chap2_(introduction to signals

CONTENTS§ Definition and Classification of Signals§ Mathematical Model of Ideal Signals§ Linear Convolution§ Discrete-time Signals§ Discrete-time Convolution

ECT 2036: CIRCUITS & SIGNALS

Chapter 2: Introduction to Signals

1. DEFINITION & CLASSIFICATION OF SIGNALS

§ 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 Signalsb) Even & Odd Signalsc) Deterministic & Stochastic Signalsd) Energy & Power Signals

CONTINUOUS & DISCRETECONTINUOUS & DISCRETE

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 & DISCRETECONTINUOUS & DISCRETE

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

Continuous-time signal f(t) Sampling at T=1

Discrete-time signal f[n]

§ 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. :

For instance, let T = 1Then:

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)

EVEN & ODD SIGNALSEVEN & ODD SIGNALS§ 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) signalf (t ) = f (-t ) for all t

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

EVEN & ODD SIGNALSEVEN & ODD SIGNALS

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

[ ])()(21

)()(21

)()()(

tftftf

tftftfwhere

tftftf

−−=

Example 1.1: Even & Odd

§ Find the odd and even components of continuous-time signal,

Solution:

532 64)( ttttf ++=

[ ] ( )

[ ] ( ) 53532532

2532532

64646421

)()(21

646421

)()(21

tttttttttftftf

ttttttttftftf

+=++−++=−−=

=−−+++=−+=

§ Find the odd and even components of the signal f(t).

½[(f(t)+f(-t)]

½[(f(t)-f(-t)]

3 3, 2 2

( ) 2 40,

t tf t

elsewhere

+ − ≤ ≤=

=(3/4) t

§ Find the even and odd components of the following signal:

-3 0 3t

fo(t)0.5

-3 0 3t

1, 0 3( )

f telsewhere

EVEN & ODD SIGNALSEVEN & ODD SIGNALS

§ 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.

§ a) Is sin(ωt) an odd or even signal?Let f(t) = sin(ωt)

sin(ωt) = - sin(-ωt) for all tf(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.

sin(ωt+(π/6))

-π/6 5π/6 11π/6

0.5cos(ωt)

0.866sin(ωt)

-0.866

Original signal is neither even nor odd

DETERMINISTIC & STOCHASTICDETERMINISTIC & STOCHASTIC§ 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 itsoccurrence (random signals).e.g.: throwing of die, noise generated in an electronic circuit, unwanted disturbances in the atmosphere, etc.

Deterministic

# on a die thrown

T1 T2 T3

Value at T4? Value=Randombetween 1 to 6

Stochastic

T4T1 T2 T3 T4

All values at t are well defined

§ Deterministic

§ Stochastic (Random Signal)

Totally predictable

ENERGY & POWERENERGY & POWER

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

Average Power, P

Total Energy, E

f[n]f(t)Signal

Discrete-TimeContinuous-Time

2lim ( )T

E f t dt→∞

= ∫2lim [ ]

E f n→∞

21lim ( )

P f t dtT→∞

= ∫ ∑−=

∞→=

P ][21

Example 1.5: Energy & PowerExample 1.5:

Energy & Power§ A signal f(t) is defined as .

Determine whether f(t) is an energy or power signal.

E=∞ (the value is indefinite). ∴ f(t) is not an energy signal.

( ) sin( ) ; f t t tω= −∞< < ∞

2 2lim ( ) lim sin ( )T T

T TT T

E f t dt t dtω→∞ →∞

− −

= =∫ ∫1 1 sin(2 )

lim [1 cos(2 )] lim2 2 2

tt dt t

ω→∞ →∞−−

= − = − ∫1 2sin(2 )

lim 22 2T

ωω→∞

= − = ∞

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

2 21 1lim ( ) lim sin ( )

T TT T

P f t dt t dtT T

ω→∞ →∞

− −

= =∫ ∫

8)2sin(2

2)2sin(2

)]2cos(1[21

∞→

−∞→ ∫

Example 1.6: Energy & PowerExample 1.6: Energy & Power§ Determine whether f(t) is an energy signal or a power

signal or neither.

