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EECE 359 - Signals and Communications: Part 1 Spring 2014

Introduction1

Signals are generated by a wide variety of

sources and contain information about the

behaviour or nature of the source. Example

signals are text, speech, music, video, sensor

data, stock market index value, etc. Often we

are interested in processing signals to extract

information of interest. In general, systems

process signals and produce other signals.

Example of processed signals are equalized

or distorted music!

Communication

channel Source Transmitter

Sink Receiver

Simplified block diagram of a communication system

The purpose of a communication system is to

convey messages (information) from a source1These slides are based on lecture notes of Professors L.

Lampe, C. Leung, and R. Schober

SM 1


to a destination (sink). Examples of sources

and sinks: person, computer, memory, etc.

Communication is typically done using signals

that are carefully designed to result in

an efficient and cost-effective information

transfer.

The medium through which the signals are

sent is referred to as channel and can

cause distortion and introduce other types

of impairments, e.g. noise, fading.

A high level perspective

Common communication channels are

inherently analog in nature, e.g., telephone

wires, coaxial cable, optical fiber, infrared,

cellular, satellite, and power line channels.

Why are digital communication systems

preferred over their analog counterparts?

SM 2


Many trade-offs are involved in the design of

digital communication systems. The designer

tries to minimize the cost while meeting all

performance specifications.

System considerations include: cost,

bandwidth, data rate, power consumption,

security delay, error performance, and

subjective quality.

To send information digitally, we often need

to convert analog signals to digital form.

How can we do this?

Use sampling and quantization.

• How often should we sample?

• How finely should we quantize?

SM 3


Example: Compact Disc (CD) system

Each one of the left and right channels is

sampled at a rate of 44,100 times per second.

Why this sampling rate?

We will later discuss Sampling Theorem which

will shed light on the answer.

Typically, each of the samples is converted

into digital form using a 16-bit analog-to-

digital (A/D) converter or ADC.

This results in a 1.41 Mbps data stream.

A coding technique (in this case, CIRC or

cross-interleaved Reed-Solomon code) is used

to encode this 1,411,200 bps data stream

which results in a 1.88 Mbps encoded stream.

Such coding allows for error correction.

SM 4


Typical digital data rate requirements:

Single-page fax 10 – 50 kB8 x 10 inch color image 1 – 8 MBVoice 8 – 32 kbpsAudio (MP3) 32 – 384 kbpsVideo (H.261 conferencing) 64 kbps – 1.5 MbpsHD Video (MPEG-2, MPEG-4) 2 – 50 Mbps

SM 5


Important events in development of communicationsystems

1838 Telegraph (Cooke and Wheatstone)1871 Telephone “Caveat” Some believe Antonio

Meucci (not A.G. Bell) was the inventor ofthe talking telegraph or telephone.

1900 Marconi sends wireless signal across Atlantic.1920 Beginning of radio broadcasting.1936 First public B/W TV broadcast.1951 First public color TV broadcast.1957 First earth satellite, Sputnik I.1962 First communication satellite, Telstar I.1966 Principles of fibre optic communications

published (Kao and Hockham).1973 Birth of Internet.1979 First-generation cellular phone service.1985 Fax machines gain popularity.1990’s HDTV, second-generation cellular systems.2000’s Third-generation cellular systems, satellite

radio, “anytime, anywhere, multimediacommunications”.

2010’s Online social networks, smart phones, LTE,wireless sensor networks (WSNs).

Note the increasingly rapid pace of innovations.

SM 6


Continuous-time (CT) and discrete-time

(DT) signals – Section 1.1

Notation

CT: x(t), y(t), or other functions of a

variable t where t is a real-valued independent

variable; t often but not always refers to time;

e.g., it could be a spatial variable.

DT: x[n], y[n], or other functions of a variable

n where n is an integer-valued independent

variable. A DT signal is not defined for non-

integer values of n.

A DT signal may result from a source that

is inherently discrete time, e.g., daily closing

index of the TSE; it may also result from the

sampling of a CT signal.

