MM3FC Mathematical Modeling 3LECTURE 1

Times

Weeks 7,8 & 9.Lectures : Mon,Tues,Wed 10-11am,

Rm.1439Tutorials : Thurs, 10am, Rm. ULT.

Clinics : Fri, 8am, Rm.4.503Dr. Charles Unsworth,Department of Engineering Science, Rm. 4.611

Tel : 373-7599 ext. 2461Email : c.unsworth@auckland.ac.nz

The Big PlanAim of course

Understand the ‘Discrete Fourier Transform’ (DFT) .

We are going to do this through simple filter theory

1) Learn how to represent signals mathematically.2) Learn about FIR filters, impulse & frequency

response.3) The z-transform, how it helps in the design of filters.4) Learn how to design lowpass, highpass, bandpass

and nulling filters.5) Show how the DFT can be represented as a bank of

filters.

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)

Sinusoids

The most general mathematical formula for a sinusoid

x(t) = A cos(2π f t + φ) 1.1

• x(t) = amplitude at time (t)• A = maximum amplitude of sinusoid,

• f = frequency = 1/T in Hz, where T = period in secs.

• φ = phase shift in rads.• ω = angular frequency = 2πf in rads/sec

Example 1 : Determine A, w, f, φ & T of the following signals :

A) x(t) = 12cos(12t +

B) y(t) = 16sin(200t –

• 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

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.

• 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.

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)

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

NOTE **

This will give us 40 points per period.

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 statesIn 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)

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 =

1-

Re(z)

Im(z)

y

x

z

0

Re(z)

Im(z)

y

x Re(z)

Im(z)

y

x

Cartesian Form

z = x + jy

z r

θ

Polar formx = r cosθ y = r sinθ

• The polar form is very clumsy, better to make use of Euler’s famous formula.

(y/x)tan=θ

y+x=r1-

22

Euler’s formula :

ejcos + jsin1.4

Now, from cartesian coords :

z = x + jy

From polar coordinates : x = rcosθ & y = rsinθ

z = rcos+ rsin

Complex exponential representation of a sinusoid

z = rej1.5

• In this form all the rules of indices hold :

1je

ee

e

1 ee

eee ,

,

• Now s’ppose we have : z1 = r1ejα & z2 = r2ejβ

z3 = z1.z2 = r1ejα. r2ejβ

= r1r2 ejα ej

β = r1r2 ej(α

+

β)

Multiplying 2 complex numbers or exponentials wemultiply magnitude & addthe phases

… 1.6

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)

x(t) = Re { Ae j(2πft + φ) } = Re { Ae j(2πft )ejφ } = Ae j(2πft )ejφ

A sinusoid is the product

of 2 exponentials

Sometimes we just drop the Re{ }, for convienience.

Example 4 : Write in complex exponent form :

A) x(t) = 3cos(50πt + π/6)

B) y(t) = 2sin(2πt – π/3)

Hence, determine :

C) x(t)y(t)

D) x(t)/y(t)

Using Inverse Euler Formulas

jφ-jwt-jφjwt

φ)+-j(wtφ)+j(wt

ee2

A+ee

2

A=x(t)∴

2

e+eA = φ)+Acos(wt

2j

e-e=sinθ ,

2

e+e=cosθ

-jθjθ-jθjθ

The equation for a sinusoid x(t) = A cos(wt + φ) becomes :

… 1.8

… 1.9

• We now have two representations of a sinusoid.

Im(z)

Re(z)

Im(z)

Re(z)

Im(z)

A

φ

+wt+wt

-wt

A/2

A/2

+

1) As a rotating phasor with amplitude (A) and frequency (w).

2) Two counter rotating phasors of amplitude (A/2) and frequencies (w) & (-w)

OR φ-φ

Example 6 : Express as an additive linear combination of exponentials :A) x(t) = 3cos(3πt + π/12)

B) y(t) = 5sin(24 πt – π/7)

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.

• 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

}eeA Re{+A=

)+t fcos(2A+A=x(t)

kk f2jφN

1=kk0

N

1=kkk0

∑

∑

t

k

π

φπ

...} ),,-fe2

A(),f,e

2

A(),,-fe

2

A(),f,e

2

A(,0),{(A

}ee 2

A Re{+}ee

2

A Re{+A=x(t)

2jφ-2

2jφ2

1jφ-1

1jφ1

0

fj2-jφ-N

1=k

kfj2jφN

1=k

k0

2211

kkkk ∑∑ tt ππ

… 1.10

… 1.11

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.

)},-fe2

A(),f,e

2

A(),,-fe

2

A(),f,e

2

A(,0),{(A

}ee 2

A Re{+}ee

2

A Re{+A=x(t)

2jφ-2

2jφ2

1jφ-1

1jφ1

0

fj2-jφ-2

1=k

kfj2jφ2

1=k

k0

2211

kkkk ∑∑ tt ππ

… 1.12

kjφ-kk e

2

A= |X| .. 1.13

Example 7 : Plot the Magnitude Spectrum for the continuous signal. x(t) = 10 + 14cos(200πt – π/3) + 8cos(500πt + π/2)

Apply inverse Euler Formulax(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|

0 100 250-100-250 f(Hz)

|10|

|7ej π/3| |7e-j π/3|

|4ej π/2||4e-j π/2|

Two-sidedMagnitudeSpectrum

MA

GN

ITU

DE

|Xk|