)(tforetfor

tf tα

Energy, E:

If α is negative, E = -1/2αà f(t) is an energy signal.If α is positive, E = ∞ à f(t) is not an energy signal.

lim ( ) limT T

E f t dt e dtα

→∞ →∞−

= =∫ ∫2 2

1 1lim lim 1

T Te eα α

α α→∞ →∞ = = −

Power, P:

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

1 1lim ( ) lim

P f t dt e dtT T

→∞ →∞−

= =∫ ∫2 2

1 1 1lim lim 1

T Te e

T Tα α

α α→∞ →∞

= = −

Negative αà Energy signal

tPositive αà Neither energy signal nor power signal

)()( tuetf tα=

ENERGY & POWERENERGY & POWER

§ 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 MODEL OF IDEAL SIGNALS

§ We will discuss mathematical model of the following signals:a) Sinusoidal Signalb) Exponential Signalc) Unit Step Functiond) Unit Pulse Functione) Unit Impulse Functionf) Approximation of signals by Impulse Functions

SINUSOIDAL SIGNALSINUSOIDAL SIGNAL

( )( )θω

sin2sin

2sin)(

angle phaseperiod

amplitude signal

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

EXPONENTIAL SIGNALEXPONENTIAL SIGNAL

Decaying Exponential Signal Growing Exponential Signal

τ/)( tAetf −= ( )τ/1)( teAtf −−=

Arbitrary Exponential Signal:

τ/)()( tfinalinitialfinal eVVVtf −−+=

A V initial=A

V initial=0V final=0

V final=A

Example 2.1:Exponential Signal

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

5exp(-t/τ)

Approaches 0

5[1-exp(-t/τ)](i)

Decaying

Growing

Approaches 5

UNIT STEP FUNCTIONUNIT STEP FUNCTION

Unit Step Delayed Unit Step

tu 1( )

u t Tt T≥

− = <

Example 2.2: Unit Step

§ Sketch the waveform of

)2()()( −+= tututf

§ Draw the exponential signals described by:i) 5exp(-t/τ)u(t) and ii) 5[1-exp(-t/τ)]u(t)

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

Example 2.3: Unit StepExample 2.3: Unit Step

5exp(-t/τ)u(t)

Approaches 0

5[1-exp(-t/τ)]u(t)(i) (ii)

Approaches 5

Note that the functions are now zero for t < 0.

UNIT PULSE FUNCTIONUNIT PULSE FUNCTION

Unit Pulse Delayed Unit Pulse

∆>∆≤≤∆−

∆−<=∆

2/02/2/1

∆+>∆+≤≤∆−

∆−<=−∆

)2/(0)2/()2/(1

)2/(0)(

UNIT PULSE FROM UNIT STEPSUNIT PULSE FROM UNIT STEPS

Construction of unit pulse from step signals

)2/()2/()( ∆−−∆+=∆ tututP

UNIT IMPULSE FUNCTIONUNIT IMPULSE FUNCTION

Unit Impulse

( ) 0 0

Area t dt

−∞

= =∫

δ=∆

→∆

Area = ∆×1/∆ =1

let ∆approach 0

Property 1: Property 4:

Property 2: Property 5:

Property 3:

( ) ( ) (0)f t t dt fδ∞

−∞

)()()( TfdtTttf =−∫∞

∞−

)()( tt δδ =−

)( tt δβ

βδ =

)()()()( TtTfTttf −=− δδ

Properties of Unit Impulse Function

Unit Impulse Function: Properties 1 & 2Unit Impulse Function: Properties 1 & 2

f(t) δ(t-T)

δ(t-T)=0, t≠T

f(t)δ(t-T)f(T)

f(t)δ(t-T)=f(T)δ(t-T)

f(t) δ(t)

δ(t)=0, t≠0

f(t)δ(t)

f(t)δ(t)=f(0)δ(t)When T=0

§ Property 2

§ Property 3

Unit Impulse Function:Properties 2 & 3

( ) ( ) (0) ( ) (0) ( ) (0)f t t dt f t dt f t dt fδ δ δ∞ ∞ ∞

−∞ −∞ −∞

= = =∫ ∫ ∫14243

( ) ( ) ( ) ( ) ( ) ( )

f t t T dt f T t T dt f T t T dt

δ δ δ∞ ∞ ∞

−∞ −∞ −∞

− = − = −

∫ ∫ ∫14243

Property 1

Example 2.4: Unit ImpulseExample 2.4: Unit ImpulseEvaluate the following integrals:a) b)

c) d) cos(2 ) (2 1)t t dtπ δ∞

−∞

−∫

sin(2 ) ( 50)t t dtπ δ∞

−∞

−∫( )2 (2 )t t dtδ

−∞

( )53 ( 3)t t dtδ

−∞

+ +∫

Solution:a)

