n] h[n] Discrete-Time FIR filter … (2.6)

21. Example 9 : Take our 3-point running averager again. y[n] = 1/3{ x[n] + x[n+1] + x[n+2] It’s filter coefficients are : b k = { 1/3, 1/3, 1/3 } for k = 0, 1 & 2. Essentially, all we do is take the filter co-efficients and mutiply them by 1 at the place in the sequence they occur ! 1313 1313 1313

22. Thus, by passing a unit impulse through ‘any’ filter we determine the pure response of the filter for a unit input. Example 10 : An order 10 FIR filter is defined below. Write down the impulse response of the filter. Expand the equation and plot its impulse response.

23. The Unit Delay Another trivial system that has great power is the ‘unit delay’. It ‘shifts’ or ‘delays’ a sytem by an amount (n 0 ). Such that : y[n] = x[n – n 0 ] When n 0 = 1, the system is called a ‘unit delay’. In a plot of y[n] the values of x[n] occur one time interval after they do in the input. … (2.7)

24. Example 11 : A system produces a delay of 3. a) Write down the difference equation of the system. b) Calculate the output y[n] of the system in a difference table. c) Plot the input and ouput of the delay 3 system. d) Write down the impulse response equation of the system e) Plot the impulse response of the system It has filter coefficients arew : b k = { 0, 0, 0, 1 }. NOTE ** : Don’t be fooled that this has only 1 coefficient. It has order M=3 and has length L =M+1 = 4. But coefficients 0,1,2 are weighted to zero. a) The difference equation is : y[n] = 0.x[n] + 0.x[n-1] + 0.x[n-2] + 1.x[n-3] = x[n-3] b) The differnce table is : n n 7 x[n] 0 0 024 6 42 0 0 0 0 y[n] 0 0 0 00 0 2 4 6 4 2 0

25. c) The input and output from the delay 3 system. d) For the impulse response. Just replace y[n] with h[n] and x[n] with  [n]. Thus, y[n] = x[n-3]  h[n] =  [n-3] 1 n=3 0 n  3 e) The impulse response of the delay 3 system. {