MM3FC Mathematical Modeling 3 LECTURE 1

1. MM3FC Mathematical Modeling 3LECTURE 1 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. The Big Plan Aim of course Understand the ‘Discrete Fourier Transform’ (DFT) . We are going to do this through simple filter theory • Learn how to represent signals mathematically. • Learn about FIR filters, impulse & frequency response. • The z-transform, how it helps in the design of filters. • Learn how to design lowpass, highpass, bandpass and nulling filters. • Show how the DFT can be represented as a bank of filters.

3. This LectureWhat are we going to cover & Why ? • Sinusoidal representation of continuous signals. (because sinusoids make up all signals) • Complex exponentials & phasors. (to simplify the addition & multiplication of sinusoids) • Spectrum Representation. (to represent frequency content of a mixed signal)

4. Sinusoids The most general mathematical formula for a sinusoid x(t) = A cos(2π ft + φ) 1.1 • x(t) = amplitude at time (t) • A = maximum amplitude of sinusoid, • f= frequency = 1/Tin Hz, where T = period in secs. • φ = phase shift in rads. • ω = angular frequency = 2πf in rads/sec

5. Example 1 : Determine A, w, f, φ & T of the following signals : • x(t) = 12cos(12pt + p/3) • y(t) = 16sin(200pt – p/9)

6. Can only do two things to the sinusoid of a set frequency : • x(t) = A cos(wt + φ) • 1) We can magnify it in amplitude, A. • 2) We can shift it in phase, φ. • SHIFTING SIGN CONVENTION • If we want to shift the sinusoid to the : • Right (in the positive direction) then φ is -ve • Left (in the negative direction) then φ is +ve • IMPORTANT SINE/COSINE RELATIONSHIP • sin(wt + φ) = cos(wt + (φ - π/2f)) 1.2 • cos(wt + φ) = sin(wt + (φ + π/2f)) 1.3

7. 3) In Signal Analysis, we always change sines to cosines using these relations. • (This is to make the math easy when we deal with complex numbers, later.) • Example 2 : Change sin(wt + π/3) to cosine form. • Plot both the sine and cosine result and verify using the shifting properties. • We see sin(wt + π/3) is sine shifted π/3 in the –ve direction. • This moves the maxima of sin(wt) at π/2 to the new position (π/2 - π/3) = π/6 from the origin.

8. From the sine/cosine relations • sin(wt + π/3) = cos(wt + (π/3 - π/2)) • = cos(wt - π/6) • Thus, cos(wt - π/6) is cos shifted π/6 in the +ve direction. • This moves the maxima cos(wt) at 0 to the new position π/6 from the origin. • Hence. Both the sine and cosine maximas lie on top of each other.

9. Example 3 : What is the Amplitude, frequency and phase of the equivalent sine/cosine. A) x(t) = 12cos(200πt + π/3) B) y(t) = 16sin(160πt – π/9)

10. NOTE ** This will give us 40 points per period. Here is the MATLAB code used to plot : x(t) = 10 cos(2π 1000 t + π/2 ). • A = 10; f = 1000; phi = pi/2; • T = 1/f; • t = -2*T : T/40 : 2*T; • x = A*cos(2*pi*f*t + phi); • plot(t,x) • title('Sinusoid: x(t) = 10 cos(2*pi*1000*t + pi/2)'); • xlabel('Time (sec)'); • grid on

11. Continuous & Discrete Signals • Real world signals are “continuous ” in time. • Any recorded signal is said to be “discrete ” in time. • It is impossible to collect every time sample of a real world signal. • The “trick” is to “sample” the data to collect enough points such that the signal is accurately represented. Shannon’s sampling theorem states In order to recover a signal of frequency (f) We have to sample at a minimum frequency of (2f). (Namely, we must have a minimum of 2 points/period)

12. Im(z) y x Re(z) Complex Exponentials & Phasors • Believe it or not !! … Analysis & manipulation of sinusoids is “ greatly simplified ” by using complex exponentials. • First let’s review complex numbers. z = x + j y , (z) is a the complex number. (x) is the real part of (y) is the imaginary part of z (j) is an imaginary number = z 0

13. Im(z) Im(z) y y x x Re(z) Re(z) r z θ Cartesian Form z = x + jy Polar form x = r cosθ y = r sinθ • The polar form is very clumsy, better to make use of • Euler’s famous formula.

