© Arizona State University EEE 203, Prof. A.

Papandreou-Suppappola

Signals and Systems I–EEE 203

Lecture Notes 2b
Importance of processing signals in the frequency domain

– Provides distribution of signal energy over a range of frequencies

– Enables efficient processing for linear time-invariant systems

frequency domain
y ptq “ xptq ‹ hptq ÐÝÝÝÝÝÝÝÝÝÑ Y pjΩq “ X pjΩq H pjΩq

– faster implementation in frequency by multiplication instead of convolution in time

– easier to design filters such as bandpass filter in figure

© Arizona State University EEE 203, Prof. A. Papandreou-Suppappola

1
A. Fourier transform definition (Sections 4.1-4.7)

– Used Fourier series for periodic signals; use continuous-time Fourier transform (FT) for aperiodic signals

– Definition of FT of signal xptq

ż8
X pjΩq “ xptq e´jΩt dt
´8

where Ω is continuous frequency in radians per second

– Inverse FT (IFT)
ż8
1
x p tq “ X pjΩq ejΩt dΩ
2π ´8

expansion of signal xptq in terms of complex sinusoids

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2
B. Convergence of Fourier transform

FT exists provided signal xptq satisfies Dirichlet conditions

ż8
a) xptq must be absolutely integrable over a period ñ |xptq| dt ă 8
´8

b) xptq must have a finite number of maxima and minima (extrema) over a finite interval

c) xptq must have a finite number of discontinuities over a finite interval and discontinuities must be finite

Example: xptq with no FT, xptq = e2 t uptq

ż8 ż8 ż8 8

|xptq| dt “ |e2 t uptq| dt “ e2 t dt “ 0.5 e2 t  “ 8

´8 ´8 0 0

Example: xptq with FT, xptq = e´2 t uptq

ż8 ż8 ż8 8

|xptq| dt “ |e´2 t uptq| dt “ e´2 t dt “ ´0.5 e´2 t  “ 0.5

0
´8 ´8 0

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C. Examples of FTs
´ ¯
(a) Rectangular pulse: xptq = 1.5 upt ` 4q ´ upt ´ 4q (not periodic)

x p tq
2

t
´4 ´2 2 4

To compute FT

ż8 ż4 4
´jΩt ´jΩt ´1.5 ´jΩt ´1.5 ´ ´jΩ4 jΩ4
¯
X pjΩq “ x p tq e dt “ 1.5 e dt “ e “ e ´e
´4 jΩ  jΩ
´8 ´4

ejΩ4 ´ e´jΩ4
ˆ ˙
3 3 sin p4 Ωq
“ “ sin p4 Ωq “ 12 “ 12 sincp4 Ωq, @Ω
Ω 2j Ω 4Ω

sin pΩq
Note: Continuous-time sinc function: sincpΩq =
Ω

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4
FT pair (equation)
´ ¯
FT sin p4 Ωq
xptq “ 1.5 upt ` 4q ´ upt ´ 4q ÐÑ X pjΩq “ 12 “ 12 sincp4 Ωq, @Ω
4Ω

To sketch X pjΩq, zero-crossing occur at 4 Ω = k π, k = ˘1, ˘2, . . .

– zero-crossings are the values of Ω for which X pjΩq = 0 ñ Ω = k π {4

&'(%) *$

!"#$%&'()%*#+,- $!
"

%
$! ! ! $!
# # "
" " "

© Arizona State University EEE 203, Prof. A. Papandreou-Suppappola

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FT pair (graph)

&'(%) *$
x p tq
2

!"#$%&'()%*#+,- $!
"
1

%
t #
$!
#
! ! $!
" " " "
´4 ´2 2 4
FT
Ð
Ñ

Observations:

– Finite duration in time Td = 8 and infinite bandwidth in frequency

where bandwidth the range of frequencies for which X pjΩq exists (ending frequency - starting frequency)

– Mainlobe: bandwidth between k = ˘1 zero-crossings

90% of energy of X pjΩq lies in the main lobe

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FT pair (general)
´ ¯
FT sin pT Ωq
x p tq “ A u p t ` T q ´ u p t ´ T q Ð
Ñ X pjΩq “ 2 A “ 2 A T sincpT Ωq, @Ω
Ω

'()&* $"%

,(+*

$! ! ! $! &
#" " # #
+ " " " "

./0",#'$ $" !"#$%&'()%*#+,- $!

