Chap6StudentSolutions

# Then it follows that cs 133 nf and rs1 10 k the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: as illustrated in Figure S33. Rf R s1 R s2 Cs + vi (t) - + vo(t) Figure S33 Pre-emphasis filter The source impedance is 1 1 Rs1 Rs2 + j 2πf + j 2πfCs j 2πfCsRs1Rs2 + Rs1 RR CsRs2 = s1 s2 Zs ( f ) = = 1 1 j 2πfCs ( Rs1 + Rs2 ) + 1 Rs1 + Rs2 j 2πf + Rs1 + Rs2 + j 2πfCs Cs ( Rs1 + Rs2 ) The numerator provides the zero of the source impedance (the pole of the overall gain) and the denominator provides the pole of the source impedance (the zero of the overall gain). So we want the gain-pole location to be set by 1 = 2π × 12000 CsRs2 and the gain-zero location to be set by 1 = 2π × 6000 . Cs ( Rs1 + Rs2 ) There is no unique solution so let’s arbitrarily set Rs2 = 10 kΩ. Then it follows that Cs = 1.33 nF and Rs1 = 10 kΩ . The overall gain is Solutions 6-26 M. J. Roberts - 8/16/04 Rf H( f ) = − Rs1Rs2 Rs1 + Rs2 1 j 2πf + CsRs2 1 j 2πf + Cs ( Rs1 + Rs2 ) = −Rf Rs1 + Rs2 Rs1Rs2 1 Cs ( Rs1 + Rs2 ) 1 j 2πf + CsRs2 j 2πf + At low frequencies, H( f ) = − Rf . Rs1 To make the low-frequency gain one, set R f = Rs1 = 10 kΩ . 34. One problem with causal CT filters is that the response of the filter always lags the excitation. This problem cannot be eliminated if the filtering is done in real time but if the signal is recorded for later “off-line” filtering one simple way of eliminating the lag effect is to filter the signal, record the response and then filter that recorded response with the same filter but playing the signal back through the system backward. Suppose the filter is a single-pole filter with a transfer function of the form, H( jω ) = 1 ω 1+ j ωc , where ω c is the cutoff frequency (half-power frequency) of the filter. (a) What is the effective transfer function of the entire process of filtering the signal forward, then backward? (b) What is the effective impulse response? (a) The impulse response of the filter is h( t) = F −1[H( jω )] . The response of the filter on the first pass through the filter forward is y1 ( t) = x( t) ∗ h( t) in the time domain or Y1 ( jω ) = X( jω ) H( jω ) in the frequency domain. The response of the filter on the second pass through the filter backward is...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online