Chap2_(introduction To Signals

ECT 2036: 2036: ECT CIRCUITS & & SIGNALS SIGNALS CIRCUITS Chapter 2: Introduction to Signals CONTENTS § Definition and Classification of Signals § Mathematical Model of Ideal Signals § Linear Convolution § Discrete-time Signals § Discrete-time Convolution

1. DEFINITION DEFINITION & & CLASSIFICATION CLASSIFICATION 1. OF SIGNALS SIGNALS OF § SIGNAL: A function of one or more variables and it carries qualitative as well as quantitative information of a physical event. § Single variable function: most common is the time variable, t (continuous-time) or n (discrete-time) -> mathematically denoted by f(t) or f[n] respectively. § Classifications: a) Continuous & Discrete-Time Signals b) Even & Odd Signals c) Deterministic & Stochastic Signals d) Energy & Power Signals

CONTINUOUS & & DISCRETE DISCRETE CONTINUOUS Continuous-time Signal -Signal is defined for all time instants -Occurs naturally in any physical process -We use a round bracket ( ) to denote the function: f( ) -Commonly used: f(t) or f(τ) for single variable

-f(t) = t, f(t) = t2, f(t) = sin(t), etc. -Example: voice signal

CONTINUOUS & & DISCRETE DISCRETE CONTINUOUS Discrete-time Signal -Signal is defined only at discrete values of time variable -We use a square bracket [ ] to denote the function: f[ ] -Commonly used: f[n] or f[k] for single variable

-f[n] = δ[n], f[n] = u[n], f[n] = 2n, etc. -Example: sampled voice signal

§ Some discrete-time signals can be obtained from continuous-time signals by sampling operation mathematical given by: f[n] = f(nT) , n = 0, ± 1, ± 2, ± 3, … where T is the sampling period. :

Continuous-time signal f(t) For instance, let T = 1 Then: n = -1; f[-1] = f(-1) n = 0; f[0] = f(0) n = 1; f[1] = f(1) n = 2; f[2] = f(2) n = 3; f[3] = f(3) :

Sampling at T=1

Discrete-time signal f[n]

EVEN & & ODD ODD SIGNALS SIGNALS EVEN § EVEN signals: signals which are symmetric with respect to the vertical axis. Mathematically described by: f(t) = f(-t) § ODD signals: signals which are anti-symmetric with respect to the origin. Mathematically described by: f(t) = - f(-t)

Even (symmetric) signal f (t ) = f (-t ) for all t

Odd (anti-symmetric) signal f (t ) = - f (-t ) for all t

EVEN & & ODD ODD SIGNALS SIGNALS EVEN • Any arbitrary signal, f(t) can be expressed as a sum of even and odd components:

f (t ) = f e (t ) + f o (t ) where

1 f e (t ) = [ f (t ) + f (−t )] 2 1 f o (t ) = [ f (t ) − f (−t )] 2

Example 1.1: 1.1: Example Even & & Odd Odd Even § Find the odd and even components of continuous2 3 5 f ( t ) = t + 4 t + 6 t time signal, Solution:

(

)

(

)

1 1 2 f e (t ) = [ f (t ) + f (−t )] = t + 4t 3 + 6t 5 + t 2 − 4t 3 − 6t 5 = t 2 2 2 1 1 2 f o (t ) = [ f (t ) − f (−t )] = t + 4t 3 + 6t 5 − t 2 + 4t 3 + 6t 5 = 4t 3 + 6t 5 2 2

Example 1.2: 1.2: Example Even & & Odd Odd Even § Find the odd and even components of the signal f(t). ½[(f(t)+f(-t)] =1.5 ½[(f(t)-f(-t)] 3 3  + t , −2 ≤ t ≤ 2 f (t ) =  2 4  0, elsewhere =(3/4) t

Example 1.3: 1.3: Example Even & & Odd Odd Even § Find the even and odd components of the following signal: f(t) 1 1, 0 < t < 3 -3

0

3

fo(t)

fe(t)

0.5 -3

t

f (t ) =  0, elsewhere

0.5 0

3

t

-3

0

3

t

EVEN & & ODD ODD SIGNALS SIGNALS EVEN § Bear in mind that any arbitrary signals can be expressed as sum of odd and even components. Look at Examples 1.1 to 1.3, the main signal f(t) itself is neither even nor odd but when decomposed, has odd fo(t) and even fe(t) parts. § If the signal f(t) itself is already an even signal, then its even part is exactly the same as itself, f(t)=fe(t), the odd part being zero. § If the signal f(t) itself is already an odd signal, f(t)=fo(t), then its odd part is exactly the same as itself, the even part being zero. § Thus, a signal can be purely even, purely odd or neither even nor odd.

