Question 14.15: Determine the frequency response of the circuit shown in Fig......

We let the input voltage v_{s}  be a sinusoid of amplitude 1 V and phase 0° .  Figure 14.51 is the schematic for the circuit. The capacitor is rotated 270° counterclockwise to ensure that pin 1 (the positive terminal) is on top. The voltage marker is inserted to the output voltage across the capacitor. To perform a linear sweep for  1 < f < 100 Hz with 50 points, we select Analysis/Setup/AC Sweep, DCLICK Linear, type 50 in the Total Pts box, type 1 in the Start Freq box, and type 1000 in the End Freq box. After saving the file, we select Analysis/Simulate to simulate the circuit. If there are no errors, the PSpice A/D window will display the plot of V(C1:1), which is the same as V_{o}   or  H(ω) = V_{o} /1 ,   as shown in Fig. 14.52(a). This is the magnitude plot, since V(C1:1) is the same as VM(C1:1). To obtain the phase plot, select Trace/Add in the PSpice A/D menu and type VP(C1:1) in the Trace Command box. Figure 14.52(b) shows the result. By hand, the transfer function is

H(ω) = \frac{V_{o}}{V_{s}} = \frac{1,000}{9,000  +  jω8}

or

  H(ω) =  \frac{1}{9  +  j16π  ×  10^{-3}}

showing that the circuit is a lowpass filter as demonstrated in Fig. 14.52. Notice that the plots in Fig. 14.52 are similar to those in Fig. 14.3 (note that the horizontal axis in Fig. 14.52 is logrithic while the horizontal axis in Fig. 14.3 is linear.)