FT "
Ð
Ñ

2π
Duration is time: 2 T Mainlobe width in frequency:
T

As T increases, duration in time increases but mainlobe bandwith decreases (inverse relationship)

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(b) Exponential signal: xptq = e´at uptq (also see Ex. 4.1 in book)

FT 1
xptq “ e´at uptq, a ą 0 Ð
Ñ X pjΩq “
a ` jΩ

To show FT computation

ż8 ż8 8
´1  1
X pjΩq “ e´at uptqe´jΩt dt “ e ´tpjΩ`aq
dt “ e ´tpjΩ`aq 
“ , aą0
0 a ` jΩ  a ` jΩ
0
´8

In general, FT X pjΩq = RpjΩq ejΦpjΩq is complex ñ plot magnitude RpjΩq and phase ΦpjΩq
ˇ ˇ ˇ ˇ ˇ ˇ dˆ ˙2 ˆ ˙2
ˇ 1 ˇ ˇ a ´ jΩ ˇ ˇ a ´ jΩ ˇ a ´ Ω
RpjΩq “ ˇˇ ˇ“ˇ ˇ“ˇ ˇ“ `
a ` jΩ ˇ ˇ pa ` jΩqpa ´ jΩq ˇ ˇ a2 ` Ω2 ˇ a2 ` Ω 2 a2 ` Ω 2
d
a2 ` Ω 2 1
“ “ ?
p a2 ` Ω 2 q 2 a2 ` Ω 2

´Ω { a 2 ` Ω 2
ˆ ˙
ΦpjΩq “ arctan “ ´arctanpΩ{aq
a {a 2 ` Ω 2

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Plot magnitude and phase

1
X pjΩq “ RpjΩq ejΦpjΩq, RpjΩq “ ? , ΦpjΩq “ ´arctanpΩ{aq
a2 ` Ω 2

magnitude spectrum RpjΩq = |X pjΩq| phase spectrum ΦpjΩq = ă

q X pjΩq


 







Figures from Example 4.1 in book ( with ω = Ω radians per second

?
Note: When magnitude drops by ?1 = 2
from its maximum 1{a, then obtain 3 dB bandwidth equal to 2 a
2 2
?
corresponds to ´20 log10 p 2{2q = 3 dB

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(c) Impulse function: xptq = δ ptq

ż8 ż8
1
δ ptqe´jΩt dt e´jΩ0
✿
✘
✘
X pjΩq “ “✘✘✘ δ ptq dt “ 1
´8 ´8

FT
ñ δ p tq Ð
Ñ 1, @Ω

Shifted impulse function xptq = δ pt ´ t0 q

ż8 ż8
X pjΩq “ δ pt ´ t0 qe´jΩt dt “ e´jΩ t0 δ pt ´ t0 q dt “ e´jΩ t0
´8 ´8

FT
ñ δ p t ´ t0 q Ð
Ñ e´jΩ t0 , @Ω

FT
(d) Complex exponential: ejΩ0t Ð
Ñ 2π δ pΩ ´ Ω0 q (derivation when cover FT properties)

Table 4.2 (book) of FT pairs (also Handout 2 in Module 2)

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D. FT of periodic signals

Recall Fourier series (FS) of a periodic signal xptq with fundamental frequency Ω0

ÿ
x p tq “ αk ejkΩ0 t
k

Obtain FT of periodic signal as

FT
ÿ ÿ
x p tq “ αk ejkΩ0t Ð
Ñ X pjΩ0q “ 2π αk δ pΩ ´ Ω0 kq
k k

FT
using FT pair ejΩ0 t Ð
Ñ 2π δ pΩ ´ Ω0 q and also that FT is a linear transformation

ñ to find FT of a periodic signal, find FS coefficients αk and fundamental frequency Ω0

© Arizona State University EEE 203, Prof. A. Papandreou-Suppappola

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Example: xptq = cos pΩ0tq with FS xptq = 0.5 ejΩ0 t ` 0.5 e´jΩ0t

Since α1 = α´1 = 0.5, and Ω0 is fundamental frequency, then FT

´ÿ ¯
FT
xptq “ cos pΩ0tq Ð
Ñ X pjΩ0q “ 2π αk δ pΩ ´ Ω0 kq “ π δ pΩ ´ Ω0 q ` π δ pΩ ` Ω0q
k

