Lecture 6-9 (Signal & Spectra) Page 1 P.Kanungo & G.Routray Department of Electronics & Telecommunication Engineering C.V.Raman College of Engineering-752054 Lecture Note (Lecture-6-9) After reading this lesson, you will learn about Fourier series expansion (Trigonometric and exponential) Properties of Fourier Series Response of a linear system Normalized power in a Fourier expansion Power spectral density Effect of transfer function on PSD 1. FOURIER SERIES EXPANSION: French Mathematician J.B.J. Fourier found that any arbitrary periodic signal can be represented with an infinite series of sinusoids with fundamental frequency and harmonically related frequency (ω 0 =fundamental frequency, n ω 0 = n th harmonic frequency, where n=1, 2, 3, 4… N). Fourier analysis is used to analysis the steady state response of a network and frequency analysis of signals. Periodic Function: A function is said to be periodic with a time period ‘T’ if it satisfies the relation f(t±T) =f(t). A numbers of such periodic signals are shown in the fig. below. Thus a periodic function repeats itself after every T seconds. -2T -T +T Fig.1 Fig.2 There are two forms of Fourier series (i) Trigonometric Fourier series (ii) Exponential Fourier Series F(t) F(t) T -T
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Lecture 6-9 (Signal & Spectra) Page 1P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
Lecture Note(Lecture-6-9)
After reading this lesson, you will learn about Fourier series expansion (Trigonometric and exponential) Properties of Fourier Series Response of a linear system Normalized power in a Fourier expansion Power spectral density Effect of transfer function on PSD
1. FOURIER SERIES EXPANSION:
French Mathematician J.B.J. Fourier found that any arbitrary periodic signal can be represented with an infinite series of sinusoids with fundamental frequency and harmonically related frequency (ω0=fundamental frequency, n ω0 = nth harmonic frequency, where n=1, 2, 3, 4… N).Fourier analysis is used to analysis the steady state response of a network and frequency analysis of signals.
Periodic Function:
A function is said to be periodic with a time period ‘T’ if it satisfies the relation f(t±T) =f(t). A numbers of such periodic signals are shown in the fig. below. Thus a periodic function repeats itself after every T seconds.
-2T -T +T
Fig.1 Fig.2
There are two forms of Fourier series
(i) Trigonometric Fourier series(ii) Exponential Fourier Series
F(t)F(t)
T-T
Lecture 6-9 (Signal & Spectra) Page 2P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
Trigonometric Fourier series:
The trigonometric Fourier series for an arbitrary periodic function f(t) is given by
)1(]sincos[)( 001
0
tnbtnaatf nn
n
Where an’s and bn’s are known as the Fourier series coefficients.
Key roots (Formulae)
''0sin0
0 nallfordttnT
0,00
0 mwhendttmCosT
nmallfordttmCostnSinT
,0.. 0
0
0
nmfordttmSintnSinT
0.. 0
0
0
nmfordttmCostnCosT
0.. 0
0
0
02.0
0
2
nforTdttnSin
T
2.0
0
2 TdttnCosT
Evaluation of Fourier series co-efficient:
This involves two operations
(i) Evaluation of coefficients a0, an and bn
(ii) Truncating of infinite series after a finite number of terms so that f(t) is represented within allowable error.
Lecture 6-9 (Signal & Spectra) Page 3P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
Evaluation of a0:
]sincos[)( 001
0 tnbtnaatf nn
n
Integrating both sides within one time period
dttnbtnadtadttfT T T
nn
n .]sincos[).(0 0 0
001
0
NB: Integration of a sinusoidal function within limit 0-T, (where T is the time period of the given function) is zero.
)2(..........)(1
0)(
0).(
0
0
0
0
00
0
T
T
TT
dttfT
a
Tatdtf
tadttf
a0 is known as the average value of the function or D.C. value of the function.