SM 7


Graphical representations of signals

Continuous-time

What is the unit step response of the RC

circuit shown below?

Input Output

R

C

SM 8


Discrete-time

Signal energy and power

CT: The total energy in a possibly complex-

valued signal x(t) over the time interval

t1 ≤ t ≤ t2 is defined as

E∆=

∫ t2

t1

|x(t)|2 dt

where |x(t)| denotes the magnitude of x(t).

SM 9


The average power is defined as

P∆=

E

t2 − t1.

Example: Determine the total energy and

average power of the signal x(t) = 1 − e−t

over the time interval (0, 5).

SM 10


DT: The total energy in a possibly complex-

valued signal x[n] over the time interval

n1 ≤ n ≤ n2 is defined as

E∆=

n2∑

n=n1

|x[n]|2 .

The average power is defined as

P∆=

E

n2 − n1 + 1.

Example: Determine the total energy and

average power of the signal x[n] = n over

the time interval [0, 10].

SM 11


Question: How do we define the total energy

and average power for signals over an infinitely

long time interval?

Answer: By taking limits in the above

definitions.

CT:

E∞∆= lim

T→∞

∫ T

−T

|x(t)|2 dt

P∞∆= lim

T→∞

1

2T

∫ T

−T

|x(t)|2 dt .

DT:

E∞∆= lim

N→∞

N∑

n=−N

|x[n]|2

P∞∆= lim

N→∞

1

2N + 1

N∑

n=−N

|x[n]|2 .

SM 12


Classification of signals

1. Finite total energy, i.e., E∞ < ∞ . This

implies that P∞ = 0.

2. Finite average power, i.e., 0 < P∞ < ∞ .

This implies that E∞ = ∞ .

3. Infinite average power, i.e., P∞ = ∞ . For

example, x[n] = n,−∞ < n < ∞ .

Transformation of the independent

variable – Section 1.2

Consider the signal shown in below.

t

x(t)

1

101–

A sample signal x(t)

SM 13


Let us sketch the signals obtained when we

change the independent variable as follows.

t

x(−t)

1

101–2– 2

t

x(t−1)

1

101–2– 2

t

x(2t)

1

101–2– 2

SM 14


t

x(2t−1)

1

101–2– 2

More generally, to obtain x(αt − β), we first

delay x(t) by β and then perform the time

scaling/reversal by α.

Equivalently, to obtain x(

α[

t− βα

])

, we first

time scale/reverse by α and then delay by βα.

Periodic signals – Section 1.2.2

CT: x(t) is periodic with period T > 0 if

x(t+ T ) = x(t) for all values of t .

The smallest value of T for which this holds

is called the fundamental period, T0.

SM 15


Example: cosω0t is periodic with

fundamental period T0 =2πω0

.

This signal is related to the CT complex

exponential signal, ejω0t, via Euler’s formula

ejθ = cos θ + j sin θ .

DT: x[n] is periodic with period N > 0 if

x[n+N ] = x[n] for all values of n .

The smallest value of N for which this holds

is called the fundamental period, N0.

Example: DT complex exponential signal,

ejω0n .

Note: There are important differences

between the CT and DT complex exponential

signals.

SM 16


1. Unlike the CT case, the rate of oscillation

of ejω0n does not increase monotonically

with ω0.

Reason: ej(ω0+2π)n = ejω0n ej2πn︸︷︷︸=1

= ejω0n.

This means that the DT exponential signal

at frequency ω0+2π is identical to that at

frequency ω0. See Fig. 1.27.

2. DT complex exponential signal, ejω0n, is

periodic (with period N) only if 2πω0

is a

rational number.

Reason: For a periodic signal, we must

have

ejω0(n+N) = ejω0n

i.e., ejω0N = 1

⇒ ω0N = m · 2π, m an integer

i.e.,2π

ω0=

N

m.

