The frequency content of a signal is obtained via the

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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 Square-Root 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 Square-Root 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 roll-off 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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