Signals and Systems 2

```SIGNALS AND SYSTEMS
‫احسان احمد عرساڻي‬
Chapter 2
S.
No.
Topic
3
Sinusoidal and complex exponential
signals
Singularity function signals and their
properties
Signal energy and signal power
4
Orthogonal signals
1
2
5
Signal representation by Generalised
Fourier Series
Continuous and discrete-time
No. of
Lecture
s
2
2
2
2
2
Sinusoidal Signals

The only natural wave
 The
only natural
oscillation
 The Simple Harmonic
motion
xt   A cos2ft   
 All other waves are
composite waves and
combinations of
several sine waves
Sinusoidal Signals

Euler’s theorem
A
sinusoidal signal can
be represented using
complex exponentials
e j  cos   j sin 
e  j  cos   j sin 
e j  e  j  2 cos 
e j  e  j  2 j sin 
e j   e  j
cos  
2
j
e e
sin  
2j
 j
The decaying sinusoid



cos(2πfot+&Oslash;)
e(-t/ )
e(-t/ ) cos(2πfot+&Oslash;)
10
5
0
-5
-10
-5
-4
-3
-2
-1
0
1
2
3
4
5
Singularity Function Signals
Unit Impulse
1
0.8
Impulse
 Abrupt
amplitude

0.6
0.4
change
0.2
0
-2
-1.5
-1
-0.5
0
0.5
time (seconds)
1
1.5
2
1
1.5
2
1
1.5
2
Unit Step u(t)
1
Step
amplitude

0.8
 Constant
0.6
0.4
0.2
0
-2
-1.5
-1
-0.5
0
0.5
time (seconds)
Unit Ramp r(t)
2
Ramp
 Linear
change/constant
rate of change
amplitude

1.5
1
0.5
0
-2
-1.5
-1
-0.5
0
time (second)
0.5
Impulse ‫اِمپلس‬
4
  0.125
3.5
3
delta(t)
2.5
  0.25
2
1.5
  0.5
1
0.5
0
-2
-1.5
-1
-0.5
1
x(t )  
0

0
t
0.5
1

for t   
for elsewhere
1.5
2
Impulse ‫اِمپلس‬
8
  0.125
7
6
5
4
  0.25
3
  0.5
2
1
0
-1
-0.8
-0.6
-0.4
-0.2
0
1  t
1 

x(t )    


0
0.2




0.4
0.6

for t   

for elsewhere
0.8
1
Impulse ‫اِمپلس‬
4
  0.125
3.5
3
x(t)
2.5
2
1.5
1
  0.25
  0.5
0.5
0
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
time (seconds)
t
1 
x(t ) 
e
2
0.4
0.6
0.8
1
Properties of impulse (t)


(-t)= (t)
(at)= (t)
Time-shifted impulse
t 0
1
 t   
0 elsewhere
(t- )

Unit Impulse
1
0.8
amplitude

0.6
0.4
0.2
0
-2
-1.5
-1
-0.5
0
0.5
time (seconds)
1
1.5
2
Properties of impulse (t)



(-t)= (t)
(at)= (t)
Time-shifted impulse

(t- )
at  0
1
 at   
0 elsewhere
t 0
1
 at   
0 elsewhere
t 0
1
 t   
0 elsewhere
t  0
1
  t   
0 elsewhere
t 0
1
  t   
0 elsewhere
Properties of impulse (t)
t 0
1
 t   
0 elsewhere
t 1
1
 t  1  
0 elsewhere
1 t  1  0
 t  1  
0 elsewhere
1 t  1  0
 t  1  
0 elsewhere
u(t)
1
0.5
0
-1.5
-1
-0.5
0
0.5
1
1.5
-1
-0.5
0
0.5
1
1.5
-1
-0.5
0
0.5
1
1.5
0.5
0
-1.5
1
u(t+1)
t  1
1
 t  1  
0 elsewhere
u(t-1)
1
0.5
0
-1.5
Properties of impulse (t)
t 0
1
 t   
0 elsewhere


t0
1
 t dt  0 elsewhere
t
t
  t dt  ut 

Impulse doesn’t have a
specific amplitude
It has a specific
Properties of impulse (t)
xt  t   x0 t 
xt  t  t0   xt0  t  t0 
t
 xt  t   x0 for t  0

t
 xt  t  t   xt  for t  t
0

0
0
1 t  1
ut  1  
0 t  1
1 t  0
u t   
0 t  0
1
u(t)
1  t  1
u t  1  
0  t  1
0.5
0
-3
-2
-1
0
1
2
3
-2
-1
0
1
2
3
-2
-1
0
1
2
3
0.5
0
-3
1
u(-t-1)
1 t  1
u t  1  
0 t  1
u(t-1)
1
0.5
0
-3
1  t  0
u t   
0  t  0
1 t  0
u t   
0 t  0
1
u t  1  
0
t 1  0
t 1  0
1 t  0
u t   
0 t  0
1 t  1
u t  1  
0 t  1
u(t)
1
0.5
0
-3
-2
-1
0
1
2
3
-2
-1
0
1
2
3
-2
-1
0
1
2
3
u(-t)
1
0.5
0
-3
u(-t+1)
1
0.5
0
-3
Ramp Signal r(t)
t 0
 At
Ar t   
 0 otherwise
