Course Outline (Tentative). Fundamental Concepts of Signals and Systems Signals Systems Linear Time-Invariant (LTI) Systems Convolution integral and sum Properties of LTI Systems … Fourier Series Response to complex exponentials Harmonically related complex exponentials … - PowerPoint PPT Presentation
Course Outline Course Outline (Tentative) (Tentative) Fundamental Concepts of Signals and Systems Signals Systems Linear Time-Invariant (LTI) Systems Convolution integral and sum Properties of LTI Systems … Fourier Series Response to complex exponentials Harmonically related complex exponentials … Fourier Integral Fourier Transform & Properties … Modulation (An application example) Discrete-Time Frequency Domain Methods DT Fourier Series DT Fourier Transform Sampling Theorem Laplace Transform Z Transform
Discrete-Time Frequency Domain Methods DT Fourier Series DT Fourier Transform Sampling Theorem
Laplace Transform Z Transform
Chapter IIIChapter IIIFourier Series
Fourier Series Fourier Series Representation of Periodic Representation of Periodic SignalsSignals
The objective is to represent complicated signals as linear combination of basic functions, i.e.,
so that if the response of the LTI system to φk(t) is known, then the response to x(t) can be expressed as the weighted sum of these responses. Try to select basic functions φk(t) such that the response to φk(t) is
kφk(t)
φk(t) are the eigenfunctions of LTI systems
k are the eigenvalues of LTI systems
Hence, the output is:
( ) ( )k kk
x t a t
( ) ( )k k kk
y t a t
Response of LTI Systems to Complex Exponentials
Why complex exponential? The most basic periodic signal with a well-defined frequency
Complex exponentials are eigenfunctions of LTI systems Response of an LTI system to a complex exponential is the same complex
exponential with only change in amplitude CT: est H(s) est DT: zn H(z) zn
Response is scaled version of the input with H(s) or H(z)
H(s) and H(z) are complex amplitude factors as functions of complex variables s and z
Response of LTI Systems to Complex Exponentials
To find H(s) consider a CT LTI system with h(t). For x(t)= est
( - ) -( ) ( ) ( - ) ( ) ( )s t st sy t h x t d h e d e h e d
For a DT LTI system with h[n], for x[n]= zn ( ) [ ] k
k
H z h k z
(show as an exercise!)
y(t) = H(s) est , where
-( ) ( ) sH s h e d
Response of LTI Systems to Response of LTI Systems to Complex ExponentialsComplex Exponentials
Hence,
Bottomline: H is known (depends on impulse response)
So, if we can express x as a linear combination of complex exponentials (i.e., find ak), we can write the output in terms of a and H! (No convolution)
We mostly use s=jω and z = ejω
( ) ( ) ( )
[ ] [ ] ( )
k ks t s tk k k
k k
n nk k k k k
k k
x t a e y t a H s e
x n a z y n a H z z
Fourier Series of CT Periodic Fourier Series of CT Periodic SignalsSignals
Consider a periodic signal,
Then, the fundamental frequency:
the fundamental period: minimum positive nonzero value of T
Two basic periodic signals:
sinusoid : periodic complex exponential :
tTtxtx for )()(2
ooT
ttx 0cos)( tjetx 0)(
Both periodic with and fundamental frequency of ω0o
oT2
Fourier Series of CT Periodic Fourier Series of CT Periodic SignalsSignals
The set of harmonically related complex exponentials:
Each of φk is periodic with fundamental frequency that is an integer multiple of ω0. All have a common period T0.
... 2, 1, 0,k ; )( 0 tjkk et
Now, take sk=kjω0
k
tjkkeatx 0)(
Periodic with T0
k=0 DC or constant term
is the Fourier series representation of x(t).
1st harmonic (fundamental component) periodic with T0
1k
2nd harmonic with fundamental period
2k20T
Need to find ak !!