SM 17


Demystifying Figure 1.27 of textbook

Observations:

1. Slowly-varying (low-frequency) DT sinusoidal

sequences have ω0 close to 0,±2π,±4π, . . .;

Rapidly-varying (high-frequency) DT sinusoidal

sequences have ω0 close to ±π,±3π, . . .

2. The sequences cos(π−θ)n and cos(π+θ)n

are identical, e.g., Fig. 1.27(c) and (g)

correspond to θ values of 3π/4.

Analytic proof of the second observation

Recall that

cos(α+ β) = cosα cosβ − sinα sin β

cos(α− β) = cosα cosβ + sinα sin β.

SM 18


Thus,

cos(π − θ)n = cos(πn− θn)

= cosπn cos θn+ sinπn sin θn

= (−1)n cos θn.

Similarly,

cos(π + θ)n = cos(πn+ θn)

= cosπn cos θn− sinπn sin θn

= (−1)n cos θn.

A graphical viewpoint

0 1 2 3 4 5 6 7 8 9 10

−1

−0.5

0

0.5

1

t (seconds)

cos (0*pi*t) cos (0.5*pi*t)cos (pi*t) cos (1.5*pi*t)cos (2*pi*t)

SM 19


Even and odd signals

Defn: A signal x(t) or x[n] is said to be even

if

CT : x(−t) = x(t)

DT : x[−n] = x[n].

Defn: A signal x(t) or x[n] is said to be odd

if

CT : x(−t) = −x(t)

DT : x[−n] = −x[n].

Examples

x(t) = t2 is even, as is x[n] = n2.

x(t) = t3 is odd, as is x[n] = n3.

x(t) = t+ 1 is neither even nor odd.

SM 20


Facts

1. A function which is both even and odd has

to be equal to 0.

2. The sum of two even functions is even.

3. The sum of two odd functions is odd.

4. Any signal x(t) (or x[n]) can be written in

a unique way as the sum of an even signal

and an odd signal, i.e.,

x(t) = xe(t) + xo(t)

where the even part is given by

xe(t) =1

2[x(t) + x(−t)]

and the odd part is given by

xo(t) =1

2[x(t)− x(−t)].

SM 21


Proof of uniqueness

Let

x(t) = xe,1(t) + xo,1(t)

and

x(t) = xe,2(t) + xo,2(t).

Then,

xe,1(t)− xe,2(t) = xo,2(t)− xo,1(t)

The LHS is even and the RHS is odd.

Therefore, both sides are identically zero.

Hence, xe,1(t) = xe,2(t) and xo,1(t) =

xo,2(t).

SM 22


Two important signals: Unit impulse and

unit step functions – Section 1.4

Discrete-Time:

Unit impulse or unit sample

δ[n] =

{0, n 6= 0

1, n = 0.

n

1

10

1–3– 2

δ n[ ]

32–

Unit step sequence

u[n] =

{0, n < 0

1, n ≥ 0.

n

1

10

1–3– 2

u n[ ]

32–

SM 23


Fact:

δ[n] = u[n]− u[n− 1]

u[n] =n∑

m=−∞

δ[m]

u[n] =

∞∑

k=0

δ[n− k].

Fact:

x[n]δ[n] = x[0]δ[n]

x[n]δ[n− n0] = x[n0]δ[n− n0]

SM 24


Continuous-Time:

Unit step function

u(t) =

{0, t < 0

1, t > 0.

t

1

10

1–3– 2

u t( )

32–

Unlike its DT counterpart, the unit impulse

function is not a proper mathematical

function. We can view it as the derivative,

δ∆(t), of an approximation, u∆(t), to u(t) as

the width ∆ → 0. Note that the area under

SM 25


δ∆(t) is constant and is equal 1.

t

1

0

u∆ t( )

∆

t

1∆---

0

δ∆ t( )

∆

t

1

0

δ t( ) δ∆ t( )∆ 0→lim=

Sampling property of δ(t): For a “smooth”

function x(t),

x(t)δ(t− t0) = x(t0)δ(t− t0).

