In Fig. 23-15, calculate the attenuation, in decibels, at the following frequencies: (a) 0 Hz; (b) 1.592 kHz; (c) 15.92 kHz. (Assume that Vin = 10 Vp-p at all frequencies.)

Question

In Fig. 23-15, calculate the attenuation, in decibels, at the following frequencies: (a) 0 Hz; (b) 1.592 kHz; (c) 15.92 kHz. (Assume that [latex]V_{in} = 10 V_{p-p} [/latex] at all frequencies.)

Accepted Answer

a. At 0 Hz, [latex]V_{out} = V_{in} = 10 V{p-p} [/latex], since the capacitor C appears as an open. Therefore,

[latex]N_{dB}=20 \log \frac{V_{out}}{V_{in}}[/latex]

[latex]=20 \log \frac{10 V_{p-p}}{10 V_{p-p}} [/latex]

[latex]=20 \log 1 [/latex]

[latex]= 20 imes 0 [/latex]

= 0 dB

b. Since 1.592 kHz is the cutoff frequency [latex]f_c, V_{out} [/latex] will be [latex]0.707 imes V_{in} or 7.07 V_{p-p} [/latex]. Therefore

[latex]N_{dB}=20 \log \frac{V_{out}}{V_{in}} [/latex]

[latex]=20 \log \frac{7.07 V_{p-p}}{10 V_{p-p}} [/latex]

[latex]= 20 \log 0.707 [/latex]

[latex]= 20 imes (-0.15) [/latex]

=-3 dB

c. To calculate [latex]N_{dB} [/latex]at 15.92 kHz, [latex] X_C and Z_T [/latex] must first be determined

[latex]X_C=\frac{1}{2\pi fC} [/latex]

[latex]=\frac{1}{2 imes \pi imes 1.592 kHz imes 0.01 \mu F} [/latex]

[latex]=1 k\Omega [/latex]

[latex]Z_T=\sqrt{R^2+X^2_C}[/latex]

[latex]=\sqrt{10^2 k\Omega +1^2 k\Omega }[/latex]

[latex]= 10.05 k\Omega [/latex]

Next,

[latex]N_{dB}=20 \log \frac{X_C}{Z_T} [/latex]

[latex]=20 \log \frac{1 k\Omega}{ 10.05 k\Omega}[/latex]

[latex]= 20 \log 0.0995 [/latex]

= 20(-1)

= -20 dB

Question 23.6: In Fig. 23-15, calculate the attenuation, in decibels, at th...

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