Example 1.4: 1.4: Example Even & & Odd Odd Even § a) Is sin(ωt) an odd or even signal? Let f(t) = sin(ωt) sin(ωt) = - sin(-ωt) for all t f(t) = - f(-t) ∴ It is an odd signal. § b) Is cos(ωt) an odd or even signal? Let f(t) = cos(ωt) cos(ωt) = cos(-ωt) for all t f(t) = f(-t) ∴ It is an even signal.

§ c) Is sin(ωt+(π/6)) an odd or even signal? sin(ωt+(π/6)) = sin(π/6)cos(ωt) + cos(π/6)sin(ωt) = 0.5cos(ωt) + 0.866sin(ωt) ∴ sin(ωt+(π/6)) is neither an even nor an odd signal. 0.5cos(ωt) 0.5

sin(ωt+(π/6)) 1 0.5 -π/6

5π/6 11π/6

0 ωt

=

Even π

ωt

2π

-0.5

+

0.866sin(ωt) 0.866

Odd

-1 Original signal is neither even nor odd

0 -0.866

π

2π

ωt

DETERMINISTIC & & STOCHASTIC STOCHASTIC DETERMINISTIC § DETERMINISTIC SIGNALS: The signals whose characteristics are well known and completely specified for all instants of time. e.g.: sine wave f(t)=sin(t), linear function f(t)=mt+c § STOCHASTIC SIGNALS: The signals whose characteristics are not fully known before its occurrence (random signals). e.g.: throwing of die, noise generated in an electronic circuit, unwanted disturbances in the atmosphere, etc.

sin t

Deterministic

T1 T2

# on a die thrown Stochastic 4 Value at T4? All values at t are 3 Value=Random well defined 2 between 1 to 6 t t T3 T4 T1 T2 T3 T4

§ Deterministic

§ Stochastic (Random Signal)

Totally predictable

ENERGY & & POWER POWER ENERGY Continuous-Time

Discrete-Time

Signal

f(t)

f[n]

Total Energy, E

T

Average Power, P

E = lim

T →∞

∫

2

f (t )dt

−T

1 P = lim T →∞ 2T

T

∫

−T

E = lim

N →∞

N

∑

2

f [n]

n =− N

1 2 f (t )dt P = lim N →∞ 2 N

N

∑

f 2 [ n]

n=− N

§ Energy Signal if: 0 < E < ∞ (finite positive value ) § Power Signal if: 0 < P < ∞ (finite positive value )

Example 1.5: 1.5: Example Energy & & Power Power Energy § A signal f(t) is defined as f (t ) = sin(ω t ) ; − ∞ < t < ∞. Determine whether f(t) is an energy or power signal. T

E = lim

T →∞

∫

−T T

T

f 2 (t )dt = lim

T →∞

2 sin ∫ (ωt )dt

−T

T

1 1  sin(2ωt )  = lim ∫ [1 − cos(2ωt )]dt = lim t −  T →∞ T →∞ 2 2 2 ω  −T −T

1 2sin(2ωT )  = lim  2T − =∞  T →∞ 2 2ω   E=∞ (the value is indefinite). ∴ f(t) is not an energy signal.

1 P = lim T →∞ 2T 1 = lim T →∞ 2T

T

∫

−T

1 f (t )dt = lim T →∞ 2T 2

T

T

2 sin ∫ (ωt )dt

−T

1 ∫−T 2 [1 − cos(2ωt )]dt

2 sin( 2ωT )    2T − 2ω  1 2 sin( 2ωT )  1 = lim  − =  T →∞ 2 8ωT   2 1 = lim T →∞ 4T

P=1/2 (the value is definite and positive). ∴ f(t) is a power signal.

Example 1.6: 1.6: Energy Energy & & Power Power Example § Determine whether f(t) is an energy signal or a power signal or neither.