FT
For Ω = π {4, FT pair is cos ppπ {4qtq Ð
Ñ π δ pΩ ´ π {4q ` π δ pΩ ` π {4q, and plot shown below

π π
X pjΩq

0
´π {2 ´π {4 0 π {4 π {2
Ω

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E. Fourier transform properties (see Table 4.1)

FT FT
Considering TF pairs x1 ptq Ð
Ñ X1 pjΩq and x2 ptq Ð
Ñ X2 pjΩq

[P1] Linearity

FT
a x 1 p tq ` b x 2 p tq Ð
Ñ a X1 pjΩq ` b X2 pjΩq

ż
satisfied since X pjΩq = xptq e´jΩt dt and integration is linear

FT
Ñ e´jΩ t0 , then
Example: since δ pt ´ t0 q Ð

FT
Ñ 5 ` 3 ej2 Ω ´ e´j3 Ω
5 δ p tq ` 3 δ p t ` 2 q ´ 2 δ p t ´ 3 q Ð

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[P2] Time-shift : shift in time results in multiplication by complex sinusoid in frequency

FT
Ñ Y pjΩq “ e´jΩ t0 X pjΩq
y p tq “ x p t ´ t0 q Ð

ñ magnitude of FT remains unchanged, i.e., |Y pjΩq| “ |X pjΩq|

ż8 ż8
To prove property: Y pjΩq = y ptq e´jΩt dt = xpt ´ t0 q e´jΩt dt
´8 ´8

Letting τ = t ´ t0 , ñ t = τ ` t0 and dt = dτ

ż8 ż8
Y pjΩq “ xpτ q e´jΩpτ `t0q dτ “ e´jΩt0 xpτ q e´jΩτ dτ “ e´jΩt0 X pjΩq
´8 ´8

Example: Find FT of y ptq = e´2 t e4 upt ` 2q

FT 1
Note that xptq = e´2t uptq Ð
Ñ X pjΩq =
2 ` jΩ
and from unit step upt ` 2q, y ptq can be expressed in terms of xpt ` 2q ñ must also have shift in e´2 t term

y ptq “ e´2 t e4 upt ` 2q “ e´2 pt`2q e4 e4 upt ` 2q “ e8 xpt ` 2q

FT 8 j2 Ω 8 ej2 Ω
Ð
Ñ Y pjΩq “ e e X pjΩq “ e
2 ` jΩ

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[P3] Modulation: multiplication by complex sinusoid in time results in shift in frequency

FT
y ptq “ ejΩ0 t xptq Ð
Ñ Y pjΩq “ X pj pΩ ´ Ω0 qq

FT
Ñ e´jΩ t0 X pjΩq (duality)
Note similarity in structure with time shift property: xpt ´ t0 q Ð

[P4] Scaling time axis: scale in time by a results in scale in frequency by 1{a

FT 1 ´ Ω¯
y p tq “ x p a tq Ð
Ñ Y pjΩq “ X j
| a| a

Example: Find the FT of y ptq = e´6 t uptq

Since up3tq = uptq, and using FT of xptq = e´2 t uptq,

FT 1 1 1 1
y ptq “ e´6 t uptq “ e´3p2 tq uptq “ xp3 tq Ð
Ñ Y pjΩq “ X pjΩ{3q “ “
3 3 2 ` j p Ω {3 q 6 ` jΩ

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[P5] Scaling frequency axis (dual)

FT
y p tq “ | a | x p t{ a q Ð
Ñ Y pjΩq “ X pjaΩq

[P6] Time-reversal y ptq = xp´tq (similar to scaling with a = ´1

FT
y ptq “ xp´tq Ð
Ñ Y pjΩq “ X p´Ωq

ñ time reversal results in frequency reversal

Example: Find FT of y ptq = e2 t up´tq

Considering xptq = e´2 t uptq, then

FT 1
y ptq “ e2 t up´tq “ xp´tq Ð
Ñ Y pjΩq “ X p´jΩq “
2 ´ jΩ

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[P7] Duality
FT
Using duality, if known one FT pair, xptq Ð
Ñ X pjΩq, then can obtain a second FT pair

Let z pΩq = X pjΩq (some function of Ω)