5
4
r(t)
3
2
1
0
-2
-1
0
1
2
time (seconds)
3
4
5
Shifted Ramp Signal r(t- )
t 
 At   
Ar t     
otherwise
 0
r(t+1)
r(t-1)
r(t)
 At    t    0
Ar t     
otherwise
 0
4
2
0
-2
4
-1
0
1
2
3
4
5
-1
0
1
2
3
4
5
-1
0
1
2
3
4
5
2
0
-2
6
4
2
0
-2
Scaled Ramp Signal r(at)
t 0
 Aat
Ar at   
 0 otherwise
at  0
 Aat
Ar at   
 0 otherwise
r(t)
10
5
r(2t)
0
-2
10
0
1
2
3
4
5
-1
0
1
2
3
4
5
-1
0
1
2
3
4
5
5
0
-2
10
r(0.5t)
-1
5
0
-2
Time Reversed Ramp Signal r(-t)
t  0
 At
Ar  t   
otherwise
 0
t0
 At
Ar  t   
otherwise
 0
5
4
r(-t)
3
2
1
0
-5
-4
-3
-2
-1
0
time (seconds)
1
2
3
4
5
Signal Energy and Signal Power
Oppenheim




Page 5 to 7
No examples
Continuous and Discrete
No concepts explained
Carlson




Page 70 to 74
3 examples
Continuous only
Concepts explained
Signal Energy



It has no specific unit
Its unit depends on the
unit of the x(t) itself
It is the concept of
energy carried in a
signal irrespective of
the system
lim T
2


Ext   Ex 
x
t
dt

T   T
lim N
2


Exn  Ex 
x
n

N   N
Example


lim 1 T 2
ER 
v t dt

T   R T
A buzzer has the
resistance of 20Ω
A rectanguar pulse is
applied to the buzzer
t 5
vt   12 
volts
 4 
Compute the Energy
consmed by the buzzer
Joules
12
10
voltage (volts)

8
6
4
2
0
0
1
2
3
4
time (seconds)
5
6
ER  28.8 Joules
7
8
A rectangular pulse
Plot the following
rectangular pulse
 t 1 
xt   2

 4 
2.5
2
1.5
x(t)

1
0.5
0
-0.5
-3
-2
-1
0
1
time (seconds)
2
3
4
5
Example continued
lim T 2
2


Ev 
v
t
dt
v
s

T   T
lim T 2
2


Ei 
i
t
dt
A
s

T   T
3

7
Ei   i t dt   i t dt   i t dt
2

3
2
2
3
7

7
Ei   0dt   0.6 dt   0dt A 2s
2

3
Ei  0.36t  3
7
7
A s Ev   v t dt   v t dt   v 2 t dt v 2s

3
2
3
As
Ei  0.367  3 A s
7

7
Ev   0dt   12 dt   0dt v 2s
2

3
7
2

7
2
Ev  144t  3
2
Ei  1.44 A2s
3
2
7
v 2s
Ev  1447  3 v2s
Ev  1444 v2s
Ev  576 v 2s
Signal Power



It has no specific unit
lim 1 T
2


Pxt   Px 
x
t
dt

T   2T T
Its unit depends on the
unit of the x(t) itself
It is the concept of
N
lim
1
2






P
x
n

Px

x
n

power carried in a
N   2N  1 N
signal irrespective of
the system
Energy Signal and Power Signal
Energy Signal

A signal with finite
energy is an Energy
Signal
Power Signal

 0&lt;Ex&lt;∞

Note
 Px=0
 Finite
average energy
divided by infinite time
A signal with finite
power is a Power
signal
 0&lt;Px&lt;∞

Note
 Ex=
∞
 Finite average power
multiplied by infinite
time
vt   e u t 
An Example
3t
Energy
Power
T



Ev  limT    e 3t u t  dt
2
T

Ev   e 6t dt
0
1 6t 
Ev   e
0
6
1
Ev   e    e 0
6
1
Ev 
6

The average Power will
be zero, b/c average
energy is finite
1
0.8

0.6
0.4
0.2
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
Periodic Signals







Cannot be an Energy signal
Each cycle/period contains a finite energy
Total Energy is the sum of energy in all the cycles
There are inifite number of cycles
Therefore, total energy becomes infinite
But Engergy per unit time (Power) remains finite
Therefore, all the Periodic Signals are Power Signals
Neither Energy Signal nor Power Signal

Examples
 Infinite
Energy and
Zero Power
0.1
u t  1
 Eg. xt   t
1
0.8
0.6
0.4
0.2
0
-2
 Infinite
Engery and
Infinite Power
 Eg.

