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Chap6StudentSolutions

# J roberts 81604 py 1 2 k sinc 4 64 f c

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Unformatted text preview: − f c − 250 k ) + δ ( f + f c − 250 k ) rect 4 2B 8 k =−∞ [ where “B” is the bandwidth of the LPF. ∞ f k rect ∑ sinc δ ( f − f c − 250 k ) 2 B k =−∞ 4 1 Y( f ) = ∞ 8 f k + rect ∑ sinc δ ( f + f c − 250 k ) 2 B k =−∞ 4 1 k Y( f ) = ∑ sinc δ ( f − f c − 250 k ) + 4 8 f c + 250 k < B k sinc δ ( f + f c − 250 k ) 4 f c − 250 k < B ∑ Solutions 6-16 M. J. Roberts - 8/16/04 Py = 1 2 k ∑ sinc + 4 64 f c + 250 k < B k sinc 2 4 f c − 250 k < B ∑ Signal Power 0.1 2000 f c Signal Power 0.1 2000 fc 2000 fc Signal Power 0.1 23. A signal, x(t) is described by x( t) = 500 rect (1000 t) ∗ comb(500 t) (a) If x(t), is the excitation of an ideal lowpass filter with a cutoff frequency of 3 kHz, plot the excitation , x(t) and the response, y(t) on the same scale and compare. Fourier transform the excitation, to yield X. Write the transfer function, H, of the ideal filter as a rectangle function. Form the transform of the response, Y, from X times H. Recognize it as a finite summation of impulses. Inverse transform the impulses in pairs to form y, and graph y. y(t) 1 -2 ms 2 ms t This looks like the partial sums in the discussion of convergence of the CTFS because, mathematically, the same thing is happening. (b) Similar to (a) 24. Determine whether or not the CT systems with these transfer functions are causal. The test for causality is that a causal system has an impulse response that is zero fall time, t < 0 . (a) H( jω ) = 2 jω (b) H( jω ) = Solutions 6-17 10 6 + j 4ω M. J. Roberts - 8/16/04 (c) H( jω ) = 4 25 − ω + j 6ω 2 = 4 ( jω + 3) 2 + 16 F → Using e− at sin (ω 0t ) u ( t ) ← ω0 ( jω + a )2 + ω 02 h( t) = e −3 t sin( 4 t) u( t) (d) H( jω ) = (f) H( jω ) = 4 25 − ω + j 6ω 2 Causal e jω jω + 9 45 − ω 2 + j 6ω (e) H( jω ) = (g) H( jω ) = 4 25 − ω 2 + j 6ω e − jω 49 49 + ω 2 25. Determine whether or not the DT systems with these transfer functions are causal. (a) H( F ) = [rect (10 F )...
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