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A band-pass filter has two cutoff frequencies, oa and ob. Suppose that oa is quite a bit
Chapter 16, Problem P16.3-5(choose chapter or problem)
A band-pass filter has two cutoff frequencies, \(\omega_n\) and
\(\omega_{\mathrm{b}}\). Suppose that \(\omega_{\mathrm{a}}\) is quite a bit smaller than \(\omega_{\mathrm{b}}\), say \(\omega_{\mathrm{a}}<\omega_{\mathrm{b}} / 10\). Let \(H_{\mathrm{L}}(s)\) be a low-pass transfer function having a cutoff frequency equal to \(\omega_{\mathrm{b}}\) and \(H_{\mathrm{H}}(s)\) be a high-pass transfer function having a cutoff frequency equal to \(\omega_{\mathrm{a}}\). A band-pass transfer function can be obtained as a product of low-pass and high-pass transfer functions, \(H_{\mathrm{B}}(s)=H_{\mathrm{L}}(s) \cdot H_{\mathrm{H}}(s)\). The order of the band-pass filter is equal to the sum of the orders of the low-pass and high-pass filters. We usually make the orders of the low-pass and high-pass filter equal, in which case the order of the band-pass is even. The pass-band gain of the bandpass filter is the product of pass-band gains of the low-pass and high-pass transfer functions. Obtain the transfer function of a fourth-order band-pass filter having cutoff frequencies equal to \(100 \mathrm{rad} / \mathrm{s}\) and \(2000 \mathrm{rad} / \mathrm{s}\) and a pass-band gain equal to 4.
Questions & Answers
QUESTION:
A band-pass filter has two cutoff frequencies, \(\omega_n\) and
\(\omega_{\mathrm{b}}\). Suppose that \(\omega_{\mathrm{a}}\) is quite a bit smaller than \(\omega_{\mathrm{b}}\), say \(\omega_{\mathrm{a}}<\omega_{\mathrm{b}} / 10\). Let \(H_{\mathrm{L}}(s)\) be a low-pass transfer function having a cutoff frequency equal to \(\omega_{\mathrm{b}}\) and \(H_{\mathrm{H}}(s)\) be a high-pass transfer function having a cutoff frequency equal to \(\omega_{\mathrm{a}}\). A band-pass transfer function can be obtained as a product of low-pass and high-pass transfer functions, \(H_{\mathrm{B}}(s)=H_{\mathrm{L}}(s) \cdot H_{\mathrm{H}}(s)\). The order of the band-pass filter is equal to the sum of the orders of the low-pass and high-pass filters. We usually make the orders of the low-pass and high-pass filter equal, in which case the order of the band-pass is even. The pass-band gain of the bandpass filter is the product of pass-band gains of the low-pass and high-pass transfer functions. Obtain the transfer function of a fourth-order band-pass filter having cutoff frequencies equal to \(100 \mathrm{rad} / \mathrm{s}\) and \(2000 \mathrm{rad} / \mathrm{s}\) and a pass-band gain equal to 4.
ANSWER:Step 1 of 5
A band-pass transfer function can be calculated as a product of low-pass and high-pass transfer functions. The order of the band-pass filter is equal to the sum of the orders of the low-pass and high-pass filters.
Because the order of the band-pass filter is equal to the sum of the orders of the low-pass and high-pass filters, and needed band-pass filter has 4th order, we choose low-pass and high-pass filter with second order.