Energy, E:

0 f (t ) =  αt e

for t < 0 for t ≥ 0

T

T

E = lim

T →∞

∫

−T

f (t )dt = lim ∫ e dt 2

T →∞

T

2α t

0

1 2α t 1 2αT e − 1 = lim e = lim T →∞ 2α T →∞ 2α 0 If α is negative, E = -1/2α à f(t) is an energy signal. If α is positive, E = ∞ à f(t) is not an energy signal.

Power, P:

1 P = lim T →∞ 2T 1 = lim T →∞ 2T

T

∫

−T

T

1 2α t f (t )dt = lim e dt ∫ T →∞ 2T 0 2

T

1  1 2α t  2α T  e − 1  2α e  = Tlim  →∞ 4α T  0

If α is negative, P = 0 à f(t) is not a power signal. If α is positive, P = ∞ à f(t) is not a power signal.

f (t ) = eαt u (t ) t Negative α à Energy signal

t

Positive α à Neither energy signal nor power signal

ENERGY & & POWER POWER ENERGY § Observations: – Bounded periodic signals are power signals (Example 1.5). – Bounded decreasing signals are energy signals (Example 1.6). – Unbounded growing signals are neither energy nor power signals (Example 1.6). – Energy signal: E is finite and nonzero, P is zero. – Power signal: E is infinite, P is finite and nonzero. – Neither energy nor power signal: E and P infinite. Bounded: the amplitude of the signals has definite upper and lower limits. If one limit is indefinite, it is unbounded.

2. MATHEMATICAL MATHEMATICAL MODEL MODEL 2. OF IDEAL IDEAL SIGNALS SIGNALS OF § We will discuss mathematical model of the following signals: a) Sinusoidal Signal b) Exponential Signal c) Unit Step Function d) Unit Pulse Function e) Unit Impulse Function f) Approximation of signals by Impulse Functions

SINUSOIDAL SIGNAL SIGNAL SINUSOIDAL f1(t) f2(t)

 2π t  f ( t ) = A sin  +θ   T  = A sin (2 π ft + θ ) = A sin (ω t + θ ) A = signal amplitude where T = period θ = phase angle

f(t) = 5 sin (ωt) f1(t) = 5 sin (ωt + π/2) f2(t) = 5 sin (ωt - π/2)

EXPONENTIAL SIGNAL SIGNAL EXPONENTIAL A

V initial=A V final=A V final=0

Decaying Exponential Signal

f (t ) = Ae − t /τ Arbitrary Exponential Signal:

V initial=0 Growing Exponential Signal

(

f (t ) = A 1 − e − t /τ

)

f (t ) = V final + (Vinitial − V final )e

− t /τ

Example 2.1: 2.1: Example Exponential Signal Signal Exponential Draw the exponential signals described by: i) 5exp(-t/τ) and ii) 5[1-exp(-t/τ)]

§

5exp(-t/τ)

(i)

5[1-exp(-t/τ)]

Growing

Decaying 5

Approaches 5

5 3.16 Approaches 0

1.84 0

(ii)

t=τ

t

0 t=τ

t

UNIT STEP STEP FUNCTION FUNCTION UNIT

Unit Step

1 t ≥ 0 u (t ) =  0 t < 0

Delayed Unit Step

1 t ≥ T u (t − T ) =  0 t < T

Example 2.2: 2.2: Example Unit Step Step Unit § Sketch the waveform of

f (t ) = u (t ) + u (t − 2)

Example 2.3: 2.3: Unit Unit Step Step Example Draw the exponential signals described by: i) 5exp(-t/τ)u(t) and Note that the functions are ii) 5[1-exp(-t/τ)]u(t) now zero for t < 0. 5exp(-t/τ)u(t) 5[1-exp(-t/τ)]u(t) (i) (ii)

§

5

5 3.16 Approaches 0

1.84 0 §

Approaches 5

t=τ

t

0 t=τ

t

Unit step function can be used in converting an arbitrary function into a right-sided function.