yt   r t 

Impulse signal
-1
0
1
2
3
4
5
Convolution

Mathematically:

‫رياضتا‬

‫جڏهن ته هڪ فرضي ويريئيبل آهي‬

x(t ) * h(t ) 
 x( )h(t   )d

Where is a dummy variable
 It is integral of one function as weighted by the
other function
‫ جنهن جو وزن ٻيو‬،‫اهو هڪ فنڪشن جو اهڙو انٽيگرل آهي‬
.‫فنڪشن متعين ڪري ٿو‬

Procedure ‫طريقه ڪار‬

x(t ) * y (t ) 
 x( ) y(t   )d




Replace t with in x(t) and y(t)
‫جاء تي رکو‬
ِ ‫ جي‬t ‫ ۾‬y(t) ‫ ۽‬x(t) ‫فنڪشن‬
Time-reverse the function y( ) to find y(- )
‫ لهو‬y(- ) ‫ کي وقت جي محور تي ابتو ڪري‬y( ) ‫فنڪشن‬
Shift the function y(- ) by t units to find y(t- )
‫ لهو‬y(t- ) ‫ ايڪا سيري‬t ‫ کي‬y(t- ) ‫فنڪشن‬



An Example ‫ڪ مثال‬
ُ ‫ِه‬
0.5e 0.5t
h(t )  
 0
for t  4 

for t  4
1
x(t )  
0
for t  0

for t  0 
x(t)
1
0.5
0
-1
0
1
2
3
4
t (sec)
5
6
7
8
9
0
1
2
3
4
t (sec)
5
6
7
8
9
h(t)
0.6
0.4
0.2
0
-1
for   4 

for   4
h(t-tau)
h(-tau)
x(tau)
1
x( )  
0
0.5e 0.5
h( )  
 0
for   0

for   0 
1
0.5
0
-6
-4
-2
0
2
4
6
8
0.4
0.2
0
-6
-4
-2
0
2
4
6
8
0.4
0.2
0
-6
-4
-2
0
2
4
6
8
t
0.5e 0.5(t  )
h(t   )  
 0
for   t 

for   t

Case I: t&lt;4

Case 2: t&gt;4
There’s no overlap


x(t ) * y (t ) 
 x( ) y (t   )d

xt * y (t ) 
 00.5e
4

xt * y (t )  0
 0.5 t 

d
There’s overlap
that increases
logarithmcally
t


x t  * y (t )   1 0.5e  0.5t   d
4
t
x t  * y (t )  0.5 e  0.5t  d
4
x t  * y (t )  0.5e
 0 .5 t
t
e
0.5
d
4
x t  * y (t )  e  0.5t e 0.5

t
4
x t  * y (t )  e  0.5t e 0.5t  e 2

Convoltion of x(t) and y(t)
xt * y (t )  1  e 2e 0.5t
1
x(t)*y(t)
0.8
0.6
0.4
0.2
0
-2
0
2
4
6
t (sec)
8
10
12
14
An Exercise Question ‫ڪ مشقي سوال‬
ُ ‫ِه‬
Find
‫لهو‬
𝑦(𝑡) = 𝑎(𝑡) ∗ 𝑏(𝑡)
Exercise 4.12 (Carlson)
1
for t  1 
b(t )  

for t  1
0
 et
a(t )  
0
for t  1

for t  1
a(t)
0.4
0.2
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-4
-3
-2
-1
0
1
2
3
4
