Chap6StudentSolutions

J roberts 81604 x t sin2t cos20t find the phase

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Unformatted text preview: CTFT’s. x t(t) yt (t) = x r(t) m 1 cos(2πfct) yd (t) LPF yf (t) cos(2πfct) Similar to Exercise 18. 20. An RC lowpass filter with a time constant of 16 ms is excited by a DSBSC signal, Solutions 6-14 M. J. Roberts - 8/16/04 x( t) = sin(2πt) cos(20πt) . Find the phase and group delays at a the carrier frequency. The transfer function of the RC lowpass filter is H( jω ) = A A = . 1 + jωτ 1 + j 0.016ω The phase of the transfer function is φ ( jω ) = − tan −1 (ωτ ) = − tan −1 (0.016ω ) . The carrier frequency is 10 Hz. Therefore φ ( j 20π ) = − tan −1 (0.016 × 20π ) = 0.788 and the phase delay is − φ ( j 20π ) 0.788 = = 0.01254 or 12.54 ms . The derivative of the ωc 20π phase shift function is d (φ ( jω )) = − 1 + (τ )2 . dω ωτ Evaluating this derivative at the carrier frequency we get 0.016 d − (φ ( jω )) = τ 2 = 1 + (0.016 × 20π )2 = 7.95 ms . dω ω =ω c 1 + (ω cτ ) 21. A pulse train, p( t) = rect (100 t) ∗ 10 comb(10 t) is modulated by a signal, x( t) = sin( 4πt) . Plot the response of the modulator, y( t) , and the CTFT’s of the excitation and response. Similar to Exercise 18. ∞ n y( t) = sin( 4πt) ∑ rect 100 t − 10 n =−∞ Y( f ) = ∞ j∞ k k sinc δ ( f + 2 − 10 k ) − ∑ sinc δ ( f − 2 − 10 k ) ∑ 10 10 20 k =−∞ k =−∞ 22. In the system below, let the excitation be Solutions 6-15 M. J. Roberts - 8/16/04 x( t) = rect (1000 t) ∗ 250 comb(250 t) and let the filter be ideal, with unity passband gain. Plot the signal power of the response, y( t) , of this system versus the sweep frequency, f c , over the range, 0 < f c < 2000 for a LPF bandwidth of (a) (b) and 5 Hz 50 Hz (c) 500 Hz. Multiplier xsh(t) x(t) LPF y(t) cos(2πfc t) x sh ( t) = [rect (1000 t) ∗ 250 comb(250 t)] cos(2πf c t) ∞ n x sh ( t) = cos(2πf c t) ∑ rect 1000 t − 250 n =−∞ 1 f f 1 X sh ( f ) = sinc ∗ δ ( f − fc ) + δ ( f + fc ) comb 1000 250 2 1000 [ ∞ 1 f X sh ( f ) = sinc ∑ δ ( f − 250 k ) ∗ δ ( f − f c ) + δ ( f + f c ) 1000 k =−∞ 8 [ X sh ( f ) = 1∞ k ∑ sinc 4 δ ( f − f c − 250k ) + δ ( f + f c − 250k ) 8 k =−∞ [ 1 ∞ k f Y( f ) = ∑ sinc δ ( f...
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