UNIT PULSE PULSE FUNCTION FUNCTION UNIT

Unit Pulse

0 t < −∆ / 2  P∆ (t ) = 1 − ∆ / 2 ≤ t ≤ ∆ / 2 0 t > ∆ / 2 

Delayed Unit Pulse

0 t < (T − ∆ / 2)  P∆ (t − T ) = 1 (T − ∆ / 2) ≤ t ≤ (T + ∆ / 2) 0 t > (T + ∆ / 2) 

UNIT PULSE PULSE FROM FROM UNIT UNIT STEPS STEPS UNIT

Construction of unit pulse from step signals

P∆ (t ) = u (t + ∆ / 2) − u (t − ∆ / 2)

UNIT IMPULSE IMPULSE FUNCTION FUNCTION UNIT

let ∆ approach 0

P∆ (t ) lim = δ (t ) ∆ →0 ∆ Area = ∆×1/∆ =1

Unit Impulse δ (t ) = 0 t ≠ 0 Area =

∞

∫ δ (t )dt = 1

−∞

Properties of Unit Impulse Function Property 1:

f (t )δ (t − T ) = f (T )δ (t − T ) Property 2: ∞

∫

f (t )δ (t )dt = f (0)

−∞

Property 3: ∞

∫ f (t )δ (t − T )dt = f (T )

−∞

Property 4:

δ (−t ) = δ (t ) Property 5:

1 δ ( βt ) = δ (t ) β

Unit Impulse Impulse Function: Function: Properties Properties 11 & & 22 Unit δ(t-T)

f(t)

0

f(t)δ(t-T)

=

X

t

T

f(T)

t

f(t)δ(t-T)=f(T)δ(t-T)

0

t

T

f(t)

=

δ(t-T)=0, t≠T

f(t)δ(t)

0

t

δ(t)

X

f(0) t

δ(t)=0, t≠0 0

t

When T=0 f(t)δ(t)=f(0)δ(t)

Unit Impulse Impulse Function: Function: Unit Properties 22 & & 33 Properties § Property 2 ∞

∫

f (t )δ (t )dt =

−∞

∞

∫

−∞

∞

f (0)δ (t )dt = f (0) ∫ δ (t )dt = f (0) −∞ 1 424 3

§ Property 3 ∞

∫

−∞

f (t )δ (t − T )dt =

Property 1 ∞

∫

−∞

1 ∞

f (T )δ (t − T )dt = f (T ) ∫ δ (t − T )dt −∞ 14243 1

= f (T )

Example 2.4: 2.4: Unit Unit Impulse Impulse Example Evaluate the following integrals: ∞ 5 a) b) ( t + 3) δ (t + 3)dt

∫

c)

−∞ ∞

∫ ( t + 2 ) δ (2t )dt

d)

−∞

−∞

∫ sin(2π t ) δ (t − 50)dt

−∞ ∞

∫ cos(2π t )δ (2t − 1)dt

−∞

Solution: ∞ 5 a) ( t + 3) δ (t + 3)dt =

∫

∞

∞

∫

f (t )δ (t + 3)dt

−∞

= f (−3) = ( −3 + 3) = 0 5

b)

∞

∞

∫ sin(2π t ) δ (t − 50)dt = ∫

−∞

Property 3

f (t ) δ (t − 50)dt

−∞

= f (50) = sin(2π × 50) = 0 Property 3

c)

∞

∫ ( t + 2 ) δ (2t )dt =

−∞

=

∞

1  t + 2 δ ( t ) ∫−∞ ( )  2  dt ∞

∫

Property 5

f (t )δ (t )dt = f (0) = ( 0 + 2 )

−∞

d)

∞

∞

−∞

−∞

1 = 1 Property 2 2

∫ cos(2π t )δ (2t − 1)dt = ∫ cos(2π t )δ [2(t − 1/ 2)]dt ∞

1  = ∫ cos(2π t )  δ (t − 1/ 2)  dt 2  −∞ =

∞

∫

Property 5

f (t )δ (t − 1/ 2)dt

−∞

= f (1/ 2) = cos(2π ×1/ 2)

1 1 =− 2 2

Property 3

APPROXIMATION OF OF SIGNALS SIGNALS BY BY APPROXIMATION IMPULSE FUNCTIONS FUNCTIONS IMPULSE

f (t ) ≅

∞

∑ Tf (kT )δ (t − kT )

k = −∞

This equation is the impulse approximation of the signal f (t). Valid iff the pulse width is very small à sampling frequency is very high (higher than frequency of signal f (t))

Signal Approximation Approximation by by Impulse Impulse Signal f(t)

