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ƒ r<br />

27.051 kHz. Clearly, therefore, the frequency at which the phase angle<br />

is zero and the total impedance appears resistive is less than the frequency<br />

at which the output voltage is a maximum.<br />

Electronics Workbench<br />

The results of Example 20.9 will now be confirmed using Electronics<br />

Workbench. The network of Fig. 20.36 will appear as shown in Fig.<br />

20.47 after all the elements have been placed as described in earlier<br />

chapters. In particular, note that the frequency assigned to the 2-mA ac<br />

current source is 100 kHz. Since we have some idea that the resonant<br />

frequency is a few hundred kilohertz, it seemed appropriate that the<br />

starting frequency for the plot begin at 100 kHz and extend to 1 MHz.<br />

Also, be sure that the AC Magnitude is set to 2 mA in the Analysis<br />

Setup within the AC Current dialog box.<br />

FIG. 20.47<br />

Using Electronics Workbench to confirm the results of Example 20.9.<br />

For simulation, the sequence Simulate-Analyses-AC Analysis is<br />

first selected to obtain the AC Analysis dialog box. The Start frequency<br />

is set at 100 kHz, and the Stop frequency at 1 MHz; Sweep<br />

type is Decade; Number of points per decade is 1000; and the Vertical<br />

scale is Linear. Under Output variables, node number 1 is selected<br />

as a Variable for analysis followed by Simulate to run the program.<br />

The results are the magnitude and phase plots of Fig. 20.48. Starting<br />

with the Voltage plot, the Show/Hide Grid key, Show/Hide Legend<br />

key, and Show/Hide Cursors key are selected. You will immediately<br />

note under the AC Analysis cursor box that the maximum value is<br />

95.24 V and the minimum value is 6.94 V. By moving the cursor until<br />

we reach 95.24 V (y1), the resonant frequency can be found. As shown<br />

COMPUTER ANALYSIS ⏐⏐⏐ 927

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