5
b(t)
1
0.5
0
-5
( t  )
e
a(t   )  
0
( t  )

e
for t    1  a(t   ) 


for t    1 
0
for   1  t 

for   1  t 
a(-tau)
0.4
0.2
b(tau)
a(t-tau)
0
-5
0.4
-4
-3
-2
-1
0
1
-4
-3
-2
-1
0
1
-4
-3
-2
-1
0
1
2
3
4
5
2
3
4
5
2
3
4
5
0.2
0
-5
t
1
0.5
0
-5

Case 1: 0 ≤ 𝑡 ≤ ∞
e
t 
y (t ) 
d
t 
e
 d
1
t 1
 
y (t )  e
e 
y (t )  0  e 
 

y (t )  e
e

y (t )  0  e
y (t )   e t 

y (t )   e
t 

t 
1
t 1
t 
t  t 1
t 1
t 1
t  t 1
y (t )  e
Case 2: −∞ ≤ 𝑡 ≤ 0


y (t ) 

y (t )  e t 1
1
y(t)=a(t)*b(t)
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Test Question

‫امتحاني سوال‬
Find the convolution of 𝑥(𝑡) and ℎ(𝑡).
‫ جو ڪنووليوشن لهو‬ℎ(𝑡) ‫)𝑡(𝑥 ۽‬
x(t )  2 cos(2t )
e  t
h(t )  
0
for t  0 

elsewhere




0
0
y t    e  e j 2 t  d   e  e  j 2 t  d


0
0
y t    e  e j 2t e  j 2 d   e  e  j 2t e j 2 d

y t    h xt   d

y t   2  e u   cos2 t   d

y t   e
y t   2  e cos2 t   d

 e j 2 t    e  j 2 t   
y t   2  e 
d
2


0
y t   e
j 2t


e
0
 1 j 2 
d  e
 j 2t

0
0
j 2t

e
0
0


y t   e j 2t  e  e  j 2 d  e  j 2t  e  e j 2 d



0
h   e  u  
xt     2 cos2 t   


y t    e  e j 2 t    e  j 2 t   d
e
for t  0
ht   

0
elsewhere


xt   2 cos2t 
t

 1 j 2 
e
d

0
 1 j 2 
d  e
 j 2t

 1 j 2 
e
d

0
y t   e j 2t
y t   e j 2t
y t   e j 2t
y t   e j 2t






1
1
e  1 j 2  0  e  j 2t
e  1 j 2 
 1  j 2 
 1  j 2 
1
1
e   e 0  e  j 2t
e   e 0
 1  j 2 
 1  j 2 
1
1
0  1  e  j 2t
0  1
 1  j 2 
 1  j 2 
1
1
 e  j 2t
1  j 2 
1  j 2 


1
1






1
1
j 2t
 j 2t
1 j 2 
1 j 2 
y t   e 
e
e
e






1

j
2

1

j
2









1
1
y t   e j 2t 
e 0   e  j 2t 
e0 
2
2
 1  4

 1  4




1
1
 j 2t 
y t   e j 2t 

e



2
2
1

4

1

4





  e j 2t  e  j 2t 
1
1
j 2t
 j 2t

y t  
e
e

2
2
2
2
1  4
1  4  

2
y t  
cos2t 
2
1  4




0
From Exercise ‫مشق مان‬

Find and sketch
convolution of the
function x(t) with itself.