…

…

-2T –T 0 T 2T f(-T)PT(t+T) f(0)PT(t) f(T)PT(t-T)

f(-2T)PT(t+2T)

…

-2T

+

+

+ -T

f (t ) ≅

∞

∑

k =−∞

0

T

f (kT ) PT (t − kT )

f(2T)PT(t-2T)

…

+

2T

Signal Approximation Approximation by by Impulse Impulse Signal f(t)

f(t) Let the pulse width be very small so the unit pulse becomes unit impulse

-2T –T 0 T 2T ∞ PT (t − kT ) ≅ ∑ δ (t − kT ) ∑ T k =−∞ k =−∞ ∞

T

T→very small

∞

∞

∞

k=−∞

k=−∞

k=−∞

f (t) ≅ ∑ f (kT)PT (t −kT) = ∑ f (kT)T [ PT (t −kT)/T] ≅ ∑Tf (kT)δ(t −kT)

3. LINEAR LINEAR CONVOLUTION CONVOLUTION 3. Input-output model of a linear time invariant continuous-time system

Input

Output Transform operator

3. LINEAR LINEAR CONVOLUTION CONVOLUTION 3. Single Impulse

Impulse response

•If input x(t)=δ(t), then output y(t)=h(t) is called the impulse response. y(t)=G[x(t)]=G[δ(t)]=h(t)

3. LINEAR LINEAR CONVOLUTION CONVOLUTION 3. •Let x(t) be signal approximated using impulses: y (t ) = G[ x (t )]  ∞  ≅ G  ∑ Tx(kT )δ (t − kT )   k =−∞  =

x(t ) ≅

∞

∑ Tx(kT )δ (t − kT )

k =−∞

∞

∑ Tx(kT )G [δ (t − kT )] → Linearity Property

k =−∞

=

∞

∑ Tx(kT )h(t − kT ) → Time invariant Property

k =−∞

Linearity Property: G[kx(t)]=kG[x(t)]; k=constant

Time invariant Property: If G[δ(t)]=h(t);

then G[δ(t-τ)]=h(t-τ); τ=constant

3. LINEAR LINEAR CONVOLUTION CONVOLUTION 3. Approximation by Impulses

Output response

y (t ) =

∞

∑ Tx(kT )h(t − kT )

k =−∞

Let τ=kT, as T à 0, τ à 0, Σ à ∫, T à dτ

y (t ) =

∞

∫ x(τ )h(t − τ )dτ

−∞

y (t ) = x(t ) * h(t ) Convolution Integral

Properties of Linear Convolution x(t ) * h(t ) =

∞

∫ x(τ )h(t − τ )dτ

−∞

h(t ) * x(t ) =

∞

∫ h(τ ) x(t − τ )dτ

−∞

EVALUATION OF OF EVALUATION CONVOLUTION INTEGRAL INTEGRAL CONVOLUTION There are two widely known procedures for evaluation of convolution integral: 1) Graphical Evaluation §

This method is very useful when mathematical models of the signals are not available.

2) Numerical Evaluation Performed by approximation of both the functions by finite-width pulse trains.

GRAPHICAL EVALUATION EVALUATION GRAPHICAL § 1. 2. 3. 4. 5. 6.

Steps involved in the graphical evaluation of two ∞ signals à y(t) = h(t)*x(t) y (t ) = ∫ h(τ ) x(t − τ )dτ Plot h(τ) and x(τ). −∞ Obtain x(-τ). = h(t ) * x(t ) Shift x(-τ) by t to get x(t-τ)= x(-(τ - t)). Multiply h(τ) and x(t-τ). Area under h(τ)x(t-τ) is equal to convolution of two signals, y(t). Repeat steps 3 - 4 for various values of t = t1, t2, t3, …. Plot y(t), the convolution.

1.

same as flip or mirror at the y-axis 2.

shift forward if t is positive, backward if negative 3. multiply both signals, then compute the area

5. Repeat

4.

6.

