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Unformatted text preview: Frequency Domain Analysis
Signals and ﬁltering can also be described in the frequency
domain.
The frequency content of a signal is obtained via the Fourier
Transform.
X (f ) = ∞ x (t )e−j 2πft dt . −∞ Convolution in the time domain corresponds to multiplication in the
frequency domain and thus
y (t ) = x (t ) ∗ h(t ) ⇔ Y (f ) = H (f )X (f ).
One useful relation between the frequency domain and time
domain is Parseval’s Theorem
∞
−∞ EECS 455 (Univ. of Michigan) ∗
x1 (t )x2 (t )dt = ∞
−∞ Fall 2012 ∗
X1 (f )X2 (f )df .
September 7, 2012 40 / 174 Lecture Notes 2 Linear Systems Ex. 1: Rectangular Filtering of Rectangular Pulses
Motivation: This is the simplest form of modulation. A single data bit
is transmitted by sending either a positive pulse to represent a 0 or a
negative pulse of duration T to represent a 1. The receiver decides
which bit was transmitted by ﬁltering the received signal with a ﬁlter
matched to the transmitted signal and sampling the ﬁlter output. x (t ) = h(t ) = pT (t ), t,
0≤t ≤T
t
(2 − T )T , T ≤ t ≤ 2T . 0,
otherwise
sin(π fT ) −j πfT
X (f ) = H (f ) = T sinc(fT )e−j πfT = T
e
.
π fT
Y (f ) = H (f )X (f ) = T 2 sinc2 (fT )e−j 2πfT .
y (t ) = h(t ) ∗ x (t ) = ΛT (t ) = EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 41 / 174 Lecture Notes 2 Linear Systems Ex. 1: Rectangular Filtering of Rectangular Pulses
Time Domain Frequency Domain
0 X(f) x(t) 1
0.5 40 0
0.5
2 0 2 60
4 4 time H(f) h(t) 1 0 2 4 2 0
frequency 2 4 2 0
frequency 2 4 0
Y(f)=X(f)H(f) y(t)=x(t)*h(t) 2 20 60
4 4 time
1
0.5
0
0 2 4 time EECS 455 (Univ. of Michigan) 0
frequency 40 0 0.5
2 2 0 0.5 0.5
2 20 Fall 2012 20
40
60
4 September 7, 2012 42 / 174 Lecture Notes 2 Linear Systems Example 2: Square Root Raised Cosine Filtering of
SquareRoot Raised Cosine Pulses x (t ) = h(t ) = X (f ) = H (f ) = sin(π (1 − α)t /T ) + 4αt /T cos(π (1 + α)t /T )
.
π [1 − (4αt /T )2 ]t /T √
T
2 [1 T, − sin(π T (f  −
0, 1
2T )/α)], 1− α
2T
+
≤ 12Tα 0 ≤ f  ≤ 1− α
2T ≤ f 
otherwise. α is called the rolloff. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 43 / 174 Lecture Notes 2 Linear Systems Example 2: Square Root Raised Cosine Filtering of
SquareRoot Raised Cosine Pulses
y (t ) = Y (f ) = T
2 [1 sin(π t /T ) cos(απ t /T )
.
π t /T 1 − 4α2 t 2 /T 2 T,
− sin(π T (f  −
0, 1
2T )/α)], −
0 ≤ f  ≤ 12Tα
+
≤ f  ≤ 12Tα
otherwise. 1− α
2T The parameter α is called the rolloff factor and is between 0 and 1.
The (absolute) bandwidth is W = (1 + α)/2T . Notice that the output is
zero at multiples of T except at t = 0. Note also that
1+α
2T
−
− 12Tα
EECS 455 (Univ. of Michigan) Y (f )df = 1
Fall 2012 September 7, 2012 44 / 174 Lecture Notes 2 Linear Systems Raised Cosine Pulses and Filtering
Time Domain Frequency Domain
1.5 X(f) x(t) 1
0.5 0.5 0
0.5
4 2 0
time 2 0
1 4 H(f) h(t) 1 2 0
time 2 0.5 1 0.5 0
frequency 0.5 1 0.5 0
frequency 0.5 1 1 0
1 4 1.5
Y(f)=X(f)H(f) y(t)=x(t)*h(t) 0
frequency 0.5 0 1
0.5
0
0.5
4 0.5 1.5 0.5 0.5
4 1 2 EECS 455 (Univ. of Michigan) 0
time 2 4 Fall 2012 1
0.5
0
1 September 7, 2012 45 / 174 Lecture Notes 2 Linear Systems Aside: Evaluaton of h(t )
h (t ) = sin[π (1 − α)t ] + 4αt cos[π (1 + α)t ]
π t [1 − (4αt )2 ] This impulse response can be evaluated at t = −1/(4α), 0, 1/(4α) using L’Hospitals rule. First note that both the numerator
and denominator are clearly 0 at t = 0. Applying L’Hospitals rule at t = 0 we get
lim h(t ) t →0 =
=
= lim cos[π (1 − α)t ]π (1 − α) + 4α cos[π (1 + α)t )] − 4αt sin[π (1 + α)t ]π (1 + α)
π [1 − (4αt )2 ] + π t [−2(4α)2 t ] t →0 π (1 − α) + 4α
π
(1 − α) + 4α
π . Consider now, the case where t = ±1/(4α). Applying L’Hospitals rule again we get
lim t →±1/(4α) h (t ) =
=
=
= lim t →±1/(4α) cos[π (1 − α)t ]π (1 − α) + 4α cos[π (1 + α)t )] − 4αt sin[π (1 + α)t ]π (1 + α)
π [1 − (4αt )2 ] + π t [−2(4α)2 t ] cos[π (1 − α)/(4α)]π (1 − α) + 4α cos[π (1 + α)/(4α)] − sin[π (1 + α)/(4α)]π (1 + α)
π/(4α)[−8α]
π
π
π
cos[ 4α − π ]π (1 − α) + 4α cos[ 4α + π ] − sin[ 4α + π ]π (1 + α)
4
4
4
−2 π
π
π
π
cos[ 4α − π ]π (1 − α) + 4α cos[ 4α + π ] − sin[ 4α + π ]π (1 + α)
4
4
4 EECS 455 (Univ. of Michigan) −2 π
Fall 2012 September 7, 2012 46 / 174 Lecture Notes 2 Linear Systems Aside: Evaluation of h(t )
Let C = cos[π/(4α)] and S = sin[π/(4α)]. We will use the following facts
π 4 ) cos(a) cos(π/4) − sin(a) sin(π/4) K [C − S ] = cos(a) cos(π/4) + sin(a) sin(π/4) = cos(a − π ) K [cos(a) − sin(a)] = 4 =
= cos(a + K [cos(a) + sin(a)] =
4 ) K [C + S ]
sin(a) cos(π/4) + cos(a) sin(π/4)
K [cos(a) + sin(a)] = π =
= sin(a + K [C + S ] where K = cos(π/4). Consider the numerator only. = π
π
π
π
π
π
− ]π (1 − α) + 4α cos[
+ ] − sin[
+ ]π (1 + α)
4α
4
4α
4
4α
4
K {[C + S ]π (1 − α) + 4α[C − S ] − [C + S ]π (1 +...
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