SM 26


Continuous-time and discrete-time systems

– Section 1.5

CT systemx t( ) y t( )

Short-hand: x t( ) y t( )→

DT systemx n[ ] y n[ ]

Short-hand: x n[ ] y n[ ]→

Interconnection of systems

1. Series or cascade

2. Parallel

SM 27


3. Hybrid – Series and parallel

4. Feedback

SM 28


System properties – Section 1.61. Linearity

Defn (CT): If x1(t) → y1(t) and x2(t) →

y2(t), then for any constants a, b we have

ax1(t) + bx2(t) → ay1(t) + by2(t).

An important result: For a linear system,

a 0 input always results in a 0 output.

Example: System whose I/O relationship is

given by y(t) = tx(t).

SM 29



given by y[n] = x2[n].

Example: System whose I/O relationship

is given by y[n] = 2x[n] + 1.

SM 30


2. Time-invariance

Defn (DT): A system is time-invariant if

a time-shift in the input signal results in an

identical shift in the output signal, i.e.,

If x[n] → y[n], then x[n− n0] → y[n− n0].

A similar definition holds for CT systems.


given by y[n] = x2[n].

SM 31



given by y[n] = nx[n].

3. Memoryless: The output at any time

depends only on the input at that time.

Example: Voltage across and the current

through a resistor.

Example: Cumulative amount of rainfall in

April in Vancouver as a function of the date

of the month.

SM 32


4. Causal: The output at any time depends

only on the input values at that time or

previous times (system is non-anticipative).

Example: Any physical system in which the

independent variable is time.

Example:

y[n] =1

3{x[n− 1] + x[n] + x[n+ 1]}.

5. Stable If the input is finite, then the output

is also finite.

SM 33


Example: In the RC circuit shown below,

the input is a voltage across the input

terminals and the output is the voltage

across the output terminals.

Input Output

R

C

Example: In the same RC circuit, the input

is a constant current and the output is the

voltage across the output terminals.

SM 34


Linear time-invariant (LTI) systems –

Chapter 2

We will derive the output of DT LTI

systems. We will show that the input-output

relationship of such a system is completely

characterized by the unit impulse response of

the system.

Discrete-time LTI systems – Section 2.1

Convolution sum

Let h[n] denote the unit impulse (sample)

response of an LTI system, i.e.,

δ[n] → h[n].

Then, by the TI property,

δ[n− k] → h[n− k] (1)

We note that any signal x[n] can be written

SM 35


as

x[n] =

∞∑

k=−∞

x[k]δ[n− k]. (2)

To explain why (2) is true, we consider the

following.

Example:

n

1

10

1–3– 2

x n[ ]

32–

2

3

For the signal in the above figure, we have

x[n] = δ[n+ 1] + 2δ[n] + 3δ[n− 1] + δ[n− 2]

which verifies (2).

Using (1) and the scaling property, we can

SM 36


write

x[k]δ[n− k] → x[k]h[n− k].

Finally using the additive property , we have

∞∑

k=−∞

x[k]δ[n− k] →∞∑

k=−∞

x[k]h[n− k].

Note that h[n− k] is a time-reversed version

of h[k] delayed by n, i.e., if n is positive, we

shift the time-reversed version to the right.

x n[ ] y n[ ]LTIh n[ ]

The output in the above figure can be

expressed as

y[n] =

∞∑

k=−∞

x[k]h[n− k]

SM 37


We say that y[n] is the convolution of x[n] and

h[n]; this relationship is commonly written as

y[n] = x[n] ∗ h[n].

Example:

n

1

10

1– 2

x n[ ]

32–

n

1

10

1– 2

h n[ ]

32–

We want to determine the output of the above

LTI system, i.e.,

y[n] = x[n] ∗ h[n].

