Question

please answer all questions and show all steps

3. Draw the magnitude characteristic of the Bode plot of the following transfer function: Hsout20(0.1s + 1 Vin S(s1)(0.01s +1) a) ldentify the poles and zeros b) Sketch the magnitude plot. The y-axis should be in dB. The x-axis should logarithmic, but labeled in rad/sec. Use MATLAB to draw the complete Bode plot and check your answer. If the input current is uln(t)-2 cos(2t + 30%), what is the output voltage. Use your MATLAB Bode plot to answer this. c) d)

Answer 1

Ans)

a) from the given transfer function putting s=jw

$H(jw)=\frac{20(j\frac{w}{10}+1)}{jw(j\frac{w}{1}+1)(j\frac{w}{100}+1)}$

poles are roots of denominator polynomial i.e

s=0,s=-1,s=-100  

zeroes are roots of numerator

s=-10

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b)

For magnitude plot we find asymptodes first comes pole at contributes -20dB/decade slope gain at w=0.1 rad/sec is

$G_{w=0.1}=20log_{10}\frac{20}{w}=20log_{10}\frac{20}{0.1}=46.02\: \: dB$

Next comes pole at w=1 rad/s contributes -20dB decade net slope is

-20dB/decade-20dB/decade=-40dB/decade

Next comes it contributes slope of +20dB/decade ,now net slope of the line to be drawn is -40dB/decade+20dB/decade=-20dB/decade

Next comes pole at w=100 rad/s contributes -20dB decade net slope is

-20dB/decade-20dB/decade=-40dB/decade

using the above details magnitude plot is

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c)

The matlab code for bode plot given below ,convolution command used for multiplication of polynomials in the denominator of transferfunction

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d)

Given input from this the gain and phase at w=2 rad/s is marked in the bode plot above

$Gain=10^{\frac{12.9}{20}}=4.416$

$Gain=4.416$

$phase=-144^{o}$ from plot using this output voltage is

$v_{out}(t)=gain*v_{IN}(t)\angle -144^{o}=4.416*2cos(2t+30^{o}-144^{o})\: \: V$

$\mathbf{v_{out}(t)=8.83cos(2t-114^{o})\: \: V}$