Evaluation of an
Multiplying tn 0cos in Equation (1) and integrating both sides over a period 0-T, we will get
T T T
nn
n tnbtnatnCosdttnCosadttnCostf0 0 0
001
0000 ]sincos[...)(
(3) .)(2
020.)(
cos.sincos.cos...)(
0
0
0
0
01 0
001
0
0
00
0
0
tdtnCostfTa
TatdtnCostf
dtntnbdttntnadttnCosatdtnCostf
T
n
n
T
n
T
nn
T
o
n
TT
Evaluation of bn
Multiplying tnSin 0 in Equation (1) and integrating both sides over a period 0-T, we will get
Lecture 6-9 (Signal & Spectra) Page 4P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
T T T
nn
n tnbtnatnSindttnSinadttnSintf0 0 0
001
0000 ]sincos[...)(
(4) .)(2
020.)(
sin.sinsin.cos...)(
0
0
0
0
01 0
001
0
0
00
0
0
tdtnSintfTb
TbtdtnSintf
dtntnbdttntnadttnSinatdtnSintf
T
n
n
T
n
T
nn
T
o
n
TT
Types of Symmetry:
(i) Even symmetry(ii) Odd symmetry(iii) Half wave symmetry(i) Even symmetry: If a function satisfies the relation f(t)=f(-t) it is said to be even
symmetryExamples:
(ii) Odd symmetry: If a function satisfies the relation f(t)=-f-(t) then it is an odd symmetry function.
f(t) f(t)
f(t)
Cos θ is an even symmetry function
T T
T-T
-T-T
Lecture 6-9 (Signal & Spectra) Page 5P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
Examples:
(iii) Half wave symmetry: If a function satisfies the relation f(t)=-f(t+T/2), then it is a half wave symmetry functionExample:
Evaluation of Fourier Coefficients in Symmetry Conditions
(i) Even symmetry: Evaluation of a0:
2/
0
0
2/
2/
2/0
0 )(1
)(1
)(1
)(1 T
T
T
T
T
dttfT
dttfT
dttfT
dttfT
a
Replace t by –t in the first term
2/
0
0
2/
0 )(1
))((1 T
T
dttfT
dttfT
a
f(t) is even symmetry function, so f(t)=f(-t)
f(t)f(t)
Sin θ is an odd symmetry function
T
T-T-T
-T/4T/4
3T/4
f(t)
f(t)
TT/2
-T-T/2
Fig: Example os another half wave symmetry
The amplitude is taken as A
Take t=T/4
f(t)=f(T/4)=A
f(T/4+T/2)=f(3T/4)=-A
Similarly
f(t-T/2)=f(-T/4)=-A
So Sine functions are half wave symmetry
Lecture 6-9 (Signal & Spectra) Page 6P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
2/
0
2/
0
2/
0
0 )(2
)(1
)(1 TTT
dttfT
dttfT
dttfT
a (5)
Evaluation of an:
2/
0
0
0
2/
0
2/
2/
0
0
0
cos)(2
cos)(2
cos)(2
cos)(2
T
T
T
T
T
n
tdtnwtfT
tdtnwtfT
tdtnwtfT
tdtnwtfT
a
Replace t with –t in the first term
2/
0
0
0
2/
0 cos)(2
))(cos()(2 T
T
tdtnwtfT
dttnwtfT
f(t) is even symmetry function, so f(t)=f(-t)
(6) cos)(4
cos)(2
))(cos()(2
2/
0
0
2/
0
0
2/
0
0
T
TT
tdtnwtfT
tdtnwtfT
dttnwtfT
Evaluation of bn:
2/
0
0
0
2/
0
2/
2/
0
0
0
sin)(2
sin)(2
sin)(2
sin)(2
T
T
T
T
T
n
tdtnwtfT
tdtnwtfT
tdtnwtfT
tdtnwtfT
b
Replace t with –t in the first term
2/
0
0
0
2/
0 sin)(2
))(sin()(2 T
T
tdtnwtfT
dttnwtfT
f(t) is even symmetry function, so f(t)=f(-t)
(7) 0
sin)(2
))(sin()(2 2/
