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Unformatted text preview: EE 562a Homework Solutions 8 December 1, 2009 1 1. (a) If the input to the system is δ D ( t ), then the input to the integrator is δ D ( t ) δ D ( t τ ). Since the integral of a Dirac delta is a step function, h ( t ) = U( t ) U( t τ ) = rect t τ/ 2 τ . Since we only need the magnitude squared of the frequency response we can time shift the impulse response without changing the answer. (why?)  H ( f )  2 = F rect t τ/ 2 τ 2 = [ τ sinc( τf )] 2 , where sinc( z ) = sin( πz ) / ( πz ). The maximum of this spectrum occurs at f = 0, so max f  H ( f )  2 =  H (0)  2 = τ 2 . To integrate the sincsquared it is easiest to use Parseval’s theorem: Z ∞∞  H ( f )  2 df = Z ∞∞  h ( t )  2 dt = Z τ dt = τ It follows that for this system B N = 1 2 τ . (b) Using complex impedance methods, this simple voltage divider has frequency response H ( f ) = 1 / (i2 πfC ) R + 1 / (i2 πfC ) = 1 1 + (i2 πf ) RC . The corresponding squaremagnitude is  H ( f )  2 = 1 [1 + (i2 πf ) RC ][1 (i2 πf ) RC ] = 1 1 + (2 πfRC ) 2 . The maximum of this spectrum again occurs at DC. max f  H ( f )  2 =  H (0)  2 = 1 . There are several ways to evaluate the integral of  H ( f )  2 . One possibility is to use the fact that the Cauchy probability density a π [ a 2 + x 2 ] integrates to 1 over x for any positive parameter a . Z ∞∞  H ( f )  2 df = Z ∞∞ 1 1 + (2 πfRC ) 2 df = 1 2 πRC Z ∞∞ 1 2 πRC 1 (2 πRC ) 2 + f 2 df = 1 2 RC . could also find the inverse transform F 1 { H ( f )  2 } = 1 2 RC e  τ  RC . Evaluting at τ = 0 yields the same answer as found before. Whatever the method, it follows that the noise bandwidth for this system is B N = 1 4 RC . Note that if R is in Ohms and C is in Farads, then B N is in Hertz. 2 EE 562a Homework Solutions 8 December 1, 2009 (c) If we call the input to system n ( u,t ) and the output x ( u,t ), then S x ( f ) =  H ( f )  2 S n ( f ) =  H ( f )  2 N o 2 . So the average power in the output is R x (0) = Z ∞∞ S x ( f ) df = N o 2 Z ∞∞  H ( f )  2 df = N o B N max f  H ( f )  2 . Since there is no delta function at f = 0, this process has no DC component and hence its mean is zero. So we have R x (0) = K x (0) = V ar { x ( u,t ) } . 2. (a) Answer for a sinusoid of random phase: i. Note that the signal x ( u,t ) is a linear combination of eigenfunctions of a LTI system, the eigenfunction indices (frequencies) being ± f 1 . If these indices are close enough to f o these signals are both passed by the resolution filter with eigenvalue 1, i.e., they are undistorted. y ( u,t ) = x ( u,t ) = A sin(2 πf 1 t + φ ( u )) ,  f 1 f o  < B res / 2 , otherwise y 2 ( u,t ) = A 2 sin 2 (2 πf 1 t + φ ( u )) = A 2 2 [1 cos(2 π (2 f 1 ) t + 2 φ ( u ))] ,  f 1 f o  < B res / 2 , otherwise The second harmonic (at frequency 2 f 1 ) in p 2 ( u,t ) is rejected by the video filter, and so p ( u,t ) = A 2 2 ,  f 1 f o  < B res / 2 , otherwise ii. In this caseii....
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 Fall '07
 ToddBrun
 Fourier Series, Normal Distribution, Probability theory, probability density function, power spectral density

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