Question

The main advantage of Bode plot is to

• A. Show complex conjugate zeros
• B. Show complex conjugate poles
• C. Calculate the constant gain
• D. Convert multiplicative factors into additive factors

Hint:

Bode plots use logarithmic scales for frequency and magnitude (in dB).
Logarithms convert multiplication into addition (\(\log(AB) = \log A + \log B\)) and division into subtraction (\(\log(A/B) = \log A - \log B\)).
This allows for easy sketching of frequency response by summing the asymptotic contributions of individual poles and zeros.

Answer 1

The Correct Option is D

A Bode plot consists of two graphs: 1. Magnitude (in decibels, dB) vs. frequency (on a logarithmic scale). 2. Phase (in degrees or radians) vs. frequency (on a logarithmic scale). The transfer function \(G(j\omega)\) is often expressed as a product of factors (e.g., constant gain, poles, zeros, quadratic terms). \( G(j\omega) = K \frac{\prod (1 + j\omega/z_i)}{\prod (1 + j\omega/p_j)} \frac{\prod (\text{quadratic zero factors})}{\prod (\text{quadratic pole factors})} (j\omega)^{\pm N} \) When taking the magnitude in dB: \( |G(j\omega)|_{dB} = 20 \log_{10} |G(j\omega)| \) \( = 20 \log_{10} |K| + \sum 20 \log_{10} |1 + j\omega/z_i| - \sum 20 \log_{10} |1 + j\omega/p_j| + \dots \) The logarithm converts the multiplication of magnitudes into a sum of logarithmic terms (dB values). Similarly, for phase: \( \angle G(j\omega) = \angle K + \sum \angle (1 + j\omega/z_i) - \sum \angle (1 + j\omega/p_j) + \dots \) The phase of a product/division of complex numbers is the sum/difference of individual phases. Thus, a major advantage of Bode plots is that the contributions of individual factors (poles, zeros, gain) can be plotted as straight-line approximations (asymptotes) and then added graphically (or their dB values added) to get the overall response. This conversion of multiplicative factors in the transfer function magnitude to additive factors in the dB magnitude plot (and additive phase contributions) simplifies the construction and analysis of the frequency response. (a) & (b): Bode plots can represent systems with complex conjugate poles/zeros, but this isn't their "main advantage" in the sense of a unique simplifying feature compared to other plots. (c) Calculate the constant gain: The constant gain \(K\) is one factor, and its contribution \(20\log_{10}|K|\) is easily found, but this isn't the main broad advantage. (d) "Convert multiplicative factors into additive factors": This is the core reason Bode plots (using logarithmic scales for magnitude and frequency) are so useful. \[ \boxed{\text{Convert multiplicative factors into additive factors}} \]