( ) = V s ( jω ) = 2 kω, a = 4, R s. = 500 nf Draw a Bode diagram of the magnitude and phase of the frequency. Let R p. response H jω. V in.

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1 Let R p = 2 kω, a = 4, = 6 kω, = 500 nf Draw a Bode diagram of the magnitude and phase of the frequency response H jω = V s ( jω ) ( jω ). V in The secondary impedance is Z s ( jω ) = R / jω s = +/ jω jω +.

2 The secondary impedance reflects through the ideal transformer to an equivalent primary impedance of Z p V p ( jω ) = Z s ( jω ) / a 2 = ( jω ) = = Z p ( jω ) R p + Z p ( jω ) V in jω / a 2 / a 2. The primary voltage is then jω + = jω R p + R p + / a 2 / a 2 jω + R R p + s / a 2 jω + V in jω V in ( jω )

3 V p V s ( jω ) == / a 2 V in jω jω R p + R p + / a 2 ( jω ) = av p ( jω ) V s ( jω ) = H( jω ) = / a jω R p + R p + / a 2 / a V in jω jω R p + R p + / a 2 = ar p jω + R + R / p s a2 R p

4 Putting the frequency response into a form that illustrates the simple system components, H( jω ) = ar p jω + R + R / p s a2 R p = R = s ar p + / a jω Frequency Independent R + R / p s a2 Gain R p C s Single-Real-Pole System. The single real pole is at jω = jω 395.8

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6 Let C = 20 nf, R = 20 kω, C 2 = 00 nf, R 2 = 5 kω and K = 8. Draw a Bode diagram of the magnitude and phase of the frequency response H jω = V out ( jω ) ( jω ) V in of this Sallen-Key bandpass filter.

7 R +/ jωc + jωc 2 + G 2 V out ( jω ) = KV x jω V x ( jω ) + ( jωc 2 + G 2 ) ( jωc 2 + G 2 ) R +/ jωc K R +/ jωc V in V x V out ( jω ) ( jω ) ( jω ) ( jωc 2 + G 2 )V out ( jω ) = 0 = R +/ jωc V in 0 ( jω )

8 = V out ( jω ) ( jω ) = H jω V in + ( jωc 2 + G 2 ) R +/ jωc R +/ jωc K 0 + ( jωc R +/ jωc 2 + G 2 ) ( jωc 2 + G 2 ) K

9 H( jω ) = H( jω ) = R C 2 K R +/ jωc + ( jωc 2 + G 2 ) K ( jωc 2 + G 2 ) R +/ jωc jωk / K ( jω ) 2 + jω / R C 2 ( K ) +/ R 2 C 2 +/ R C +/ R R 2 C C 2

10 H( jω ) = R C 2 jωk / K ( jω ) 2 + jω / R C 2 ( K ) +/ R 2 C 2 +/ R C +/ R R C C 2 2 Substituting in numerical values H( jω ) = 57.4 jω ( jω ) jω Poles at s = ± j Underdamped

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12 A signal generator generates a constant-amplitude, variablefrequency analog sinusoid that is sampled at a rate of khz to form the input signal for this bandpass digital filter. Find the frequency of the signal generator that corresponds to the center frequency of the filter. Also find the effective -3dB bandwidth of the filter by finding the signal frequencies at which the power of the output signal is half of its value at the center frequency.

13 This filter's block diagram is drawn in Direct Form II so the transfer function is H( z) = 2.38z z z 4.989z z 2.675z response is H( e jω ) = and the frequency 2.38e j 4Ω 24.77e j 2Ω e j 4Ω.989e j 3Ω e j 2Ω.675e jω

14 Squared Magnitude of the Frequency Response.

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16 The digital center frequency is Ω = radians/sample. That corresponds to a signal generator frequency of ω = 999 radians/second or 59 Hz. The -3dB points are at Ω = radians/sample Hz Ω =.24 radians/sample Hz for a half-power bandwidth of about 39. Hz.