14. Euler’s formula : ejq = cosq + jsinq 1.4 Now, from cartesian coords : z = x + jy From polar coordinates : x = rcosθ & y = rsinθ z = rcosq + rsinq Complex exponential representation of a sinusoid z = rejq 1.5

15. a b a + b = e e e , 1 - b = e b e , a e a - b = e b e 2p = j e 1 • In this form all the rules of indices hold : … 1.6 • Now s’ppose we have : z1 = r1ejα& z2 = r2ejβ Multiplying 2 complex numbers or exponentials we multiply magnitude & add the phases z3 = z1.z2 = r1ejα. r2ejβ = r1r2 ejαejβ = r1r2 ej(α+β)

16. How do we express a sinusoid in exponential form ? The most general mathematical formula for a sinusoid x(t) = A cos(2πft + φ) Now consider the expansion of : A e j(2πft + φ) = Acos(2πft + φ) + jAsin(2πft + φ) A sinusoid is the product of 2 exponentials x(t) = Re { Ae j(2πft + φ)} = Re { Ae j(2πft )ejφ } = Ae j(2πft )ejφ Sometimes we just drop the Re{ }, for convienience.

17. Example 4 : Write in complex exponent form : • x(t) = 3cos(50πt + π/6) • y(t) = 2sin(2πt – π/3) • Hence, determine : • C) x(t)y(t) • D) x(t)/y(t)

18. Using Inverse Euler Formulas … 1.8 The equation for a sinusoid x(t) = A cos(wt + φ) becomes : … 1.9

19. Im(z) Im(z) Re(z) • We now have two representations of a sinusoid. +wt +wt -wt Im(z) A A/2 φ OR φ + -φ Re(z) A/2 • As a rotating phasor • with amplitude (A) • and frequency (w). 2) Two counter rotating phasors of amplitude (A/2) and frequencies (w) & (-w)

20. Example 6 : Express as an additive linear combination of exponentials : • x(t) = 3cos(3πt + π/12) • y(t) = 5sin(24 πt – π/7)

21. The Spectrum • The way we are use to observing a signal is by viewing its time-course. • This is known as the “time domain representation” of the signal. • The spectrum is a graphical representation of the frequency content of the sum of sinusoids in a signal. • Known as the “frequency domain representation” of the signal. • This visual form allows us to see the relationships between the different frequency components and their relative amplitudes quickly and easy.

22. A spectrum can be produced from “additive linear combination” of a constant and (N) sinusoids with different frequency, amplitude and phase. Using Inverse Euler formulae … 1.10 … 1.11

23. NOTE** • Observe what happens for k=1,2. • For each (fk) the we have ‘complex conjugate pair’ that represents the sinusoidal component contributing at frequency (fk) . • So we could write this as : • {(X0,0), (X1,f1), (X1*,-f1), (X2,f2), (X2*,-f2), …} • (2N+1) frequency components with Magnitude’s • (X0) is a DC component that can be expressed as complex exponential signal with a frequency of zero. • The ‘Magnitude Spectrum’ of the signal x(t) is a plot of |Xk| vs. f. … 1.12 .. 1.13

24. Two-sided Magnitude Spectrum |10| |7ej π/3| |7e-j π/3| MAGNITUDE |Xk| |4e-j π/2| |4ej π/2| 0 250 f(Hz) -250 -100 100 Example 7 : Plot the Magnitude Spectrum for the continuous signal. x(t) = 10 + 14cos(200πt – π/3) + 8cos(500πt + π/2) Apply inverse Euler Formula x(t) = 10 + 7e-j π/3 e j2π(100)t + 7ej π/3 e -j2π(100)t + 4ej π/2 e j2π(250)t + 4e-j π/2 e -j2π(250)t Thus, {(10,0),(7e-j π/3, 100),(7ej π/3, -100),(4ej π/2, 250),(4e-j π/2, -250)} • Draw each frequency as a vertical line of length of magnitude |Xk|