SM 38


k

1

10

1– 2

x k[ ]

32–

k

1

10

1– 2

h k[ ]

32–

k

1

10

1– 2

h k–[ ]

32–

k

1

10

1– 2

h n k–[ ]

n

2–

n 1–

SM 39


n

1

10

1– 2

y n[ ]

32– 4 5

2

MATLABr is a high-level language and

environment for numerical computations

for variety of applications including signal

processing. It can be used to compute y[n].

A simple MATLABr script, convol.m, for

this purpose and the result of running the

script is shown below.

SM 40


% convol.m

% Matlab script file to evaluate the % convolution of two signals

x=[1 1 1] % index of first element is 1

h=[1 1]

y=conv(x,h)

Result from running above MATLAB script

EDU» clear allEDU» convol

x =

1 1 1

h =

1 1

y =

1 2 2 1

SM 41


A more complicated example:

x[n] =

{1, 0 ≤ n ≤ 3

0, otherwise

and

h[n] =

{0.5n, 0 ≤ n ≤ 1

0, otherwise.

n

1

10

1– 2

x n[ ]

32–

n

1

10

1– 2

h n[ ]

32–

4 5

0.5

4

SM 42


% convol1.m

% Matlab script file to evaluate the

% convolution of two signals

x=[1 1 1 1] % index of first element is 1

h=[1 0.5]

y=conv(x,h)

Result from running above MATLAB script

>> clear all>> convol1

x =

1 1 1 1

h =

1.0000 0.5000

y =

1.0000 1.5000 1.5000 1.5000 0.5000

SM 43


k

1

10

1– 2

x k[ ]

32–

k

1

10

1– 2

h n k–[ ]

32–

4 5

0.5

4

for n = −1

n

1

10

1– 2

y n[ ]

32– 4 5 6

2

SM 44


Continuous-time LTI systems – Section 2.2

In a manner analogous to the DT case, it

can be shown that the output y(t) of a CT

LTI system with input x(t) and unit impulse

response h(t), as depicted in the following

figure, is given by the convolution integral,

i.e.

y(t) =

∫ ∞

−∞

x(τ)h(t − τ) dτ

x t( ) y t( )LTIh t( )

We say that y(t) is the convolution of x(t) and

h(t); this relationship is commonly written as

y(t) = x(t) ∗ h(t).

SM 45


Example:

t

1

0.50

0.5– 1

x t( ) rect t( )=

1.51–

t

1

0.50

0.5– 1

h t( ) rect t 0.5–( )=

1.51–

We want to determine the output of the LTI

system with x(t) and h(t) as shown in the

above figure, i.e.

y(t) = x(t) ∗ h(t).

To help in performing the convolution, x(τ)

and h(t− τ) are plotted below.

SM 46


1

0.50

0.5– 1

x τ( )

1.51–

1

0

h t τ–( )

τ

τt

t 1–

We need to evaluate

y(t) =

∫ ∞

−∞

x(τ)h(t− τ) dτ .

• Case 1: t ≤ −0.5

SM 47


• Case 2: −0.5 ≤ t ≤ 0.5

• Case 3: 0.5 ≤ t ≤ 1.5

• Case 4: t ≥ 1.5

The output y(t) is sketched below.

t

1

0.50

0.5– 1

y t( )

1.51–

SM 48


Example: x(t) = e−tu(t) is input to an

LTI system with unit impulse response h(t) =

u(t). Find the output y(t).

As in the previous example, we first plot x(τ)

and h(t− τ) below.

1

0

x τ( )

1

0

h t τ–( )

τ

τt

1

eτ–u τ( )

Sketch the output, y(t), in the figure below.

SM 49


t

1

0 1

y t( )

1–

• Case 1: t ≤ 0

• Case 2: t ≥ 0

SM 50


Properties of LTI systems – Section 2.2

Note that these properties are exhibited by

both DT and CT systems.

1. Commutative property

x[n] ∗ h[n] = h[n] ∗ x[n].

This is easily shown by a change of

variables.

The commutative property implies that

y2[n] = y1[n] in the following figure.

