3 gaussian filtering of a rectangular pulse x t pt t

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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: Spread-spectrum 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 out-of-band 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 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 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 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 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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