Example 3.1: 3.1: Example Graphical Linear Linear Convolution Convolution Graphical § Evaluate the convolution of f1(t) and f2(t) at t = 0 and t = 1.

y (t ) = f1 (t ) * f 2 (t ) =

∞

∫

f1 (τ ) f 2 (t − τ )dτ

−∞

At t = 0

At t = 1 f1(τ)

f2(-τ)

f1(τ) τ

×

Area f1(τ). f2(-τ)=0

τ

y (0) = 0

f2(1-τ)

τ

×

Area f1(τ). f2(1-τ)=1/2

τ

y (1) = 1/ 2

NUMERICAL EVALUATION EVALUATION NUMERICAL § Let’s approximate both functions by finite-width impulse trains: n y (t ) = lim T ∑ f1 (kT ) f 2 (t − kT ) T →0

k =0

§ At t = nT, T = sampling period, n = positive integer n

y (nT ) = lim T ∑ f1 (kT ) f 2 (nT − kT ) T →0

k =0

Numerical evaluation form of convolution

Example 3.2: 3.2: Example Numerical Linear Linear Convolution Convolution Numerical § Perform numerical convolution of the following functions with sampling time T = 0.1 for three consecutive sampling instants, n = 0, 1, 2. n

f1 (t ) = t u (t ) 2

y (nT ) = lim T ∑ f1 (kT ) f 2 (nT − kT ) T →0

f 2 (t ) = u (t )

k =0

n

y (0.1n) = 0.1∑ f1 (0.1k ) f 2 (0.1n − 0.1k )

§ At n = 0,

k =0

0

y (0) = 0.1∑ f1 (0.1k ) f 2 (−0.1k ) k =0

= 0.1{ f1 (0) f 2 (0)} = 0 0

1

At n = 1,

1

y (0.1) = 0.1∑ f1 (0.1k ) f 2 (0.1 − 0.1k ) k =0

= 0.1{ f1 (0) f 2 (0.1) + f1 (0.1) f 2 (0)} = 0.1{0 + 0.01} = 0.001

0

1

0.01

1

At n = 2, 2

y (0.2) = 0.1∑ f1 (0.1k ) f 2 (0.2 − 0.1k ) k =0

0

= 0.1{ f1 (0) f 2 (0.2) + f1 (0.1) f 2 (0.1) + f1 (0.2) f 2 (0)}

1 = 0.1{0 + 0.01 + 0.04}

= 0.005

0.01 1

0.04 1

Interpretation of of Numerical Numerical Evaluation Evaluation of of Convolution Convolution Interpretation Graphically using using Example Example 3.2 3.2 Graphically

§ From Example 3.2 f1(τ)= τ 2u(τ)

f2(τ)=u(τ)

τ

T=0.1

τ 1 0.04 0.01 0 0.1 0.2 0.3

0 0.1 0.2 0.3

Interpretation of of Numerical Numerical Evaluation Evaluation of of Convolution Convolution Interpretation Graphically using using Example Example 3.2 3.2 Graphically f1(τ)= τ 2u(τ) 0.04 0.01 0 0.1 0.2 0.3

f2(-τ)=u(-τ) X τ

0 0.1 0.2 0.3

= τ

0

f1(τ)= τ 2u(τ) 0.04 0.01

1

y(0)= f1(τ)f2(-τ) =0

f2(0.1-τ)= u(0.1-τ) X τ

1

=

y(0.1)=f1(τ)f2(0.1-τ) =0.01x0.1

0.1

τ

=0.001

Interpretation of of Numerical Numerical Evaluation Evaluation of of Convolution Convolution Interpretation Graphically using using Example Example 3.2 3.2 Graphically f1(τ)= τ 2u(τ) 0.04 0.01 0 0.1 0.2 0.3

f2(0.2-τ)= u(0.2-τ) X τ

1

= 0.2

τ

y(0.2)=f1(τ)f2(0.2-τ) =0.01x1+0.04x1 =0.005

§ The Numerical Evaluation of Convolution performs by approximation of both the functions by finite-width pulse trains at T=0.1.

4. DISCRETE-TIME DISCRETE-TIME SIGNALS SIGNALS 4. § Some basic sequences – Unit step: u[n]={…0 1 1 1 1 1 1 1 1 1 1…} The arrow indicates the origin, n = 0

1, or u[n] =  0,

n≥0 n<0

4. DISCRETE-TIME DISCRETE-TIME SIGNALS SIGNALS 4. – Unit sample/impulse: δ[n]={…0 1 0…} The arrow indicates the origin, n = 0 or

1, δ [ n] =  0,

n=0 n≠0

4. DISCRETE-TIME DISCRETE-TIME SIGNALS SIGNALS 4. § Both discrete-time functions can also be used without arrows to indicate the origin: – Unit step: u[n]={…0 1 1 1 1 1 1 1 1 1 1…} – Unit sample/impulse: δ[n]={…0 1 0…}

§ In such a case, the first nonzero value at the left hand side of the sequence denotes the position of the origin.