for  3  t  3
elsewhere
1
xt   
0
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-10
-8
-6
-4
-2
0
2
4
6
8
10
Convolution Exercise
xt   xt  
1
3
 x xt   d
t 3
0.9
0.8
0.7
0.6
xt   xt  
3
 1d
t 3
xt   xt   
0.5
0.4
0.3
0.2
3
t 3
xt   xt   3   t  3
xt   xt   3  t  3
xt   xt   6  t
0.1
0
-10
-8
-6
-4
-2
3 3 t
0
2
4
3
6− 𝑡
𝑥 𝑡 ∗𝑥 𝑡 =ቊ
0
6
8
3t
𝑡≤6
𝑡&gt;6
10
𝑡 &lt; −6
−6 ≤ 𝑡 &lt; 0
𝑡=0
0&lt;𝑡≤6
𝑡&gt;6
𝑛𝑜 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
𝑛𝑜 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
𝑡=0
𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
൞0 &lt; 𝑡 ≤ 6 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
𝑡 &gt;6
𝑛𝑜 𝑜𝑣𝑒𝑟𝑙𝑎𝑝
6
5
4
3
2
1
0
-10
-8
-6
-4
-2
0
t
2
4
6
8
10
Properties of Convolution

Commutative

Associative
f t * g t   g t * f t 
f t *g t * ht    f t * g t * ht 

Distributive
f t *g t   ht   f t * g t   f t * ht 
Discrete-time convolution
xn   1
hn   3
1
4
3
1
1  2  3
1
2
4
2
3
yn 
yn  xn* hn  ?
yn 
2

 xmhn  m
n  
M 1
 xmhn  m
n   ( L 1)
4
3
x[n]
2
1
0
-1
-2
-3
-4
-3
-2
-1
1
0
time index n
2
3
4
5
4
x[n]
2
0
-2
4
h[n]
2
0
-2
h[-n]
4
2
0
-2
-5
-4
-3
-2
-1
0
1
time index (n)
2
3
4
5
xn   1
2
3
h n   4
2
1




1
4
3
 3  2 1
1
2
3
x[m] = {0 1 2 -3 1 4 3 -1 2}
h[-m] = {4 2 1 -3 -2 -1 3 0 0}
Sum = 0&times;4+1&times;2+2&times;1-3&times;-3+1&times;-2+4&times;-1+3&times;31&times;0+2&times;0
Sum=0+2+2+9-2-4+9-0+0=16
Impulse Response

System’s zero-state
response to a unit
impulse applied at
t=0.
Represented by h(t)
Unit Impulse
1
0.8
System
amplitude

0.6
0.4
0.2
0
-2
-1.5
-1
-0.5
0
0.5
time (seconds)
 t 
1
1.5
2
h(t)
Causality Revisited
ℎ(𝑡) = 0 𝑓𝑜𝑟 𝑡 &lt; 0
 ℎ(𝑡 − 𝜏) = 0 𝑓𝑜𝑟 𝑡 − 𝜏 &lt; 0
 ℎ(𝑡 − 𝜏) = 0 𝑓𝑜𝑟 − 𝜏 &lt; −𝑡
 ℎ 𝑡 − 𝜏 = 0 𝑓𝑜𝑟 𝜏 &gt; 𝑡

y t  





x

h
t


d



y t  
t
 x ht   d

Function Approximation

A series of basis signals to
approximate a complex
signal over an approximation
interval
N
xˆ t    X nn t 
n 1
A
DC value (constant)
 A constant slope (ramp)

Output for each basis signal
is computed separately
 Sum
of individual Outputs
estimates the output
corresponding to the complex
input
 This is called superposition
 System characteristics
Approximation Criteria
Minimising the:
 maximum error

area under the error
Energy of the error signal
over the approximation
interval

minmax xˆ t   xt  
t2




min xˆ t   xt  dt 


t1

t2




2
min xˆ t   xt  dt 


t1

Example

The approximation interval is from 0 to 2
5
x(t)
4
t2+1
3
2
1
0
-1
-0.5
0
0.5
1
1.5
time (seconds)
2
2.5
3
x(t)
5
phi1(t)
0
-1
-0.5
0
0.5
1
1.5
2
2.5
3
-0.5
0
0.5
1
1.5
2
2.5
3
-0.5
0
0.5
1
1.5
2
2.5
3
1
phi2(t)
0.5
0
-1
2
1
0
-1
time (seconds)
N
xˆ t    X nn t 
For N=1
n 1
xˆt   X 11 t 
1   X 11 t   t  1 dt
t2
2
2
t1
2
1   X 1  t  1 dt
2
2
0
28
206
1  2 X  X 1 
3
15
2
1
d1
28
 4 X1 
dX1
3
d 1
0
dX1
28
4 X1 
0
3
28
4 X1 
3
28 7
X1 

12 3
128
1 
45