3 ( 3) ( ) ( 3)

( 3) 3 3 0

t t dt f t t dt

δ δ∞ ∞

−∞ −∞

+ + = +

= − = − + =

∫ ∫

sin(2 ) ( 50) ( ) ( 50)

(50) sin(2 50) 0

t t dt f t t dt

π δ δ

∞ ∞

−∞ −∞

− = −

= = × =

∫ ∫

Property 3

( ) ( )

12 (2 ) 2 ( )

1 ( ) ( ) (0) 0 2 1

t t dt t t dt

f t t dt f

∞ ∞

−∞ −∞

−∞

= = = + =

∫ ∫

Property 5

cos(2 ) (2 1) cos(2 ) [2( 1/ 2)]

1 cos(2 ) ( 1/ 2)

( ) ( 1/ 2)

t t dt t t dt

t t dt

f t t dt

π δ π δ

∞ ∞

−∞ −∞

−∞

− = −

∫ ∫

∫1 1

(1/ 2) cos(2 1/ 2)2 2

f π= = × = −

Property 2

Property 5

Property 3

APPROXIMATION OF SIGNALS BY IMPULSE FUNCTIONS

∑∞

−∞=

−≅k

kTtkTTftf )()()( δ

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 by ImpulseSignal Approximation by Impulse

f(-T)PT(t+T)

f(0)PT(t)

f(T)PT(t-T)

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

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

+ + + +… …

( ) ( ) ( )Tk

f t f kT P t kT∞

=−∞

≅ −∑

-2T –T 0 T 2T

… …

Signal Approximation by ImpulseSignal Approximation by Impulse

[ ]( ) ( ) ( ) ( ) ( )/ ( ) ( )T Tk k k

f t f kT P t kT f kT T P t kT T Tf kT t kTδ∞ ∞ ∞

=−∞ =−∞ =−∞

≅ − = − ≅ −∑ ∑ ∑

-2T –T 0 T 2T

T→very small

( )( )T

P t kTt kT

∞ ∞

=−∞ =−∞

−≅ −∑ ∑

Let the pulse width be very small so the unit pulse becomes unit impulse

3. LINEAR CONVOLUTION3. LINEAR CONVOLUTION

Input-output model of a linear time invariant continuous-time system

Transform operatorInput Output

•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)

Single Impulse

Impulse response

( ) [ ( )]

( ) ( )

( ) ( ) Linearity Property

( ) ( ) Time invariant Property

y t G x t

G Tx kT t kT

Tx kT G t kT

Tx kT h t kT

=−∞

≅ −

= − →

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

•Let x(t) be signal approximated using impulses:

( ) ( ) ( )k

x t Tx kT t kTδ∞

=−∞

≅ −∑

( ) ( ) ( )k

y t Tx kT h t kT∞

=−∞

= −∑

Let τ=kT, as Tà 0, τà 0, Σà ∫, Tà dτ

( ) ( ) ( )

( ) ( )* ( )

y t x h t d

y t x t h t

τ τ τ∞

−∞

Convolution Integral

Approximation by Impulses

Output response

Properties of Linear Convolution

( )* ( ) ( ) ( )

x t h t x h t d

h t x t h x t d

τ τ τ

−∞

EVALUATION OF CONVOLUTION INTEGRAL

§ There are two widely known procedures for evaluation of convolution integral:

1) Graphical EvaluationThis method is very useful when mathematical models of the signals are not available.

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

GRAPHICAL EVALUATIONGRAPHICAL EVALUATION

§ Steps involved in the graphical evaluation of two signals à y(t) = h(t)*x(t)

1. Plot h(τ) and x(τ).2. Obtain x(-τ).3. Shift x(-τ) by t to get x(t-τ)= x(-(τ - t)).4. Multiply h(τ) and x(t-τ). Area under h(τ)x(t-τ) is

equal to convolution of two signals, y(t).5. Repeat steps 3 - 4 for various values of t = t1, t2,

t3, ….6. Plot y(t), the convolution.