SM 51


x n[ ] y1 n[ ]LTIh n[ ]

h n[ ] y2 n[ ]LTIx n[ ]

Illustrating a consequence of the commutative

property

2. Distributive property

x[n] ∗ {h1[n] + h2[n]}

= x[n] ∗ h1[n] + x[n] ∗ h2[n].

This equation states that convolution

distributes over addition.

One interpretation of this result is that the

outputs of the two systems shown in the

SM 52


figure below are identical, i.e. ya[n] =

yb[n].

x n[ ]

y1 n[ ]LTIh1 n[ ]

y2 n[ ]LTIh2 n[ ]

+ya n[ ]

LTIh1 n[ ] h+

2n[ ]

x n[ ] yb n[ ]

Illustrating a consequence of the distributive property

3. Associative property

x[n] ∗ {h1[n] ∗ h2[n]}

= {x[n] ∗ h1[n]} ∗ h2[n]. (3)

Equation (3) states that the order in

SM 53


which the convolution is performed is

unimportant. We can thus write x[n] ∗

h1[n] ∗ h2[n] without ambiguity.

One interpretation of this result is that the

outputs of the two systems shown in the

figure below are identical, i.e. ya[n] =

yb[n]. Note that ya[n] is the RHS of (3)

whereas yb[n] is the LHS of (3).

x n[ ] LTIh1 n[ ]

LTIh2 n[ ]

ya n[ ]

LTIh1 n[ ]∗h

2n[ ]

x n[ ] yb n[ ]

Illustrating a consequence of the associative property

Applying the commutative property

followed by the associative property to the

second system in the above figure, we have

the result shown in the figure below.

SM 54


x n[ ] LTIh2 n[ ]

LTIh1 n[ ]

yd n[ ]

LTIh2 n[ ]∗h

1n[ ]

x n[ ] yc n[ ]

Re-ordering of cascade of LTI systems

We can therefore conclude that the unit

impulse response of a cascade of LTI

systems does not depend on the order in

which they are cascaded.

SM 55


Unit impulse responses of various types

of LTI systems

Memoryless systems – Section 2.3.4

Necessary and sufficient conditions are:

DT : h[n] = K · δ[n]

CT : h(t) = K · δ(t).

Invertible systems – Section 2.3.5

A system is invertible if different inputs

produce different outputs. This means that

we can recover the input signal exactly from

the output signal.

Fact: An invertible LTI system has an LTI

inverse. (Problem 2.50)

Given an invertible LTI system with unit

impulse response h[n] or h(t), the unit

impulse response h1[n] or h1(t) of the inverse

SM 56


system must satisfy

DT : h[n] ∗ h1[n] = δ[n]

CT : h(t) ∗ h1(t) = δ(t).

Example:

h[n] = δ[n− n0] and h1[n] = δ[n+ n0].

Causal systems – Section 2.3.6


DT : h[n] = 0 for n < 0

CT : h(t) = 0 for t < 0.

Example:

h[n] = δ[n]− δ[n− 1].

SM 57


Stable systems – Section 2.3.7


DT :∞∑

n=−∞

|h[n]| < ∞.

The above equation states that the unit

impulse response is absolutely summable.

CT :

∫ ∞

−∞

|h(t)| dt < ∞.

The above equation states that the unit

impulse response is absolutely integrable.

Example:

h[n] = δ[n]− δ[n− 1] is stable.

h[n] = u[n] corresponds to an accumulator

and is not stable.

SM 58


Summary

Property Discrete-time LTI system Continuous-time LTI system

Memoryless h[n] = δ[n] h(t) = δ(t)Invertibility h[n] ∗ hinv[n] = δ[n] h(t) ∗ hinv(t) = δ(t)Causal h[n] = 0, for n < 0 h(t) = 0, for t < 0

Stability∞∑

n=−∞|h[n]| < ∞

∞∫

−∞|h(τ)| dτ < ∞

Unit Step-Response of LTI Systems –

Section 2.3.8

It can be used as an alternative to impulse

response and uniquely describes the behaviour

of LTI systems.