Manipulating Sequences q

Shifted unit step

1, u[n − 1] =  0, q

n ≥1 n <1

Shifted unit sample/impulse 1, n = 1 δ [n − 1] =  0, n ≠ 1 1, n = 2 δ [n − 2] =  0, n ≠ 2

q

Unit sample/impulse using unit step

1, n ≥ 0 u[n] =  0, n < 0

1, n ≥ 1 u[n − 1] =  0, n < 1

δ [n] = u[n] − u[n − 1]

q

Unit step as sum of unit sample/impulse

∞

u[n ] = δ [n ] + δ [n − 1] + δ [n − 2] + δ [n − 3] + ... = ∑ δ [n − i ] q

Sample/impulse of arbitrary amplitude

i =0

x2

x0 x1 x-1

x[n ] = ... + x−1δ [n + 1] + x0δ [n ] + x1δ [n − 1] + x2δ [n − 2] + ... =

∞

∑ x δ [n − i ]

i = −∞

i

Example 4.1: 4.1: Example Discrete Sequences Sequences Discrete § Express the following signals graphically: (i) x[n] = u[n] − u[n − 5] (ii) x[n]={…0 5 0…} (iii) x[n]={…0 0 5 0…} (iv) x[n] = δ [n] + 2δ [n − 2] − δ [n − 3]

§ Solution: (i) x[n] = u[n] − u[n − 5]

-

(ii) x[n]={…0 5 0…}

(iii) x[n]={…0 0 5 0…}

=

(iv) x[n] = δ [n] + 2δ [n − 2] − δ [n − 3]

+ =

+

5. DISCRETE-TIME DISCRETE-TIME 5. CONVOLUTION CONVOLUTION § For a LTI discrete-time system, the discrete-time convolution is given mathematically by:

y [ n] =

∞

∑ x [k ]h [n − k ]

k =−∞

§ If the LTI system is causal, then: n

y [ n] = ∑ x [ k ] h [ n − k ] k =0

§ By causal system, we mean that the input and output signals of the system start from t≥0.

Example 5.1: 5.1: Example Discrete-Time Convolution Convolution Discrete-Time

§ Perform convolution mathematically for the following sequences at n = 3. f1 [ n ] = nu [ n ] n

§ For n = 3 y [3 ] =

0 (1/3)3

1 f2 [ n] =   u [ n] 3

n

y [ n ] = ∑ f1 [k ] f 2 [ n − k ] k =0

Causal signals 1 (1/3)2

2 (1/3)1

3

∑ f [ k ] f [3 − k ] k =0

1

2

= f 1 [ 0 ] f 2 [3 ] + f 1 [1] f 2 [ 2 ] + f 1 [ 2 ] f 2 [1] + f 1 [3 ] f 2 [ 0 ] 2

1 1 = 0 +   + 2 + 3 3 3 = 3.777

3

(1/3)0

5. DISCRETE-TIME DISCRETE-TIME 5. CONVOLUTION CONVOLUTION § All the properties for continuous-time convolution still hold for discrete-time convolution. § Graphical method can be performed on discretetime signals to obtain discrete-time convolution. Same procedures as in continuous-time case.

GRAPHICAL EVALUATION EVALUATION GRAPHICAL § 1. 2. 3. 4. 5. 6.

Steps involved in the graphical evaluation of two ∞ signals à y[n] = h[n]*x[n] y [n] = ∑ h [k ] x [n − k ] Plot h[k] and x[k]. k =−∞ = h[n]* x[ n] Obtain x[-k]. Shift x[-k] by n to get x[n-k]= x[-(k - n)]. Multiply h[k] and x[n-k]. Area under h[k]x[n-k] is equal to convolution of two signals, y[n]. Repeat steps 3 - 4 for various values of n = n1, n2, n3, …. Plot y[n], the convolution.

Example 5.2: 5.2: Example Discrete-Time Convolution Convolution Discrete-Time § Repeat Example 5.1, now using graphical convolution.