17 A signal x( t) = 4 cos( 2000πt)cos( 200πt) is sampled at its Nyquist rate. What is the signal power of the resultant discrete-time signal x[ n]? The Nyquist rate is 2200 samples/second. Therefore x[ n] = 4 cos( 2π ( 5 /)n)cos( 2π ( / 22)n) = 2 cos 2π ( 9 / 22)n = N 0 = 2 = 2 x[ n] 2 = 4 22 n= N 0 n= 22 n= 22 cos 2 ( 2π ( 9 / 22)n) + cos 2 πn + cos πn 2 cos 2π ( 9 / 22)n + 2cos 2π 9 / 22 + cos πn ( n)cos πn cos 2 ( 2π ( 9 / 22)n) + cos 2 ( πn) + 2 cos( 2π ( 9 / 22)n)cos( πn) n= 22 n= 22 =0 n= 22 { } = 6 = 2 ( / 2 )

18 Alternate Solution: A signal x( t) = 4 cos( 2000πt)cos 200πt What is the signal power of the resultant discrete-time signal x[ n]? The Nyquist rate is 2200 samples/second. Therefore x[ n] = 4 cos( 2π ( 5 /)n)cos( 2π ( / 22)n) X F is sampled at its Nyquist rate. = δ ( F 5 /) + δ ( F + 5 /) δ ( F / 22 ) + δ ( F +/ 22 ) = δ ( F / 2) + δ ( F + 9 / 22) + δ ( F +/ 2) X F + δ F 9 / 22 =δ F+/2 = 2δ ( F / 2) + δ ( F 9 / 22) + δ ( F + 9 / 22) X F = = 6 =δ F /2

19 A signal x( t) = 4 cos( 2000πt)cos( 200πt) is sampled at twice its Nyquist rate. What is the signal power of the resultant discrete-time signal x[ n]? The Nyquist rate is 2200 samples/second. Therefore x[ n] = 4 cos( 2π ( 5 / 22)n)cos( 2π ( / 44)n) = 2 cos 2π ( 9 / 44)n = N 0 = = x[ n] 2 = 4 44 n= N 0 n= 44 n= 44 cos 2 ( 2π ( 9 / 44)n) + cos 2 πn / 2 + cos πn / 2 2 cos 2π ( 9 / 44)n + 2 cos 2π 9 / 44 + cos πn / 2 ( n)cos πn / 2 cos 2 ( 2π ( 9 / 44)n) + cos 2 ( πn / 2) + 2 cos( 2π ( 9 / 44)n)cos πn / 2 n= 44 n= 44 n= 44 = {( / 2 )44 + ( / 2 )44 } = 4 It can be shown that this answer is the same for any sampling rate that is greater than the Nyquist rate. Why is this signal power less than the previous signal power?

20 A signal x( t) = 4 cos( 2000πt)sin( 200πt) is sampled at twice its Nyquist rate. What is the signal power of the resultant discrete-time signal x[ n]? The Nyquist rate is 2200 samples/second. Therefore x[ n] = 4 cos( 2π ( 5 / 22)n)sin( 2π ( / 44)n) = 2 cos 2π ( 9 / 44)n = N 0 = = x[ n] 2 = 4 44 n= N 0 n= 44 n= 44 sin 2 ( 2π ( 9 / 44)n) + sin 2 πn / 2 + sin πn / 2 2 sin 2π ( 9 / 44)n + 2sin 2π 9 / 44 + sin πn / 2 ( n)sin πn / 2 sin 2 ( 2π ( 9 / 44)n) + sin 2 ( πn / 2) + 2 sin( 2π ( 9 / 44)n)sin πn / 2 n= 44 { } = 4 = ( / 2 )44 + ( / 2 )44 n= 44 n= 44 It can be shown that this answer is the same for any sampling rate that is greater than the Nyquist rate. This is the same as the previous signal power.

21 A signal x( t) = 4 cos( 2000πt)sin( 200πt) is sampled at its Nyquist rate. What is the signal power of the resultant discrete-time signal x[ n]? The Nyquist rate is 2200 samples/second. Therefore x[ n] = 4 cos( 2π ( 5 /)n)sin ( 2π ( / 22)n) = 2 sin 2π ( 9 / 22)n = N 0 = 2 x[ n] 2 = 4 22 n= N 0 n= 22 sin 2 2π ( 9 / 22)n = 2 n= 22 + sin πn sin 2π ( 9 / 22)n + sin πn Why is this signal power less than the previous two signal powers? =0 2