( ) ( ) ( )

( )* ( )

y t h x t d

h t x t

τ τ τ∞

−∞

same as flip or mirror at the y-axis

shift forward if t is positive, backward if negative

multiply both signals, then compute the area

5. Repeat

§ Evaluate the convolution of f1(t) and f2(t) at t = 0 and t = 1.

Example 3.1: Graphical Linear Convolution

At t = 0 At t = 1

( ) ( )* ( )

( ) ( )

y t f t f t

f f t dτ τ τ∞

−∞

= −∫

(0) 0y =Area f1(τ). f2(-τ)=0 (1) 1/ 2y =Area f1(τ). f2(1-τ)=1/2

f1(τ) f2(-τ) f1(τ) f2(1-τ)

NUMERICAL EVALUATIONNUMERICAL EVALUATION

§ Let’s approximate both functions by finite-width impulse trains:

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

( ) lim ( ) ( )n

y t T f kT f t kT→

= −∑

( ) lim ( ) ( )n

y nT T f kT f nT kT→

= −∑

Numerical evaluation form of convolution

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

§ At n = 0,

)()()()(

tutftuttf

(0) 0.1 (0.1 ) ( 0.1 )

0.1{ (0) (0)} 0k

y f k f k

Example 3.2: Numerical Linear Convolution

( ) lim ( ) ( )

(0.1 ) 0.1 (0.1 ) (0.1 0.1 )

y nT T f kT f nT kT

y n f k f n k

At n = 1,

At n = 2,

{ }{ }

1 2 1 2

(0.1) 0.1 (0.1 ) (0.1 0.1 )

0.1 (0) (0.1) (0.1) (0)

0.1 0 0.01

y f k f k

f f f f=

{ }{ }

1 2 1 2 1 2

(0.2) 0.1 (0.1 ) (0.2 0.1 )

0.1 (0) (0.2) (0.1) (0.1) (0.2) (0)

0.1 0 0.01 0.04

y f k f k

f f f f f f=

0 1 0.01 1

0 10.01 1 0.04 1

Interpretation of Numerical Evaluation of Convolution Graphically using Example 3.2

§ From Example 3.2f1(τ)= τ 2u(τ)

f2(τ)=u(τ)

τT=0.1

0 0.1 0.2 0.3

0.010.04

0 0.1 0.2 0.3

f1(τ)= τ 2u(τ)

y(0)= f1(τ)f2(-τ)

f2(-τ)=u(-τ)

f2(0.1-τ)= u(0.1-τ)

y(0.1)=f1(τ)f2(0.1-τ)

=0.01x0.1

=0.001

0 0.1 0.2 0.3

0.010.04

f1(τ)= τ 2u(τ)

τ0 0.1 0.2 0.3

0.010.04 X

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

f1(τ)= τ 2u(τ)

f2(0.2-τ)= u(0.2-τ)

1y(0.2)=f1(τ)f2(0.2-τ)

=0.01x1+0.04x1

=0.0050 0.1 0.2 0.3

0.010.04

4. DISCRETE-TIME SIGNALS4. DISCRETE-TIME SIGNALS

§ 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, 0[ ]

≥= <

– Unit sample/impulse: δ[n]={…0 1 0…}

The arrow indicates the origin, n = 0

1, 0[ ]

§ 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, 1[ 1]

≥− = <

q Shifted unit sample/impulse

1, 1[ 1]

− = ≠

1, 2[ 2]

− = ≠

1, 0[ ]

≥= <

[ ] [ ] [ 1]n u n u nδ = − −

q Unit sample/impulse using unit step

1, 1[ 1]

≥− = <

q Unit step as sum of unit sample/impulse

q Sample/impulse of arbitrary amplitude

∑∞

−=+−+−+−+=0

][...]3[]2[]1[][][i

innnnnnu δδδδδ

∑∞

−∞=− −=+−+−++++=

ii inxnxnxnxnxnx ][...]2[]1[][]1[...][ 2101 δδδδδ

§ Express the following signals graphically:(i)

(ii) x[n]={…0 5 0…}

(iii) x[n]={…0 0 5 0…}

Example 4.1: Discrete Sequences

]5[][][ −−= nununx

]3[]2[2][][ −−−+= nnnnx δδδ

§ Solution:(i)

(ii) x[n]={…0 5 0…}

(iii) x[n]={…0 0 5 0…}

]5[][][ −−= nununx

(iv) ]3[]2[2][][ −−−+= nnnnx δδδ

5. DISCRETE-TIME CONVOLUTION

§ For a LTI discrete-time system, the discrete-time convolution is given mathematically by:

§ If the LTI system is causal, then:

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

[ ] [ ] [ ]k

y n x k h n k∞

=−∞

= −∑

[ ] [ ] [ ]0

y n x k h n k=

= −∑

§ Perform convolution mathematically for the following sequences at n = 3.

