Question

.A network has a voltage gain of:

H(ω) = 10 / (1+jω)(10+jω)
Sketch the Bode plot for the gain.

Answer 1

To create the Bode plot for the given voltage gain, we plot an initial gain of 20 dB, decreasing at -20 dB/decade after ω=1 and at -40 dB/decade after ω=10, with the phase reaching -90 degrees at high frequencies - characteristic of a two-pole low pass filter.

Step-by-step explanation:

The given network has a voltage gain expressed as H(ω) = 10 / (1+jω)(10+jω). To sketch the Bode plot for this gain, we need to examine the magnitude and phase of the gain as a function of angular frequency (ω). The given gain expression can be considered as two first order terms in the denominator having corner frequencies at ω=1 and ω=10 respectively.

To construct the Bode plot:

1. We start by plotting a gain of 20*(log(10)) or 20 dB at very low frequencies.
2. At the first corner frequency ω=1, the gain begins to decrease at a slope of -20 dB/decade.
3. After the second corner frequency ω=10, the gain decreases further at a slope of -40 dB/decade (combination of both the first and second order terms).
4. For the phase plot, each first order term contributes -45 degrees at its corner frequency, leading to -90 degrees at high frequencies.

The Bode plot for this gain will show the characteristic of a two-pole low pass filter.