Finding X1 that gives the least error
e1
15
10
5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
3
3.5
4
4.5
-3
de1/dX1
x 10
X=2.333
Y=0.00
5
0
-5
0
0.5
1
1.5
2
2.5
X1
For N=2
N
xˆ t    X nn t 
n 1
xˆt   X11 t   X 22 t 
 2   X 11 t   X 22 t   t  1 dt
t2
2
2
t1
 2   X 1  X 2t  t  1 dt
2
2
2
0
8 2
28
 2  2 X  X 2  4 X 1 X 2  X 1  12 X 2  12
3
3
2
1
d 2
16
 4 X 1  X 2  12
dX 2
3
d 2
28
 4 X1  4 X 2 
dX1
3
1
X1 
3
X2  2
8
2 
45
Finality of coefficients

The coefficient X1 also changes
7
X1 
3



1
X1 
3
This doesn’t demonstrate finality of coefficients
Need to recompute the already computed coefs
Orthogonal basis signals can assure finality of
coefficients
Mutually Orthogonal basis signals
phi1(t)
1
0.5
0
-1
-0.5
0
0.5
-0.5
0
0.5
1
1.5
2
2.5
3
1
1.5
time (seconds)
2
2.5
3
phi2(t)
1
0
-1
-1
 2   X 11 t   X 22 t   t  1 dt
t2
2
2
t1
 2   X 1  X 2 t  1  t  1 dt
t2
2
2
t1
d 2
28
 4 X1 
0
dX1
3
d 2
4
8
 X2   0
dX 2 3
3
7
X1 
3
X2  2
Orthogonality

Signals &Oslash;1(t) and &Oslash;2 (t)
are said to be orthogonal to each other if:
n
t n t m t dt   0
1
t2
n  0
nm
nm

Generalised Fourier
Series is a weighted
sum of orthogonal basis
signals that approximates a signal over the
time interval t1&lt; t &lt; t2
by minimising the integral square error εN
over the interval
t2
 N   xt   xˆ t 2 dt
t1
 N   x 2 t   2 xt xˆ t   xˆ 2 t dt
t2
t1
N
2
N
N


 
 
 2

   x t   2 xt  X nn t    X nn t  dt
 n 1
  n 1
 
t1 


N
 2
N
 N
 N

   x t   2 xt  X nn t    X nn t   X mm t  dt
 n 1
  n 1
  m1

t1 
t2
t2
t2
N
t2
t1
n 1
t1
N
M
t2
 N   x 2 t dt  2 X n  xt n t dt  X n X m  n t m t dt
n 1 m 1
t1
t2
N
t2
t1
n 1
t1
N
M
t2
 N   x 2 t dt  2 X n  xt n t dt  X n X m  n t m t dt
n 1 m 1
t2


2
2
 N   x t dt    2 X n  xt n t dt  X n n 
n 1 

t1
t1

t2
N
1
 N   x t dt   n 
n 1
 n
t1
t2
N
2
1
  n 
n 1
 n
N

t xt n t dt
1

t2

t xt n t dt  X n 
1

t2
2
2
t1
t2
 xt  t dt
1
Xn 
n
n
t1
1
 N   x t dt   n 
n 1
 n
t1
t2
N
2
t2
N
 N   xt  dt   n X n
2
2
2
n 1
t1
X1 

t xt n t dt
1

t2
1
t2
 xt  t dt
1 t
1
1
X2 
1
2
t2
 xt  t dt
2
t1
t2
N
 N   xt  dt   n X n
2
n 1
t1
 N  0 as N  
t2
 xt 
t1
2

dt   n X n
n 1
2
2

Parseval’s theorem
 The
total energy
contained in 𝑁 basis
signals approach the
total energy in the signal
𝑥(𝑡) when 𝑁 → ∞
The Test Question

For the system equations shown below, describe the
characteristics of the corresponding systems. Are they causal,
linear, and/or time-invariant? Give sound reasons.
1






y
n

2
x
n

x
n

1

xn  2
1.
2
1
2. yn  2 xn  nxn  1  xn  3
2
1
2








y
n

2
x
n

x
n

2

xn  3
3.
2
1
4. yn  2 xn  xn  1  2n xn  1
n
2
5. yn  2 xn  xn  1  xn  1
2
1.
System 1



Causal, b/c the output doesn’t depend on the future
inputs
Time invariant, b/c the input or output coefficients
aren’t function of the time index n
Linear, b/c all the input and output coefficients are
linear
```