FT
x p tq Ð
Ñ z pΩq “ X pjΩq
FT
y p tq “ z p tq Ð
Ñ Y pjΩq “ 2πxp´Ωq
1 FT
g p tq “ z p´tq Ð
Ñ GpjΩq “ xpΩq
2π
 
where z ptq “ z pΩq and x p Ω q “ x p tq 
 
Ω Ñ t t Ñ Ω

Example

FT FT sin pΩ T1 q
box Ð
Ñ sinc x p tq “ u p t ` T 1 q ´ u p t ´ T 1 q Ð
Ñ z pΩq “ X pjΩq “ 2
Ω
FT sin pt Ω1q FT ´ ¯
sinc Ð
Ñ box z p tq “ 2 Ð
Ñ 2πxp´Ωq “ 2π upΩ ` Ω1 q ´ upΩ ´ Ω1 q
t

Note: up´Ω ` Ω1q ´ up´Ω ´ Ωq = upΩ ` Ω1 q ´ upΩ ´ Ω1 q

(see Example 4.5: find the inverse FT of a box)

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 1
Example: Find the FT of rptq =
3´jt

FT 1
Using duality and the FT pair xptq = e´3t uptq Ð
Ñ X pjΩq “
3`jΩ

FT
Ñ RpjΩq “ 2π xpΩq “ 2π e´3Ω upΩq
rptq “ z p´tq Ð

1 FT
Resulting FT pair: Ñ 2πe´3Ω upΩq
Ð
3´jt

Example: Find the FT of rptq = ejΩ0t

FT
Ñ z pΩq “ X pjΩq “ e´jΩt0 , then
Using xptq = δ pt ´ t0 q Ð

FT
rptq “ z p´tq “ ejΩ0 t Ð
Ñ RpjΩq “ 2πxpΩq “ 2πδ pΩ ´ Ω0q

FT
Resulting FT pair: ejΩ0 t Ð
Ñ 2πδ pΩ ´ Ω0 q

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[P8] Differentiation in time results in multiplication by frequency variable in frequency

d FT
y p tq “ x p tq Ð
Ñ Y pjΩq “ jΩ X pjΩq
dt

Example: Find the FT of y ptq = sin pΩ0tq

FT
´ ¯ d
Using cos pΩ0tq ÐÑ π δ pΩ ´ Ω0q ` δ pΩ ` Ω0q , and since cos pΩ0tq “ ´Ω0 sin pΩ0tq, then
dt
1 d FT Ω Ω ´ ¯
ñ y ptq “ sin pΩ0tq “ ´ x p tq Ð Ñ Y pjΩq “ ´j X pjΩq “ ´j π δ pΩ ´ Ω0 q ` δ pΩ ` Ω0 q
Ω0 dt Ω0 Ω0
Ω0 ´ ´Ω ¯ π´ ¯
0
“ ´j πδ pΩ ´ Ω0 q ´ j πδ pΩ ` Ω0q “ δ pΩ ´ Ω0 q ` δ pΩ ` Ω0 q
Ω0 Ω0 j
FT π
´ ¯
sin pΩ0tq Ð Ñ δ pΩ ´ Ω0 q ` δ pΩ ` Ω0 q
j

FT d
[P9] Dual property of P8: differentiation in frequency y ptq “ t xptq Ð
Ñ Y pjΩq “ j X pjΩq
dΩ
FT 1
Example: Find the FT of y ptq = t e´2t uptq using xptq = e´2t uptq Ð
Ñ X pjΩq = ,
2 ` jΩ

´2t FT d d ´ 1 ¯ 1 1 1
y p tq “ t e u p tq “ t x p tq Ð
Ñ Y pjΩq “ j X pjΩq “ j “ ´j 2 “ “
dΩ dΩ 2 ` jΩ p2 ` jΩq2 p2 ` jΩq2 4 ´ Ω2

FT 1
t e´at uptq Ð
Ñ , aą0
pa ` jΩq2

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[P10] Integration in time
żt
FT 1
y p tq “ xpτ q dτ Ð
Ñ Y pjΩq “ X pjΩq ` πX p0qδ pΩq
´8 jΩ

Example: Find the FT of y ptq = uptq

żt
FT
Using δ ptq Ð
Ñ 1 and also uptq = δ pτ q dτ , then
´8
żt
FT 1 1
y p tq “ u p tq “ δ pτ q dτ Ð
Ñ Y pjΩq “ X pjΩq ` πX p0qδ pΩq “ ` πδ pΩq
´8 jΩ jΩ

