Chap6StudentSolutions

This can be seen both in the transfer function

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Unformatted text preview: lutions 6-9 M. J. Roberts - 8/16/04 The circuit is a highpass filter because at low frequencies the response approaches zero and at high frequencies it approaches K. This can be seen both in the transfer function formula and in the physical nature of the circuit connection itself. At zero frequency no current flows through the capacitors and therefore not current flows through resistor, R2 . Therefore the voltage at the input of the amplifier is zero making the response voltage zero. At high frequencies the capacitor impedances become practically zero, making the excitation voltage and the voltage at the amplifier input equal. Therefore the transfer function must be K at high frequencies. (b) C1 R1 R2 K + vx(t) vi (t) + vo(t) C2 - - Similar to (a) 14. Show that this system has a highpass frequency response. y(t) x(t) ∫ Write the differential equation from the block diagram. You should get d d y( t) + y( t) = x( t) dt dt Fourier transform both sides and solve for the ratio of Y to X. 15. Draw the block diagram of a system with a bandpass frequency response using two integrators as functional blocks. Then find its transfer function and verify that it has a bandpass frequency response. Lowpass cascaded with highpass. Find the transfer function of both stages (lowpass and highpass) and multiply the transfer functions. Solutions 6-10 M. J. Roberts - 8/16/04 Y( jΩ) , and sketch the frequency response of each X( jΩ) of these DT filters over the range, −4π < Ω < 4π . 16. Find the transfer function, H( jΩ) = (a) x[n] y[n] D y[ n ] = x[ n ] − x[ n − 1] H ( jΩ ) = 1 − e − j Ω (b) Similar to (a) (c) y[n] x[n] D D 1 − e − jΩ H( jΩ) = 1 + e − jΩ (d) x[n] D y[n] D Let z be the output of the left-hand summer. Then y[ n ] = x[ n ] − x[ n − 1] + z[ n ] z[ n ] = x[ n ] − z[ n − 1] Take the DTFT of both equations, eliminate Z and solve for the ratio of Y to X. H ( jΩ ) = 2 − e− j 2 Ω 1 + e− jΩ Solutions 6-11 M. J. Roberts -...
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