Bode Plot Numericals — ECT (IOE)

1. 2082 Baishakh Q4b [8 marks]

Q. Draw the approximate Bode plot of the given transfer function:

Solution

Quadratic numerator: $s^2+5s+25 = 25\!\left[1+\tfrac{s}{5}+\!\left(\tfrac{s}{5}\right)^{\!2}\right]$, so $\omega_n = 5$, $2\zeta\omega_n=5 \Rightarrow \zeta = 0.5$.

Corner frequencies: 5(+), 10(−), 30(−)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{50/3}{(j\omega)^2}\right|_{\omega=0.1} = 20\log\!\left(\dfrac{16.67}{0.01}\right) = 64.44$ dB

Phase plot

Starting phase $= 0° + (-90°)\cdot 2 = -180°$   (constant $K_B>0$ plus double pole at origin)

2. 2081 Bhadra Q4b [2+8 marks]

Q. Find and plot poles/zeros; draw asymptotic Bode plot of:

Solution

Quadratic: $s^2+5s+225 \Rightarrow \omega_n=15$, $2\zeta\omega_n=5 \Rightarrow \zeta=\tfrac{1}{6}\approx 0.167$.

$K_B = 5000/2025 \approx 2.469$, so $20\log K_B \approx 7.85$ dB.

Corner frequencies: 1(+), 9(−), 15(−)

Starting frequency $\omega_0 = 0.01$ rad/s (well below the smallest corner).

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{2.469}{j\omega}\right|_{\omega=0.01} = 20\log\!\left(\dfrac{2.469}{0.01}\right) = 47.85$ dB

Phase plot

Starting phase $= 0° + (-90°) = -90°$

3. 2081 Baishakh Q5b [8 marks]

Q. Draw the asymptotic Bode graph of:

Solution

$s^2+21s+20 = (s+1)(s+20)$. The other quadratic is irreducible: $\omega_n=10$, $2\zeta\omega_n=2 \Rightarrow \zeta=0.1$ (sharp resonance).

Corner frequencies: 1(−), 2(+), 10(−), 20(−)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{1}{j\omega}\right|_{\omega=0.1} = 20\log(10) = 20$ dB

Phase plot

Starting phase $= -90°$

4. 2080 Bhadra Q5b [8 marks]

Q. Plot the frequency response as asymptotic Bode plot:

Solution

Already in standard Bode form. Quadratic: $\omega_n=50$, $2\zeta = 0.6 \Rightarrow \zeta = 0.3$. $K_B=15 \Rightarrow 20\log(15)=23.52$ dB.

Corner frequencies: 2(−), 10(+), 50(−)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{15}{j\omega}\right|_{\omega=0.1} = 20\log(150) = 43.52$ dB

Phase plot

Starting phase $= -90°$

5. 2080 Baishakh Q4b [8 marks]

Q. Sketch the Bode Plot for:

Solution

Quadratic: $\omega_n=\sqrt{50}\approx 7.07$, $2\zeta\omega_n=3 \Rightarrow \zeta \approx 0.212$.

Corner frequencies: 7.07(−), 10(+)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{6}{j\omega}\right|_{\omega=0.1} = 20\log(60) = 35.56$ dB

Phase plot

Starting phase $= -90°$

6. 2079 Bhadra Q4b [8 marks]

Q. Draw the asymptotic Bode plot for:

Solution

Quadratic: $\omega_n=15$, $2\zeta\omega_n=2 \Rightarrow \zeta=\tfrac{1}{15}\approx 0.067$ (very sharp peak).

$K_B = 1/9 \approx 0.111 \Rightarrow 20\log(1/9) = -19.08$ dB.

Corner frequencies: 10(+), 15(−), 20(−)

Starting frequency $\omega_0 = 0.01$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{1/9}{j\omega}\right|_{\omega=0.01} = 20\log(11.11) = 20.92$ dB

Phase plot

Starting phase $= -90°$

7. 2079 Baishakh Q4b [8 marks]

Q. Define frequency response; draw the Bode plot of:

Solution

Corner frequencies: 2(−), 20(−)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log|K_B|_{\omega=0.1} = 20\log(1) = 0$ dB

Phase plot

Starting phase $= 0°$

8. 2078 Kartik Q4b [5 marks]

Q. Draw the Bode log-magnitude and phase plots for:

Solution

$K_B = 1.5 \Rightarrow 20\log(1.5) = 3.52$ dB.

Corner frequencies: 1(−), 2(−), 3(+)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{1.5}{j\omega}\right|_{\omega=0.1} = 20\log(15) = 23.52$ dB

Phase plot

Starting phase $= -90°$

9. 2078 Bhadra Q5b [8 marks]

Q. Draw the asymptotic Bode plot for:

Solution

Quadratic: $\omega_n=4$, $2\zeta\omega_n=4 \Rightarrow \zeta=0.5$ (no resonant peak above asymptote).

Corner frequencies: 2(+), 4(−), 5(−)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{0.5}{j\omega}\right|_{\omega=0.1} = 20\log(5) = 13.98$ dB

Phase plot

Starting phase $= -90°$

10. 2076 Chaitra Q5b [8 marks]

Q. Draw the asymptotic Bode plot for:

Solution

Corner frequencies: 1(−), 5(+), 10(−), 20(−)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{0.05}{j\omega}\right|_{\omega=0.1} = 20\log(0.5) = -6.02$ dB

Phase plot

Starting phase $= -90°$

11. 2076 Ashwin (Back) Q5a [8 marks]

Q. Draw the asymptotic Bode Plot for:

Solution

Quadratic: $\omega_n=10$, $2\zeta\omega_n=2 \Rightarrow \zeta=0.1$.

$K_B = 5/2000 = 0.0025 \Rightarrow 20\log K_B = -52.04$ dB.

Corner frequencies: 1(−), 5(+), 10(−), 20(−)

Starting frequency $\omega_0 = 0.1$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{0.0025}{j\omega}\right|_{\omega=0.1} = 20\log(0.025) = -32.04$ dB

Phase plot

Starting phase $= -90°$

12. 2075 Chaitra Q5b [8 marks]

Q. Draw the asymptotic Bode plot for:

Solution

Quadratic: $\omega_n=8$, $2\zeta\omega_n=3.2 \Rightarrow \zeta=0.2$.

$K_B = 128/32 = 4 \Rightarrow 20\log(4) = 12.04$ dB.

Corner frequencies: 0.5(−), 2(+), 8(−)

Starting frequency $\omega_0 = 0.01$ rad/s.

Magnitude plot

Starting magnitude $= 20\log\left|\dfrac{4}{j\omega}\right|_{\omega=0.01} = 20\log(400) = 52.04$ dB

Phase plot

Starting phase $= -90°$