0
0
2/
0
0
TT
tdtnwtfT
dttnwtfT
(ii) Odd symmetry: Evaluation of a0:
2/
0
0
2/
2/
2/0
0 )(1
)(1
)(1
)(1 T
T
T
T
T
dttfT
dttfT
dttfT
dttfT
a
Replace t by –t in the first term
Lecture 6-9 (Signal & Spectra) Page 7P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
2/
0
0
2/
0 )(1
))((1 T
T
dttfT
dttfT
a
f(t) is odd symmetry function, so f(t)=-f(-t)
0)(1
)(1 2/
0
2/
0
0 TT
dttfT
dttfT
a (8)
Evaluation of an:
2/
0
0
0
2/
0
2/
2/
0
0
0
cos)(2
cos)(2
cos)(2
cos)(2
T
T
T
T
T
n
tdtnwtfT
tdtnwtfT
tdtnwtfT
tdtnwtfT
a
Replace t with –t in the first term
2/
0
0
0
2/
0 cos)(2
))(cos()(2 T
T
tdtnwtfT
dttnwtfT
f(t) is odd symmetry function, so f(t)=-f(-t)
(9) 0
cos)(2
))(cos()(2 2/
0
0
0
2/
0
T
T
tdtnwtfT
dttnwtfT
Evaluation of bn:
2/
0
0
0
2/
0
2/
2/
0
0
0
sin)(2
sin)(2
sin)(2
sin)(2
T
T
T
T
T
n
tdtnwtfT
tdtnwtfT
tdtnwtfT
tdtnwtfT
b
Replace t with –t in the first term
2/
0
0
0
2/
0 sin)(2
))(sin()(2 T
T
tdtnwtfT
dttnwtfT
f(t) is odd symmetry function, so f(t)=-f(-t)
(10) sin)(4
sin)(2
))(sin()(2
2/
0
0
2/
0
0
0
2/
0
T
T
T
tdtnwtfT
tdtnwtfT
dttnwtfT
Lecture 6-9 (Signal & Spectra) Page 8P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
(iii) Half wave symmetry:
Evaluation of a0:
2/
0
0
2/
2/
2/0
0 )(1
)(1
)(1
)(1 T
T
T
T
T
dttfT
dttfT
dttfT
dttfT
a
Replace (t+T/2) by λ in the first term ;t=λ-T/2 and dt=dλ
TT
dttfT
dTfT
a0
2/
0
0 )(1
)2/(1
f(t) is half wave symmetry function, so f(λ-T/2)=-f(λ)
0)(1
)(1 2/
0
2/
0
0 TT
dttfT
dtfT
a (11)
Evaluation of an:
2/
0
0
0
2/
0
2/
2/
0
0
0
cos)(2
cos)(2
cos)(2
cos)(2
T
T
T
T
T
n
tdtnwtfT
tdtnwtfT
tdtnwtfT
tdtnwtfT
a
Replace (t+T/2) by λ in the first term ;t=λ-T/2 and dt=dλ
2/
0
0
2/
0
0 cos)(2
))2/(cos()2/(2 TT
tdtnwtfT
dTnwTfT
f(t) is half wave symmetry function, so f(λ-T/2)=-f(λ)
2/
0
0
2/
0
00 cos)(2
)2/cos()(2 TT
tdtnwtfT
dTnwnwfT
2/
0
0
2/
0
0 cos)(2
)cos()(2 TT
tdtnwtfT
dnnwfT
oddnwhennw
evennwhennwnnw
0
00 cos
cos)cos(
Lecture 6-9 (Signal & Spectra) Page 9P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
So
2/
0
0cos)(4
0T
n oddnfortdtnwtfT
evennfor
a (12)
Evaluation of bn:
2/
0
0
0
2/
0
2/
2/
0
0
0
sin)(2
sin)(2
sin)(2
sin)(2
T
T
T
T
T
n
tdtnwtfT
tdtnwtfT
tdtnwtfT
tdtnwtfT
b
Replace (t+T/2) by λ in the first term ;t=λ-T/2 and dt=dλ
2/
0
0
2/
0
0 sin)(2
))2/(sin()2/(2 TT
tdtnwtfT
dTnwTfT
f(t) is half wave symmetry function, so f(λ-T/2)=-f(λ)
2/
0
0
2/
0
00 sin)(2
)2/sin()(2 TT
tdtnwtfT
dTnwnwfT
2/
0
0
2/
0
0 sin)(2
)sin()(2 TT
tdtnwtfT
dnnwfT
oddnwhennw
evennwhennwnnw
0
00 sin
sin)sin(
So
2/
0
0sin)(4
0T
n oddnfortdtnwtfT
evennfor
b (13)
The Fourier series expansion for a function having half wave symmetry contains only the odd harmonics frequencies.SUMMARY: (i) Even Symmetry