Unit step response = s(t) or s[n] = system

response to input u(t) or u[n]

Relationship:

s(t) = u(t) ∗ h(t) = h(t) ∗ u(t) =

t∫

−∞

h(τ) dτ

s[n] = u[n] ∗ h[n] = h[n] ∗ u[n] =n∑

k=−∞

h[k]

SM 59


andh(t) = ds(t)

t = s(t)

h[n] = s[n]− s[n− 1]

Example:

RC circuit

• Impulse response: h(t) = 1RC e−t/(RC)u(t)

• Step response: s(t) = 1RC

t∫

0

e−τ/(RC) dτ

= 1RC (−RC) e−τ/(RC)

∣∣∣∣∣

t

0

= (1− e−t/(RC))× u(t)

SM 60


Systems described by differential and

difference equations – Section 2.4

Many systems and physical phenomena can

be modelled by linear constant-coefficient

difference (when the system is discrete time)

or differential (when the system is continuous

time) equations.

In general to solve difference or differential

equations we need to know the initial

conditions (e.g., initial energy of the system).

• General N th order linear constant-

coefficient differential equation

N∑

k=0

akdky(t)

dtk=

M∑

k=0

bkdkx(t)

dtk

• Solution:

y(t) = yh(t) + yp(t)

SM 61


– Homogeneous solution

N∑

k=0

akdkyh(t)

dtk= 0

– Particular solution

any solution yp(t) for given x(t)

– Condition of initial rest: If x(t) = 0 for

t < t0 then

y(t)∣∣t−0

=dy(t)

dt

∣∣∣t−0

= . . . =dN−1y(t)

dtN−1

∣∣∣t−0

= 0

(corresponds to causal LTI system)

• Differential equation + initial condition

specify system output completely

SM 62


Example 1:

• First–order differential equation

dy(t)

dt+ 2y(t) = x(t)

with input

x(t) = Ke3t for t > 0 K ∈ R

• Solution

– Particular solution:

Hypothesis: yp(t) = Y e3t for t > 0

⇒ 3Y e3t + 2Y e3t = Ke3t

⇒ Y = K5

⇒ yp(t) =K5 e

3t for t > 0


dy(t)

dt+ 2y(t) = 0

SM 63


Hypothesis: yh(t) = Aest

⇒ Asest + 2Aest = 0

⇒ s = −2

⇒ yh(t) = Ae−2t for t > 0

SM 64


– Combination

y(t) = Ae−2t +K

5e3t for t > 0

– Condition of initial rest

y(t) = 0 for t < 0 since x(t) = 0 for t < 0

No impulse introduced at t = 0 ⇒

y(0) = 0

⇒ A+K

5= 0

– Complete solution

y(t) = yh(t) + yp(t)

=K

5(e3t − e−2t) for t > 0

Linear Constant-Coefficient Difference Equations

SM 65


• General N th order linear constant-

coefficient difference equation

N∑

k=0

aky[n− k] =M∑

k=0

bkx[n− k]

• Solution

y[n] = yh[n] + yp[n]


N∑

k=0

aky[n− k] = 0

SM 66


– Particular solution

any solution yp[n] for given x[n]

– Condition of initial rest: If x[n] = 0 for

n < n0 then

y[n] = 0 for n < n0

(corresponds to causal LTI system)

• Alternatively: rewrite difference equation

y[n] =1

a0

(M∑

k=0

bkx[n− k]−N∑

k=1

aky[n− k]

)

and solve for successive values of n ≥ n0

Example 2:

1. Assume nonrecursive difference equation

y[n] = x[n]− x[n− 1]

SM 67


Impulse response of underlying LTI system

obtained for

x[n] = δ[n]

⇒ y[n] = h[n] = δ[n]− δ[n− 1]

Since h[n] = 0 for n > 2 the underlying

system is referred to as finite impulse

response (FIR) system.

