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Unformatted text preview: )} = N K {[C + S ][π (1 − α) − π (1 + α)] + 4α[C − S ]} = cos[ = K {[C + S ][−2απ ] + 4α[C − S ]} = 2K α{C (2 − π ) − S (2 + π )} EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 47 / 174 Lecture Notes 2 Linear Systems Aside: Evaluation of h(t ) Thus lim t →1/(4α) h (t ) =
=
= EECS 455 (Univ. of Michigan) 2K α{C (2 − π ) − S (2 + π )}
−2 π
K α{C (2 − π ) − S (2 + π )}]}
−π
K α{C (1 − Fall 2012 2
π ) − S (1 + 2
π )}]} September 7, 2012 48 / 174 Lecture Notes 2 Linear Systems Ex. 3: Gaussian Filtering of a Rectangular Pulse
x (t ) = pT (t )
1
exp{−t 2 /(2σ 2 )}
h(t ) = √
2πσ
The parameter σ is related to the 3dB bandwidth B by
σ2 = ln(2)
.
π2 B t
t −T
y (t ) = Φ( ) − Φ(
)
σ
σ
where
x Φ(x ) =
−∞
EECS 455 (Univ. of Michigan) 1
2
√ e−u /2 du .
2π
Fall 2012 September 7, 2012 49 / 174 Lecture Notes 2 Linear Systems Ex. 3: Gaussian Filtering of a Rectangular Pulse In the frequency domain
X (f ) = T sinc(fT )e−j πfT
H (f ) = exp{−2π 2 σ 2 f 2 } Y (f ) = X (f )H (f ). This is a noncausal ﬁlter. In practice a delay must be added to make
the ﬁlter implementable. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 50 / 174 Lecture Notes 2 Linear Systems Ex. 3: Gaussian Filtering of a Rectangular Pulse
20
X(f) x(t) 1
0 40
60 1
0 1 2
time/Tb 0.4 3 80 4 0 2 4
f 6 8 0 2 4
f/f3 6 8 0 2 4
f/f3 6 8 H(f) h(t) 20
0.2 40
60 0
4 2 0
time/Tb 0.4 2 80 4 Y(f) y(t) 20
0.2 40
60 0
2 0 2 4 time/Tb EECS 455 (Univ. of Michigan) Fall 2012 80 September 7, 2012 51 / 174 Lecture Notes 2 Linear Systems Example 4: Spreadspectrum signals In this example the basic pulse shape has much larger bandwidth.
The pulse shape consists of a sequence of shorter pulses (called
chips).
The ﬁlter is the time reverse (and delayed) version of the pulse.
Notice that the output lasts for 2T seconds and is zero at time 0
and time 2T .
Notice also that the output is a piecewise linear function of time. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 52 / 174 Lecture Notes 2 Linear Systems Motivation As in the ﬁrst example a transmitter can send a 0 by sending the
T second waveform shown below and send a 1 by sending the
same waveform but with opposite polarity.
The receiver ﬁlters the signal (to remove outofband noise).
The ﬁlter is matched to the transmitted signal (with a time
reversal).
The receiver decides 0 is sent if the ﬁlter output at time T is larger
than 0.
Otherwise the receiver decides 1. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 53 / 174 Lecture Notes 2 Linear Systems Filtering Spread Spectrum Pulses
Time Domain Frequency Domain
20
X(f) x(t) 1
0 60 1
2 40 0 2 80
50 4 time 0 0 2 80
50 4
Y(f)=X(f)H(f) y(t)=x(t)*h(t) time
1
0
1
2 50 0 2 4 0
frequency 50 20
40
60
80
50 time EECS 455 (Univ. of Michigan) 0
frequency 40
60 1
2 50 20
H(f) h(t) 1 0
frequency Fall 2012 September 7, 2012 54 / 174 Lecture Notes 2 Linear Systems Filtering Spread Spectrum Pulses
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 55 / 174 Lecture Notes 2 Linear Systems Multiuser System Now consider two users which have different basic signal waveforms.
Consider a ﬁlter matched to the basic signal of the ﬁrst user. The
output due to the signal of the ﬁrst user alone is shown as is the output
due the second user alone. If these users both transmitted
simultaneously then the output would be the sum of the two outputs. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 56 / 174 Lecture Notes 2 Linear Systems Filtering Spread Spectrum Pulses
User 1 User 2
1
x2(t) x(t) 1
0
1 1
0 1 2
time 3 4 0 1 2
time 3 4 1 2
time 3 4 1 2
time 3 4 1
h(t) h(t) 1
0
1 0
1 1 2
time 3 4 1
0
1
0 0
y2(t)=x2(t)*h(t) 0
y1(t)=x1(t)*h(t) 0 1 EECS 455 (Univ. of Michigan) 2
time 3 4 Fall 2012 1
0
1
0 September 7, 2012 57 / 174 Lecture Notes 2 Linear Systems Filtering Spread Spectrum Pulses
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 58 / 174 Lecture Notes 2 Linear Systems Sampling a Causal Filter
t=T
x (t ) h (t )
y (T ) If we assume that the ﬁlter is sampled at time T and that the ﬁlter is causal
h(t ) = 0, t < 0 (h(t − α) = 0, α > t ) then
t y (t ) =
−∞ h(t − α)x (α)d α. If the ﬁlter has ﬁnite response, say for T seconds then
t y (t ) =
t −T h(t − α)x (α)d α. If the desired signal is the output sampled at time T .
T y (T ) =
0
EECS 455 (Univ. of Michigan) h(T − α)x (α)d α.
Fall 2012 September 7, 2012 59 / 174 Lecture Notes 2 Linear Systems Correlator This can be implemented with a correlator as shown below x (t ) × T
0 y (T ) h(T − t ) EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 60 / 174 Lecture Notes 2 Linear Syste...
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