T
dttfT
a0
0 )(2
cos)(4 2/
0
0T
n tdtnwtfT
a
bn=0
(ii) Odd Symmetry
00 a
Lecture 6-9 (Signal & Spectra) Page 10P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
0na
sin)(4 2/
0
0T
n tdtnwtfT
b
(iii) Half wave symmetry
00 a
evennforba nn 0,0
oddnforsin)(4
oddnforcos)(4
2/
0
0
2/
0
0
T
n
T
n
tdtnwtfT
b
tdtnwtfT
a
If any function contains (i) Even symmetry and half wave symmetry
oddnforcos)(8
oddnforcos)(4
0
00
4/
0
0
2/
0
0
0
T
T
n
n
n
tdtnwtfT
tdtnwtfT
a
evennfora
ba
(ii) Odd symmetry and half wave symmetry
oddnforsin)(8
oddnforsin)(4
0
00
4/
0
0
2/
0
0
0
T
T
n
n
n
tdtnwtfT
tdtnwtfT
b
evennforb
aa
Exponential Fourier series:
Let v(t) is a periodic signal then according to Fourier series
1
00 sincos)(n
non tnwbtnwaatv
Lecture 6-9 (Signal & Spectra) Page 11P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
10
10
22
22
00
0000
n
nntjnwnntjnw
n
tjnwtjnw
n
tjnwtjnw
n
jbae
jbaea
j
eeb
eeaa
Let nnn v
jba
2, and
nnnn vv
jba
2
Where vn* is the complex conjugate of vn, and a0=v0.
Then
n
tjnwn
n
tjnwn
tjnwn evevevvtv 000
10)(
Where vn is the exponential Fourier series Coefficients.
Ttjnw
T
TTnn
n
dtetfT
dttnwjtnwtfT
tdtnwtfT
jtdtnwtfT
jbav
0
0
00
0
0
0
0
0)(1
]sin)[cos(1
sin)(2
cos)(2
2
1
2
We can represent v(t) as
1 00
2cos)(
nnn T
ntCCtv
Where C0,Cn and Φn are related to a0,an,bn as
n
nn
nnn
a
b
baC
aC
1
22
00
tan
Cn is also known as spectral amplitude i.e Cn is the amplitude of the spectral
component nn tnfC 02cos at frequency nf0.
Fourier series Frequency Spectrum: The plot of the amplitude of frequency component vs the frequency known as the discrete frequency spectrum or line spectrum. The frequency spectrum consists of discrete lines. The length of the line represents the amplitude of the corresponding frequency component.Phase Spectrum: The plot of the phase of the frequency component vs the frequency is known as phase spectrum. Phase spectrum is a odd function.
Lecture 6-9 (Signal & Spectra) Page 12P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
Sampling Function: The sampling function is defined as x
xxSa
sin)( . The function
is shown in the following figure.
2. FOURIER SERIES PROPOTIES:
If the Fourier series representation of v(t) given as
n
Tntjnevtv 0/2)( then the following
properties are satisfied by the signal(i) Time shift: Fourier series representation of v(t+τ), is
n
Tntjn
n
Ttnjn evevtv 00 /2/)(2 ')(
Where 0/2' Tnjnn evv
(ii) Time inversion: Fourier series representation of v(-t) is given by
n
Tntjn
n
Ttnjn evevtv 00 /2/)(2)(
i.e. the magnitude of vn remains constant, phase is shifted by 1800. In trigonometric representation an remains constant but bn becomes negative.
