Sample of Industrial Attachment Report in Civil Engineering

Submitted by: Submitted by

Views: 494

Words: 2692

Pages: 11

Category: Science and Technology

Date Submitted: 09/28/2014 12:58 AM

Report This Essay

Transfer Functions and Bode Plots

Transfer Functions

For sinusoidal time variations, the input voltage to a filter can be written ¤ £ vI (t) = Re Vi ejωt

where Vi is the phasor input voltage, i.e. it has an amplitude and a phase, and ejωt = cos ωt + j sin ωt. A sinusoidal signal is the only signal in nature that is preserved by a linear system. Therefore, if the filter is linear, its output voltage can be written ¤ £ vO (t) = Re Vo ejωt where Vo is the phasor output voltage. The ratio of Vo to Vi is called the voltage-gain transfer function. It is a function of frequency. Let us denote Vo T (jω) = Vi We can write T (jω) as follows: T (jω) = A (ω) ejϕ(ω) where A (ω) and ϕ (ω) are real functions of ω. A (ω) is called the gain function and ϕ (ω) is called the phase function. As an example, consider the filter input voltage ¤ £ vI (t) = V1 cos (ωt + θ) = Re V1 ejθ ejωt The corresponding phasor input and output voltages are Vi = V1 ejθ Vo = V1 ejθ A (ω) ejϕ(ω) It follows that the time domain output voltage is i h vO (t) = Re V1 ejθ A (ω) ejϕ(ω) ejωt = A (ω) V1 cos [ωt + θ + ϕ (ω)]

This equation illustrates why A (ω) is called the gain function and ϕ (ω) is called the phase function. The complex frequency s is usually used in place of jω in writing transfer functions. In general, most transfer functions can be written in the form T (s) = K N (s) D (s)

where K is a gain constant and N (s) and D (s) are polynomials in s containing no reciprocal powers of s. The roots of D (s) are called the poles of the transfer function. The roots of N (s) are called the zeros. As an example, consider the function T (s) = 4 s/4 + 1 s/4 + 1 =4 s2 /6 + 5s/6 + 1 (s/2 + 1) (s/3 + 1) 1

The function has a zero at s = −4 and poles at s = −2 and s = −3. Note that T (∞) = 0. Because of this, some texts would say that T (s) has a zero at s = ∞. However, this is not correct because N (∞) = 0. 6 Note that the constant terms in the numerator and denominator of T (s) are both unity. This is...