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Unformatted text preview: Transfer functions for each type of circuit are of the form: H ω j ω RC 1 j ω RC H ω ω RC 1 ω 2 R 2 C 2 Generate this plot. Some example code for MATLAB is provided below: 10 1 10 2 10 3 10 4 10 5 10 6 10 7 ! 140 ! 120 ! 100 ! 80 ! 60 ! 40 ! 20 f (Hz) H (dB) f=logspace(1,7,1000); w=2*pi*f; semilogx(f, 20*log10(w*982*560e12./sqrt(1+w.^2*982^2*560e12^2)), ''); hold; semilogx(f, 20*log10(w*72*560e12./sqrt(1+w.^2*72^2*560e12^2)), ''); semilogx(f, 20*log10(w*10*560e12./sqrt(1+w.^2*10^2*560e12^2)), ''); To calculate cutoff frequencies algebraically, use the fact that for these firstorder type filters, ω c = 1 / RC . Be sure to divide by 2 π to find an answer in regular frequency, not angular frequency: ω c , E A 1 2 π 982 Ω 560pF 289kHz ω c , E B 1 2 π 982 Ω 560pF 3 . 95MHz ω c , E C 1 2 π 982 Ω 560pF 28 . 4MHz EECS 314 Fall 2007 HW 07 Solutions Problem 1 Calculate all the output signals using the above transfer function magnitude, each frequency (multiplied by 2 π ) and each resistance: 50 kHz 500 kHz 5 MHz EA EB EC 170 mV 866 mV 998 mV 12.7 mV 126 mV 785 mV 1.76 mV 17.6 mV 173 mV Well above the cutoff frequency, the maximum transfer function magnitude remains the same for each filter. The decay rate for each filter is also the same (20 dB per decade frequency). Each filter has a different cutoff frequency, so their Bode plots are shifted from each other on the frequency axis. EECS 314 Fall 2007 HW 07 Solutions Problem 1 Problem 2 Solution 1kHz Solution The impedance of C and L are = 1 = 1 2 ∙ 1000 ¡ (125 ∙ 10 − 9 ) = − 1273.24 Ω = = 2 ∙ 1000 ¡ (12.5 ∙ 10 − 3 ) = 78.54 Ω Now we can treat the capacitor and inductor as we have treated resistors in the past. Add the capacitor in series with the 5R resistor to form Z 1 and the inductor in series with the R resistor to form Z 2 . 1 = 5 + = 625 − 1273.24 Ω 2 = + = 125 + 78.54 Ω Find the total current in the circuit, and then use current division to find the currents I C and I L . Combine Z 1 and Z 2 in parallel. = 1  2 = 625 − 1273.24 ¡ 125 + 78.54 ¡ 625 − 1273.24 + 125 + 78.54 = 133.22 + 65.46 Ω ⟶ 148.43 ∠ 26.17° Now the total current is found in phasor form using Ohm’s law. = 125 ∠ 0° 148.43 ∠ 26.17° = 0.8421 ∠ ( − 26.17°) Use current division to find I C and I L , = + ∙ = 125 + 78.54 625 − 1273.24 + 125 + 78.54 ∙ [0.8421 ∠ − 26.17° ¡ ] = 0.0881 ∠ 63.85° = + ∙ = 625 − 1273.24 625 − 1273.24 + 125 + 78.54 ∙ ¢ 0.8421 ∠ − 26.17° ¡£ = 0.8467 ∠ ( − 32.14°) Use Ohm’s Law to find V C and V L , = ∙ = −...
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This note was uploaded on 04/10/2008 for the course EECS 314 taught by Professor Ganago during the Fall '07 term at University of Michigan.
 Fall '07
 Ganago

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