Fourier Series of CT Periodic Fourier Series of CT Periodic SignalsSignalsExampleExample
Consider a periodic signal
3
3
2)(k
tjkkeatx
31 ,
21 ,
41 ,1 3322110 aaaaaaa
)(31)(
21)(
411)( 664422 tjtjtjtjtjtj eeeeeetx
1 2( ) 1 cos 2 cos 4 cos 62 3
x t t t t
Fourier Series of CT Periodic Fourier Series of CT Periodic SignalsSignals
If x(t) is real x*(t)=x(t)
0
0
* *
*
( ) ( ) jkw tk
k
jkw tk
k
x t x t a e
a e
kk aa * *
k ka a and for real periodic signals
Alternative form of Fourier Alternative form of Fourier SeriesSeries
Another form:
][)( 00
10
tjkk
k
tjkk eaeaatx
][ 00
10
tjkk
k
tjkk eaeaa
1
00Re2
k
tjkkeaa
1
)(0
0Re2)(k
tkjk
keAatx
1
00 )cos(2k
kk tkAa
kjkk eAaLet
k k kLet a B jC
1
000 ]sincos[2)(k
kk tkCtkBatx
Determination of Fourier Series Determination of Fourier Series of CT Periodic Signalsof CT Periodic Signals
Assume that a given periodic signal x(t) has a FS representation Find FS coefficients, ak of:
k
tjkkeatx 0)(
tjne 0Multiply both sides with and integrate over one period (from 0 to T=2/ω0)
T
k
tjntjkk
Ttjn eeadtetx
00
000)(
k
Ttnkj
k
Ttjn dteadtetx
0
)(
0
00)(
T T T
tnkj tdtnkjtdtnkdte0 0 0
00)( )sin()cos(0
Determination of Fourier Series of Determination of Fourier Series of CT Periodic SignalsCT Periodic Signals
0( )
0
, 0,
Tj k n t T k n
e dtk n
T
tjnn dtetx
Ta
0
0)(1
we can take integral over any interval of length T
k
tjkwkeatx 0)(
T
tjkk dtetx
Ta 0)(1
synthesis equation analysis equation
ak : FS coefficients, spectral coefficients
T
dttxT
a )(10a0 : DC component , average of signal
Determination of Fourier Series Determination of Fourier Series of CT Periodic Signalsof CT Periodic SignalsExampleExample
2 ,0
,1)(
1
1
TtT
Tttx
Consider the periodic square wave with fundamental period T (ω0=2T)
Find FS coefficients!
x(t)
-T -T/2 -T1 t- - -- - -
T1 T/2 T
T
tjkk dtetx
Ta 0)(1
1
1
10
21 T
T TTdt
Ta (average, DC
value!)
Determination of Fourier Series Determination of Fourier Series of CT Periodic Signalsof CT Periodic SignalsExampleExample
1
1
0
1
1
0
0
1.11 ,0 T
T
tjkT
T
ktjk e
Tjkdte
TakFor
jee
Tkee
Tjk
TjkTjkTjkTjk
221 1010
1010
00
TkTk
0
10 )sin(2
0kfor )sin(2 100
kTkaTFor k
Convergence of Fourier Convergence of Fourier SeriesSeries
FS expansion is possible if x(t) is a periodic function satisfying Dirichlet conditions:
T
dttx )(
C2) x(t) has a finite number of maxima and minima within any period.
C3) x(t) has a finite number of discontinuities within a period.
C1) Over any period x(t) is absolutely integrable, i.e.,
Convergence of Fourier Convergence of Fourier SeriesSeriesExamplesExamples
1) period(with signal periodic as 1t0 ,1)( )1 t
tx
x(t)
- - --1-2 1 2
- - -t
Not absolutely integrable over period!
Violates C1!
Convergence of Fourier Convergence of Fourier SeriesSeriesExampleExample
1 period with 1t0 ),2sin()( )2 t
tx
- meets C1 !- violates C2!
1 2
- - -
3)
violates C3!
- - -
x(t)
Convergence of Fourier Convergence of Fourier SeriesSeries
Complex exponentials are continuous for . Henceis continuous as well.
t k
tjkwkeatx 0)(
What if x(t) has discontinuity at t=t0 ?
If Dirichlet conditions are satisfied x(t) is equal toat every continuity point of x(t).
k
tjkwkeatx 0)(
k
tjkwkeatx 0)( At discontinuities, equals to the average of the values
on
either side of the discontinuity, i.e., ,)( 0
tx )( 0tx
if x(t) is discontinuous at t=t0,
)()(21
000 txtxea
k
tjkwk
Remarks:Remarks:
Properties of Fourier Properties of Fourier SeriesSeries
1) Even and Odd Functions: if a function x(t) is even, x(t)=x(-t)
k
tjkk
k
tjkk
k
tjkk eaeaeatxtx 000)()(
ak=a-
k
Hence,
1
001
0 cos2 )( 00
kk
k
tjktjkk tkaaeeaatx
if x(t) is odd
ak=-a-k , a0=0
1
0sin2)(k
k tkajtx
Properties of Fourier Properties of Fourier SeriesSeries
2) Time Shifting: if x(t) has ak as FS coefficients, i.e., ( )FS
kx t a
k
tjkkeatx 0)(
k
tjktjkk
k
ttjkk eeaeattx 00000 )(
0 )(
00)( 0tjk
k
FS
eattx
Properties of Fourier Properties of Fourier SeriesSeries
3. Time Reversal: ( ) , then (- )FS FS
k kif x t a x t a (prove it as exercise !)
4. Differentiation:
0( ) ( ) , then
FS FS
k kdx tif x t a jk a
dt
k
tjkkea
dtd
dttdx
0)( … (prove it as exercise !)
Consider the cascade (series) connection of two LTI systems, whose impulse responses are
and
Evaluate the output signal y[n] corresponding to the input signal