§ For n = 3

[ ] [ ]

f n nu n

f n u n

[ ] [ ] [ ]1 20

y n f k f n k=

= −∑

[ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]

1 2 1 2 1 2 1 2

0 3 1 2 2 1 3 0

1 10 2 3

3 33.777

y f k f k

f f f f f f f f=

= + + +

Example 5.1: Discrete-Time Convolution

(1/3)3

1 (1/3)2 2 (1/3)1

3 (1/3)0

Causal signals

5. DISCRETE-TIME CONVOLUTION

§ All the properties for continuous-time convolution still hold for discrete-time convolution.§ Graphical method can be performed on discrete-

time signals to obtain discrete-time convolution. Same procedures as in continuous-time case.

GRAPHICAL EVALUATIONGRAPHICAL EVALUATION

§ Steps involved in the graphical evaluation of two signals à y[n] = h[n]*x[n]

1. Plot h[k] and x[k].2. Obtain x[-k].3. Shift x[-k] by n to get x[n-k]= x[-(k - n)].4. Multiply h[k] and x[n-k]. Area under h[k]x[n-k] is

equal to convolution of two signals, y[n].5. Repeat steps 3 - 4 for various values of n = n1,

n2, n3, ….6. Plot y[n], the convolution.

[ ] [ ] [ ]

[ ]* [ ]k

y n h k x n k

h n x n

=−∞

§ Repeat Example 5.1, now using graphical convolution.

§ At n= 3

§ Find mathematically, the discrete time convolution of the two signals below.

[0] 6; [1] 13; [2] 10;[3] 10; [4] 11; [5] 2

6 [ ] 0

y y yy y yn y n

= = == = − =

≥ ⇒ =

[ ] [ ] [ ]0

y n f k h n k=

= −∑Causal signals

f[k] h[k]

§ Repeat Example 5.3, now using graphical convolution

Example 5.4: Discrete-Time ConvolutionExample 5.4: Discrete-Time Convolution

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

h[1-k]

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

h[2-k]

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

f[k] f[k] f[k]

h[3-k]

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

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

h[4-k]

h[5-k]

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

f[k] f[k] f[k]

§ For the two discrete signals defined below:

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

]6[][][ −−= nununx

]5[]2[][ −−−= nununh

(ii) y[n] = x[n]*h[n]

654-3 -2 -1 0 321

10981 2 3 4 765

]6[][][ −−= nununx

]5[]2[][ −−−= nununh

CONVOLUTION TABLECONVOLUTION TABLE

Index 0 + 0 = 0

Start of convolutionsequence, y[0]

CONVOLUTION TABLECONVOLUTION TABLE

§ 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 isy[1+1] = y[2].

§ Repeat Example 5.1 using convolution table.

Index 0 + 1 = 1

Start of convolutionsequence, y[1]For n = 3, compute y[3]

y[3]Σ(3, 2/3, 1/9) = 34/9 = 3.777

Example 5.6: Convolution Table

§ Repeat Example 5.3 using convolution table.

y[0]=6 y[3]=-3+15+(-2)=10y[1]=3+10=13 y[4]=-5+(-6)=-11y[2]=9+5+(-4)=10 y[5]=2

ΣIndex(0,0) = 0

Start of convolutionsequence, y[0]

Example 5.7:Convolution Table

f[3]h[2]=2f[3]h[1]=-5f[3]h[0]=-3f[3]=-1f[2]h[2]=-6f[2]h[1]=15f[2]h[0]=9f[2]=3f[1]h[2]=-2f[1]h[1]=5f[1]h[0]=3f[1]=1f[0]h[2]=-4f[0]h[1]=10f[0]h[0]=6f[0]=2h[2]=-2h[1]=5h[0]=3

SummarySummary

§ 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

chap2_(introduction to signals

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