(iii) Time scaling:
n
Tntjn
n
Tatnjn evevatv '/2/)(2 00)(
Where T0’=T0/a . i.e. vn remains constant but shifts to a new frequency na/T0. If the signal is compressed in time domain (a>1) it is expanded in frequency domain; and if it is expanded in time domain (a<1) then compressed in frequency domain.
(iv) Time derivative:
n n
Tntjn
Tntjn
n
Tntjn evevTnj
dt
evd
dt
tdv00
0/2/2
0
/2
')/2()()(
-π 0 π 2π
Lecture 6-9 (Signal & Spectra) Page 13P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
Where vn’=j2πnvn/T0
(v) Integration:
n
Tntjn
n
Tntjn
n
tTntj
n
t
eveTnjvdtevdttv 000 /2/20
0
/2
0
')/2/()(
Where )/2/(' 0Tnjvv nn
3. APPLICATIONS OF FOURIER SERIES EXPANSION:
(i) Response of a Linear System: When a sinusoidal excitation is applied to a linear system the response of the system is similarly sinusoidal, i.e., sinusoidal waveform preserves the wave shape. The relationship of the response to the excitation is characterized by the relation of input -output amplitude and phase.Let the input to the linear system be the spectral component
tjwn
Tntjnni
nevevwtv 0/2),( (3i-1)
The output vo(t,wn) is related to the input vi(t,wn) by a complex transfer function
)(|)(|)( nwjnn ewHwH (3i-2)
The output is
)]([)(0 |)(||)(|),()(),( nnnn wtwj
nntjw
nwj
nninn evwHevewHwtvwHwtv (3i-3)
The physical input (vip(t)) is the sum of the spectral component and its complex conjugate. i.e.
)Re(2),( * tjwn
tjwn
tjwn
tjwn
tjwnnip
nnnnn evevevevevwtv (3i-4)
The corresponding physical output is
tjwnn
tjwnnnop
nn evwHevwHwtv *)()(),( (3i-5)
Since the output is +ve and real the two terms in (3i-5) must be complex conjugate. Hence H(wn)=H*(-wn) So |H(wn)|=|H(-wn)| and θ(wn)=- θ(-wn). i.e. |H(wn)| is an even function and θ(wn) is an odd function.
Hence the output of the system can be expressed as
FilterH(wn)vi(t,wn) Vo(t,wn)
Lecture 6-9 (Signal & Spectra) Page 14P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
)(2
cos|)(|)0()()(01
0/2
00
nnnn
nn
Tntjnn w
T
ntCwHCHevwHtv
(ii) Normalized Power in a Fourier Expansion:
Consider two terms of the Fourier series expansion (Fundamental and the first harmonics)
2
021
01
4cos
2cos)('
T
tC
T
tCtv
The normalized power S’ of v’(t) is
2/
2/
22
212
0
0
022
)]('[1
'T
T
CCdttv
TS
By extension the normalized power associated with the entire Fourier series is
1
2
1
22
01
22
0 222 n
n
n
n
n
n baa
CCS
N.B: The power and normalized power are associated with the real waveforms not with the complex waveforms.
For exponential Fourier series the normalized power is due to the product terms
*/2/2 00nnnn
Tntjn
Tntjn vvvvevev
Total normalized power is
n
nnvvS *
In complex representation, the power associated with a particular frequency nf0=n/T0 is not associated with the spectral component at nf0 and –nf0, rather the combination of the spectral component. Thus the power is
vnvn*+ v-nv-n
*=2 vnvn*
Lecture 6-9 (Signal & Spectra) Page 15P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
(iii) Power Spectral Density (PSD):
Normalized power dS(f) at the frequency f in a range df is dfdf
fdSfdS
)()(
dS(f)/df is called the normalized power spectral density G(f).