§ At n= 3

x

=

Example 5.3: 5.3: Example Discrete-Time Convolution Convolution Discrete-Time § Find mathematically, the discrete time convolution of the two signals below. f[k]

n

y [ n] = ∑ f [ k ] h [ n − k ] k =0

Causal signals

h[k]

y[0] = 6; y[1] = 13; y[2] = 10; y[3] = 10; y[4] = −11; y[5] = 2 n ≥ 6 ⇒ y[n] = 0

Example 5.4: 5.4: Discrete-Time Discrete-Time Convolution Convolution Example § Repeat Example 5.3, now using graphical convolution f[k]

h[-k] 5

f[k]

f[k]

h[1-k] 5

h[2-k]

3

3 k

-2 n=0: y[0]=2x3=6

5

-2

3 k

n=1: c[1]=2x5+1x3=13

-2

k

n=2: y[2]=2x(-2)+1x5+3x3=10

f[k]

f[k]

f[k]

h[3-k] 5

h[4-k]

h[5-k] 5

5 3 -2

3

3 k

-2

k

n=3: n=4: y[3]=1x(-2)+3x5+(-1)x3=10 y[4]=3x(-2)+(-1)x5=-11

-2

k

n=5: y[5]=(-1)x(-2)=2

Example 5.5: 5.5: Example Discrete-Time Convolution Convolution Discrete-Time § For the two discrete signals defined below:

x[n] = u[n] − u[n − 6] h[ n ] = u [ n − 2 ] − u [ n − 5 ]

(i) Sketch the two sequences. (ii) Compute the convolution graphically.

(i)

1

x[n] = u[n] − u[n − 6] -3

-2

-1

0

1

2

3

4

5

6

3

4

5

6

1

h[n] = u[n − 2] − u[n − 5] -3

(ii)

-2

-1

0

1

2

3

y[n] = x[n]*h[n]

2 1

0

1

2

3

4

5

6

7

8

9

10

11

CONVOLUTION TABLE TABLE CONVOLUTION Index 0 + 0 = 0 Start of convolution sequence, y[0] Σ y[0] Σ y[1] y[2]

Σ

y[3]

Σ y[4]

Σ

CONVOLUTION TABLE TABLE CONVOLUTION § Note that the top entry and left entry does not necessary start with h[0] or f[0]. § The rule is that, we start the entries using the first nonzero value of h[n] or f[n]. § For instance: if we start with h[1] and f[0], then y[1+0] = y[1] is the first nonzero value of the convolution. If h[1] and f[1], then it is y[1+1] = y[2].

Example 5.6: 5.6: Example Convolution Table Table Convolution § Repeat Example 5.1 using convolution table. Index 0 + 1 = 1 Start of convolution sequence, y[1] For n = 3, compute y[3]

y[3]

Σ(3, 2/3, 1/9) = 34/9 = 3.777

Example 5.7: 5.7: Example Convolution Table Table Convolution § Repeat Example 5.3 using convolution table. ΣIndex(0,0) = 0 Start of convolution sequence, y[0]

f[0]=2 f[1]=1 f[2]=3 f[3]=-1

h[0]=3 f[0]h[0]=6 f[1]h[0]=3 f[2]h[0]=9 f[3]h[0]=-3

y[0]=6 y[1]=3+10=13 y[2]=9+5+(-4)=10

h[1]=5 f[0]h[1]=10 f[1]h[1]=5 f[2]h[1]=15 f[3]h[1]=-5

h[2]=-2 f[0]h[2]=-4 f[1]h[2]=-2 f[2]h[2]=-6 f[3]h[2]=2

y[3]=-3+15+(-2)=10 y[4]=-5+(-6)=-11 y[5]=2

Summary Summary § At the end of this chapter, you should understand: – Concept of signals and its common classifications • Able to define, draw, compare, compute and give examples

– Mathematical model of ideal signals (continuous) and discrete-time signals • Able to formulate signal functions as well as graphically drawn the signals, perform signal operations mathematically and graphically, able to differentiate continuous- and discrete-time signals

– Linear convolution (continuous) and discrete-time convolution • Able to perform convolution operations mathematically and graphically, able to differentiate continuous- and discrete-time convolutions

§ Next Chapter: Application of Fourier Series