FT 1
u p tq Ð
Ñ ` πδ pΩq
jΩ

[P11] Dual property of P10

żΩ
1 FT
´ x p tq ` π x p 0 q δ p tq Ð
Ñ X pjφq dφ
jt ´8

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[P12] Multiplication of two signals in time

FT 1
g p tq “ x p tq r p tq Ð
Ñ GpjΩq “ X pjΩq ‹ RpjΩq
2π

Example: Amplitude modulation (AM) in communications

amplitude A, transmit signal xptq, carrier frequency Ωc (to modulate xptq to appropriate frequency band)
¸
FT A A´
´ ´ ¯ ¯
A xptq cos pΩc tq Ð
Ñ X pjΩq ‹ π δ pΩ ´ Ωc q ` δ pΩ ´ Ωc q “ X pj pΩ ´ Ωc qq ` X pj pΩ ` Ωc qq “
2π 2

Convolution with δ pΩ ´ Ωc q ‹ X pjΩq = X pj pΩ ´ Ωc qq

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[P13] Convolution in time (dual to P11)

FT
y ptq “ xptq ‹ hptq Ð
Ñ Y pjΩq “ X pjΩq RpjΩq

Example: Find the FT of a triangular pulse

Recall: convolution of two boxes

%&'( %&'( )&'(

"
! "

! "#$ "#$ ' ! "#$ "#$ ' !" " '

FT 2
Recall: FT of a box bptq = upt ` a{2q ´ upt ´ a{2q Ð
Ñ B pjΩq “ Ω
sin pΩ a{2q

Using property

FT 4
xptq “ bptq ‹ bptq “ |t| pupt ` aq ´ upt ´ aqq Ð
Ñ X pjΩq “ B pjΩq B pjΩq “ 2
sin2 pΩ a{2q
Ω

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22
Example: Linear time-invariant (LTI) with impulse response hptq and frequency response H pjΩq (Sect. 4.5)
ż
FT
y ptq “ xptq ‹ hptq “ xpt ´ τ q hpτ q dτ Ð
Ñ Y pjΩq “ X pjΩq H pjΩq

!"#$ & #
%"#$

*"()$ +"()$
'"()$

To find y ptq: (a) obtain FTs X pjΩq and H pjΩq

(b) multiple them to obtain Y pjΩq = X pjΩq and H pjΩq;

ż
1
(c) obtain inverse FT y ptq = 2π Y pjΩq ejΩt dΩ

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23
Filtering: processing to remove frequencies that are not of interest

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Example: Signal xptq = 2 sin p2π50 tq ` sin p2π250 tq duration 1 s and frequencies 50 Hz and 250 Hz
Design lowpass filter (LPF) with frequency response H pjΩq) and cuttoff 150 Hz

fs=1e3;t=0:1/fs:1; % Sampling frequency and time axis

x=[1 2]*sin(2*pi*[50 250]’.*t); % Generate 50 Hz and 250 Hz sinusoids
N=2ˆ10;Xf=fftshift(fft(x,N)); % Obtain Fourier transform of x(t)
freq=linspace(-fs/2,fs/2,N); % Frequency axis
lpFilt=designfilt(’lowpassiir’,’FilterOrder’,8, PassbandFrequency’,...
150,’PassbandRipple’,0.2, ’SampleRate’,fs); % Design LPF, cutoff 150 Hz
y=filter(lpFilt,x); % Apply filter to process x(t) and obtain y(t)
Yf=fftshift(fft(y,N)); % Obtain Fourier transform of y(t)
fvtool(lpFilt) % GUI to view magnitude response | H(j Omega)| of lowpass filter
Magnitude Response
1

0.9

0.8

0.7

0.6
Magnitude

0.5

0.4

0.3

0.2

0.1

0
-500 -400 -300 -200 -100 0 100 200 300 400 500
Frequency (Hz)

© Arizona State University EEE 203, Prof. A. Papandreou-Suppappola

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 FT
xptq “ 2 sin p2π50 tq ` sin p2π250 tq Ð
Ñ
´ ¯ ´ ¯
X pjΩq “ j π δ pΩ ` 2π50q ´ δ pΩ ´ 2π50q ` j π δ pΩ ` 2π250q ´ δ pΩ ´ 2π250q `