The power in the range df at f is G(f)df. The power in the range f1-f2 is
2
1
)()( 21
f
f
dffGfffS
And power in the range –f1to -f2 is
1
2
)()( 12
f
f
dffGfffS
-nf0 -3f0 -2f0 -f0 0 f0 2f0 3f0 nf0
Sn
|V0|2
|Vn|2
|V-n|2
|V-1|2 |V1|
2
A two sided power spectrum
-2f0 -f0 0 f0 2f0 3f0
|v1|2
|v0|2
Sf
The sum S(f) of the normalized power in all spectral components from f=-∞ to ∞
Lecture 6-9 (Signal & Spectra) Page 16P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
The quantities in above two equations have no physical significance but the total powers in the real frequency range f1-f2 have physical significance, and the power is given as
2
1
1
2
)()(|)||(| 21
f
f
f
f
dffGdffGfffS
To find the power spectral density, differentiate S(f). But in between the harmonics G(f)=0. So at harmonics G(f) gives an impulse of strength equal to the jump in S(f). Hence
n
n nffvfG )()( 0
2
(iv) Effect of Transfer function on PSD:
Let vi(t) is the input to a filter having psd Gi(f). If vin is the spectral amplitude of the input signal then
n
ini nffvfG )()( 0
2
Where
2/
2/
/2
0
0
0
0)(1
T
T
Tntjiin dtetv
Tv
Let the output is vo having spectral amplitude von, then the corresponding psd is
n
on nffvfG )()( 0
2
0
And
2/
2/
/2
0
0
0
0)(1
T
T
Tntjoon dtetv
Tv
If H(f) is the transfer function of the filter then the input and output spectral amplitudes are
related as von=H(f)vin ; Hence |von|2=|H(f)|2|vin|2
Substituting in the equation for G0(f) above we have G0(f)=Gi(f)|H(f)|2.Assignments: 1. Find the Fourier series expansion for the following wave forms(i)
-T 0 T t
T/2
Lecture 6-9 (Signal & Spectra) Page 17P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
(ii)
(iii)
(iv)
(v)
2. Determine the Fourier expansion for the following signals
a. 1)( ntnforetx nt
b. tttx 5.2coscos)(
c. tftx 02cos)( (full wave rectifier output)
d. tftftx 00 2cos2cos)( (half wave rectifier output)
3. Show that for real x(t)
-3T/4 -T/4 0 T/4 3T/4
v
-v
-T -T/2 0 T/2 T
A
-A
-T -T/2 0 T/2 T
1
-1
-6 -4 -2 0 2 4 6 8 10
-1
1
Lecture 6-9 (Signal & Spectra) Page 18P.Kanungo & G.Routray
Department of Electronics & Telecommunication EngineeringC.V.Raman College of Engineering-752054
tnfatxandtnfaa
tx none 000 2sin)(,2cos
2)(
Where xe(t) and xo(t) denote the even and odd parts of x(t)
2
)()()(
2
)()()(
txtxtxand
txtxtx oe
4. Let x(t) and y(t) be two periodic signals with period T0, and xn and yn denotes the Fourier series coefficients of these two signals. Show that
n
nn yxdttytxT
**
0
)()(1
5. Show that for all periodic physical signal that have finite power, the coefficients of the Fourier series expansion xn tend to zero as n .
6. A periodic triangular waveform v(t) is defined by
22
2)(
Tt
Tfor
T
ttv and v(t±T)=v(t)
Calculate the fraction of the normalized power of this waveform which is contained in its first three harmonics.
7. Find G(f) for the following voltagesa. An impulse train of strength I and period Tb. A pulse train of amplitude A, duration τ=I/A, and period T
8. Plot G(f) for a voltage source represented by an impulse train of strength I and period nT for n= 1,2,10, infinity. Comment on this limiting result.
9. Gi(f) is the power spectral density of a square wave voltage of peak-to-peak amplitude 1 and period 1. The square-wave is filtered by a low-pass RC filter with 3dB frequency 1. The output is taken across the capacitor
a. Calculate Gi(f)b. Find Go(f)
10. (a) A symmetrical square-wave of zero mean value, peak-to-peak voltage 1 volt, and period 1 sec is applied to an ideal low-pass filter. The filter has a transfer function |H(f)|=1/2 in the frequency range -3.5≤f≤3.5 Hz, and H(f)=0 elsewhere. Plot the power spectral density of the filter output(b)What is the normalized power of the input square wave? What is the normalized power of the filter output?
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