2. Consider recursive difference equation

y[n]−1

2y[n− 1] = x[n]

or

y[n] = x[n] +1

2y[n− 1]

Impulse response of underlying LTI system

(Condition of initial rest: y[n] = 0 for

SM 68


n < 0)

y[0] = x[0] +1

2y[−1] = 1

y[1] = x[1] +1

2y[0] =

1

2

y[2] = x[2] +1

2y[1] =

1

4· · ·

y[n] = x[n] +1

2y[n− 1] =

(1

2

)n

⇒ impulse response h[n] = (1/2)nu[n]

Since h[n] has an infinite duration, the

underlying system is referred to as infinite

impulse response (IIR) system.

SM 69


Fourier series representation of periodic

signals – Chapter 3

We have seen that a signal can generally be

represented as a linear combination of shifted

impulse (or sample) functions.

We will show that a signal can also

be represented as a linear combination

of complex exponential functions, provided

certain conditions are satisfied.

Why is this useful? Section 3.2

Reason is due to this important fact: The

output of an LTI system due to a complex

exponential input is the same complex

exponential multiplied by a (possibly complex)

gain factor.

We say that complex exponentials are

eigenfunctions of LTI systems; the gain

factors are termed eigenvalues. (cf.

SM 70


eigenvectors and eigenvalues of a matrix)

We now prove the following.

Continuous-time: If x(t) = est is input to a

LTI system with impulse response h(t), the

output y(t) is

y(t) = H(s)est

where

H(s) =

∫ ∞

−∞

h(τ) e−sτ dτ

Proof:

SM 71


Example:

Suppose y(t) = x(t− 1) and x(t) = ej2πt.

Example:

Suppose y(t) = x(t−1) and x(t) = cos 2πt+

cos 3πt.

SM 72


Discrete-time: If x[n] = zn is input to a

LTI system with impulse response h[n], the

output y[n] is

y[n] = H(z)zn

where

H(z) =∞∑

k=−∞

h[k] z−k

Proof:

SM 73


Fourier series representation of CT

periodic signals – Section 3.3

Let x(t) be a periodic signal with fundamental

period T . Then if certain conditions

(Dirichlet, pp. 197-200) are satisfied, we can

represent x(t) as a Fourier series (FS) (an

infinite sum of complex exponentials):

x(t) =

∞∑

k=−∞

ak ejk(2πT )t

where the (possibly complex) Fourier

coefficients {ak} are given by

ak =1

T

∫ T2

−T2

x(t) e−jk(2πT )t dt,

k = 0,±1,±2, . . .

The above equations are referred to as the

synthesis and analysis equations respectively.

SM 74


Section 3.4 provides a good discussion of the

convergence of the FS representation.

Notes:

1. Using Euler’s relationship, we can also

express the FS representation as an infinite

sum of sine and cosine terms.

2. Given x(t), we can determine {ak}∞k=−∞;

conversely, we can reconstruct x(t) from

{ak}∞k=−∞.

{ak}∞k=−∞ give the frequency-domain

description of the signal and are called its

spectral coefficients.

3. A periodic signal x(t) of fundamental

period T has components at frequencies

0,±2πT ,±4π

T , . . ., i.e. at multiples of the

fundamental frequency ω0 =2πT or f0 =

1T .

SM 75


The component at freq nf0 is called the

nth harmonic .

4. Conjugate symmetry property: If x(t) is

real, then a−k = a∗k.

Proof:

As a result,

|a−k| = |a∗k| = |ak|

and ∠a−k = ∠a∗k = −∠ak .

SM 76


Thus, when x(t) is real, the amplitudes

of the spectral coefficients have even

symmetry whereas their phases have odd

symmetry .

Example: Periodic square wave shown below.

1

T10T1– T

2---

x t( )

T2---–

tTT–

Periodic square wave

We would like to determine its Fourier or

spectral coefficients {ak}∞k=−∞.

SM 77

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