$"# !"
Sum of 50 and 250 Hz sinusoids in time in frequency
1000

900

800

700

600
magnitude

500

400

300

200

100

0
-500 -400 -300 -200 -100 0 100 200 300 400 500
!"
frequency, Hz
"# !"#

© Arizona State University EEE 203, Prof. A. Papandreou-Suppappola

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Time and frequency domain signal representations, before and after LPF

Filtering removed the 250 Hz signal component

Sum of 50 and 250 Hz sinusoids in time After lowpass filtering with cutoff 150 Hz in time
3 1

2
0.5
1
amplitude

amplitude
0 0

-1
-0.5
-2

-3 -1
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
time, s time, s

Sum of 50 and 250 Hz sinusoids in time in frequency After lowpass filtering with cutoff 150 Hz in frequeny
1000 500

800 400
magnitude

magnitude
600 300

400 200

200 100

0 0
-500 0 500 -500 0 500
frequency, Hz frequency, Hz

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27
Systems characterized by linear constant coefficient differential equations (DE) (Sect. 4.7)

d d
LTI system whose input and output satisfy DE a0 y ptq ` a1 y ptq = b0 xptq ` b1 xptq (stable, initially at rest)
dt dt

To find the impulse response hptq:

Step 1: take FT of both sides of the DE (if differentiate in time then multiply with Ω in frequency)

d d FT
a 0 y p tq ` a 1 y ptq` “ b0 xptq ` b1 xptq Ð
Ñ a0 Y pΩq ` a1 j Ω Y pjΩq “ b0 X pΩq ` b1 j Ω X pjΩq
dt dt

FT Y pjΩq
Step 2: From y ptq = xptq ‹ hptq Ð
Ñ Y pjΩq = X pjΩq H pjΩq, solve for frequency response: H pjΩq =
X pjΩq
` ˘ ` ˘ Y pjΩq b0 ` jb1 Ω
from Step 1, Y pjΩq a0 ` ja1 Ωq = X pjΩq b0 ` jb1 Ωq ñ H pjΩq = =
X pjΩq a0 ` ja1Ω

Step 3: Obtain inverse FT (FT) of H pjΩq

b0 ` jb1 Ω IFT
H pjΩq “ ÐÑ hptq
a0 ` ja1 Ω

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 d d
Example: LTI stable system as DE 2 y ptq = 3 xptq ` xptq ´ y ptq; find impulse response hptq
dt dt

Using FT properties

d d FT
2 y p tq “ 3 x p tq ` x p tq ´ y p tq Ð
Ñ 2Y pjΩq “ 3X pjΩq ` jΩX pjΩq ´ jΩY pjΩq
dt dt

ñ Y pjΩqp2 ` jΩq “ X pjΩqp3 ` jΩq

Y pjΩq 3 ` jΩ 3 1
ñ H pjΩq “ “ “ ` jΩ
X pjΩq 2 ` jΩ 2 ` jΩ 2 ` jΩ

´d ¯
´2t ´2t
ñ hptq “ 3 e u p tq ` e uptq “ 3 e´2t uptq ´ 2 e´2tuptq “ e´2t uptq
dt

P[14] Parseval’s relation: FT preserves signal energy Ex

ż ż
2
Ex “ |xptq| dt “ |X pjΩq|2 dΩ

© Arizona State University EEE 203, Prof. A. Papandreou-Suppappola

29
Some additional FT properties

[P15] Conjugation in time and symmetry properties

FT
x ˚ p tq Ð
Ñ X ˚ p´jΩq

FT
Also, recall xp´tq Ð
Ñ X p´jΩq

Using these two properties Note the following types of signals

FT
real signal : xptq “ x˚ ptq Ð
Ñ X pjΩq “ X ˚ p´jΩq conjugate symmetric in frequency
FT
imaginary signal : xptq “ ´x˚ ptq Ð
Ñ X pjΩq “ ´X ˚ p´jΩq conjugate antisymmetric
FT
even signal : xptq “ xp´tq Ð
Ñ X pjΩq “ X p´jΩq even in frequency
FT
odd signal : xptq “ ´xp´tq Ð
Ñ X pjΩq “ ´X p´jΩq odd in frequency
FT
real and even signal : xptq “ xp´tq “ x˚ ptq Ð
Ñ X pjΩq “ X p´jΩq “ X ˚ p´jΩq real and even

ñ X pjΩq “ X ˚pjΩq

where conjugate symmetric implies that real part is even and imaginary part is odd

© Arizona State University EEE 203, Prof. A. Papandreou-Suppappola

30