This preview shows page 1. Sign up to view the full content.
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.
C1 R1 R2 K + vx(t) vi (t) + vo(t)
C2 - - Similar to (a)
14. Show that this system has a highpass frequency response.
Write the differential equation from the block diagram. You should get
y( t) + y( t) = x( t)
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
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 -...
View Full Document
- Spring '09